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HUGHES AIRCRAFT COMPANY

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HUGHES

HUGHES AIRCRAFT COMPANY AEROSPACE GROUP

SPACE SYSTEMS DIVISION

EL SEGUNDO CALIFORNIA

QUARTERLY TECHNICAL REPORT

PERIOD I JULY - 30 SEPTEMBER 1968

ELECTRIC THRUSTER POWER CONDITIONER

CONTRACT NO 952297



QUT Zl 1)38

W JgYMuldoon Assistant Project Manager

J A Robertson ontract Administrator

This work was performed for the Jet Propulsion Laboratory

California Institute of Technology as sponsored by the National Aeronautiis and Space Administration under Contract NAS 7-100

This report contains information prepared by the

Hughes Aircraft Company under JPL subcontract

Its ontent is tot necessarily endorsed by the

Jet Propulsion Laboratory California Institute of

Technology or the National Aeronautics and Space

Administration

ABSTRACT

Quarterly report on JPL Contract 952297 for a 20 C Electric

Thruster Power Conditioner and Support Equipment presents circuit

and physical design description efficiency analysis reliability

analysis and test data taken on preliminary breadboards

Report discusses techniques of high-efficiency regulation and

control ie single inverter pulse-width modulation staggeredshy

phase inverter modulation line-regulator modulation magnetic

modulation Design goal efficiency is 93 percent Preliminary

tests suggest this is feasible-

Techniques of analog thruster loop control are discussed also

overload trips standby and active redundancy

Report also discusses physical design approach for self-radiating

cooling

TABE OF CONTENTS

Page

INTRODUCTION 1

TECHNICAL DISCUSSION 2

1 General 2

a Objectives 2

b System Description 2

2 Pulse-Width Modulation Techniques 6

a Single Inverter Modulation 6

b Staggered Phase Modulation

12

-d Magnetic Modulation I

c Line Regulator Modulation i

3 Cathode RMS Current Regulatorl

4 D C Voltage Regulation i

a Arc Supply is

b Accelerator Supply 21

c Screen System 21

5 System Control Techniques 22

a General 2

b Command Controls 24

c Overload-Trips 2f

d Analog Controls 2E

e Standby Controls 2E

6 Test Data 41

a Low Voltage Group 41

b Power Inverter 41

c Cathode Supply 41

d Function Generator 41

e Components 41

f Analog Controls 41

iii

Page

7 Efficiency Analysis 62

a Screen Inverter 62

b Arc Inverter 66

c Acceletator Inverter 67

d Cathode Inverter 68

e 5 KHz Inverter 70

f 5 KHz Line Regulator 70

g Accelerator Line Regulator 72

h Arc Rectifier Filter 73

i High Voltage Filter 73

j Magnetic Modulator 74

k Summary of Losses 75

8 Reliability Analysis 76

9 Physical Design 99

a General 99

b Support Structure 100

c Module Assemblies 100

d Miscellaneous Considerations 105

10 Power Conditioner Support Equipment - Test Console 112

a General 112

b FunctionalCircuit Description 113

c HardwarePackaging Description 117

d Energy Needs and Losses 117

11 Calorimeter 134

12 List of Drawings 136

INTRODUCTION

The objective of this contract is the design and fabrication of a power

conditioning and control system for the JPL 20 cm electric thruster which

requires approximately 3 KW of conditioned power with prime power from a

solar array ranging from 40 volts out to 80 volts out

The requirements of 1 regulation on all outputs and a wide range of

control combined with a target efficiency of 93 and a weight at eight

pounds per kilowatt necessitates the use of pulse-width modulation on all

supplies The large step-up for high voltage dc supplies and large

step-down for high current low voltage ac supplies requires the use

of dc to ac inverters

To satisfythe requirements of high efficiency and low weight inverters

must operate at relatively high frequency The choice of transistors

with the high speed switching required for high efficiency is limited to

20 amp rating usable at f0 amp for reliability At 40 volt low line the

resultant limit is 400 watts per inverter ( paralleling transistors is

costly in efficiency due to need for equalizing techniques) Hence in one

supply requiring an output of 2000 volts at I amp (2 K-W) the total power

must be obtained from a group of inverters For high efficiency with wide

range of regulation and control to avoid the losses associated with separate

modulation and inversion pulse-width modulated inverters are used

In this first three month period in the contract principal efforts were in

the design and development of the subsystems and circuits required by the

system The results of this development effort are described in this

report with emphasis on technology rather than description of the detailed

embodiment of the techniques described although detailed circuits are

presented to illustrate the reduction to practice of these techniques

TECHNICAL DISCUSSION

1 General

a Objectives

The broad objectives are to design a system providing the outputs

shown in Tables I and II with -prime power from a solar array varying from

40 V to 80 V such as obtained on a Jupiter mission (40 V at earth 80 V

at Jupiter) The 6ystem weight should not exceed 235 pounds and efficiency

should not be less than 92 Reliability for a 10000 hour flight should

exceed 96

Start-up and shut-down shall be commanded by pulse signals of

20 V to 31 V of 20 milliseconds duration at 75 mA

Control of thruster screen current shall be by an analog signal with

5 volts for I amp continuously variable to 05 amps at 0 volts control

Telemetry shall be supplied as indicated in Table II at anormalized

full-scale output of 5 volts at a source impedance of 10K ohms or less

b System Description

All voltages delivered by the Power Conditioning System delineated

in Table I are derived from inverters operating at two synchronized

frequencies ie 125 KHz and 5 KHz Individual inverters deliver power

in the range of 100 to 300 watts Where higher power is required such as

the screen system at 2 KV and 1A total power is derived from the series

output of 300 watt inverters

By restricting the maximum output per inverter to 300 watts it is

possible to use high speed transistors and thereby retain high efficiency

at reasonably high frequency such as 125 KHz permitting low weight and

high efficiency transformers Further advantages from the modularization are

adaptability to fractional redundancy for high reliability and adaptability

to staggered phasing to minimize ripple from pulse-width regulation The

staggered phase technique also minimizes the peaking of the screen ntput

-filters at no load

The 5 KHz inverter frequency is used for ac heater loads where a

higher frequency would result in excessive line and load inductive impedance

4 MacRatings Nominal Ratings Range of Frequency v ii-i - shy

ply spl R 1 fk Control2) 4 RippLIN P u i 5) Output RippleGroup Supupply Type

No Name = =I JIi i -

S V A A V A W 7 A kHz kHz

S Magnet Mantfld Htr DC F 19 085 09 15 067 105 10(I) 5 --- --- 10

3 Cath Htr AC V 5 40 45 45 35 160 Loop --- 10 - 40 5

7 Nautr Htr AC V 12 34 4 12 28 35 Loop 03- 34 5

300V 20mA 2 at 30V 8 Neutr Kpr DC F 0o5 10 050 5 10(E) 002-05 --- 10

5mA 30V 5 at lOVr (4- - -n--si - - - - - shy-

2 Vaporizer AC V 10 20 205 55 11 6 Loop --- 05 - 12 5

-5-1S 7 t36O 2 7 4 Arc DC V 2mA - 8 345 6 210 1O(z) 2 2-7 --- 30

fl7V -i--i

5 Beam DC V 2050 10 105 2000V 10 2kw 10(E) 5 05-10 --- 30

2050 [13)6 Accel DO P 0105 2000V 001- 20 10(g) 5at --- -30

03 amp

1) Output V - Variable F - Fixed

2) Current Limits or overload trips - Exact values to be specified by the manufacturer

3) Current stays at this level for less than 10 minutes at very low rep rate

4) Starting Characteristicsa (150V N to 36V at 20 ma) 5) Current varies as nmetion of engine loop control See Figures below

5-29-68 5aa L3-4 7amp 91 4o 8 2

15 EE

Supply I Name

440~-5VmTRange

Current

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Voltage

High Accuracy Part T Rane

Current Voltage

Shutdown Levels

(Note 1 Low High

I

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0 - 3A

10 shy 45A

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5

6

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0 - IA

0 - 2OmA

0 - 4A

0 - IA

1700 - 2100

1700 - 2100

-

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05 -IA

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1950 - 2050

1950 - 2050

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t2 1 0 0 V

2

2100V

Neutr Emission 0 - IA

ff a- 5V Output

Areingamp 10 AreMmn

NTZ It At these livels P vil be ordered to shut-down

Table IIS Telemetry Data and

at relatively high currents and low-voltages required It is also used for

low power loads such as Magnet-Supply and Neutralizer Keeper to minimize

circuit complexity

The 125 KHz inverter frequency is used for dc-to-dc converters

at high power (300 watts) for reasons stated above

Referring to Functional Block Diagram Drawing X3188131 regulated dc

voltage of 35 volts is derived from the chopper line regulator which

regulates output within one percent over the rangeof 40 to 80 volt line

A master oscillator serves as the chopping frequency source for the regulator

and as the master oscillator for all pulse-width-regulated inverters Also

derived from the oscillator is synchronized 10 KHz for the 5 KHz heater inverter

and 10 KHz for the cathode heater pulse-width-regulator

The 5 KHz heater inverter supplies square-wave voltage to magneticallyshy

regulated supplies for the Magnet (VI) vaporizer-heater (V2) neutralizer

heater (V7) and neutralizer keeper (V8) supplies Additionally it supplies

regulated dc voltages for driven inverters base drive (cathode arc accelerat

and screen) and for miscellaneous logic and control functions

The cathode arc and screen inverters all use pulse-width modulated

base drive for regulation with prime power derived from the 40 to 80 volt

line

The accelerator inverter is a driven inverter not modulated but

regulated by a chopper line regulator in series with the line and controlled

by feedback from the accelerator supply output This technique was chosen

rather than a modulated inverter in view of the resonant ringing with high

step-up between output filter choke and transformer distributed capacity

the choke necessitated by the modulated inverter By regulating (controlling)

input dc inverter output is square-wave and requires only a small capacity

for filtering minimizing output ringing Such ringing is of course highly

undesirable at the 2 KV level of the output

As mentioned above in discussing the advantages of the modular system

the screen system inverters are stagger-phased The phase shifting required

is accomplished in the oscillator section since this section includes the

master oscillator used as the reference frequency for phase-shifting The

screen inverters are phase-shifted 18 of a half-cycle apart with all inverter

5

dc outputs (unfiltered) in series- All inverters are modulated under control

of feedback from the total screen output voltage In event of failure of one

inverter other inverters advance On time to compensate

The cathode arc and accelerator inverters consist of an Operate

and Standby inverter in each supply the Standby being switched on in

event of a failure of the Operate inverter

The Control Section pr6vides the control and telemetry interface with

the vehicle provides the required control link between interdependent supplies

(minor loop gain and transfer function) provides the necessary automatic

sequencing and time delays in turn-on monitors the system for overload

trips controls recycle time senses inverter failures and controls standby

switching It also contains the system master oscillator and phase shifter

Details of the individual circuits will be discussed in later sections

of this review

2 Pulse-Width Modulation Techniques

a Single Inverter Modulation

Referring to System Diagram Drawing X3188131cathode and arc systems

shown use a single operating inverter with a standby inverter Each inverter

is pulse-width modulated for control andor regulation The cathode inverter

operates at 5 KHz and the arc inverter at 125 KHz

In the following discussion reference should be made to schematic of

arc inverter Dwg X3188109

The mode of inverter modulation chosen will maintain accurate balance

on pulse-width of alternate half cycles of carrier thereby minimizing ratcheting

of output transformer from DC unbalance restricting this unbalance to that

due to unequal saturation or unequal storage time of output transistors

Modulation is achieved as follows a modulation carrier of twice

the output frequency (square-wave) is integrated (R26 Cl) to obtain a

triangular wave-form The triangular wave-form is summed with the difference

between output feedback and a reference bias the net difference being applied

to a comparitor amplifier Q15 A small droop error signal will swing the

comparitor trigger from off to on for full half-cycle (see following figure)

6

-II

i7fl iY- -S

Modifications in the conventional drive utilizing a single transformer

ere necessary to accommodate a pulse width modulated drive In the single

ransformer drive circuits a pulse would not be transformed perfectly due to

eakage inductances The secondary overshoots in the negative-direction and

n an ECU application turns the off transistor on for a short period of time

his effect is undesirable and can be overcome by 1) using two drive

ransistors (turn-on and turn-off) directly connected to each of two power

witches (eliminates the transformer) and 2) using two transformers Alternate I

as thoroughly investigated utilizing a variety of designs The major disadshy

antages were power consumption circuit complexity and high failure rate

omponents (power transistor drive)

The circuit chosen is essentially two energy storage converters one

riving each side of the ECU Several advantages result from this configurashy

ion ie low failure rate components (signal transistor drive) low

omplexity and power savings The natural kickback of the transformer is

seful in supplying the needed turnoff bias and with separate transformers

oes not affect the off transistor Also the transformer coupling permits

mpedanee match from a 35 volt drive source to the -2volt base drive The basic

ower drive stage for one-half the ECU is shown in Figure 2-1 The drive pulse

or Q1 of Fig 2-1 is shown and the resulting voltage and current waveforms

re shown for each part of the circuit

Referring to Fig 2-1 upon application of power to the base of QI the

I collector voltage drops to near zero impressing the supply voltage across

he TI primary The voltage appearing at the secondary turns Q2 on During

he conduction period the collector current increases as shown The initial

tep in current is the reflected base current When base drive is removed from

allet 2r F-

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Qi the stored energy in the transformer causes the voltages to reverse and

CRI clamps the primary Since TI is center tapped the voltage developed at

Qi is twice the supply voltage and the voltage on the secondary is just the

reverse of its previous value The reverse base current is limited by CR2

and Q2 base impedances and when measured was found to be 2 amperes peak

The transistor is thus not subjected to high reverse base-emitter voltages and

a low impedance path is provided to remove the stored charge The commutation

(fly-back) current at turn-off must be large enough to supply the turn-off curshy

rent demand of Q2 when the transformer reverses The peak design collector

current for 100 percent duty cycle is 04 amperes and at 25 percent would be

015 amperes which when reflected to the secondary would be 195 amperes

The system is designed to operate a no less than 50 percent duty cycle

thus adequate turn-off current is assured

A damping network is required to limit the overshoot voltage at Q1

The damping current is seen in Fig 2-i The line current is a composite of

the collector and transformer fly-back currents and is also shown A sketch

of flux density is also given

The drive pulse for Ql is generated by a 932 gate Part of the

gate is used to and the flip-flop output and pulse amplifier output The two

gates are capacitance coupled as shown to prevent destruction of Q3 or Q4 due

to loss of drive signal (Dwg X3188109)

This circuit was breadboarded and optimized with a measured base drive

power of 15 watts per side or 30 watts total drive power for 300 watts out

(10 percent) which is within the design goal for the stage The total weight

of the two drive transformers is 28 grams also within the design goal on weigh

The variable pulse-width output of the comparitor amplifier (see

schematic) is summed with the output from a flip-flop triggered at 2f on

a nand gate Ql Q2 for each phase of the carrier The output of each nand

gate is invertered in a power gate coupling to the driver amplifier Q3 Q4

The output transistor chosen Q5 Q6 is the Solitron SDT 8805 rated

at 325 volts Vcex 300 volts Vcesfst and 20 amps with a minimum beta of 15

10 amps From the published data the expected minimum beta at 8 amps (252

watts out) is 20 required a circuit beta of 15 for good saturation at 8 amps

10

The switching time for the SDT 8805 is less than microsecond

resulting in good switching efficiency at 125 KHz The storage time will be

about I to 2 microseconds which requires compensation when full-on to prevent

overlap of output transistors and resultant output current spikes and high

switching dissipation The compensation method chosen derives a hold-off

bias from an output transformer winding T5 such that the drive on the

lion-coming transistor is held off until the current in the off-going

transistor starts to collapse

Diodes CR39 to CR40 are used to clamp the output transistors to a

voltage close to double line voltage The clamp operates through the tight

coupling of the bifilar-primary the induced voltage due to collapsing current

being-clamped on the opposite side of primary to the voltage on the line

capacitor through the clamping diodes

The operating frequencies of 125 and 5 KHz were chosen as compatible

with a master oscillator and consistent with maximum allowable frequency of

5 KHz for heaters (line drop at low voltage high current and engine heater

inductance) and a frequency for DC-DC converters compatible with high

efficiency and minimum size and weight A discrete jump in size of available

cores for output transformers suggests 125 KHz since the next smaller core

is too small for 252 watts at 15 KC and the core chosen permits operation at

125 KHz at 300 watts hence higher efficiency than that at 15 KHz (higher

switching losses)

The output transformer to be used with the inverter will be a ferrite

cup-core Indiana CF 217-05 material with a 2000C Curie point Tests have

verified that this core will operate at 1400c without saturation at 2 kilogauss

operating flux density a temperature well above the expected maximum of 950C

Core loss was verified as being within the expected 1 of output power Copper

loss was found to be significantly higher (-50) than the DC value due~to skin

effect An approximation analysis using super-position of harmonic currents

verified the measured loss as expected Use of Litz wire multi-filar or

foil windings would result in lower space factor hence is not justified

The choice of ferrite for the output transformer was made on the basis

of low core loss at the frequency desired comparable to that obtained with

1i

nickel-iron I mil tape cores plusthe tolerance in a cup-core of some

dc unbalance inherent in the inverter The residual air-gap in the

cup-core is sufficient to permit a significant dc without excessive

ratcheting A further advantage is the ideal heat-transfer to the mounting

inherent in the cup-core shape


Referring to Dwg X3188131System Block Diagram and X3188105 Screen

Inverter S6hematic the screen system delivering 2000 volts at 1 amp uses

eight (8) inverters with dc outputs in series each inverter delivering

250 volts or 250 watts

The modulation circuitry used in the screen inverter is similar to

that described abovefor the cathode and arc inverter

To minimize ripple reduce size of ripple filter and improve no-load

to full load regulation each inverter is driven at 125 KHz with phase

displacement of 18 of 12 cycle between inverters resulting in a ripple

frequency of 200 KHz and a maximum peak to peak ripple of 500 volts at 2000

volts unfiltered ie plusmn 250 volts

Dwg No X3188121-2 Master Oscillator and Phase Shifter indicates

the method for obtaining the staggered phase drive for the screen inverters

Fig 2-2 is the timing diagram for this circuit

The digital staggered phase generator (SCPG) is designed to provide

two sets of 125 KHz square waves Each set consists of eight separate

square waves separated by exactly 18 of cycle The two sets of staggered

sqdare waves are then logically combined to produce 16 modulated pulses which

are buffered to provide current drive for the inverter transistors in the ECUs

The 125 KHz staggered square waves are produced by shifting a 125 KHz

square wave through a seven bit shift register at a 200 KHz clock rate The

200 KHz square wave is given for reference The 125 KHz waveform is the

output from the second 16 bit counter in Channel A Channel B is the delayed

channel The outputs from the first four bits of the shift register in

Channel A are shown as 2A 3A 4A

Channel B is a duplicate of Channel A The Channel B counters are

reset at the trailing edge of the pulse width modulator output The pulse

12

width modulator is of standard triangle wavezero crossing type The butput

of the Channel A counter is integrated t6 obtain the triangle waveform The

ramp compared with the control voltage produces the PWM waveform shown

By reseting the B counters at the end of the PWM pulse the

count is effectively delayed by one-half the PWM pulse width By anding

the two shift register outputs a delayed pulse equal in width to the PWM

pulse is obtained for each phase of the system The two required waveforms

for each half of the inverters are shown as El and 2 Fl and 2 Cl and 2

and HI and 2

Since the counters change state only on the fall of the clock pulse

the phase difference between the two counters and hence the shift registers

cat be only increments of one clock pulse This determines the accuracy to

which the pulse width can be controlled To achieve I percent regulation

the clock rate must be at least 200 times the inverter frequency since each

one-half cycle must be controlled to within one perdent Since integer powers

of two are available from binary counters a factor of 256 is used The

oscillator frequency is thus 32 MHz

The square-wave outputs of the phase shifter are anded on the inputs

at the screen inverter input gates to produce the controlled pulse width

The remaining inverter circuitry following the input gates is

identical to that described for the arc inverter except in the output transshy

former rectifier

c Line Regulator Modulation

Referring to System Block DiagramDwgX3188131 it will be noted that

two (2) subsystems the 5 KHz inverter and associated low voltage group and

the accelerator supply use a line regulator

In the case of the 5 KHz inverter and its associated low voltage

supplies the latter use individual load regulators (magnetic-amplifiers) since

the low power required of these supplies permits a lower efficiency and

higher per-unit weight to obtain minimum complexity (high reliability) Hence

the burden of regulation on the line regulator is that of line variation only

The accelerator supply uses its associated line regulator for both

line and load regulation to permit a square-wave inverter for the high

voltage step-up required in the inverter and thereby minimize ringing of

output transformer and simplify filtering at high voltage

The line regulator circuit5 drawing no 3188119 is identical f03

both applications differing only in the voltage sense connection

The regulator is a synchronized switching regulator As a switching

regulator it converts 80 VDC to 40 VDC line variation to a regulated or

controlled nominal 35 VDC with an efficiency of approximately 941 When R4

remote feedback is connected to ground the filter regulated output power

is connected to the local feedback point 10 KHz signal (S6) is Applied to

the modulation signal input and the line power connected to the line power

input the line regulator regulates dc voltage out

The line regulator can also be utilized withremote feedback Such

is the case of the accelerator regulator The connection for this mode of

operation are as follows regulated 35 VDC (from 5 KHz line regulator) to

low voltagereference 10 KHz S6 signal to input modulation signal regulated

output to accelerator inverter input and line power to input power

Circuit Operation Description

Time Zero apply power and S6 10 KHz square wave to proper inputs

Time One current flows through R9 into the base of Q2 Q2 turns Q3

on Q3 turns Q4 on

Time Two the voltage across C4-(S7) rises to 35 VDC and is fed back

to the local feedback connection Note remote feedback is grounded

Time Three the feedback current flows through RI achieving proper

zener current through CRI and CR2 therefore supplying Al with a reference

voltage and power for operation S6 square wave is passed through a low pass

filter and converted into a 10 KHz triangle wave (S8) S7 and SS are summed

together through R5 and R14 to the summing input of Al (Als) Al is utilized

in a high gain switching amplifier configuration

Time Four the voltage at Al summing point equals the reference voltage

Time Five the output of Al switches to plus 18 VDC

Time Six Ql turns on and Q2 Q3 Q4 turn off

Time Seven the current continues to flow through LI therefore CR7

conducts and the voltage at the junction of L and CR7 is minus 7 VDC hence

voltage at the junction of RII and C3 is a minus 7 VDC

I

Time Zero One Two etc equals sequential timing not real time

14

Time Eight CR3 becomes forward biased and charges 03 to +35 VDC

through-Rll

Time Nine the voltage at the sumning point of Al decreases below the

reference voltage

Time Ten the output of Al returns to zero and Qi turns off

Time Eleven Q2 Q3 and Q4 conducts and the voltage at the juntion

point of C3 and RIl increases to1tine voltage plus 35 VDC

Time Twelve the voltage across C3 discharges through RII RIO Q2

Q3 and Q4 therefore creating an efficient drive The voltage at the junction

of R9 andRlO-is line voltage plus 16 V peak Q2 is in supe saturation

CR5 is back biased and Q3 is in hard saturation The voltage across Q4 is

Q3 hard saturation voltage plus Q4 base to emitter voltage

This completes one cycle of regulation

d Magnetic Modulation

Reference to System Block Diagram X318813 and Magnetic Modulator

Schematic Dwg No X3188123 will show that three supplies use magnetic

modulation for load regulation These are 1) Magnet Supply 2) Vaporizer

Supplyi and 3) Neutralizer Heater Supply (the Neutralizer Keeper Supply is

simply reactance current limited)

The magnet supply regulates dc load current while the vaporizer

and neutralizer heater regulate ac load current (RMS regulation) However

the mode of modulation is similar in all 3 systems differing only in current

sense circuits Therefore the following explanationof the magnet modulator

will substantially apply to the other supplies

The magnet supply is required to furnish a regulated dc current at

085 amps at a maximum voltage of 19 volts and is referenced to the screen

+ 2 KV supply

The circuit used here is that of a conventional doubler mag-amp

in series with the primary of the output transformer Current is sensed by

a current transformer in the primary circuit with a peak sensitive filter on

the rectified output of the current transformer

15

The load inductance supplieg the necessary filtering of the pulse-width

modulated output current at 10-KHz fundamental with rich harmonics Consequently

the output current is substantially constant while the average ac current

varies substantially with load impedance While the wave-form of primary

current is substantially exponential rather than square-wave it has been

determined experimentally that a peak current detector provide the necessary

accuracy in current regulation and is a substantially less complex hence

more reliable method than other techniques which might be used on the output

floating at 2 KV (Magnetic transductor Hall effect devices etc)

The current feedback signal is nulled with a reference signal at the

differential transistor The resulting collectorcurrent due to null unbalance

is used to control the mag-amp

The bias winding insures that the mag-amp is driven close to saturation

in absence of a null unbalance signal thus eliminating the control range

which might otherwise be lost due to residual reset

Telemetry is supplied through an isolation resistor permitting a

short in telemetry with only small effect on regulated output (approx 1)

7 volt zener acts asa clamp on possible transients

To minimize effect of temperature on regulation and telemetry the

current sense rectifier is supplied with 25 volts full scale thereby

minimizing effect of diode drop variation with temperature The 25 volts is then

scaled down to 5 volts for control and telemetry

This circuit wil hold substantially constant current into a shortshy

circuited load Fortunately the exponential character of the output wave-form

due to finite saturated inductance in mag-amp and leakage inductance in

transformer resdlts in a diminishing peak output voltage for small firing

angles and corresponding limited peak current into a short This is of

course desirable to-prevent excessive current in inverter

As noted above the modulation of the vaporizer and neutralizer heater

supplies is similar to that of the magnet however with differences in current

sensing

For the vaporizer and neutralizer heater supplies two (2) modes of

control are required 1) constant RMS current and 2) 100 control from on

to off when under control of engine loops

16

For RMS regulation an RMS aproximation circuit was chosen as more

reliable than thermal sensing - This consists essentially of a hybrid peak

average filter on the rectified output of the current transformer (An averaging

filter will indicate acurrent less than the RMS value a peak filter will

indicate a current greater than the RMS value) Since the line regulator qill

hold the input square-wave voltage constant to the magnetic modulator the

latter must only regulate for load changes including however a shorted

output

3 Cathode RMS- Current Regulator

See Drawing No X3188113 for schematic As indicated above in the discussion

of single inverter modulation the cathode supply uses a pulse-width modulated

driven inverter The output of this inverter therefore will with a line

voltage varying from 40 to 80 volts be required to vary from nearly full on

pulse-width to a narrow pulse width with the peak voltage varying 2 to 1

With the large step-down from line voltage to 5 volts at 40 amps to the

load ahd the need for a remote transformer at the thruster to avoid excessive

linedrops the transfoenar leakage inductance for the combined inverter autoshy

transformer and remote transformer results in a significantly inductive load

This -inductive load will therefore result in a current waveform which is

exponential with a time constant significant relative to the half-period It

is this waveform then which must be modulated to maintain constant RMS

current as the line voltage varies

The detection circuit chosen for EMS control is an approximation circuit

rather than atrue RIS thermal sensing circuit since it offers higher

reliability greater independenceof ambient temperature andfaster response

(the latter is an important consideration in engine loop stability)

Referring to schematic drawing no X3188113 the network which provides the

RMS approximation detector consists of the dual series R-C shunts across the

current transformer T6 in combination with hybrid peak-average rectifier-filter

CR19 CR20 RIO and C13

The output of the circuit shown closely tracks the desired function relating

RMS current value to peak current value as a function of pulse width Fig 3-1

is a plot of-the analytical value of the desired functions

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As the line voltage increases from 40 to 80 volts the inverter peak

output voltage and peak current increases proportionately for a given load

For a constant RMS value the pulse-width required will decrease and this

decrease will be forced by the feedback to the pulse-width modulator

As the pul-se-width narrows the degree of shunting contributed by the currentshy

transformer RC shunt loads will increase reducing the voltage across the

transformer secondary Thus the rate-of-change of detection output as a

function of pulse-width may be chosen to closely fit the desired function

Tests of this network at 25 amps and 40 amps load have verified a regulation

accuracy of 1 between 40 and 80 volt line Similarly short circuit current

is held to within 2 of regulated nominal

Current telemetry for this circuit is derived from the detector emittershy

follower in parallel with the current feedback Isolation resistors in telemetry

circuit permit shorting of telemetry without shifting controlled current more

than 1

Three modes of operation are required of this circuit I) regulated at

40 amps in a preheat condition 2) regulated at 25 amps in a preheatshy

standby condition and 3) controlled between 10 and 40 amps as a function of arc current ie when arc current exceeds a set-point value cathode

current will be cut back rapidly from 40 amps to 10 amps or less Hencej in

normal engine-loop operation the cathode current will stabilize at some value

between 40 amps and 10 amps such as to maintain the desired arc current

The control for the above indicated functions is derived from a function

generator located in the control module described elsewhere This control

is shown as an input signal command reference on schematic X3188113 The

circuit is mechanized to provide 40 amps with 0 volts command and cut-off at

+5 volts

The basic inverter circuit drawing X3188113 has been described under

paragraph 2 (b) above Single Inverter Modulation

4 DC Voltage Regulation

a Arc Supply

See Drawing X3188109 and Drawing X3188117 for schematics of this

circuit

This circuit uses the pulsezwidth modulated inverter discussed above

As in the cathode circuit discussed above the arc circuit uses an

operate and standby inverter with a aonmon output transformer

The arc supply must provide a starting boost of 150 volts at

no load dropping to 36 volts at 20 ma and regulated at 36 volts for line

and load up to 7 amps Also the supply is to be tripped off for overloads

in V4 (arc) V5 (screen) or V6 (accelerator) Telemetry is required for

both current and voltage with high calibration accuracy telemetry for current

between 2 and 7 amps and voltage between 34 and 36 volts

The supply is floating at the screen potential of +2 KV

in Drawing X3188117 the starting boost is provided through currentshy

limiting reactor Ll to rectifier CR4 When load increases to 20 ma voltage

from boost circuit will drop below 36 volts and load current will be drawn

from main rectifier CRI CR2 through filter L2 Cl

In Drawing X3188109 voltage telemetry and feedback are derived from

output transformer through CR19 CR20 and associated averaging filter since

average AC voltage is linear proportional to DC output voltage except for

load regulation in rectifier and filter The latter is compensated by load

current feedback sunned with voltage feedback Change in output due to

temperature change in rectifier and choke will not be compensated but should be

less than for 500C variation This technique is less complex than a

mag-amp voltage sensor or Hall effect voltage sensor and is expected to be

as accurate

In Drawing X3188117 current sensing is obtained with current transshy

former TI located in load secondary There will be a lack of linear cbrrelashy

tion between AC current and DC load current due to pulse-width modulation

Assuming perfect transformation and perfect square-waves a linear correlation

would exist between DC current and peak AC current For exponential wave

forms however this is not true A quasi-peak detector filter gives adequate

correlation

Two (2) rectifier-filter circuits are shown with T1 transformer These

are necessary to provide positive telemetry and negative current feedback

the latter necessary for compensation summing with positive voltage feedback

20

b Accelerator Supply

See Drawings X3188125 X3188107 and X3188115 for schematics of this

supply

This circuit will use a driven square-wave inverter

Regulation will be obtained by controlling the DC voltage input to

the inverter By this technique transformer will put out square waves

requiring only capacitor filtering for switching period and minimizing voltage

ringing which is high with a low-pass filter on output

The line regulator used here is identical with that described above

for the 5 KHz system with feedback from accelerator output instead of

from line regulator output

The accelerator supply uses one operating inverter and one standby

inverter with DC outputs from transformer rectifier in series thus providing

a redundant transformer rectifier

A winding on the output transformer of the operating inverter senses

failure of this inverter The control module on sensing failure will

switch the standby inverter on by applying a gating ON signal to the 946

gates in the inverter base circuit while removing the gating signal from

the 946 gate of the operating inverter

Referring to Drawing X3188115 output of accelerator inverter

rectifier is filtered with low-pass filter Ll Cl Choke Ll also provides

arc suppression Bleeder resister Rl R2 R3 provides voltage and current

sense for regulation telemetry and overload trip

c Screen System

See DrawingsX3188131 System Block Diagram X3188105Screen Inverter

Schematic andX3188115 High Voltage Filter Schematic

The screen system uses eight pulse-width modulated inverters phased

18 of a half-cycle apart with rectified outputs in series feeding a cormmon

L-C low-pass filter Frequency of inverters is 125 KHz resulting in a

ripple frequency of 200 KHz with a peak-to-peak unfiltered ripple of 500 volts

maximum at high line

21

With the relatively high frequency ripple the size and weight of the

output filter is small for the-high power out (2 kilowatts) A ferrite

cup-core choke (42 mm diameter) is sufficient weighing only 180 grams The

output capacitor is a small (005 mfd) mica type chosen for small sze and

tolerance of high frequency ripple

The output filter is important in its contribution to high-voltage

arc suppression Since the output capacitor may be relatively small for

ripple suppression the intensity of arc drawn from the capacitor is

significantly less than that from the larger capacitor which would be required

with a low-frequency ripple

The output filter choke is also sized to provide arc suppression by

limiting the-current which may be drawn from the inverters before overload

trip shuts the inverters down

The staggered phase technique has a further advantage relative to the

size of the output filter choke This is the relatively low peak ripple

unfiltered ie 250 volts Thus at no load or very light load where

choke is no longer critical permitting output voltage to peak the maximum

rise in output voltage due to peaking will be 250 volts during a step-load

transient before the closed-loop pulse modulation can reduce the pulse-width

and drop the voltage

Referring to Drawing X318811 high voltage filter schematic output

voltage of the screen system is sensed with a tapped bleeder resistor

providing both voltage feedback for regulation and votage telemetry at 5

volts full scale (telemetry is isolated in control module)

Current is sensed with a resistor in the return to provide an overcurrent

trip in the control module where isolation amplification and inversion is also

provided for telemetry

The technique of pulse-width modulation for regulating output voltage

has been described before under discussions of single inverter modulation

and staggered phase modulation

22

5 System Control Techniques

a General

See Drawing X3188121-l Control Module Block Diagram Control

system accepts both high level digital command pulses and a low level analog

command processes and stores the commands and controls all functions of the

power conditioner The six input commands are listed below

1 On I - turns on power to heaters to warm power conditioner

2 On LI - turns on Group I power supplies and starts the cathode

preheat sequence

3 On III - turns on Group II power supplies

4 Off I - turns off the vaporizer supply

5 Off II - turns off power conditioner heaters and all supplies

Resets all memory prior to application of power to input bus

6 IBref - Used as reference in controlling the vaporizer and as

an input to the function generator to produce a reference for controlling

the cathode

The command pulses are 20 ms or greater in duration and are zero

volts when not present transitioning to 20 to 31 VDC during the pulse period

The analog control voltage is a zero to +5 volt signal capable of driving a

10 k ohm load

The control module functions are itemized below

1 Command memory

2 Overload trip and recycling

A Slow turn on

B Magnet delay

3 Standby switching

A Arc

B Cathode

C Accelerator

23

4 Function generation

5 Control amplifiers

A Cathode

B Vaporizer

C Neutralizer heater

6 Preheat sequencing

b Command Controls

Command Interface Circuit (See Figure 5-1)

All relays are initialized by a pulse command at the Off II

input When a On I command is received Ki is set which closes the circ-

to a PC preheat switch The relay contacts are not heavy enough to carry

the heater current KI is reset by applying an Off II command An On I-I

command sets K2 and also pulses the reset on the one minute timer which

initializes the sequencer and the one minute timer The line regulator is

turned on which supplies power to the 5KHz inverter which in turn supplies

power to the Group I supplies The power for the timer and sequencer comes

from the inverter and is expected to be available before the end of the On II

command The On III command sets K3 which allows the Group II supplies to

come on and switches the cathode supply set point to 40A so that the cathode

amplifier can control over the 10 to 40 A range The Off I command resets

K4 which supplies a bias to the vaporizer controller reducing the vaporizer

current to near zero

One Minute Timer (See Figure 5-2)

The one minute timer is used to control the timing of the preheat

sequence and the sampling period of the overload trip counter The timer is

mechanized using a unijunction oscillator operating at 267 Hz which is

counted down using a monolithic 4 bit counter and shaped using a monostable

multivibrator The output is a short pulse once per minute

Preheat Sequencer (See Figure 5-3)

The preheat sequencer utilizes a monolithic four bit ripple counter

for timing the 13 minute preheat sequence Zl-A allows the one minute pulses

to clock Z3 until 13 pulses have occurred The flip-flop consisting of Z2-B

24

and ZI-C is initially set so that the output to the cathode controller is at

+5 volts (Z2-B at +5v) After 10 minutes the inputs to ZI-B are true and

the flip-flop changes state allowing the cathode current to go to 40A Three

minutes later the flip-flop is reset by Z2-A and the cathode current is reshy

turned to 25A When the input from K3 is grounded the output drops to zero

allowing the cathode current to rise to 40A

c Overload Trips

Overload Trip and Integration (See Figure 5-4)

The trip and integration circuit detects and counts overloads

provides a telemetry signal which represents the total number of overloads

at any moment and detects a solar panel undervoltage When an overload

occurs Z1 is triggered which clocks the overload counter and starts the

trip sequence through Z3-A As the total number of trips increase the output

of Z8 steps in a positive direction until either a once-per-minute reset pulse

occurs 12 overloads are counted or the solar panel bus drops below 36 volts

at which time the power conditioner is shutdown The ripple counter consisting

of Z4 through 7 will probably be replaced by an MC939 for convenience of

packaging and reliability

Trip Timing Circuit (See Figure 5-5)

The trip timing circuit times the off period of the Group II

spplies The circuit consists of two monostable multivibrators constructed

of discrete components due to the long time constants required The overload

trip and integration circuit triggers the first one-shot which gates off all

Group II supplies and the magnet At the end of the trip period the second

one-shot is triggered and the Group II supplies are turned back on The

magnet supply is turned on at the end of the second one-shot pulse which

lasts one second

Trip Shutdown Circuit (See Figure 5-6)

The trip shutdown circuit provides slow turn-on for the accelerashy

tor Arc and Screen supplies to prevent high voltage rate of change and proshy

vides buffering between the supplies and the control module When the trip

occurs Ql turns on turning Q2 on which turns Q3 and Q4 on clamping the

output to zero At the end of the trip) Cl charges up slowly causing the

output to rise slowly The output is used as the reference supply for the

Arc Screen and Accelerator supplies The vaporizer is shut off by applying

25

a turn-off bias to the input through CR9 Q5 keeps the magnet off by

pulling current through the turn-off winding of the mag-amp The input

fromrelay K3 is released from ground at an on III command allowing

the vaporizer accelerator arc and screen supplies to come on

d Analog Controls

Analog Control Amplifiers (See Figures 5-7 5-8 5-9)

The system consists of three major control loops each of which will

be discussed separately

Neutralizer Heater - Neutralizer Keeper

This control function requires the neutralizer heater current to be

reduced from the preheat level of 28 amps to zero when the neutralizer

keeper voltage drops two volts below a preselected reference The current

reduction is to begin from the point at which the NK voltage equals the

present reference Under present requirements this level is between 10 and

20 volts The NK supply rides on a bias of up to + 30 volts A differential

amplifier is used to sense the voltage difference The amplifier will be

scaled so that a NK voltage of 0 - 30 volts will be 0 to 5 volts at the

amplifier output

The controller requires a zero to 5 volt signal to control the outshy

put current from full on to off An intermediate amplifier is mechanized

which provides the function below

Enhc = 15 2Enhref - inks -Fi Ench 0

Enhc=O0 2Enhref - Enks

Where

Enhc is neutralizer heater control voltage

Enref is the reference voltage

Enks is the scaled neutralizer keeper voltage

53 is a bias used to allow Enhref to be a zero to 5 volt signal

The neutralizer heater control amplifier schematic includes a bias

supply consisting bf a temperature compensated zener diode This reference

is used in the other two amplifiers The diode at the output of the amplishy

fier is to keep the output from going negative

26

Vaporizer

The vaporizer is controlled by comparing a screen current reference

voltage to the voltage scaler representing screen current As the screen

current exceeds the preset reference level by 10 ma the vaporizer is proshy

portionally reduced from 12 amps to less than 05 amps Since the screen

current is scaled 5 volts per ampere and the reference is 0 to 5 volts for

05 to 1 ampere and the vaporizer control requires a zero to 5 volt signal

to reduce the output from 12 amps to near zero an amplifier is required

that mechanizes the following function

Evapc = -200 [Esr - Ess + 25] Evapc 0

Evapc = 0 jjsr - Ess + 25 1lt 0

Where Evapc is the vaporizer control voltage

Ear is the screen reference voltage

Ess is the screen current scales

The vaporizer control amplifier schematic shows the amplifier designed

to perform the required function To accomplish the required gain the input

resistor was of necessity less than 10ko To prevent loading of the IBref

source a voltage follower will be implemented

Cathode Heater

The cathode heater is controlled by the arc current If the arc

current exceeds a reference by 01 amperes the cathode current is cut back

proportionally from 40 to 10 amperes The reference is a 0 - 5 volt signal

from function generator The arc current is scaled at 5 volts for 8 amperes

The cathode controller requires a zero to 5 volt signal to reduce the outshy

put current from 40 to 10 amperes To accomplish the required control

function an amplifier is mechanized which provides the function given below

Eco - 80 Eas - 125 -5 Earef 1 Ecc - 0

E Earef - 125 -5 Ealtr 0Ecc + 0 Eas

8 lt Where

Ecc is the cathode controller control voltage

Eas is the scaled arc current

Earef is the reference voltage

Function Generator

The function generator converts the Beam reference to an Arc reference

following a curve provided by JPL The curve is approximated by three

27

straight lines The amplifier Fijure 5-10 consisting of Zl QI Q2 and

Q3 provides dc biasing and sufficient dynamic range to provide drive to

the breakpoint network consisting of R6 RIO R14 R20 CR1 and CR2

Amplifier Z2 corrects for amplifications factor

The breakpoint network utilizes temperature compensated diodes to

provide the required switching points The diode specification gives a

temperature coefficient at one current only At lower currents the coefficient

is higher but not significant enough to effect the required performance

The current levels through the network are selecped high enough to pass

the critical knee area quickly At the knee the zener contributions is not

significant and the knee softness causes rounding of the curve at the breakshy

point Freon freeze tests showed that no significant shift occurred in the

curve occurred for variations in temperature A curve showing the output vs

input voltage is attached The circled dots are the breakpoints and end

points required to match the actual curve

e Standby Controls - See Figure 5-11

Cathode Standby Control - See Figure 5-11

The cathode standby control circuit operates exactly as does the

Arc circuit of the Accelerator Arc standby control circuit explained below

The sense signal ts either zero volts for off or minus 5 volts for on sense

Accelerator amp Arc Standby Control - See Figure 5-12

The Accelerator and Arc standby control circuit delays the On III

command -approximately 01 seconds and turns on the standby module if the

primary module is inoperative Transistor 01 discharges and releases C2 at

the end of the On III command C2 charges to a voltage equal to VR2 and

the base to emitter drops of Q2 and Q3 at which time ZI-A output goes high

If the accelerator primary on sense is not present the Accelerator standby

drive is turned on and the Accelerator primary drive is turned off If the

Arc on sense is present when ZI-A goes high Z2-B is inhibited and Q5 cannot

turn on If however Zl-A goes high when the Arc sense is not present Z2-B

output goes low turning Q5 on and inhibiting the Z2-A gate

28

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6 Test Data

Presented herewith are test results taken on preliminary breadboards

of key circuits of the system together with data on critical components

Some subsystems such as Screen Supply require the assembly of a number

of modular circuits which have not yet been built hence cannot be

reported on at this time

a tow Voltage Group

Data is presented on the Vaporizer Neutralizer Heater

Neutralizer Keeper and Magnet Supplies showing 1) regulations vs

load (line regulation is performed by line regulator) 2) telemetry

output vs load current and 3) controlled current vs control voltage

b Modulated Power Inverter

Data-is presented on losses measured in principal elements

of a 300 watt screen inverter

c Cathode Heater Supply

Data is presented on load current regulation vs line voltage

variation for 3 values of load current

d Function Generator

Aiplot of measured output-is given for arc current reference

as a function-of commanded screen current reference

e Components

Data is given on switching speeds of power transistor used

in inverters also on PIV of screen high voltage recterfier bridge

f Analog Control Circuits

Data presented shows the roll-off characteristics of the

operational amplifiers for

1) Cathode Current vs Arc Control

2) Vaporizer Current vs Screens Control

3) Neutralizer Heater Current vs Neutralizer Keeper Control

41

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Losses in 300 Watt Inverter

Screen drive inverter circuit with 16 MR inductor and a 005pf

5000 VDG capacitor in output filter

a Power Loss in Power Transistor

Line 80 VC

Pulse width 25ps

Power output 300 W

Room temperature 820F

Power transistor beat sink temperature 1220F

AT 400F

610F rise = 1 watt

40OF 2 61 = 645 watts loss in power transistor

Drive transformer voltage 35

Drive transformer average current 70 ma

Power for drive circuit - 70 ma x 35 v = 246 w

Base drive current 125 A pulse width 24 is

VBE IV

Power VBE = 251s40ps x 125A x 1lV = 086W

Corrected power transistor power loss equals 645W - 086W 559W

b Line Voltage 40VDC

Pulse width 36ps

Power output 300 W


Heat Sink temperature 136F

T - 51OF

Power in loss in power transistor = 511 4 6-1 835 W

Power in drive = 35 V x 115 ma = 4 W

Power loss in base = 125A x I1V x 3540 P 120 W

Corrected power loss in power transistors = 835 W - 120 W = 715 W

c Summary of Transistor and Drive Loss

High tine Transistor loss in power 559 W 80 VDC Drive Power 246 W

Total 805 Watts loss

Low Line Transistor Power Loss 715 W 40 VDC Drive Power 400 W


d Output Transformer Loss

At 80V line 50 percent duty cycle output 300 V DC at 3A measured

loss (calorimeter) = 385 Watts

54

Cathode Heater Regulation

(RMS Regulator)

Line Voltage

40

45

50

55

60

65

70

75

80

lOA Load

100

100

101

100

100

100

100

100

100

25A 40A

250 401

251 402

251 402

251 403

252 404

251 404

251 405

251 404

251 404

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Power Transistor Switching Speeds

Solitron SDT 325VI01 Transistors Salitron SDT 8805 Transistors

Collector Current = 7A Collector Current = 7A

Serial Serial No On Off No On Off

31 8 7 x 10-6sec 1 2 12 x 10- 6sec

32 1 10 2 18 10

33 1 10 3 22 15

34 9 9 4 20 15

35 8 8 5 25 10

36 9 7 6 25 10

37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

40 11 8 10 25 10

41 11 8 11 30 14

42 11 8 12 20 10

43 10 9 13 25 15

44 10 8 14 17 10

45 10 8 15 20 10

46 11 9 16 25 15

47 10 29 17 22 12

48 10 7 18 25 10

49 11 9 19 22 10

2-1 10 8 20 30 15

2-2 12 8 21 30 15

2-3 11 8 22 20 15

2-4 11 9 23 15 15

24 10 10

25 25 10

26 20 08

27 20 10

28 15 08

29 15 06

30 15 06

57

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P19 l0p A RY RY BY BY

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3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

34 1160 1040 1160 1040

35 1000 960 1080 1200

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7 EFPICIENCY ANALYSIS

a Screce InverterNodule

Worst Case 300 watts out (2100 watts seven inverters)( qorst case = 7 inverters on normal = eight inverters on)

At 40-Volt line

1) Transistor saturated loss Assume 93 percent effici6ncy 95 percent duty cycle

I 300 = 87 amperes peak = 87 x 095- 83 ampere average

093 x 39 x 095

Psat IcVce sat 83 x 07 - 58 watts (two transistors)

2) Transistor switching loss Assume partially inductive load Assume is = 05 microsecond f = 125 KH3

P 29 E I t f per transistor sw mces p

= 49 E I t f for two transistors

-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419

= 190 watts (two trensistors)

3) Psat + P = 58 + 19 = 77 watts (excluding base loss)

4) Base drive loss (including modulator) (circuit P = 9)

-PBD 20 volts x 10 = 200 watts + 10 watt in modulator (estimated)

30 watts

5) Line capacitor loss (foil tantalum)

P C 041 watts ceest

6) Transformer loss (from design data)

PT = 5 watts (20 w core loss 30 w copper loss)

7) Output rectifier loss (manufacturers date)

4 diodes OSV dropdiode lOA = 32 watts

GE Bulletin 121EC for two 175 microfarad200 volts page 9

6

8) Total inverter loss at 40-volt line

Transistors 77 (65 w measured)

Bdse drive 30 (33 w measured)

Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193

L0) Efficiency = 300 = 940 percent 3193

At 8O-VQlt Line

1) Transistor saturated loss (transistor on half time)

Psat = 83 x 07 = 279 watts (two transistors)sat 2

2) Transistor switching loss

P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw

= 380 watts (two transistors)

3) Psat + Psw 290 + 380 = 687 watts

4) rbase drive = 300 watts (on half time) - 150 WaLL 2

5) Line cap PC 093 watts

6) Transformer loss Pcore loss= 200 watts (from design data)

p 30 x 05 = 15 W cop loss

- (same current as 40-volt line on half time PW)

GE Bulletin 121EC page 9

PT = 20 - 15 =-35 watts

7) Output rectifier loss PR = 32watts = 16 watts (on 12 time-same current as volume) 2



Base drive 150

Line capacitor 093

Transformer 350 (385 w measured)

Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423

10) Efficiency = 300 = 952 percent 31423

Nprmal Case 8 inverters on 2100 w =262 watts out 8

At40-VolE Line (Duty Cycle = 95 x 78 - 083)

1) Transistor saturated loss

I = 262 = 87 amps peak093 x 30 x 083 - = 87 x 083 x 72 ampsaver

Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103

= 190 watts (2 transistors)

P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64

5) Line Capacitor Loss = 041 x (76)2 = 031 W(971

6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w

Copper Loss cy Duty Cycle

P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w

7) Output Rectifier Loss

a Duty Cycle

Pr = 32 x 083 = 28 w 095

8) Total Inverter Loss at 40V line

Transistors 694 BAse Drive 262 Line Capacitor 031 Transformer 396 Rectifiers 280

1663

9) Efficiency = 250 94 2666

At 80-Volt Line (Duty Cycle - 083 = 042) 2


Ic av = 72 = 36 2

Psat = 36 x 07 =252 w


Ic peak same as 40 V line voltage = 2x

Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss

Pcore = 134 w (same as 40V line)

Pcopper = 262 x 12 = 131 w (same current 12 time)

PT = 134 + 131 = 265 w

7) Output rectifier loss

Pr = 28 x 12 = 14 w (same drop 12 time)

8) Total inverter loss at 80V line

Transistors 632 Base drive 131 Line capacity 071 Transformer 265 Rectifier 140

1239 w

9) Efficiency = 250 = 951

2624

b Arc Inverter Module

1 operate 1 standby Po = 36V x 7A = 252W

Scaling losses from screen inverter

At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382

5) Total Inverter Losses 1332 w

6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62

5) Total Inverter Losses =1060 W

6) Efficiency = 252 =966 2606

c Accelerator Inverter Nodule

1 operate 1 standby inverter POT = 200 W transient

(2000 V 01A transient)

(2000 V 001A steady state) Pos s = 20 W steady state

This inverter operates at duty cycle of 100 with a regulated input line of 35 volts

Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss

=Psat 632 x 07 = 44 W -Psw = 49 x 68 x 632 x 05 x 10 6x 125 x 103 r 12 W

PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W

3) Line capac loss = negligible (Ion ripple)

4) Transformer loss = 200 x 336 = 302 W -R2

5) Rectifiers (10 diodes in series)

PR = 01A x 08V x ID = 08 W

6) total Inverter Loss (transient)

Transistors 56 Base Drive 16 Transformer 30 Rectifiers 08

Total 110 W

B - Steady State

Ic 632 x 110 = 062 A peak = average

1) Transistor Loss = 56 x 110 = 056

2y BaseDrive Loss

3) Transformer

Pcore x 12 total of transient loss = 302 = 15 2 -2

P = 161 x i = negligiblecopper 10

T = 151

4) Rectifiers Loss = 08 x 110 = 08 W

5) Total inverter loss

Transistors 056 Base Drive 160 Transformer 151

Rectifiers 08 Total 375 W

d Cathode Inverter Module

= operate 1 standby inverter PQ 5V x 40A 200 W

40V Line (95 duty cycle)

I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss

4WPBO = 2V x 07A = shy

68 6657-CC-E REV 3-61

3) LineCapac Loss = 025W (from G- E Bulletin 121 EC)

4) Transformer Loss = 200 W x 02 4 watts (98 effic)

5) TotalInverter Loss

Transistors 425 Base Drive 140 Line Capacitor 025 Transformer 400

Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W

3) Line Capac Loss e 080 W (G E Bulletin 121 EC)

4) Transformer Loss

Since this is currentregulator same current as at 40V For same RMS voltage average voltage will be lower but assume same

same core loss

PT = 400W (same as 40V)

5) Total Inverter Loss

Transistors 287 Base Drive 070 Line Capacity 080 Transformer 400 Total 637 W

69

447-CC-EN PEV -61

e 5 KHz Inverter Module

I Operate 1 Standby Inverter

Supplies Magnet Vaporizer Neutralizer Heater Neutralizer Keeper

+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W

This Inverter is supplied from Line Regulator at 35V out (constant)

1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W

3) Line Capac Loss = negligible (low ripple)

4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w

f 5 K-z Line Regulator

This Regulator supplies power at 35V to 5 KHz Inverter and to all Pm

Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W

70 pound87-CC--EN REV 3-81

At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A

1) Power Transistor Loss

Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W

2) Choke Loss (Design Loss) = 02 x 135 = 27 W

3) Output Capac Loss (est) = 05 W

4) Line Capac Loss (est) = 05 W

5) Power Transistor Drive Loss (est) = 10 W

6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line

1) Power Transistor Loss (On 12 Time)

Psat = 256 = 128 - 2

Psw = 063x 2 = 126 W (same current)

Pt = 128 + 126 = 254 w (2 x voltage)

2) Choke Loss (same current) = 27 W

3) Output Capac Loss est = 10 W

4) Line Capac Loss est = 10 W

5) Drive Loss est = 10 W

6) Pt = 830 W (measured loss = 82 W)

71

g Accelerator Line Reulator (35 V out)

40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W

2) Choke Lobs (Design Loss) 02 x 200 = 40 W


4) Line CapacLoss (est) = 07 W

5) Drive Loss (est) = 15 W

6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W

1) Transistor Loss-= 307 x 01 = 051 W

2) Choke Loss = 40 x 01 = 004

3) Output Capac Loss (est) = 00

4) Line Capac Loss (est) = 00

5) Drive Loss (est) = 15

6) Total Loss = 205 W

7-CC-W RrY 3-61 72 85

--

80V Line

A Transient B Steady State

1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041

2) Choke Loss (design) = 40 W =- 004

3) Output Capac Loss (est) = 15 W = 000

4) Line Capac Loss (est) = 15 W = 000

5) Drive Loss (est) = 15 W = 150

6) Total Loss =125 W = 195 W

h Arc Rectifier Fitter

40V80V Line same since current remains same except divides between Main Rectifiers and Commutating Rectifier

1) Rectifier Loss = 08 x 70 = 56 w (3)

2) Choke Loss (design) = 252 x 01 = 25 W


4) Total Loss =101 W

i HV Filter

Same at 40V and 80V Line

1) Screen Bleeder = 2000V x 001 A = 2 W

2) Aceel Bleeder = 2000V x 001 A = 2 W

3) Screen Current Sens-e Resistor = I W x 1 A I W

4) Accel Current Sense Resistor = 5V x 01A = 05 W

5) Screen Choke (design) = 20 W

6) Accel Choke (design) 05 W

7) TOTAL (73) = 80

j Maanetic Modulator

At 40-volt line amp SOV line (regulated by 5 KHz Line Regulator)

1) Vaporizer

=Pout 20 Watts

Ptransformer = 02 x 20 = 040

P 02 x 20 = 040 mag-amp

Pdiodes 20 x 10 =025 O

2) Neutralizer Heater

Pout 41 Watts

PNH = 41 x 105 c-21 Watts

) Neutralizer Keeper

VAou t 180

a) Ptransformer = 18098 = 184 = 04 Watt

b) Pchoke - 1899 - 18 -- 02 watt

c) Prectifiers 05 ampere x 16 volts = 08 Watt

d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet

=Pout 20 Watts Iout = 085 amperes

a) Prectifier = 085 x 10 = 085 Watts

b) Ptransformer = 20098 - 20 = 040 Watt

P) P 040 Wattmag-amp

d) Ptotal i165 Watts

5) Ptotal at 40-volt line magnetic regulators

PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses

Module 40V Line 80V Line

1 Screen inverter - Worst Case - 71NV x (193W) (1423W)

Normal Case- 81NV x 166W 1239W

133W 1060W2 Arc Inverter shy

3 Accel Inverter - Transient 110W 110W

Normal 375W 375W

4 Cathode Inverter 990W 637W

567W 567W5 6K 3 Inverter

6 5KH3 Line Regulator 79W 83W

7 Accel Line Regulator - Transient (1197W) (125W)

Normal 205W 195W

101W 101W8 Arc Rectifier - Filter

9 High-Voltage Filter 80W 80W

10 Magnetic Modulator 48W 48W

50W 50WI1 Control Module

75

8 RELIABILITY

8a Reliability Analysis

W Mission Reliability calculated to 09702 where R(t)= r R and

mission time is 10000 hours The reliability block diagram (Fig 8l)

displays the complete system with all involved subsystems or units The

reliab4lity for each of the units was determined by the relationship

-4t R = e except for those units where redundancy practices were followed

(for which math models are noted) The set of failure rates utilized

for the analysis effort were developed by utilization of Hughes Document

R22-100DC After determination the subsystem failure rates were then

modified by use of an E-factor the Hughes Experience factor The set

ofutilized failure rates and a rationale justifying use of these

failure rates together with the E-factor follow The resultant failure

rate (k) of-each unit modified by the E-factor (0475) is rioted on the

applicable Parts List (see Section 8a4)

See Table 81 for listing of unit failure rates and reliability values

8al) The-Screen Inverters

For the purpose of this analysis the missien time of (10) 4

hours was broken into three intervals

Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000

At the start of flight seven (of the eight active) Inverters are

required At the end of each interval the requirement drops as

follows

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Interval Inverters Required at

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1 7

2 6

3 5

The reliability Probability of meeting these requirements for

each interval was determined from the formula

d = R i )n-i

edund = z) (1 - R1 2) i = c 12

where n = operating inverters

c = inverters required at end of each interval

R e-lt X = failure rate

t = interval time

The redundant screen inverter reliability calculated to 09998

8a2) Transformer Short-to-Ground Failure Mode

Since the entire power conditioner will be inoperative

if one of the output transformers of the Accelerator and The

Screen Inverters fail in failure mode short to ground and

since each transformer is exposed to seven failure modes a

conservative failure rate of 5 x 10 (onefifth of the transshy

-former failure rate of 20 x 10 9 ) has been placed in series

with the other elements for each of those transformers (see

block 14 on Figure 81)

Where those transformers are part of the redundant

or standby circuitry the failure rate has been accordingly

-modified from 20x 10 9 Hrs to 15 x 10- 9 Hrs to account

for the balance of failure modes inside the redundant portion

of the circuit

78

8a3) Assumptions and Considerations

1 All parts exhibit constant failure rate

-2 Parts within a block on the logic diagram fail

independently Failure of any part in a block (except

for several redundant Operational Amplifiers in the

Control Module) causes block failure

3 All blocks on the logic diagram fail independently

4 Mission time is 10000 hours

5 Power Conditioner temperatures are 250C (base plate) and

350C (junction) except for the Control Module where the

temperatures are 200C and 250C respectively

6 Discrete semiconductors operating in digital circuits

are derated to 10-15 of their rated power stress All

other parts (in general) are derated to 30-40of their

rated stress

7 Part failure rates reflect Hughes spacecraft experience

through 31 August 1968

8- The reliability analysis predictions do not consider

wearout failure

9 The reliability mathematical models for standby

redundancy consider that the dormant failure rates are

greater than zero

79

TABLE 81

FAILURE RATES FOR POWER CONDITIONER

PART-TYPE

Diode Switching

General Purpose

Zener

Transistor Switching

General Purpose

Power

Integrated Circuits DTL

NA741

Resistor Carbon Comp (RC)

Metal Film

Power (RW)

Precision

Capacitor Ceramic etc

Tantalum

Magnetics Transformer Choke

Mag Amplifier

Relay

Connector

Fuse

-FAILURE RATE x 10 9 HRS

(932 946 948 etc)

Active Dormant 24 10

60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119

(Issue with approval4 pAWT COuVTS date of 9-30-68)

LINE REGULATOR - Block 1

- 9Part Type Qty X(10) Total x

Resistor Carbon Comp 10- 10 100

Metal Film 2 22 44

Power W W 1 65 65

Capacitor GlassMylar etc 5 60 300

IF Tantalum 2 260 520

Diode Switching 7 24 168

Zener 2 240 480

Transistor Power 4 1200 48000

Operational Amplifier 840

Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81

Ref Dwg No X3188111 (Issue with approval date of 9-30-68)

5K HZ POWER INVERTER - Block 2

- 9 Total x (10)- 9 Quantity (10)

Part Type Active Standby Active Dormant x I X 3

Resistor Carbon Comp 6 5 10 010 60 050 50

Power W W 1 1 65 001 65 001 65

Capacitor Glass Mylar etc 1 1 60 002 60 002 60

Tantalum 6 6 260 100 1560 600 15600

Diode General Purpose 9 8 24 10 216 900 192

Transistor Power 5 4 12000 100 6000 4000 48000

Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000

Integrated Circuit 3 2 140 50 420 1000 280

Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit

X2 = Standby Unit Off

X3 = Standby Unit On

82


MAGNETIC MODULATOR - Block 3

- 9- 9 Part Type Quantity (10) Total (10)

Operational Amplifier 1 840 840

Magnetic Amplifier 3 300 900


Zener 10 240 2400

Transistor Switching 3 75 225

Capacitor Glass Mylar etc 3 60 180

Tantalum 6 260 1560

Resistor Carbon Comp 11 10 110

Metal Film 25 22 550

Power W W 4 65 260

Transforme 8 200 16D0

Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83

CATHODE PULSE MOD amp FREQ DIV - Block A4

Ref Dwg X3188113 (Issue with approval date of 93068)

Part Type Quaptity -9

X(10) -9

Total X (10)

Integrated Circuit 1 140 140


Diode Genl Purpose 2 60 120

Resistor Carbon Comp

Metal Film

4

5

10

22

40

110

Capacitor Glass Mylar etc

it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84

Ref Dwg Numbers (Arc) X3188109 and (Cathode) X3188113

(Issue with approval CATHODE INVERTER AND ARC INVERTER - Block 5 date of 93068)

Part Type Quantity x (10) - Total X (10) - 9

Active Standby Active Dormant 1 22 23


Transistor Switching 2 2 75 100 150 200 150


Diode General Purpose 18 18 60 100 108-0 1800 1080

Capacitor Glass Mylar etc 2 1 60 002 120 002 60 Tantalum 3 3 260 100 780 300 780


Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85


CATHODE HEATER LOOP - Block 6

-9 -9 Part Type Quantity X(I0) Total x(10)

Resistor Carbon CQmp 3 10 30

I Power W W 1 65 65

Diode Qenl Purpose 4 60 240


Capacitor Tantalum 2 260 520

Transformer 2 200 400

- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86

Ref Dwg X318810 (Issue with approval date of 9-30-68)

ARC PULSE MOD amp FREQ DIV - Block 7

Part Type Quantity A(10)- 9 Total (10)-9





Metal Film 4 22 88


I Tantalum 2 260 520

= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87

Ref dwg nos are X3188117 amp X3188109

(Both issued with approshyval dates of 93068)

ARC RECT FILTER - Block 8

Part Type Quantity ()0)-9 Total x(10) - 9


Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400

Capacitor Ceramic Glass etc 4 60 240

Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88

Ref Dwg X3188 107 (Issue with approval date of 93068)

ACCEL FREQ DN - Block 9

Part Type Quantity X(10) shy 9 Total X (10)-9



ResistorCarbon Comp- 1 10 10

X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89

Ref Dwg No X3188107 (Issue with approval date of 93068)

ACCEL INVERTER - Block 10

- 9- 9lO) Total x(10)Quantity Part Type

Active Dormant Active Dormant Active Dormant

Integrated Circuit 2 2 140 500 280 1000

Transistor Switching 2 2 75 100 150 200

Power 2 12002 1000 2400 2000

Diode Genl Purpose 8 7 60 100 480 700

Capacitor Glass Mylar etc 5 5 60 002 300 010

Tantalum 4 3 260 100 1040 300

Resistor Carbon Comp 4 4 10 010 40 040

Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100

Diode Assembly (10 CRs) 1 1 240 1000 240 1000

Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200

Connector I - 250 - 250 shy

5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy

Normally 200 but one of the failure modes is allocated to Block 15

90

i


HIGH VOLTAGE FILTER - Block 11

-Pait Type Quantity X(l0) Total X (10)


Power W W 2 65 130

Resistor Assy (10 Carbon 4 100 400 Comp PCs)

Capacitor Ceramic Mylar etc 4 60 240

Tantalum 1 260 260

Diode Zener 3 240 720

Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91


SCREEN INVERTER - Block 12

- 9Part Type Quantity x (10) Total X(I0)-9

Integrated Circdit 2 140 280


Power

2

2

75

1200

150

2400


Diode Assembly (2-CR Pkg) 1 48 48

Capacifor Class Mylar etc

Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92

Ref Drawing (See below)

PRELIMINARY CONTROL MODULES - Block 13

Utilized data from earlier design review

(See 2228DR 1701 Rev A pages 60 seq which lists)

= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93

SPECIAL TRANSFORNER SEORT-TO-GROUND FAILURE MODE - Block 14

If any of the Accel or Screen Inv output transformers (a total of five) fail the entire system fails We assign a failure rate to this failure mode of a highly conservative nature 5(10) - 9 based on two factors

I) The operating voltage of the involved transformers is 2 Kv Several of the transformers have been tested in a destructive hi-pot test and insulation breakdown occurred at 15 to 17 Kv the large safety factor in the amount is indicative of HV insulation supplied

2) The total failure rate for-transformers is 20(10) 9 which rate covers all the failure modes These involve short or open in each of the -hree windings insulation breakdown arcing and so on

-Thus is a product of (5 transformers)(5)(10) 9 25(10)

Block 14 X (475((250) = 19(10)-9

94

8b Rationale for Use of the Selected Failure Rates

The component part failure rates utilized for the program are

based on failure rates taken from the high reliability Minuteman

ballistic missile program somewhat modified by Hughes using engineering

judgement These failure rates (detailed in Hughes Document R22-100DC)

were originally set up for the Hughes Space Satellite Systems They

basic rates modified as necessary toare utilized for this program as

allow for the estimated environmental stresses (temperature electrical

etc) that the system is expected to encounter in flight These basic

failure rates are then adjusted as a set by the Hughes Experience Factor

(E-factor) described below

8bl) E-Factor Usage

It is Hughes practice to modify the basic set of failure

rates by a prediction factor

E = k = observed failures

7t expected failures11

based on flight experience of Hughes Satellite Systems (Syncom II

Syncom III Early Bird Intelsat II-Fl F2 F3 and F4 ATS-l 2 and 3)

Multiplication of the basic set of failure rates by the E-factor in

effect will correct any optimistic or pessimistic bias in the basic

rates This utilization of the E-factor when significant differences

between the flight experience rates and the basic rates are observed

will cause the basic failure rates to converge to true values The

technique is asympototically unbiased regardless of the validity of the

originally assumed basic failure rates The data from the HAC Space

Satellites are considered applicable to the subject program as the

mission environments are similar and long time service and essentially

free-fall operation are also involved here The Hughes Satellites

95

8b1) (Contd)

to date have accumulated over 650 million operational component

hours and 6ver 400 million component hours in the dormant stage

The current value (as of 31 August 1968) of the E-factor is 0475

8b2) Internal Temperatures

Based on design considerations and calculations which

were confirmed by experimental-tests in a vacuum chamber the

componentt mounting base plate temperature will be at +250C +

50 and-the junction temperature of semiconductors at plusmn350C + 50C

except for the Control Module where the temperatures will be +20 0C

+ 250C respectively The utilized failure rates were subjected to

(and modified by) derating curves in the aforementioned Hughes Document

R22-100DQ In order to assure stability of these temperatures the

system is designed so that the solar panels will always face the sun

while the power unit is always facing away from the sun

8b3) External Environment

The space environment to be encountered by the system

in flight is considered to be gentler than that experienced by the

presently operating Hughes Space Synchronous Satellites The only

large consideration is exposure to radiation from the Van Allen belt

The effect is judged to be very minor for the subject system whereas

the Hughes Synchronous Satellites see the radiation for several days

while in transfer orbits Even though this system could encounter solar

flares in flight these are not considered as potentially harmful as the

Van Allen belt radiation

8b4) Part Selection

The component parts used in the subject system are highshy

reliability parts selected from a JPL Preferred Parts List The specifishy

cations for these parts have extensive and stringent reliability-proof

test requirements which appear satisfactory to assure procurement of

high reliability parts for the program

96

The reliability proof tests include among others

a) qualification tests such as - soaking at temperature extremes

humidity shock hermetic seal plus-a high temperature test to

demonstrate with 60 percent assurance that a tested lots

failure rate is less than I percent per 1000 hours and b)

quality assurance tests such as - life stability environmental

plus final visual and electrical tests

8b5y Conclusions

The computed reliability of 09702 for 10000 hours is

somewhat higher than the design objective of 096 and somewhat

lower than the proposal estimate of 09826

The drop from the proposal estimate is substantially due

to increased complexity resulting from mugh tighter regulation

requirementsintroduced by JPL during the negotiation phase for

Screen and Accelerator supplies (no line regulation originally

required and5 to 10 percent load regulation required respectively

now + 10 percent for line and load) A small drop in reliability

is due to somewhat higher temperature stresses than originally

assumed in analysis for purposes of simplified analysis

The present calculation should not be considered as

representing the best effort at optimizing reliability since it

represents an interim cut at the total system Now that a

detailed analysis has been made it will be apparent which areas

deserve the most attention in reducing failure rate and a tradeshy

off study may be made on cost in efficiency andor weight

Possibly in some areas it may be desirable to substitute lower

failure rate components ie a digital gate plus discrete

transistor for an integrated circuit op-amp Also closer

97

examination of stresses should result in lower failure rates

since in many cases components are at substantially lower stress than

than those assumed in this conservative calculation

98

9 Physical Design

a e

l-Cirawing X318810-500 Tnstallation Control for oucline and mounting

s of the assembly As indicated maximum dimensions are 36 x 30 x 378

Synte is designed for mounting from four (4) edges of the assembly with one

prt point in cunter of rear face Cover is removable from the-back without disshy

ni the acmbly making accessible components and circuit connections which are

wtd up Qu rtar face of individual module plates

Jic systt consists of an assembly of twenty (20) modules mounted on an egg-

Cis structure Individual modules are removable from the front after disconnecting

fN 1ar ss connectors on rear face

ront face of assembly i electrically dead wth protruding power transistor

s aad nust cqvercd by epoxy conformal coating All other components and connections

Iubehfld module plates Thus the assembly with its cover is totally enclosed to

provide MI shielding and freedom from high voltage hazard during testing

M|hcclear front face facing space in a vehicle provides sMielding from incident

iadiation Since in a typical vehicle ipstallation the rear of the assembly would be

Mhce from radiation by the vehicle structure and other poter conditioning assewsli es

tie back cover lay be perforated to reduce weight and permit free outgassirg thereby

pre-ccting pressure build-up and transition to regions of voltage breakdown (pasehens

curve) (Although all materils are selected for low outgassing)

All external connections are made at connectors located at one end of assenbly

with all low voltage connections in one corner made with space approved sub--minaturn

rectangular connectors All high voltage connections are made at the other corner to

high voltage standoffs with screw terminals These connections are made behind a

reijovable small connection cover

The various modules used in the system are sized in frontal area for a worstshy

case thermal radiation of 03 watt per square inch all from che front face assuming

no radiation from rear face a conservative assumption since in a typical vehicle the

vehicle structure to the rear is expected to be at a lower temperature than the Power

Conditioner with sowe radiation from the rear At this level of radiation with no

solar incidence a worst-case plate temperature of 350 C may be expected providing

high reliability operation of components which with few exceptions are mnunted on the

module plates

With the design philosophy of icw Wattssquaru inch each module is therurlly

self-sufficient and will not require conduction to structuun and sharing of radiation

area between modules Thus hot spots are mipimized

99

Since high reliability is obtained by extensive use of redundant or standby

circuits the thermal design philosophy has resulted in packaging of redundaft cirshy

cuits on same module plate with operating circuits intermixing the components on

the plate to obtain uniform thermal density regardless of whidh circuit is operating

b SUPPORT STIUCTURE

Refer to HAG Drawing X3188101 Frame Module

This supporting frame forms an eggcrate structure with a structural web or

I beam running between all modules and a channel section forming the edges Roles

located in I beam webs reduce weight without significantly reducing the section

modulus In order to provide a flat surface for module mounting the structure is

formed by assembling front and back plates with module cut-outs with webs intershy

laced in eggcrate fashiof the whole being dip-brazed Material is 0040

aluminum 6061T4 chosen rather than lighter magnesium to permit dip-brazing of

thin sections A riveted magnesium structure of the same strength would be

heavier

A stress analysis has been made of the structure (See Appendix) using conshy

servative simplifying assumptions of no contribution to strength from attached

module plates and no reduction of stress due to interlacing of beams This

analysis shows that structure is more than adequate to survive vibration at launch

phase

Module mounting screws fit into steel Pem nuts riveted to frame providing

a lightweight fastening which will permit frequent replacing of screws without

damage to threads The many screws provided will result in significant damping of

the structure in vibration as well as permitting ready replacement of modules

during test or service

Riveted Pem nuts are also used along edges of structure for niounting

c NODULE ASSE1l-LIES

1 Screen Inverter (See HAG Drawing X3188104)

As indicated under General discussion module is sized for a thermal

radiation of mwatt per square inch Since this circuit will dissipate 218 watts

for 300 watts out in-worst case of one module failed out of eight or 190 watts

for normal operation an area of 705 in2 will result in 027 wattsin2 or 250 C

normally or 031 wattsin2 maximum With an operating plate temperature (maximum)

of 350 C

Principle sources of dissipation have been separated in the assembly to

minimize local hot spots ie power transistors output transformer and output

rectifiers

6S57-CC-EN RFV 3-5I 100

Large area available pormits mounting all components on base plate which is

0050 sheet magnesium for low weight and maximum thermal conductivity per weight

To obtain good thermal conductivity yet conservative insulation of

components the area used by small components is covered with a 1 mil sheet of

Kapton (H-Film) bonded to plate with epoxy adhesive Thus the component may be

placed against plate and when conformally coated with epoxy will be thermally

intimate with plate solidly anchored against vibration and sealed against moisture

with a minimum of weight

To provide good thermal interface and security from vibration transformers

and output rectifiers are bonded with a thin layer of RTV in addition to mounting

screws RTV is sealed against out-gassing by epoxy conformal coating Use of RTV

for bond will penit ready replacement of component if necessary during tests

The method shown for mounting power transistors is subject to tests in

vacuum to confirm this technique Previous practice of using beryllium-oxide washer

and indium foil is subject to question due to possible cold-flow of indium and

resultant opening up of interface The relatively low wattage dissipated per transistor

(approximately 3 watts) permits use of Kapton with its lower thermal conductivity

than Be-O (even with I mil vs 0060 Be-O) Similarly it is believed that a thin

layer of RTV in interfaces may be superior long-term to indium foil

Low-voltage connections are made through a sub-miniature rectangular conshy

nector shown at bottom edge High-voltage connections are made at top to two

stand-off screw terminals

2 H ig-Voltage Filter (See Drawing No X3188114)

This assembly consists of the Screen and Accelerator Output Filters and

and bleeder resistors with voltage and current sensing networks -Since Screen

output is at -2000 V special attention must be given to voltage isolation

The bleeder resistors RI R2 R4 and R5 are potted in epoxy in a plastic

case The cases therefore provide a convenient mounting for the mica capacitors

which are shunted across the bleederOases provides a positive thermal path for

losses in the capacitors which are flat with a good interface and also provides a

secure mounting for vibration

Low voltage components are mounted in the fashion described for the Screen

Inverter

High voltage input and output connections are made at stand-off screw terminals

with low voltage connections through a rectatgular sub-miniature connector

Magnetic components and bleeder resistor assemblies are bonded to chassis for

goud thermal interface and resistance to vibration

d5-7 CC nv 3-5 101

Entire assembly will be conforirally coated with low-outgassing epoxy

Loss on this assembly will be approximately 5 watts Thus with area of2 2a

386 in2 and 0135wattsin temperature will be between 0 C for a free body and

50 C of adjacent modules

3 5 IQlz Heater Inverter (See 11AC Drawing NoX3188110)

This assembly is typical of design of inverters with standby cirshy

cuits where the operating and Standby inverters are packaged on the same

plate making the thermal radiation area available equally to either circuit

In this case a common output transformer is switched between inverter circuits

In normal operation transistors Q6 and Q7 will be on In standby transistors Q12 and

Q13 will be on It is thus apparent that a symmetrical thermal distribution is

presentin eiLher case With a dissipation of 9 watts expected and an area of 5 2 2 0

475 in or 019 wattsin a plate temperature of 5 C would result for a free

body or close to average system temperature of -_) C when mounted Mounting of

small components pox er transistors and mignetic components is similar to that

described for Screen Inverter

4 Accelerator Inverter (See h1AC Drawing NoX3188106)

This chassis similar to the 5aiGz inverter except for two (2) output

transformers and two (2) output rectifiers (21 T4 CR8 CR15) High voltage outputs

are brought out to two (2) stand-off screw terminals To share radiation area

equally between operate and standby conditions active components are staggered

Thus in operate Q7 Q6 T3 and CR8 will be on in standby Qi3 Q12 T4 and

CR15 will be on

In a transient mode at system turn-on dissipation will be-14 watts whereas

steady-state dissipation will be less than 3 watts At 429 in2 therefore and 14 2 0

watts or 029 wattsin final value temperature would be 35 C a safe value

However since transient should not last more than 15 mfnutes and thermal time constant

will be about one hour expected rise in 15 minutes is 1560 x 063 x (35 - 0) = 550 C

rise above assumed 00 C starting conditions free body

At 3 watts or 0078 wattsin2 a final temperature of -50 C would be expected

for a free body or closo to the average temperature of the system (about 20deg C)

5 Line Regulator (51Iz and Accelerator) See BAc Drawing No X3188118)

This assembly is identical in layout for the 5kilz and Accelerator

applications

At a loss of 105 watts worst case steady-state for the 5Kz applications 2and an area of 386 in or 027 wattsin 22 tmperature will be 250 C free body

8 4 1025-CC-EN RE 3-81

Components are mounted with techniques similar to those described above

6 Arc Inverter and Cathode Inverter (See HAC Drawings X3188108 X3188112)

These inverters are similar in layout differing in only the few small

components associated with output telemetry and feedback Techniques are similar

to those described above for the 5Kqlz and Accelerator inverters where -radiating

area is shared between operating and standby inverters

Dissirationii will be approximately 12 watts for the Cathode inverter and 2 215 watts for the Arc inverter With 479 in of radiation or 025 and 031 wattsin

respectively free body temperatures would be 200 C and 350 C respectively

7 Arc Rectifier and Filter (See HAC Drawing X3188116)

This module is a unique case of mounting relatively high dissipation

components at high voltage- This applies in particular to the output rectifiers

CRI CR2 and CR3 which dissipate 7 watts at a potential of 2KV To accomplish

the firm objective of good thermal conductivity and good insulation the rectifiers

are mounted on a magnesium bracket which is bonded to an epoxy-glass insulator

in turn bonded to the chassis By providing a liberal interface area of 3 in2

between -bracket and insulator a temperature rise across interface of 60 C will

result quite acceptable for the components with a plate temperature of 250 C The

latter results from a total dissipation of 10 watts with an area of 386 in 2 or 2

026 wattsin

Choke L2 is a source of about 3 watts and is a ferrite cup core directly

heat-sunk to chassis with high-voltage insulation internally

Capacitor Cl is a tantalum type which will dissipate about 1 watt due to

high-frequency ripple losses It is mounted in a fashion similar to that used for

rectifiers

Note that a inch creepage path is maintained between high-voltage elements

and chassis a desirable minimum to prevent flash-over

Except as noted other techniques are similar to those described before

S Magnetic Modulator (See HAC Drawing X3188122)

This is an assembly of magnetically-regulated supplies ie Vaporizer

Neutralizer Heater Neutralizer Keeper and Magnet Supplies The only supply at highshy

voltage is the Magnet with its output at a reference of + 2KV Estimated total 2 2dissipation is 48 watts with a radiation area of 395 in or 012 wattsin thus

an expected plate temperature of 10 0C free body or closed to average temperature of io C mounted

103

66857-CC-EN REV 3-51

Thus the mounting area required by components is the limit in this case

rather than temperature As indicated in drawing all magnetic components the

principal source of dissipation are mounted on chassis with small components

mounted on a plate above the magnetics with a thermal path to base plate through

corner posts

The components requiring high-voltage insulation are th6 output diodes of the

Mgnet Supply Using a technique similar to that described before for the Arc

Rectifier these diodes are heat-sunk to a small magnesium plate which is bonded

to a 0032 epoxy-glass insulator in turn bonded to a supporting magnesium bracket

which donducts heat to main plate (approximately 1 watt) A one-half inch creepage

path is provided between the diode heat sinkand the bracket across the epoxy-glass

A thin 0002 H film layer insulates the diodes from the heat sink and thus

from each other The bracket also carries the high-voltage screw-terminal standshy

offs for the magnet output

Techniques for mounting magnetics and small low-voltage components are

similar to those described for other assemblies

9 Control Module (See HAC Drawing X3188120)

This module is an assembly of low power components both discrete and

micro-circuit Nevertheless all components are mounted to heat sinks with defined

conductive paths to the base plate radiator Total watts dissipated is estimated

at approximately 5 watts maximum Hence temperature of assembly will be controlled

primarily by heat conduction from adjacent modules and should be close to the average

temperature of the system ie about SO0 C

Discrete components are mounted on a 132 epoxy-glass terminal board bonded

to a magnesium heat sink All components will iest on terminal board and will be

conformally coated with low-outgassing epoxy to improve heat transfer seal against

moisture and secure against vibration during launch

Microcircuits used here will be of the flat pack type and will be mounted on

a printed circuit board with an etched copper heat sink for a good thermal path to the

main heat sink

10 Low Voltage Connection Module (See HAC Drawing X3188127)

This module will mount all low voltage connectors interfacing with vehicle

connectors ie Control and Telemetry Low Voltage Loads and Input Power Conshy

nectors are all of the Sub-Miniature rectangular type rated at 5 amps per contact

All internal connections will be made at the rear of the assembly and all

external connections on the front face Each connector used is a different size from

adjacent connectors to prevert interchange of connectors

6O5 7-CC-EN REV -51 104

Harness connectors will be provided with side access covers for maximum

convenience in routing vehicle harness from edge of assembly

11 Hji Voltaga Connection Module (See HAG Drawing X3188126)

This module will mount all high-voltage connections to vehicle harness

Connections used are high-voltage standoff screw terminal type for minimum weight

compatible with high-vbltage insulation A protective cover will be provided for

personnel protection during tests

d ISCELTANEOUS GONS DEIRATIONS

1 Thermal

System thermal design philosophy was discussed under General and

detail thermal design was discussed under each module discussion For overall 2

sunmmations Figure 1 attached gives heat loads for each module wattsin and

expected temperatures

Although not indicated in drawings an array of power resistors will be

arranged throughout the eggerate frame to be powered from vehicle control to

pre-heat assembly to -20deg C before starting up system and insure minimum-temperature

above -40 C at any time

2 Choice of Materials

Two (2) criteria have been used in selection of materials over and above

the usual criteria strength insulation etc These are 1) low outgassing and

2) compatibility with Freon to be used in Calorimeter test Choice of materials for

first criterion are based on extensive data collected by HAG Materials Laboratory

Choice of materials for the second criterion has been made on the basis of tests for

24 hours immersionf in Freon at 450 C (expected operating temperature) and previous

experience with Surveyor batteries in Freon Calorimeters These tests indicate that

RTV is incompatible hence must be sealed with epoxy (only hair-line joints must be

sealed) However nylon adhesive mylar film H film teflon and epoxy have been

verified as compatible insulations

105

S1EEN 5CREEN 5CPEFN RF__N INV INV INV INV

it I -2 3k 4

9ow 1- w 19o w k9ow -o-l WIN 2 C27W IN 027 W IN o-z2W ItN

zsdegc 25 5C -c

Sr-CRIMEN SCREEN SC-9EEN sCRkEEN INV INV INV INV

11o w 19 ow le)o w 13 0w 02- WIN 012- WIN 0 2 7 WIN2 027 W IN

asoc zswC cC2C -

ACCEL ARC KH C H 0O DE

INV INV INV INV

4 W 4) IS2WI SW OW IN 03al WIN 2ISWINL 02 wINL

I V ARC LINE RE CONTROL

swIILTR5Iw bcFMILR - RCT OSW105) MWDs L

013 WIN Z lOW02SWIN 2 0 7 WINz25C

0 1WIN22o 0 C

za00c 2s 0 ~C

i-V MAlG LINE LOW VOL-rcE TERMINaL M0)DUL ATO RFI- (ACCEL) TEMINL BO ARDk 4ampW 4 W ESOaRD

c O W OowIN

0 10 WIN Z00 C

0 10 wIN

OO 00 OW

WIN

T-TOTAL LvOS Z31uA3 Ar7S-To-AL- ArLEA= 1ofo3 w

AVJEtLUtO-rSIU= 0--Z5 AULRamp-TtAPZ 2-30 C

THERMAL MAP F 5-1 ULSTI i cLU-Tpound P -vZr- 9( 06)

d Miscellaneous Considerations (Contd)

3 Gas Density Calculation for Arc Dischtrge Versis Altitude

Problem

Find the maximum gas density that will accumulate in a vented power

conditioning housing of 47000 cm The gas being generated by the outgassing

rate of a conformal coating The vented area being a mesh screen plate with2

an atea of 6800 cm The holes in the mesh represent 40 percent of this area 2 or approximately 2700 cm

Given

The Components and Materials Laboratory at Hughes Aircraft ran some pre-shy

liminary tests on the conformal coating (used on the power conditioning comshy

ponents) with the following results

a) The outgassing molecules consist of water and various absorbed

gasses

b) The molecular weight averages 100

c) Approximately 3 x 1019 moleculescm 2 does outgas from a coating

0005 inch thick

d) Temperature of the gas is 3280 K

e) Approximately 0001 weight loss occurs in the first four hours in

space This is the maximum rate loss expected This is equivalent 6to 70 x 10- grams per second

f) The vented power conditioning housing contains 47000 cm3

g) One side of the housing is a mesh plate with a total hole area of e22700

Find

The maximum gas density developed inside the power cohditioning housing

after launch and injection in orbit Plot results on a gas density versus

altitude from the surface of the earth (Figure I)

Solution

1 Molecular velocity of the outgas is

-2 3R TVag = mM

1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23

Vag = 286 x 104 cmrsec

whez k = Boltzmanns constant

107

SDensity Calculation for Are Discharge Versus Altitute

T- absolute temperature in kelvin

m = molecular mass (grams)

vag =velocity in cmsec

2 The maximum gas density developed inside the housing R

-=K A vag

61070 x

05-x 2700 x 103 x 286 x 104

= 181 x 10 13 gramscm

3

3 x 10-

1 2 slugsft-36

where R = gas flow rate in gramssecond2

in cmA = vented area

vag = gas velocity in cmsec

K =-flow efficiency taken -as 05

Conclusion

The maximum gas density developed inside the vented housingcontaining the -12

power conditioning system is approximately 36 x 10 slugs percubic foot

This value corresponds to an altitude of 100 to 130 nautical miles on an

atmospheric versus altitude graph

Gas Pressure Calculation

Using the ideal gas law

Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104

P = 49x 10-4 dynes-cm

2

or P = 49 x 10-I0 atmospheres

or 7P - 376 r 10 shy mm Hg

where P-= pressure in dynexcm2

v = volume of 47000 in3

n = number of molecules in v 51 x 1014

k = Boltmanns constant 138 x 10-16

T abs temperature 328 0 K

108

Gs Pressure Calculation

Conclusion

Thevoltage breakdown for the power conditioning system will be above

10000 volts for 05 inch terminal spacings See Figure III

109

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10 POWER CONDITIONER SUPPORT EQUIPMENT SYSTEMS TEST CONSOLE

a General

The test console will operate test and evaluate the power conditioner during design de elopment qualification and acceptance testing

It provides the following basic functions

1 Simulated loads for the power conditioner Each of the eight power supplies are terminated in the correct impedance to operate the power conditioner and evaluate its performnce

2 Power input to the power conditioner power supplies Four power supplies provide power to the poier conditioner An array of switches applies and removes power from each group of inverters A master switch removes all power Selector switches are provided to remove power from individual inverters to simulate failures and demonstrate the abiliLy of the reserve circuits to cut in and mainshytain operation

3 Instrumentation for monitoring voltage current waveform and temperature All instrumentation consists of standard commercial equipment All pertinent voltages currents and waveforms can be selected and displayed on the appropriate drgital voltmeter ENS voltmeter andor scope Provisions for continual scanning and monitoring of 10 thermistors are provided

4 Continual telemetry monitoring All 15 telemetry outputs arc terminated and individually monitored at all times

5 The command generator will provide control conimands to the power conditioner The commands are 22 VDC positive pulses of 60 milliseconds duration (internally adjustable from 40-100 milliseconds) Any one of five command circuits can be selected Command status is displayed by lights on the status panel A 0 - 5 V analog command is provided continuously for screen reference current

6 High voltage protection of personnel is provided by electrical interlock and shorting relays in the -2 KV and + 2 KV high voltage areas within the test console

The test console block diagram displays the interconnection and flow between the above described functions

The instrumentation by selector switch monitors the voltage current and wave-form of all eight dummy loads In order to measure the relatively low voltageshyand currents which are at 2000 volts with respect to ground under normal operating conditions a meter enabling circuit has been provided This circuit allows DVM or EMS meter connection only when all high voltage has been removed The telemetry readout will provide continuous monitoring of all load voltages and currents All loads for the power conditioner are provided by the test console or transferred to the ion engine by the test console This will enable the operator to recheck the calibration of the telemetry at any time that it should become necessary during the test with the ion engine

REV 3-6s16637-CC-EN

The arc test will short any two supplies or any supply and ground by a thyratron

ischarge when so commanded There are also provisions to apply a direct short or

xternal resistance between any two supplies or any supply and ground

A main three pole circuit breaker is provided to remove all power Individual

ircuit breakers are provided for the 28V pouer supply and the arc test power nput- All instrumentation and the remaining power supplies are commercial units

ith front panel switches

FUNCTIONALCIRCUIT DESCRIPTI

Schematic diagrams of the various circuits comprising the test console are ttached

1 Control Panel

The DC power distribution consists of a main disconnect switch and five ower conditioner power disconnect switches As each switch is closed an indicator

amp will glow on the status panel indicating the presence of power

The main DC power switch is a DPDT center off switch The upper switch

esition connects the four main power supplies two in series and two series groups n parallel The lower switch position connects a9l four in parallel The relay ontacts for each pair of power supplies will carry 60 amps

All high voltage supplies are interlocked by a series string of switches

he switches are placed on all access panels to high voltage areas When a panel s removed the associated switch will open This will open the circuit to the igh voltage warning lamp and a relay that disconnects power to the high voltage upplies The meter enable switch on the high voltage panel disconnects the etering circuits on the magnet (VI) cathode (V3) and anode (V4 ) loads This ircuit is discussed in a separate section The scope digital voltmeter and

rue PNS Voltmeter selector switches are located on this panel The Arc Initiate ommand and Command Selector are located in this panel They are discussed in eparate sections A means of disconnecting the power to all active inverters to imulate failurewill be accomplished by opening the selected switch A failure n the 5 KC Cathode and Arc Inverters is simulated by disconnecting the drive ith a relay located near the power conditioner

2 Comiiand Generator

The Command is initiated bya momentary on push-button switch This pplies 28 volts across the 10K Q and 47 f time delay network With the 10K 0 esistor adjusted to 58K Q the transistor will conduct enough to energize the elay in 60 milliseconds The result is conmand circuit providing a positive DC 2 volt pulse of 60 milliseconds duration with less than a one millisecond rise and

all tine 4K resistors are provided to effectively ground the command circuits and

enerator when they are not selected or commanded The second contact (normally open)

f the relay provides a means of displaying the command status

113-Crrr-jnr $-8

Off 2 must be comanded first or no DC power will be available to the power conditioner The Off 2 status light will remain lit (after command) until the selection switch is changed As On 1 On 2 and On 3 conmmands are enabled

a status light turns on for each When Off 1 is commanded the Off I status light turns on When On 3 is commanded (this light is already on) the Off 1 light -turns off Off 2 command turns off all command status lights except Off 2 A zero to five volt DC analog command is provided for screen reference current

3 Telemetry Panel

Fifteen telemetry outputs are monitored continuously They are terminated P90K ohms This termination and the source impedance will provide a 5 volt indicashytion on the 50ga meter for a 5VDC signal Switches are provided at each meter to momentarily disconnect the meter load for more accurate readings on the DVM

4 DC power is supplied by four 40V 30A fast response power supplies connectedshyin series parallel for normal operation or in parallel for low voltage operation

5 Arc Test

A thyratron tube is used for the arc test When energized it can be disshycharged by depressing the Arc initiate momentary on switch This energizes a conshytrol relay and applies a positive voltage to the grid circuit driving the thyratron into full conduction Conduction will cease when the anode-cathode voltage drops to a very low level or the initiate switch is released which opens the anode circuit An RC network in the grid circuit limits the time the grid remains positive to less than 10 milliseconds

6 Meter Enabling Circuit

All voltage mbnitoring is done at the power conditioner andnot at the loads (except the cathode) It is necessary to protect the instrumentation from the applishycation of 2000 volts when either the screen or accelerator supplies are operating A 101 voltage divider is provided to measure 2 KV screen and accelerator voltages A meter enablingcircuit is provided to protect the instrumentation by disconnecting the instrumentation with high voltage relays Under normal operating conditions the magnet (VI) cathode (V) and anode (V4) loads are connected to the +2KV screen supply When measurements are to be made on these supplies they are disconnected shy

from the screensupply and grounded The parameter to be measured is selected by a seven position selector switch This circuit also prevents measurement at 2 KV when the power conditioner is not connected such as when the load is transferred to the ion engine This protection is required because all voltage monitoring is done direct This circuit uses the ground leg of the vaporizer supply (V2) to connect or disconnect the +28V transfer voltage A red status light indicates when the circuit is disconnected

7 -Magnet Load (Vi)

The magnet load consists of a variable 4 ohm rheostat in series with a 7 mh choke and a 185 ohm resistor The magnet load can be varied from 185 ohm cold to 225 ohm hot This load is connected to the screen supply except when the instrumentashytion is in use Provisions are made for open circuit short and arc short to any

supply and ground

14

8 Vaorizer Load (V_2

This load consists of a 5 ohm non-inductive resistor It can be open circuited shorted and arc shorted to any supply and ground

9 Cathode Load (V3)

Provisions are made to mount the output transformer supplied with the power conditioner within the test console The SERT IT vaporizer and cathode loads were combined and rearranged to form this load A breadboard was made The best circuit arrangement consisted of seven parallel strings of three series 866A tube filaments shunted by approximately 0-28 ohms The test arrangement consisted of this load a 40A current limited DC power supply and current and voltage recorders Current was constant at 40A (except for initial surge) Voltage was 18V cold(0045 ohms) 28V in 8 seconds and stabilized at 42V after 90 seconds (0105 ohms) The 0045 ohm cold will be reduced to 0035 ohm by increasing wire size and reducing lead length The center tap of the output transformer is connected to the arc load Provisions are made to short and arc short to any supply and ground

10 Arc Load (V4)

The large rheostat from SERT II anode load and 11V selector switch from the vAporizer keeper load were retained for this load The load consists of a 5-position step switch and a combination of resistors and the rheostat The load terminals are connected one to the cathode supply and the other to the screen supply via the meter enabling relay The 5 positions will provide a load of 1) open 2) _2 ma 150V (75K ohms) 3) 20 ma 30V (1800 ohms) 4) al to 29A (124 to 474 ohms) 5) 58 to 84A (57 to 43 ohms) It can be shorted and arc shorted to any supply and ground Note The plus lead will not be cormmon to the screen lead at the power conditioner provisions for this connection are made in the meter enable circuit of the test console

11 Screen Load_(V5)

Many of the resistors and the high voltage relays from the SERT II console are used in this load Most of the resistors are operated at their published maximum wattages therefore a 700 CFM exhaust fan will be added at the top of the console inmediately above this load to keep these resistors at below their rated temperatures Three resistance banks are used in parallel combinations to provide loads at 2 KV of 1) open (divider in meter circuit will draw 1 ma) 2) 01 to 058A 3) - 068A 4) 08A 5) pOS4 to 106 A A make before break selector switch is used to make selection This load can be shorted and are shorted to any other supply and ground A 0-075A current limited supply can be connected across the screen current sensing resistor in the power condition to simulate an analog load condition in the active range of the screen supply

12 Accelerator Load (V6)

This load consists of a 5-position selector switch and resistors which will provide these loads at 2KV 1) Open (divider in meter circuit will draw 1 ma) 2)

105 ma (188K) 3) 53 ma (38K) 4) _ 77 ma (26K) 5 01 to 011A (20K to 18K) This load can be shorted and arc shorted to any other supply and ground

GM 7-cc-EN FEV 3-61 115

13 Neutralizer Cathode and Neutralizer Vaporizer Load (V7) (Same as SERT II)

The neutralizer cathode and neutralizer vaporizer loads consist of seven 866 tube filaments wired in series with an additional seven series connected 866

tube filaments shunted by one 1651 lamp (5V at 06A each) This circuit is an analytical design based on data received from the breadboard of the SERT II vaporizer load circuit The mathematical results show the following

a) The initial cold resistance should be 12 ohms

b) The final (after 10 t1me constants) resistance will approach 53 ohms Note that the one time constant resistance is 38 ohms

c) One time constant will be 40 seconds

116

The specification requires a cold resistance of I ohm a one-rime constant resistance of 35 ohms and a one-time constant value of 30 seconds

14 Neutralizer Keeper Load (V8 ) (Same as SERT IT)

This load consists of a potentiometer and a high-voltage transistor Adjustshyment of the potentiometer will cause the transistor to load the supply in the range of from approximately 300V minimum to 2V with a corresponding change in current of from approximately 0 ma to 500 ma A vernier adjustment is provided for

15 Status Panel

The following indicator lamps are displayed on the status panel

Status Panel Lamps Color

a) Warning Ihigh Voltage Red b) On I Green c) On 2 Green d) On 3 Green e) Off I Green f) Off 2 Green g) Main AC Power Neon (Amber) h) Arc Test Power Neon (Amber) i) 28V Supply Power Amber j) Main DC Power Amber k) Line Regulator Power Amber 1Y Cathode Power (V3) Amber m) Arc Power (V4 ) Amber

n) Screen Power (V5) Amber o) Accelerator Power (V6 ) Amber p) -2KV Shorting Switch Closed Red q) +2KV Shorting Switch Closed Red r) Short Test On Red s) Meter enable not ready Red

c HARDWAREPACKAGING DESCRIPTION

The test console will be housed in two 6 foot rack cabinets 25 inches deep There will be a work shelf at approximately 29 inches from the floor A pictorial diagram is included Custom equipment will be placed as indicated in the pictorial diagram Location of controls is approximate and will be determined during the mechanical design phase of this program

d ENERGY NEEDS AND LOSSES

Energy requirements of the test console will be supplied by 120208 VAC 3shyphase power source Current loading will be divided between the three phases as equally as possible

Two blowers and filters will be installed at the bottom of the cabinets to cool the test console An exhaust fan will be installed over the high voltage section to adequately cool these components

6657-CC-EN REV 3-61 117

LIST OF DRAWINGS

POWER CONDITIOiMR TEST CONSOLE (BLOCK DIAGRAM) Fig 10-1i

CONTROL (SCOIFMATTC) Fig 10-2

COMMAND -ENRATOR (SCmTMATIC) Fig 10-3

TEIII-tTRY PANEL (SCHEMIATIC) Fig 10-4

ARC TEST (SCHEMATIC) Fig 10-5

METER ENABLING CIRC-IT FOR VAPORIZER ThRUSTER CATHODE Fig 10-6

MAGNET LOAD (V1 ) (SCUEt1ATIC) Fig 10-7

VAPORIZEIR LOAD (V2 ) (SCHISMATiC) leig 10-8

CATHODE LOAD (V3) (SCPEIMATIC) Fig 10-9

ARC LOAD (V4 )(SCHEMLATIC) Fig 10-10

SCRIEN LOAD (V5 ) (SCHEIINAIC) Fig 10-11

ACCEI3_RATOR LOAD (V6) (SCHEMhATIC) Fig- I0-12

NEUTRALZBR HEATER LOAD (V7 ) (SCHEIItTIC) Fig 10-13

NEUTRALIZER KEEPER LOAD (V ) (SCFEWTIC) Fig 10-14

CONSOLE DRAWING Fig 10-15

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11 Isothermal Calorimeter

A calorimeteris being constructed for the purpose of measuring the operating

efficiencies of the power conditioner The calorimetric method of measuring

operating efficiences has the inherent advantage of measuring inefficiencies

directly so that a two percent error in measuring the inefficiency for example

would lead to an error in calculating the efficiency of less than 02 percent if

the efficiency is greater than 90 percent This assumes of course that the

input power can be measured to within less than 02 percent

The calorimeter is of the isothermal type which indicates heat generation

by measuring the rate of boil off of a liquid The liquid used is one of the

Freons (usually Freon 11 or Freon TF) because of their low toxicity non-flammability

and non-conductance The item under test is placed inside the inner chamber of

the calorimeter X3182788 This chamber and the surrounding one are filled

with Freon A hole in the bottom of the inner chamber connects the Freon in the

outer chamber with that in the inner one The Freon in both chambers is maintained

at the boiling point with heaters mounted in the chambers so that any additional

heat input to the system such as that from the item being tested will be seen

as an increase in the volume of the vapor The reason for the outer chamber of

Freon is to serve as a thermal insulator for the test chamber The Freon vapor

from both chambers is condensed in a condensor and returned to the outer chamber

The vapor produced in the inner chamber goes through a small orifice in the

lid and into the vapor space ii the outer chamber before going to the condensor

A very sensitive differential pressure transducer located on the lid to the inner

chamber senses the pressure difference between the inner and outer chambers

caused by Freon vapor from the inner chamber as it passes through the small

orifice into the outer chamber

When the item under test begings to generate heat the Freon flow rate inshy

creases which causes an increased pressure drop across the orifice The increased

output from the transducer causes the heater in the inner chamber to be turned

down and a constant flow rate of Freon vapor (and therefore a constant heat

outputfrom the inner chamber) will be maintained provided that the inner chamber

heater was originally set at some value higher than the anticipated heat output

of the item under test The circuit includes a zero suppression so that this

decreased is read directly This readout obviates any data reduction The heat

generation rates can be read directly on a strip chart recorder or they can be

sent to any other data storage system

134

In addition to the heaters which are part of the feedback system the inner

chamber also contains a calibration~heater This heater is used not only as a

dummy load for calibrating but also assists in the initial warm-up It can also

be used to maintain a positive heat output when measuring endothermic reactions

such as batteries under certain circumstances

This calorimeter has been stressed for pressurization to 15 psig with a

safety factor of more than 41 This is to allow a range of operating temperashy

tures to be attained By adjusting the pressure in the range from 0 to 15 psig

for example the boiling point of Freon 11 can be varied over the temperature

range from 24 to 45 0C

The assembly drawing is shown in Figure 1 The inner chamber which is

indicated by the dotted lines has the following dimensions

Length 40 (inside)

Width 10 (Inside)

Height 35 38 (Inside)

The maximum height available for an item being tested is about 32 inches because

the chamber must not be filled completely with Freon and the item under test must

be completely submerged

The total weight of the calorimeter is estimated to be 832 pounds empty

and 2257 pounds filled with Freon 11

Power input to the feedback heater will be controlled by an SCR which

senses the ignal from the differential pressure transducer The unit is the

135

12 List of Drawings

Drawing Number

X3188100-500 X3188101 (2 sheets) X3188103 x3188104 X3188105 X3188106 X3188107 X3188109 (2 sheets) X3188110 X3188111 X3188112 X3188113 x3188114 X3188115 X3188116 X3188117 X3188118 X3188119 X3188120 (2 sheets) X3188121-2 X3188122 X3188123 (2 sheets) X3188126 X3188127 X3188131 X3182788

Title

Installation Control Frame Module Plate Mounting Screen Inverter Assembly Screen Inverter Schematic Accelerator Inverter Assembly Accelerator Inverter Schematic Arc Inverter Schematic 5 KHz Heater Inverter Assembly 5 KHz Heater Inverter Schematic Cathode Inverter Assembly Cathode Inverter Schematic High Voltage Filter Assembly High Voltage Filter Schematic Arc Rectifier Filter Assembly Arc Rectifier Filter Schematic Line Regulator Assembly Line Regulator Schematic Control Module Assembly Master Oscillator amp Phase Shifter Schematic Magnetic Modulator Assembly

Magnetic Modulator Schematic High Voltage Connection Module Assembly Low Voltage Connection Module Assembly System Block Diagram Calorimeter Assembly Mark III

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HUGHES AIRCRAFT COMPANY AEROSPACE GROUP

SPACE SYSTEMS DIVISION


QUARTERLY TECHNICAL REPORT

PERIOD I JULY - 30 SEPTEMBER 1968

ELECTRIC THRUSTER POWER CONDITIONER

CONTRACT NO 952297



QUT Zl 1)38

W JgYMuldoon Assistant Project Manager

J A Robertson ontract Administrator

This work was performed for the Jet Propulsion Laboratory

California Institute of Technology as sponsored by the National Aeronautiis and Space Administration under Contract NAS 7-100






Administration

ABSTRACT













cooling

TABE OF CONTENTS

Page

INTRODUCTION 1


1 General 2

a Objectives 2





12





a Arc Supply is


c Screen System 21


a General 2


c Overload-Trips 2f



6 Test Data 41


b Power Inverter 41

c Cathode Supply 41


e Components 41


iii

Page



b Arc Inverter 66



e 5 KHz Inverter 70









a General 99





a General 112




11 Calorimeter 134


INTRODUCTION




























1 General

a Objectives






exceed 96





















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093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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of the circuit

78















rated stress




wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

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84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

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108


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6657-CC-EN REV 3-61 117

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ABSTRACT













cooling

TABE OF CONTENTS

Page

INTRODUCTION 1


1 General 2

a Objectives 2





12





a Arc Supply is


c Screen System 21


a General 2


c Overload-Trips 2f



6 Test Data 41


b Power Inverter 41

c Cathode Supply 41


e Components 41


iii

Page



b Arc Inverter 66



e 5 KHz Inverter 70









a General 99





a General 112




11 Calorimeter 134


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1 General

a Objectives






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093 x 39 x 095





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30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

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PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

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TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

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Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

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81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










heavier





phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































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applications


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rectifiers









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Find




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1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23



107






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3

3 x 10-

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Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104


2








108


Conclusion



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10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







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ABSTRACT













cooling

TABE OF CONTENTS

Page

INTRODUCTION 1


1 General 2

a Objectives 2





12





a Arc Supply is


c Screen System 21


a General 2


c Overload-Trips 2f



6 Test Data 41


b Power Inverter 41

c Cathode Supply 41


e Components 41


iii

Page



b Arc Inverter 66



e 5 KHz Inverter 70









a General 99





a General 112




11 Calorimeter 134


INTRODUCTION




























1 General

a Objectives






exceed 96





















-filters at no load








3 Cath Htr AC V 5 40 45 45 35 160 Loop --- 10 - 40 5





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output


























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20



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21































22


a General







preheat sequence







the cathode




10 k ohm load


1 Command memory


A Slow turn on

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3 Standby switching

A Arc

B Cathode

C Accelerator

23



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24







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lasts one second









25





d Analog Controls













amplifier output






Where









26

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28





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Line Voltage

40

45

50

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60

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100

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37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

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8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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1 7

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of the circuit

78















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wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










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phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































Inverter





d5-7 CC nv 3-5 101























CR15 will be on











applications


8 4 1025-CC-EN RE 3-81



















026 wattsin





rectifiers









103

66857-CC-EN REV 3-51





corner posts

























main heat sink








6O5 7-CC-EN REV -51 104









1 Thermal




















105


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Given





gasses



0005 inch thick






Find




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1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23



107






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61070 x

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= 181 x 10 13 gramscm

3

3 x 10-

1 2 slugsft-36





Conclusion







Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104


2








108


Conclusion



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113-Crrr-jnr $-8



3 Telemetry Panel



5 Arc Test





7 -Magnet Load (Vi)


supply and ground

14



9 Cathode Load (V3)


10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







116




15 Status Panel










6657-CC-EN REV 3-61 117

LIST OF DRAWINGS
















118 - v -I1

CONDIrIONIER

FAILUREJE R ON IE

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INTRODUCTION 1


1 General 2

a Objectives 2





12





a Arc Supply is


c Screen System 21


a General 2


c Overload-Trips 2f



6 Test Data 41


b Power Inverter 41

c Cathode Supply 41


e Components 41


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b Arc Inverter 66



e 5 KHz Inverter 70









a General 99





a General 112




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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



follows

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1 7

2 6

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t = interval time















of the circuit

78















rated stress




wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










heavier





phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































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d5-7 CC nv 3-5 101























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applications


8 4 1025-CC-EN RE 3-81



















026 wattsin





rectifiers









103

66857-CC-EN REV 3-51





corner posts

























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105


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166 x 10-23



107






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3

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Conclusion







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x 1014 x 138 x 1016 x 328

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2








108


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113-Crrr-jnr $-8



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5 Arc Test





7 -Magnet Load (Vi)


supply and ground

14



9 Cathode Load (V3)


10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







116




15 Status Panel










6657-CC-EN REV 3-61 117

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40

45

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100

101

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32 1 10 2 18 10

33 1 10 3 22 15

34 9 9 4 20 15

35 8 8 5 25 10

36 9 7 6 25 10

37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

40 11 8 10 25 10

41 11 8 11 30 14

42 11 8 12 20 10

43 10 9 13 25 15

44 10 8 14 17 10

45 10 8 15 20 10

46 11 9 16 25 15

47 10 29 17 22 12

48 10 7 18 25 10

49 11 9 19 22 10

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2-2 12 8 21 30 15

2-3 11 8 22 20 15

2-4 11 9 23 15 15

24 10 10

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26 20 08

27 20 10

28 15 08

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1 1040 1000 1120 1160

2 1240 1040 1080 1080

3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

34 1160 1040 1160 1040

35 1000 960 1080 1200

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



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Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

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5945 5601

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shy


90

i





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91






Power

2

2

75

1200

150

2400




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4

2

60

260

240

520



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6

2

10

65

60

130

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Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

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92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

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module plates




99





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rectifiers

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107






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3

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2








108


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109

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Power output 300 W



AT 400F

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VBE IV




Pulse width 36ps

Power output 300 W



T - 51OF













54


(RMS Regulator)

Line Voltage

40

45

50

55

60

65

70

75

80

lOA Load

100

100

101

100

100

100

100

100

100

25A 40A

250 401

251 402

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32 1 10 2 18 10

33 1 10 3 22 15

34 9 9 4 20 15

35 8 8 5 25 10

36 9 7 6 25 10

37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

40 11 8 10 25 10

41 11 8 11 30 14

42 11 8 12 20 10

43 10 9 13 25 15

44 10 8 14 17 10

45 10 8 15 20 10

46 11 9 16 25 15

47 10 29 17 22 12

48 10 7 18 25 10

49 11 9 19 22 10

2-1 10 8 20 30 15

2-2 12 8 21 30 15

2-3 11 8 22 20 15

2-4 11 9 23 15 15

24 10 10

25 25 10

26 20 08

27 20 10

28 15 08

29 15 06

30 15 06

57





1 1040 1000 1120 1160

2 1240 1040 1080 1080

3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

34 1160 1040 1160 1040

35 1000 960 1080 1200

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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1 7

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t = interval time















of the circuit

78















rated stress




wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










heavier





phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































Inverter





d5-7 CC nv 3-5 101























CR15 will be on











applications


8 4 1025-CC-EN RE 3-81



















026 wattsin





rectifiers









103

66857-CC-EN REV 3-51





corner posts

























main heat sink








6O5 7-CC-EN REV -51 104









1 Thermal




















105


it I -2 3k 4


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INV INV INV INV





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Given





gasses



0005 inch thick






Find




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1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23



107






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61070 x

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= 181 x 10 13 gramscm

3

3 x 10-

1 2 slugsft-36





Conclusion







Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104


2








108


Conclusion



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113-Crrr-jnr $-8



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5 Arc Test





7 -Magnet Load (Vi)


supply and ground

14



9 Cathode Load (V3)


10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







116




15 Status Panel










6657-CC-EN REV 3-61 117

LIST OF DRAWINGS
















118 - v -I1

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Line capacitor 093


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Total 1423 watts

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1239 w


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4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

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= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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1 7

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of the circuit

78















rated stress




wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

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module plates




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b SUPPORT STIUCTURE










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108


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7 -Magnet Load (Vi)


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6657-CC-EN REV 3-61 117

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40

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100

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33 1 10 3 22 15

34 9 9 4 20 15

35 8 8 5 25 10

36 9 7 6 25 10

37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

40 11 8 10 25 10

41 11 8 11 30 14

42 11 8 12 20 10

43 10 9 13 25 15

44 10 8 14 17 10

45 10 8 15 20 10

46 11 9 16 25 15

47 10 29 17 22 12

48 10 7 18 25 10

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2-3 11 8 22 20 15

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27 20 10

28 15 08

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1 1040 1000 1120 1160

2 1240 1040 1080 1080

3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

34 1160 1040 1160 1040

35 1000 960 1080 1200

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




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b SUPPORT STIUCTURE










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phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































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d5-7 CC nv 3-5 101























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8 4 1025-CC-EN RE 3-81



















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rectifiers









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66857-CC-EN REV 3-51





corner posts

























main heat sink








6O5 7-CC-EN REV -51 104









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105


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Problem





Given





gasses



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Find




Solution


-2 3R TVag = mM

1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23



107






-=K A vag

61070 x

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3

3 x 10-

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Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104


2








108


Conclusion



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38 10 7 8 20 10

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43 10 9 13 25 15

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45 10 8 15 20 10

46 11 9 16 25 15

47 10 29 17 22 12

48 10 7 18 25 10

49 11 9 19 22 10

2-1 10 8 20 30 15

2-2 12 8 21 30 15

2-3 11 8 22 20 15

2-4 11 9 23 15 15

24 10 10

25 25 10

26 20 08

27 20 10

28 15 08

29 15 06

30 15 06

57





1 1040 1000 1120 1160

2 1240 1040 1080 1080

3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

34 1160 1040 1160 1040

35 1000 960 1080 1200

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



follows

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1 7

2 6

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t = interval time















of the circuit

78















rated stress




wearout failure



greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










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phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































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d5-7 CC nv 3-5 101























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applications


8 4 1025-CC-EN RE 3-81



















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103

66857-CC-EN REV 3-51





corner posts

























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105


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107






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3

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Pv= n k T

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2








108


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113-Crrr-jnr $-8



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5 Arc Test





7 -Magnet Load (Vi)


supply and ground

14



9 Cathode Load (V3)


10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







116




15 Status Panel










6657-CC-EN REV 3-61 117

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13 1030 1180 1000 1100

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16 1220 1180 1220 1140

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20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

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24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

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32 1180 1160 1080 1160

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093 x 39 x 095





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30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


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P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

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P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

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1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

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Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


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PR = 01A x 08V x ID = 08 W



Total 110 W

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I = 200 = 58A peak93 x 39 x 95

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1) Thansistor Loss

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P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

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Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







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7) TOTAL (73) = 80



1) Vaporizer

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P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

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Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

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TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

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84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

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87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

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module plates




99





b SUPPORT STIUCTURE










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rectifiers

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166 x 10-23



107






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3

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2








108


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Line Voltage

40

45

50

55

60

65

70

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100

100

101

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34 9 9 4 20 15

35 8 8 5 25 10

36 9 7 6 25 10

37 9 9 7 20 10

38 10 7 8 20 10

39 11 7 9 22 08

40 11 8 10 25 10

41 11 8 11 30 14

42 11 8 12 20 10

43 10 9 13 25 15

44 10 8 14 17 10

45 10 8 15 20 10

46 11 9 16 25 15

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48 10 7 18 25 10

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2-2 12 8 21 30 15

2-3 11 8 22 20 15

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2 1240 1040 1080 1080

3 1140 1040 1020 1200

4 1240 1100 1180 1150

5 1100 1140 1080 1200

6 1140 1180 1120 1240

7 1160 1120 1160 1240

8 1120- 990 1160 1080

9 1080 1240 1140 1080

10 1160 1240 1200 1180

11 1000 1120 1180 1200

12 i000 1060 1120 1100

13 1030 1180 1000 1100

14 1210 1040 1080 1080

15 1080 1260 1300 1180

16 1220 1180 1220 1140

17 1220 1140 1260 1160

18 1140 1180 1060 1140

19 1130 1140 1120 1126

20 1040 1000 1000 1060

21 1220 1200 1180 1100

22 1100 1140 1200 1120

23 980 1160 1060 1240

24 1100 1080 1110 1220

25 1100 1110 980 1120

26 1180 1060 1120 1220

27 1060 1040 1140 1100

28 1120 1080 1200 1190

29 1190 1280 1060 1140

30 1160 1000 1040 1180

31 1040 1240 1240 1120

32 1180 1160 1080 1160

33 1100 1100 1160 1020

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At 40-Volt line



093 x 39 x 095





-x 78 x 87 x 05 x 10 6 x 125 x 103 = 419





30 watts


P C 041 watts ceest






6




Line capacitor 04

Transformer 50

Rectifiers 32

Total 193 watts

9) Input power 300 + 193 -3193


At 8O-VQlt Line




P = 49 x 156 x 87 x 05 x 10-6 x 125 x 103 sw


3) Psat + Psw 290 + 380 = 687 watts







PT = 20 - 15 =-35 watts




Base drive 150

Line capacitor 093


Rectifiers 160

Total 1423 watts

9) Input power = 300 + 1423 = 31423






Psat 72 x 07 = 504 w


-6P = 49 x 78 x 87 x 05 x 10 x 125 x 103


P3) Psat + sw = 504 + 190 694 w

4) Base Drive Loss

PBD - 30 x 083 = 262 w 095

64


6) Transformer Loss

Core Loss a B m 3E

w

PCoM 20 x (78)3 = 134 w


P 30 x 083 = 262 w wp 095

P = 134 + 262 = 396 w


a Duty Cycle

Pr = 32 x 083 = 28 w 095



1663

9) Efficiency = 250 94 2666



Ic av = 72 = 36 2

Psat = 36 x 07 =252 w



Psw = 190 x 2 = 380 w

3) Psat + Paw = 252 + 380 = 632 w

4) PBD = 262 x 12 = 131 w

2 5) PLine capac = 093 x 76 = 071 w

87

6) Transformer Loss



PT = 134 + 131 = 265 w





1239 w


2624




At 40 Volt Line

1) Transistors 252 x 694 = 668 26-2

2) Base Drive = 252 x 262 = 252 262

3) Line Capacity = 252 x 031 = 030

4) Transformer = 252 x 396 = 382


6) Effici ency = 252 = 950 2653

45-7-CC-Et REV 3-61 66

At 80 V Line

1) Transistors = 252 x 632 = 610 2

2) Base Drive 252 x 131 = 126262

3) Line Capaci-ty = 252 x 071 069 262

4) Transformer = 252 x 265 255 i62


6) Efficiency = 252 =966 2606






Ic = 200 = 632 A peak = 632 A aver 93 i 34

1) Transistor Loss


PT = 44+ 12 = 56 w

2) Base Drive Loss

PBD = 2Vx 08A = 16W




PR = 01A x 08V x ID = 08 W



Total 110 W

B - Steady State



2y BaseDrive Loss

3) Transformer



T = 151








I = 200 = 58A peak93 x 39 x 95

= 58 x 95 = 55A aver

1) Thansistor Loss

Psat = 55 x 07 = 375 W

P = 49 x 78 x 58 x 05 x 10-6 x 5 x 103 05Wsw

PTot = 375 + 05 = 425 W

2) Base Drive Loss


68 6657-CC-E REV 3-61





Total 990W

80 V Line

1) Transistor Loss

Psat = 55 x 07 = 187W 2

P = 05 x 2 = 100W Sw

PTot = 187 + 100 = 287W

2) Base Drive Loss

PBO 2 x07 x 12 = 07W


4) Transformer Loss


same core loss




69

447-CC-EN PEV -61




+ 12V + 5V

Po = 20 + 20 + 41 + 21 + 15 + 15 = 105 W


1 105 = 332 A peak = average93 x 34

1) Transistor Loss

Psat = 332 x 07 = 232 W 4 6=

Psw = -x 68 x 332 x 05 x 16-6 x 5 x 103 025 W9

Pt = 232 + 025 = 257 W

2) Basj Drive Loss

Pbd = 2 x 05 = 10 W


4) Transformer Loss

98 Effic Pt = 105 x 02 = 21 W


Transistors 257

Base Drive 100

Transformer 210

TOTAL 567 w



Inverters (10)

Po = 105 + 10 x 30 = 105 + 30 = 135 W


At 40V Line

Ic = 135 = 386 A peak35 shy

= 386 x 95 = 366 A


Psat = 366 x 07 = 256 W

46 3 -Psw 104 Hz = - x 78 x 386 x 05 x 10 x 10 x 10 = 063

Pt = 256 + 063 = 319 32 W





6) Pt = 32 + 27 + 05 + 05 + 10 = 79 W

At 80V Line


Psat = 256 = 128 - 2


Pt = 128 + 126 = 254 w (2 x voltage)






71


40 V Line

A Transient

Po-= 211 Watts

211Ic = 602 A peak =602 x 95= 575 A

1) Transistor Loss

Psat = 575 x 07 = 403 W

Psw = x 78 x 602 x 05 x 106 x 10 x 10= 104 W 9

Pt = 403+ 104 = 507 W





6) Total Loss = 507 + 40 +07 + 07 + 15 = 1197 W

B Steady State

Po = 211 W


2) Choke Loss = 40 x 01 = 004





7-CC-W RrY 3-61 72 85

--

80V Line


1) Transistor Loss

Psat = 403 = 202 W

Psw = 104 x 2 = 208 W

Pt = 410 w 041





6) Total Loss =125 W = 195 W







i HV Filter








7) TOTAL (73) = 80



1) Vaporizer

=Pout 20 Watts


P 02 x 20 = 040 mag-amp



Pout 41 Watts



VAou t 180


b) Pchoke - 1899 - 18 -- 02 watt


d) PT i 0O4 + 02 + 08 = 14 Watt

4) Magnet







PT = 105 + 21 + 14 + 165 = 480 Watts

746657-CC- EN REV 3 -61

k Sunmary of Losses

Losses






Normal 375W 375W





Normal 205W 195W





75

8 RELIABILITY



















Interval Time Hours

1 0 to 2500

2 2500 to 5000

3 5000 to 10000



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78















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greater than zero

79

TABLE 81


PART-TYPE

Diode Switching

General Purpose

Zener


General Purpose

Power


NA741


Metal Film

Power (RW)

Precision


Tantalum


Mag Amplifier

Relay

Connector

Fuse


(932 946 948 etc)


60 10

240 2shy

75 10

120 10

1206 100

140 50

840 300

10 01

22 002

65 001

39 01

60 02

260 10

200 05

300 10

1000 100

250 25

100 10

80

Ref Dwg X3188119





Metal Film 2 22 44

Power W W 1 65 65




Zener 2 240 480



Conductor 200

Fuse 200

Connector 250

(10)- 9

x = 7967

-9 Bl ck 1 X= (475)(E X) 3784 (10)

81






Power W W 1 1 65 001 65 001 65


Tantalum 6 6 260 100 1560 600 15600



Transformer 2 1 200 05 400 050 200

Relay - 2 1000 100 - 2000 2000


Fuse 2 2 100 10 200 200 200

Connector 1 - 250 - 250 - 250

9231 8703 9657

(X1) (sx2) (2x3)

Block 2 X1 = (475)(9231) = 4385(10) - 9

Block 2 X2 = (475)(8703) 413(1) - 9

B k 2 X=-(475) (9657) = 4587(10) - 9

X1 = Active Unit



82







Zener 10 240 2400



Tantalum 6 260 1560



Power W W 4 65 260


Inductor 2 200 400

Connector 1 250 250

- 9 Z = 10067(10)

Block 3 = (475)(10067) = 4782(10) - 9

83




X(10) -9

Total X (10)





Metal Film

4

5

10

22

40

110


it Tantalum

1

3

60

260

60

780

T_= 2090 -9

(10)

SBlock A 8=(475)(2090) 993(10) - 9

84











Power W W 2 2 65 001 130 002 -130

Trans former 2 2 200 050 400 100 400

Oelay - 2 1000 1000 - 2000 2000

Fuse- 2 2 100 100 200 200 200

Connector 2 250 - 250 - 250

5850 7664 7790

CE 2I) (E X2 ) (z 23)

Block 5 X1 = (475)(5850) = 2779(10)-9

Block 5 X22 = (475)(7664) = 364 (10) -

Block 4 3= (475)(7790) = 3700(10) - 9

S1 = Active Unit



85





I Power W W 1 65 65





- 9 X = 1330(10)

- 9

Block 6 X = (475) (1330) = 632(10)

86








Metal Film 4 22 88



= 1828 (10) -9

Block 7 x = (475)(1828) = 868(10)-9

87






Power 3 600 1800

Zener 2 240 480


Metal Film 5 22 110

Power W W 1 65 65



Inductor- 2 200 400


Connector 1 250 250

- 9

X= 4048(10)

Block 8 X (475)(4048) 1923(10) - 9

88







X = 270(10) - 9

- 9 - 128(10)(475)(270)Block 9 X =

89







Power 2 12002 1000 2400 2000



Tantalum 4 3 260 100 1040 300


Power W W 1 1 65 001 65 001

Transformer 2 2 150 050 300 100


Inductor 1 1 200 050 200 050

Fuse 2 2 100 100 200 200


5945 5601

(zzActive) Dorm)

Block 10 XActive = (475) (5945) = 2824(10) - 9

9 Block 10 XDorm = (475)(5601) = 266(10)

shy


90

i





Power W W 2 65 130



Tantalum 1 260 260


Inductor 2 200 400

Connector 1 250 250

= 2 2420

= 1150(10) - 9 Block 11 = (475)(2420)

91






Power

2

2

75

1200

150

2400




Tantalum

4

2

60

260

240

520



Power W W

6

2

10

65

60

130

Fuse 1 100 100

Connector 1 250 250

SBlock 12 X = (475)(5084) 2415(10) -

x = 50844(10) - 9


92





= 24959 (10)- 9 The total )

- 9 - 9Thus Block 13 = (24959 )(475)(10) = 11856(10)

93






Block 14 X (475((250) = 19(10)-9

94












8bl) E-Factor Usage

















95

8b1) (Contd)

























8b4) Part Selection






96








8b5y Conclusions





















97




98

9 Physical Design

a e
































module plates




99





b SUPPORT STIUCTURE










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phase














of 350 C



rectifiers

6S57-CC-EN RFV 3-5I 100

































Inverter





d5-7 CC nv 3-5 101























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applications


8 4 1025-CC-EN RE 3-81



















026 wattsin





rectifiers









103

66857-CC-EN REV 3-51





corner posts

























main heat sink








6O5 7-CC-EN REV -51 104









1 Thermal




















105


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Given





gasses



0005 inch thick






Find




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1 6Vag 2 = 3 x 138 x 10 x 328 x 10 2

166 x 10-23



107






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61070 x

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= 181 x 10 13 gramscm

3

3 x 10-

1 2 slugsft-36





Conclusion







Pv= n k T

x 1014 x 138 x 1016 x 328

47090 x 104


2








108


Conclusion



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113-Crrr-jnr $-8



3 Telemetry Panel



5 Arc Test





7 -Magnet Load (Vi)


supply and ground

14



9 Cathode Load (V3)


10 Arc Load (V4)


11 Screen Load_(V5)





GM 7-cc-EN FEV 3-61 115







116




15 Status Panel










6657-CC-EN REV 3-61 117

LIST OF DRAWINGS
















118 - v -I1

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