ch 2 (support) power supply circuits

8/13/2019 Ch 2 (Support) Power Supply Circuits

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ECE590: Power Supply Circuits

a. Transformersb. Rectifiers

c. Filtersd. Regulators


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Bridge (full-wave) rectifier

Calculate the Ripple Factor

TRANSFORMER RECTIFIER FILTER


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SOLUTION1. Vrms = Vp / (2) Vrms : effective value of a sinusoidal signal

Vp : peak value (amplitude) is the maximum value of a sinusoidalsignal

2. For a transformer,

The secondary voltage: N 2/N 1 = V2/V1 N2/N 1 is the secondary to primary number of coils.3. Peak bridge (full-wave) rectified voltage:

Vp(rect) = Vp(sec) 2(0.7V) Located at the input of the filter (after rectification)

4. Ripple factor: approximate peak-to-peak RIPPLE VOLTAGEat the output/approximate DC VOLTAGE at the output

Ripple voltage: V r(pp) = ILt/C = (V p(rect) /R L)/fC VDC = Vp(rect) Vr(pp) /2

Ripple Factor = V r(pp) /V DC


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Bridge (full-wave) rectifier

TRANSFORMER

Vp(sec) = 16.3V

RECTIFIER

Vp(rect) = 14.9V

FILTER

Vr(pp) = 1.13VVDC = 14.3VRipple Factor = 0.079The % ripple = 7.9%


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After rectifying & filtering will produce a DCvoltage, but with some variation (ripple).

Represented by the Ripple Factor.

Vi

Vm VDC

Vr(p-p)

t


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Jan 2012 (Q1b)


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June 2012 (Q1b)


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POWER SUPPLIES

REGULATORS


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VOLTAGE REGULATOR A voltage regulator is used to maintain a constant

output voltage.

A voltage regulator also can use dc input to provide adc voltage that not only has much less ripple voltage

but remain the same dc value even the dc inputvoltage varies or the load connected to the output dcvoltage changes.


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ZENER REGULATOR (ZENER DIODE) The characteristic drops in almost vertical manner at reverse bias potential

denoted as Vz. The current in zener diode has a direction opposite to what offorward bias diode.

Zener is designed to work in reverse bias direction without harming the diode.Two things happen when reverse breakdown voltage is reached

Diode current increase drasticallyReverse voltage across diode, V R remain constant.

A diode operated in this region will have relatively constant voltage across it,regardless the value of current through the device. A zener diode is used tooperate in reverse region and maintain constant voltage regardless thevariation current. This make zener diode a goood voltage regulator in thecircuit.


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ZENER REGULATOR (ZENER DIODE)Diode breakdown There are two types of diode brakdown1. Zener breakdown2. Avalance brakdown

(1) Zener breakdown Occur at a much lower value of reverse voltage, VR than those the

avalanhce breakdown. The heavily doping of the zener diode causesthe device to have much narrow depletion layer . As a result it takesvery little VR to cause the diode to go into breakdown. Typically 5Vor less.

Zener breakdown is a result of ionization of covalent bonds due to

high intensity electric field that inherently exist increasing thepotential across it. It will eventually yank valence electron out oftheir covalent bond, producing free electron or conduction electron.This additional carrier drift across the junction under the influence ofreverse voltage causing reverse current to increase rapidly and bringabout zener reverse breakdown.


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ZENER REGULATOR (ZENER DIODE)(2) Avalanche Breakdown

Avalanche breakdown is a result of the ionization of covalent bond byminority carriers accelerated across the reverse biased junction. This resultsin multiplication of carriers crossing the junction causing reverse current toincrease rapidly with reverse voltage (VR). Reverse breakdown in a diodeoccurs at a voltage > 5V.

Zener Equivalent circuit

r z

V Z

VZ

Zener diode Complete equivalentcircuit

Approximate equivalent


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ZENER REGULATOR (ZENER DIODE)

Vi

R

V Z

P ZM

IZ

IR

IL

R L V L

+

-

The figure shows thesimplest voltageregulator using zenerdiode.

Three condition of inputvoltage, Vi and loadresistance, R L are:

1. Vi and R L fixed

2. Vi fixed and R L variable

3. Vi variable and R L fixedVoltage regulator using zener diode


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(1) Vi and R L fixed

Vi V L

+

-

R

V Z

P ZM

IZ

IR

IL

R LVi V L

+

-

R

V Z

-

IR

IL

R L

+

a bThe simplest of Zener diode networks appears in figure (a). The applied dcvoltage is fixed, as is the load resistor. The analysis of the circuit can besolve by the following steps:

1. Determine the state of the zener diode by removing it from network as

shown in figure (b) and calculating the voltage across the resulting opencircuit.

2. Using voltage divider rule:

R R VR

VVL

iLL


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3. If VL VZ : Zener diode is ON If VL< VZ : Zener diode is OFF

4. Substitute the appropriate equivalent circuit as shown in figure below and solve forthe desired unknowns.

Vi V L

+

-

R

V Z

P ZM

IZ

IR

IL

R L

Approximate equivalent circuit of zener diode

Since the zener diode isparallel with load resistor,the voltage across parallel

element must be the same,therefore VL = V Z

Apply Kirchoffs currentlaw I R = IZ + IL

IZ = IR - IL

L

LL

R

VI

R

VV

R

VI

LiR R

Power dissipation by zener diode: P Z = V ZIZ

(1) Vi and R L fixed


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Vi VL

+

-

R

+

V

IZ

IR

IL

R L1.2k

1k

16V-

R R

VR VV

L

iLL

Voltage divider:

V73.8k 2.1k 1

)V16(k 2.1

Since V is less than V Z (8.73V < 10 V) diode is OFF .

4. Substituting the open circuit equivalent circuit will result in the same network as above.Where we find

V = V L = 8.73V V R = V i VL = 16 -8.73 =7.27V I Z = 0A P Z = I ZVZ =VZ(0A)=0W

(1) Vi and R L fixed Example


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Due to the offset voltage V Z, there is a specific range of resistor values (andtherefore load current) which will ensure that the zener is in the on state.

Too small a load resistance R L will result in a voltage V L across the loadresistor less than V Z and the zener device will be in the off state.

To determine the minimum load resistance that will turn the Zener diodeON calculate the value of R L that will result in a load voltage V L = VZ. Thatis,

R R

V RV Vz

L

i L L

Solving for R L, we have

Z

ZL

VVi

RVR mi n

(2) Vi fixed and R L varied


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a) For the network shown, determine therange of R L and I L that will result in V R,being maintained at 10V.

b) Determine the maximum wattage ratingof the diode.

R

1k

IZM =32mA

V Z =10V

IR

IZ

R L

IL+

-

Vi=50V

Solution:

a) To determine the value of RL that will turn the Zener diode on use the formula

Z

ZL

VVi

RVR min 250

40

k 10

V10V50

)V10(k 1

The voltage across the resistor R is then determined by using KVL in the input loop.

V40V10V50VVV ZiR

(2) Vi fixed and R L varied Example


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The value of IR is:

mA40k 1V40

R V

I R

R

The minimum level of I L is then determine by

mA8mA32mA40III ZMR L min The maximum R L is then determine by

k 25.1mA8

V10IVR

mi n

ma x

L

ZL

A plot of V L versus R L appear in figure (a) and for V L versus I L in figure (b)

10V

VL

0 8mA 40mAIL

10V

VL

0 250 1.25k RL

b)

(a) (b)

mW320)mA32)(V10(

IVP ZMZmax

(2) Vi fixed and R L varied Example


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Q: To determine the range of values of R L and I L that will ensure V L = VZ; i.e. Zener Diode inON mode.

R

1k

IZM =32mA

V Z =10V

IR

IZ

R L

IL+

-

Vi=50V


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Why is there a range of values? This is because of : V L = RL x IL (Ohms Law) Thus, the range of values would be:

i. RLmin x ILmaxii. RLmax x ILmin

Therefore, the objective is to find the range: themaximum and minimum values of each R L and I L that will ensure V L = VZ.


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i. RLmin x ILmax

RLmin is obtained via Voltage Divider rule RLmin = (RxVz)/(V i-Vz) = (1kx10v)/(50v-10v) = 250 ILmax = VL/RLmin = Vz/RLmin = (10v)/(250 ) = 40mASince: IR = (Vi Vz)/R = (50v 10v)/1k = 40mAThe value of I

Lmax indicates that all of the current from I

R

will flow through R L, thus leaving I Z = 0A. Therefore, I Lmax is said to be the maximum current value that could flowthrough resistor R L.


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ii. RLmax x ILmin

Since IZM is the maximum value of I Z that couldpass through the Zener Diode, then:

ILmin = IR IZM = 40mA 32mA = 8mA

RLmax = VL/ILmin = 10v/8mA = 1.25k


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For fixed value of RL, the voltage Vi must be sufficiently large to turn the Zener diode on. Theminimum turn on voltage Vi = Vimin is determined by

R R VR VV

L

iLZL where

L

ZLi

R V)R R (

V mi n

The maximum value of Vi is limited by the maximum Zener current I ZM. Since I ZM = I R-IL

LZMR III ma x

Since I L is fixed at V Z/R L and I ZM is the maximum value of I Z, the maximum Vi is defined by

ZR i

ZR i

VR IV

VVV

maxmax

maxmax

(3) Vi varied and R L fixed


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JAN 2013


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Jan 2012 (Q3a)

ch 2 (support) power supply circuits

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