References

The Fourier coefficient an (4), can bewritten as

{n 2» 1

n>kb

1. H. C. Pocklington. Proceedings, CambridgePhilosophical Society, London, England, vol. 9,1897, p. 324.

2. E. Hallen. Nova Acta Regiae SocietatisScientiarum U psaliensis, Uppsala, Sweden, seriesIV, vol. 2, no. 4, Nov. 1938.

3. Tung Chang. Technical Report no. 16, CruftLaboratory, Harvard University, Cambridge; Mass,July, 1947.

4. ANTENNA THEORY AND PRACTICE (book), S. A.Schelkunoff, H. T. Friis. John Wiley & Sons,Inc., New York, N. Y., section 13.12, 1952.

5. SUMMATION OF SLOWLY CONVERGING SERIES,I. Gumowski. Journal of Applied Physics, NewYork, N. Y., vol. 24, 19.53, p. 1068.

6. THEORY OF ANTENNAS DRIVEN FROM Two­WIRE LINE, R. King. Ibid., vol. 20, 1949, p. 832.

(n 2) [Ll - Ll -1 ]an= kb-/;b Kn+kb 11, 2 71.

Since d n vanishes for large n, the asymptoticvalue of an is given by

1 ( 2b )So Knrv~ lu -;; - 'Y-In n -

. (kb )n+l

J r(2n+2)

Finally (by simple insertion of numericalvalues), it can be shown that for k ~2.5, and

.n ~ 5, the asymptotic value of an as givendiffers negligibly from the correct value.

1 (x)nIn(x)t'oJr(n+ 1) "2

Now, using Sterling's equation to evaluatethe harmonic series, it can be shown that

Eulers constant.

Similarly, for n>kb

n-l

1 8b 1=-In-+-X

1r a 21r

f 1r E2inep(E-2jkb sin ep-1) -.- dcP+

o sm cP

1 fTr dcP- (E 2i n ep_ 1) -.-211'" 0 sm cP

1 8b 112k b

=- In --- [02n(x)+jJ2n(x)ldx-11'" a 2 0

n»l; n>kb, Knrv(~ In ~-~X11'" a 11'"

~ 1 ) j f2kbL..J2K+1 -2)0 J 2n(x)dx

o

2~ 11rL..J 2K+1

o

This result can be used conveniently to de­termine the form of K« for large N. Forn> >kb, the integral is small, vanishing inthe limit. Thus

1 8b 1Kn=-ln-+-X

7r a 27r

=i 2kbdX( -~i"E-iXSin<!>dq,)

r: r2kb

=-1/2)0 fl o(x )dx - j/2) 0 x

Jo(x)dx

It also can be shown-

j fTr (e 2j nep_1)=Ko+- E-2ikbsinep -.-- dcP7r 0 sin cP

Inserting the value for K o

n-l

x; =Ko+~L i"E-2ikb sin <!> + i(2n+l)<!>dq,

o

n-l

Kn=Ko+ LL1no

Inserting the integral expressions for d nyields

Another expression, useful for determin­ing K« for large n, can also be found. Fromthe foregoing it is seen that

1 f21r dcP 1 8b~ -- =- In -+terms O(a2/b 2 )27r 0 R(cP) 7r a

So, to the order of approximation consistentwith original integral equation

1 8b 12kbK o= - In--1/2 f2o(x )dx -

7r a 0

j/2i 2kbJo(x)dx

d n =Kn+1-Kn =fl2n+l(2kb)+jJ2n+l(2kb)

Magnetic Amplifiers

J. L. LOWRANCENONMEMBER AlEE

IN THE self-saturating type of mag­netic amplifier it is desirable to bias

the saturable reactor to a particular pointon its hysteresis loop. Higher gain andbetter linearity of the transfer functioncan be achieved in this manner. Shuntpaths around the diodes of the magneticamplifier afford a convenient means ofbiasing the saturable reactors.

The basic circuit examined in this re­port is a conventional full-wave, push­pull magnetic amplifier. An analysis ismade of the bias currents in two differentcircuit arrangements. The feedback ef­fects which are found in this type of cir­cuit are discussed and the effective feed-

Diode-Shunting •In

J. D. DOLANNONMEMBER AlEE

back factors for each circuit are deter­mined.

Various methods of obtaining bias forthe saturable reactors are shown in Fig. 1.Circuits 1 and 2 of Fig. 1 offer the advan­tage that fewer components are required.

Terms and SymbolsB = feedback coefficientE = supply voltageE = peak value of supply voltageE = average value of supply voltageHe =magnetization force required to satu­

rate the magnetic core at angle (J ofhalf-cycle, oersteds

I D = current through dummy load resistorafter both cores have sa turated

ID=average current through dummy loadresistor after both cores have sa tu­rated

I L = load current1L = average value of load current over half-

cycleIn = net control current seen by amplifierIs = current from signal sourceK e = open loop voltage gainK;' =closed loop voltage gainK t = open loop current gainK t' = closed loop current gainlm = effective magnetic path length of core,

centimetersN; = control-winding turnsN /b =feedback-winding turnsN g = gate-winding turnsRc =control-winding resistance& = dynamic foreward resistance of diodeRD = dummy load resistor

Paper 56-749, recommended by the AlEE MagneticAmplifiers Committee and approved by the AlEECommittee on Technical Operations for presenta­tion at the AlEE Summer and Pacific GeneralMeeting, San Francisco, Calif., June 25-29, 1956.Manuscript submitted March 14, 1956; madeavailable for printing May 8, 1956.

J. L. LOWRANCE and J. D. DOLAN are with theBendix Aviation Corporation, South Bend, Ind.

NOVEMBER 1956 Lotorance, Dolan-s-Diode-Shunting in Magnetic Amplifiers 619

0_1-------- CONTROL-------uCIRCUIT-S

n_...----- CONTROL-----o

n--------BIASCIRCUIT-3

RSH2

u-~-----CONTROL-------uCIRCUIT -I

o ...e------CONTROL------~u

CIRCUIT-20_---- CONTROL---~·u

CIRCUIT-4

Fig. 1. Various methods of obtaining biasfor saturable readors

Fig. 2. Saturation angle versus shunt resistor

R L = load resistorR, = signal source resistanceRsh = shunt resistorRw = gate-winding resistance of core plus

output resistance of power supplyVcr= voltage across diodeVcr = average voltage across diode over half­

cycleVg = voltage absorbed by core with respect

to gate windingr, = average value of voltage absorbed bycore with respect to gate windingduring half-cycle

Vsh = voltage across shunt resistorVSh = average voltage across shunt resistor

over half-cycle~ Vsn= change in average value of voltage

across shunt resistor over half cycleDiode dynamic resistance = slope of voltage­

current characteristic of diode afterthreshold voltage has been exceeded

Dummy load = resistors which are connectedto power supply center tap fromeach pair of diodes

Explanation of Bias Techniques

(3)

(2)

This may be written in terms of theaverage value of the supply voltage overthe half-cycle.

EVg = -(I-cos ())

2

EVg =- (I-cos ())

7l

where E is the peak voltage and 0 is theangle at which the core saturates, meas­ured from zero with respect to the begin­ning of the half-cycle. Referring to equa­tion 1, the sum of the voltages supportedby the core averaged over the completecycle is zero. Therefore the integral ofthe voltage absorbed 'by the core duringthe reset half-cycle can be expressed byequation 2.

As previously stated, for no-signal condi­tions, the angle at which the core satu­rates is a function of the bias current.In the case where shunt resistors areused, it is a function of the shunt resist-

EJ:f) . EVo= - sm eM= - (1- cos 0)1r 0 7r

Assuming negligible excitation current,the voltage supported by the core duringits gating half-cycle is equal to the inte­gral of the supply voltage over the periodbefore the core saturates. 2

( 1.)

Under conditions of no input signal thesaturation angle of each core in the self­saturating type of magnetic amplifier isa function of the bias current that flowsin its windings during the reset half­cycle. 1 This bias current can be suppliedeither by a separate winding on the coreor by allowing current to flow in the re­verse direction of the diode in the gatewinding during the reset half-cycle. Thelatter method offers certain advantagesbecause of ease in obtaining bias withlittle additional circuitry. This is thetype of circuit considered in this report.

Under .steady-state conditions, thevoltage supported by a core averaged overa complete cycle is zero.

Gating half-cycle = interval during whichsupply voltage is positive withrespect to anode of diode underconsideration

Reset half-cycle = interval during whichsupply voltage is negative withrespect to anode of diode underconsideration

Shunt or leakage resistors = resistors whichallow current to flow through gatewinding during reset half-cycle

Gate winding = winding on saturable reactorwhich is connected to power supply

rz«Jo ed()=O

10 K

30 K

50K80K

90K

620 Lowrance, Dolan-Diode-Shunting in Magnetic Amplifiers NOVEI\'IBER 1956

Analysis of Shunt Voltage

Due to the input. signal, the voltageabsorbed by each core is no longer equalto the voltages absorbed by the othercores. For d-e output from the stage,cores 1 and 4 are wound in the same sensewith respect to the control winding, andcores 2 and 3 are wound in the same sense,but opposite to cores 1 and 4. Thereforeit may be evident that the effect of con­trol signal upon cores 1 and 4 is equal andopposite to the effect upon cores 2 and 3.

Expressing the voltage absorbed by thecore in terms of the saturating anglesgives_ E _E=2(1-cos Od+IL(RD+RL)+

2R D + R LR

D(Rw+Rd)lL+

ID(RD+Rd+Rw) (18)

Adding the equations for the three timeintervals together yields the equation forthe average value of the supply voltageover the half-cycle.

E= VOl+1L(RD+RL)+ID(RD+Rw+~)+

(Rw+Rd)eR~:RL)h (14)

E= V02+ILRD+ID(RD+Rw+Rd) (15)

E= V"3+ Vcr3-1LRD-IDRD (16)

E= V"4+ Vcr4-1L(RD+RL)-IDRD (17)

After both cores' have saturated

After core 1 saturates but before core 2saturates

Ei 91

- sin OdO = V"3+ VCT3 = Vg 4+ VCr 4

7r 0

_ _ EV<,>hl= VSh2=2(1+cos O)+IDRD+1DRd (13)

It may therefore be shown by deductionthat for zero input signal conditions thetwo circuits are equivalent with respectto the bias circuit.

Analysis of Shunt Voltage withInput Signal

The preceding statement is true, how­ever, only for zero input signal conditions.The circuit will now be considered for aninput signal applied to the amplifier, overthe linear region of the transfer function.The signal is assumed to be in sense whichmakes core 1 saturate before core 2. Asin the previous cases, it is assumed thatcores 1 and 2 are positive with respect tothe power supply center tap.

CIRCUIT 1

Before core 1 saturates

.1:91~ sin Od6= VOl = Vg2

7f 0

(12)

The next step is to determine the rela­tionship between the saturating angle andthe magnetizing force. I t is possible toobtain these data by empirical methodsfor a given core material and coregeometry. 3 These data are then appli­cable to designs in which this type of coreis used.

The change in saturation angle withvarious values of shunt resistors and con­stant supply voltage is shown in Fig. 2.These data were obtained by makingmeasurements in the circuit and calculat­ing the anglefrom the foregoing equations.Other techniques such as oscilloscopemeasurements and timing devices may beused to obtain the same data. 4

An examination of circuit 2 shows thatfor zero signal conditions the equationsfor the voltage across the shunt resistorsis the same as in circuit 1.

where H 9 is the magnetizing force inoersteds and I is the magnetization cur­rent in amperes. In this case, I is the cur­rent through the shunt resistor.

The shunt resistance can be expressedin terms of the saturating angle of thecore and the supply voltage

R -R _E(1+COSO)(RD+Rd+WI2)shl- sta>:

H91tn RD+Rd+RwO.47rNG

For a given saturating angle and supplyvoltage, the value of reset magnetizingforce can be calculated from the equation

H9= O.47rNIItn (11)

(6)

(7)

(4)

_ _ _. EVcr 4 = E + I DRD- - (1 - cos 0)

2

Therefore the average voltage across thediodes over the half-cycle is_ _ _ E1,rcI3=E+IDRD--(1-cos 8)

2

After saturation

Ef7r.) sin Od{)= -IDRD+ VCr3+ Vg 3~ 0

= -IDRD+ VCr4+ V,,4 (5)

'[he voltage absorbed by the core dur­ing the half-cycle is_ E1/0 = -- (I-cos OJ. 2

_ E __V'sl l1 = 2(1+cos 8)+IDRD+IDRd (8)

_ E _V~h2= 2(1+cos 8)+IDRD+IDRd (9)

EfO:) sin OdO = VCT3+ V,,3 = VCT4+ 11Y4

- U

E _2(1+cos O)=ID(RD+Rd+Rw)

_ _ EVshl = V sh2= - (1 +cos 0) x­

2

Rd is the dynamic forward resistance ofthe diode.

The voltage across the dummy loadresistor may be written in terms of thesaturating angle (), and the equation be­comes

1D is the average value of the currentflowing in the dummy load resistor of cir­cuit 1 after both cores have saturated.

Adding the forward drop of the otherdiode being shunted by the shunt resistorduring its gating half-cycle, the voltageacross the shunt resistor becomes

Circuit 1 will be considered first.Assume that cores 1 and 2 are positive

with respect to the center tap of the powersupply and cores 3 and 4 are negative withrespect to this center tap.

Prior to saturation

ance and the voltage across the shuntresistor.

The voltage across the shunt resistorswill be analyzed in the time interval pre­ceding and the time interval following thesaturation of the cores. The first case tobe considered will be for zero inpu t signal..A.11 cores will saturate at the same angleunder this conditon.

NOVEMBER 1956 Lowrance, Dolan-Diode-Shunting in Magnetic Amplifiers 621

Fig. 3. Effect of control signal on shunt voltage

2R D+2RL ]+ID(RD+Rw+2Rd) (26)

V,:h2=IL(2RD)+lD(RD+Rw+2Rd) (27)

From equations 26 and 27, it is seen.. that the voltage across the shunt resistor

is a function of the load current and thecurrent that flows in the dummy load,after both cores have saturated. Solvingfor the difference in the shunt voltages

- - I [2RD+RLVshl- VSh2= L RD

X

(Rw+2R.!H2RL] (28)

From this equation, it is seen that thedifference or unbalance in the shunt volt­ages is a function of the load current.

CIRCUIT 2

An examination of circuit 2 shows thatthe voltage appearing across the diodesmay be expressed by the same equationsderived for circuit 1. The voltages acrossthe shunt resistors are

(29)

- - (2RD+RL)Vsh~=+ILRL+Rd R

DX

lL+RdlD+ VCT 3 (30)

Substituting equations 22 and 23 forthe diode voltages gives

Vshl= -ILRL+IDRd+lLX

[2R D+ R L ]R

D(R w+Rd)+2RD+2RL +

ID(2RD+Rw+Rd) (31)

- 1 - _RdVsh2 = + LRL+IDRd+IL R

D(2RD+RL)+

2ILRD+ID(2RD+Rw+Rd) (32)

(33)

A-Circuit 1, voltage across Rsh for no-input signalB-Circuit 2, voltage across Reh for no-input signalCo-Circuit 1, voltage across Rshl for an input signal positive with respect to dotD--Circuit 2, voltage across Rs h 2 for an input signal positive with respect to the dotE--Circuit 1, voltage across RSh 2 for an input signal positive with respect to the dotF--Circuit 2, voltage across Rehl for an input signal positive with respect to the dot

Subtracting equation 32 from 31 gives

_ - - (2RD+RL)Vshl- Vsh2 =IL R

D(Rw)

As in the previous analysis of circuit 1,it is seen that the difference in shunt volt­ages is a function of the load current.

The voltages across the shunt resistorsare

- - [2RD+RLVer 4 =IL ------- (Rw+Rd )+RD

2R D+2RL]+ID(RD+Rd+Rw) (23)

Explanation of Feedback Effect

I t is assumed that the flux that is reseton the core during the reset half-cycle isa function of the average net ampere­turns that exist during the reset half­cycle. This is true for all practical pur­poses. The net ampere-turns are thealgebraic sum of the control ampere­turns and the leakage ampere-turns. Inthe preceding mathematical analysis ofthe shunt voltage it is seen that a differ­ence in shunt current exists which is afunction of the amplifier output. A dif­ference in the reset flux of cores 1 and 2

(25)

(24)- - - (2RD+RL) -Vshl= Vcr4+ILRd RD

+IDRd

Vsh2= VCr 3+1DRd

Substituting for the diode voltages fromequations 22 and 23 gives

- .. [2RD+RLVshl=IL (Rw+2Rd)+RD

(19)

_ E _E = i( i-cos 81) -IL(R L +Rf))-

lDR D + VCT4 (21)

Solving equations 18 and 19 for cos 01

and cos O2 and, substituting in equations20 and 21, the voltages across diodes CRaand CR4 are

Lowrance, Dolan-Diode-Shunting in Magnetic Amplifiers NOVEMBER 1956

(40)J.V/b Rs+Rc1+- --KelVe R L

Ke'

Fig. 6. Analysis of output network ofam!,lifier with regard to maximum power

transfer

Is is the input signal and In is the netcontrol current to which the amplifierresponds.

Closed loop current gain is

- - - ( RL)Vshl- V sh2 =IL 2Rw + Rw RD

- Rw ­=2RwI L +- VL (38)R D

In this case, the feedback is a result ofthe impedance R w and can be minimizedby decreasing the winding resistance andthe source impedance of the powersupply.

In both circuits, there is feedback ofvoltage as a current and of current as acurrent. Fig. 4(B) shows the change ingain of a stage in circuits 1 and 2.

The feedback effect due to diode shunt­ing can be related to the conventionalfeedback equations. This is equivalentto feedback into a separate winding, inthe same winding configuration as thecontrol winding. This type of circuit isshown in Fig. 5.6 The gain expressionsare, for open loop gain

I L I LK]=- =­In Is

I L K[K/=-----=----1"lfb Nib (39)

I n+Nc I L 1+ }..TC

K]

Open loop voltage gain

R LKe=K] --Rs+Rc

Closed loop voltage gain

From this equation it is seen that bothload current and load voltage are factorsin determining the feedback in the ampli­fier. Although the two are dependentupon each other, they are separate fac­tors in the difference equation.

The difference in shunt voltages in cir­cuit 2 may be arranged in a similar man­ner

(37)

,INPUT

(8)

TRANSFER CURVERSH = 80K OHM

CIRCUIT*2

CURRENT FEEDBACKMAGNETICALLY MIXED

Fig. 5. Block diagram of amplifier withcurrent feedback, magnetically mixed

2 would indicate that there is no difference in the shunt voltages. However, anexamination of the wave form shows thatthe voltage is different on alternate half­cycles, as shown in Fig. 3(D) and (F).If VShl is larger than VSh 2 during theinterval that cores 3 and 4 are being reset,the change in leakage current will tendto urge the core farther away from satura­tion. This opposes the effect of the con­trol signal and is therefore considered tobe negative feedback.

In the next half-cycle, Vshl is largerthan Vsh2• In circuit 2 the leakage cur­rent through RSh 2 is the bias current forcore 1. The effect is the same as for theprevious half-cycle, and is also consideredto be negative feedback.

The difference in shunt voltages in cir­cuit 1 may be divided into two parts

Vshl- VSh~ =!L(2Rw+ 4Rd )+

- ( RW+2~)VL 2+---RD

50K

INPUT

(A)

TRANSFER CURVESFOR VARIOUS VALUESOF RSH

CIRCUIT -2

output = FI, input+F2 (output) (36)

This equation is of the form which com­monly describes feedback systems.'Therefore it may be said that the changein shunt voltage as a function of outputis a form of feedback.

Looking at the equation for the differ­ence in shunt voltage in circuit 1, it isseen that VSh1 is larger than VSh2 for apositive signal with respect to the dotapplied to the control winding. An in­crease in the shunt voltage Vshl will tendto urge the cores 1 and 4 farther awayfrom saturation. This is contrary to theeffect of the control signal, which tendsto urge the cores toward saturation.From this it may be deduced that the ef­fect is that of negative feedback.

An examination of circuit 2 shows thatfor the same control circuit conditionsVshl is larger than VSh2 for the half-cycleduring which cores 3 and 4 are being reset.On the next half-cycle VSh 2 will be largerthan Vsh 1• A quantitative measurementof the voltage across the shunts in circuit

Therefore

output = FI (input ampere-turns-l-zs leakagecurrent ampere-turns) (34)

~leakagecurrent ampere-turns = F2 ( ou t pu t )(35)

Fig. 4. Transfer curves. A-For various values of shunt resistance. B-For circuits 1 and 2

or cores :3 and 4 constitutes an outputvoltage from the amplifier.

On the basis of the statements in thepreceding paragraph, the following equa­tions are written

NOVEMBER 1956 Lowrance, Dolan-Diode-Shunting in Magnetic Amplifiers 623

For circuit 2

1. ON THE MECHANICS OF MAGNETIC AMPLIFIEROPERATION, R. A. Ramey. AlEE Transactions,vol. 70, pt. II, 1951, pp. 1214-23.

2. SELF-SATURATION IN MAGNETIC AMPLIFIERS,w. J. Dornhoefer. Ibid., vol. 68, pt. II, 1949, pp.835-50.

Conclusions

5. RESPONSE OF PHYSICAL SYSTEMS (book),John D. Trimmer. John Wiley & Sons, Inc.,New York, N. Y., 1950, pp. 174-77.

6. ALTERATION OF THE DYNAMIC RESPONSE OFMAGNETIC AMPLIFIERS BY FEEDBACK, R. O.Decker. AlEE Transactions, vol. 73, pt. I, 1954(Jan. 1955 section), pp. 658-65.

7. FAST RESPONSE WITH MAGNETIC AMPLIFIERS,D. G. Scorgie. Ibid., vo1..72, pt. I, 1953, pp. 741-49.

The gain of the full-wave, push-pullmagnetic amplifier is related to the satura­tion level to which the saturable reactoris biased. The design equation for deter­mining the value of shunt resistor to usein obtaining a particular saturation angle

E(1+cos (J) (RD+Ra+ Rw/2)Helm. RD+Ra+Rw

0.411"1\T"

References

3. TECHNIQUES FOR MEASURING CASCADED SELF­SATURATING MAGNETIC-AMPLIFIER PERFORMANCE,Henry Kaplan, Gerald Wolff. Ibid., vol. 73, pt. I,1954 (Jan. 1955 section), pp. 581-84.

4. OSCILLOGRAPHIC TECHNIQUES FOR THE EVALU­ATION OF MAGNETIC AMPLIFIER RESPONSE, D. L.Critchlow. Ibid., vol. 74, pt. I, Nov. 1950, pp.607-10.

The shunt resistors create a feedbackloop which introduces an appreciableamount of negative feedback into theamplifier. In instances where it is desir­able to minimize the negative feedback,circuit 2 is more appropriate than circuit1. In circuit 2, the feedback effect isa function of the winding resistance of thegate winding and the internal impedanceof the power supply. It can be mini­mized to afford higher open-loop gain.

The feedback factor B, relating thefeedback effect of the shunts on the open­loop gain of the amplifier, has beenderived. 1Jse of this factor is an aid inthe analysis of the transfer function ofthis type of amplifier.

(50)

and for circuit 2

Graphic Analysis

(N(J )(Rw)( 1)(j= - - 1+-n, Rsh y2

RD+RLRw+Rd= R D2RD+RL

R - (Rw+Rd)RD +RL- RD+Rd+Rw D

See also Fig. 6.Solving for RD and R L in terms of Rw

and R d gives

RD=y2(Rw+Ra) R L=2(Rw + Ra)

ff=(~)(;sJ[ (3+~2)XRw+(4+V2)Rd ] (49)

Substituting these values in equations45 and 48 gives for circuit 1

The effect of the control signal on theshunt voltage is shown in Fig. 3.

For zero signal conditions, the shuntvoltage in circuit 1 is equal to the shuntvoltage in circuit 2. This is shown inFig. 3(A) and (B). The sharp rise in thewave form occurs at the time the coresbecome saturated. Saturation occursat approximately 70 degrees in this par­ticular design.

Fig. 3(C) shows the increase in the volt­age across R~hl in circuit 1 when a signalvoltage is applied which is positive withrespect to the dot on the control winding.Fig. 3(E) shows the decrease in voltageacross RSh 2 for the same conditions.

Fig. 3(D) shows the increase in voltageacross RSh l in circuit 3, for a positive in­put voltage with respect to the dot of thecontrol winding. On the following half­cycle the voltage decreases across theshunt. Examination of Fig. 3(F) indi-

. cates that this wave form is identical inform to that of Fig. 3(D) and out ofphase with it by 1/2 cycle.

(45)

(48)

(42)

(46)

(41)

1+(N(J)(_1)xlVe 2Rsh

[2_R_D_+_R_~( Rw +2Rd ) +2RL]KIRJ)

(44)

Therefore for circuit

IL[2RD+RL~ VSh=-2- -----~~Rw+2Rd)+2RL (43)

K/=. (i.V(J)( 1 ) (2RD+RL)1+ .-.- R K.V( 2Rs h R D to I

(47)

An analysis of the output network ofthe amplifier with regard to maximumpower transfer yields the following equa­tions

The differential change in shunt voltageis assumed to be equally divided

1A V~h = 2(V~hl - Vsh2)

K/

In applying these equations to theshunt feedback effect, the gate windingis the counterpart of the feedback wind­ing. The change in shunt voltage dividedby shunt resistance is analogous to thefeedback current.

Making these substitutions gives

K ' I L1 .!.V(J ~VshI n +-:- -·--­

.:.v c Rsh

624 Lowrance, Dolan-Diode-Shunting in Magnetic Amplifiers NOVEMBER 1956