by s k rai electronics engineering department bkbiet, ceeri road pilani (raj.)-333 031

A LECTURE

ON

VOLTAGE REGULATORS

by

S K Rai

Electronics Engineering Department BKBIET, CEERI RoadPilani (Raj.)-333 031

EE2603-01Electronic Circuit Analysis

Q. Why DC Power Supply?

Q. Source of DC………..

Q. What is regulation? It is the process to maintain the terminal voltage as

constant even if input voltage varies or load current is varying.

Q. When regulation is required?

BKBIET, Pilani


DC Power Supply

Vm

n:1+1

VacfS

CRi

0.7

VZ

RL

IL

IB

DC Power supply is a circuit that provides a steady dc voltage obtained by rectifying the ac voltage. The following is the block diagram of a power supply stage.

Step-down transforme

r

High ac

voltage

Rectifier

Low ac voltag

e

Ripple Filter

High ripple dc voltage

Voltage regulato

r

Low ripple dc voltage

Load using dc voltage

Pure constant dc voltage

Vm

Vm

Vdc

Vdc

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n:1+1

Vac=220Vf=50Hz

Vmax

Vmax

RLC

I dc

Vdct2

t1

Vmin

Vmax

V

T

Vdc

V= - m2t2 V= m1t1VO

Ripple and dc voltage

V is higher for high Idc

voltageacoffrequencyfandR

VIwhere

3fC4

I)rms(VvoltageRipple

fC2

IV)slopefrom(

)wavetriangular(32

V)rms(voltageRipple

L

dcdc

dcr

dc

n

2VVand

voltageacoffrequencyf,R

VIwhere

fC4

IVVvoltagedc

fC2

IV)slopefrom(

2V

Vvoltagedc

acmax

L

dcdc

dcmaxdc

dcmax

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Ripple Factor

%505.0F100k1504

1

3CfR4

1V

)rms(Vr)c(

Ldc

r

3CfR4

1

3fC4V

I

V)rms(V

rfactorRipple

R

VIand

3fC4

I)rms(VhaveWe

Ldc

dc

dc

r

L

dcdc

dcr

t2t1

Vmin

Vmax

V

T

Vdc

V= - m2t2 V= m1t1VO

Example:Given the Power supply circuit shown, find (a) output dc voltage Vdc (b) output ripple voltage Vr (rms) (c) ripple factor “r”

10:1+1

Vac=220Vf=50Hz

Vmax

Vmax

RLC

I dc

Vdc1k100

F

V85.03100504k1

6.29

3fC4R

V

3fC4

I)rms(VvoltageRipple)b(

L

dcdcr

V6.2905.1

1.31V

V1.31100k1504

11V

102220

CfR41

1V

CfR4

V

102220

fC4

IVVvoltagedc)a(

dc

dcL

dc

L

dcdcmaxdc

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RC Filter in Power Supply

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

RL

t

VC1

RC2 RL

Idct

VC2

L

L1dc2dc2 RR

RVVdcforopensC

1

dcmax1dc1dc fC4

IVVfromfoundbecanV

Vdc1 Vdc2

RC2 RLt

VC1

V1(rms)

t

VC2

V2(rms)

R

X)rms(V)rms(V 2C

12 3fC4

I)rms(Vfromfoundbecan)rms(V

1

dc11

1dc

11 V

)rms(Vrripple

2dc

22 V

)rms(Vrripple

)Hz50ffor(kC56.1

5060

C3.1

X

or)Hz60ffor(kC3.1

X

kinisXandFinisCIf

222C

22C

2C2

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Series Voltage Regulators

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

RLR

VZ

Vi Vo

Unregulated power supply

Series voltage regulator

Regulated power supply

Vm

Vdc

Vdc

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RLR

VZ

Vi

1kW0.22kW

12V

50

20V

Vo=?

IZ=?

Example:Given the Series regulator circuit shown, find (a) output dc voltage Vo (b) Zener current IZ

V3.117.0127.0VV)1( Zo

mA4.36k22.0

V12V20I)2( R

mA174.36226.04.36III)4( BRZ

mA226.050

mA3.11I

mA3.11k1

V3.11I)3(

B

L

0.7V+

-

+

-

+

-

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Improved Series Regulator

0.7V

+

-

+- VZ

VB

1

IR1

VCE1+ - IL IL

Q2

IC2

IB1

IR4 R1

R2

RL

Vo

R3

Q1

R4

Vi

RL

+

-+

-

+

-

2

12Ro

21

2o2R R

R1VV

RR

RVV

7.0VV Z2R

7.0VV o1B

2B22C1

L1B

1B2C4

1Bi4R

II&I

I

IIR

VVI

L1R1RZ I1.0I&I1.0I

Analysis of the Improved Series Regulator

Z3

ZZZ3

Z2B3

Z2B

I9.07.0

R

I9.0I1.0IR

7.0

I1.0IbutR

7.0II

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IC Series Regulator

+

-

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

2

1Zo

21

2oZ2R R

R1VV

RR

RVVV

R1

Vi

R2

R3

RLVZ

Vo

VZ

VZ

Ii=0

Ii=0Because Ii=0, a small power Zener may be used to regulate the whole load current in RL

IC Series Regulator

BKBIET, Pilani


+

-

-+-

+

+-

-

+

-

+

-

R1

R2

RL

Vo

Vi

R3

VZ VZ

VZ

Ii=0

Ii=0

IL

ILRSC

Current-limiting IC Series Regulator

IO

IB1

IC2

High IL

Analysis of the Circuit

1. Vi and R3 will produce regulated voltage VZ2. Because of the Op. Amp. Property, VZ will present at the junction of R1 and R2

3. VZ=VR2 will result Vo which will be higher than VZ

4. Normal value of IL will produce normal VBE at Q2 whose collector current is small so that Io from IC will go mostly to IB1

6. Q2 collector current IC2 becomes large so that Io from IC will go mostly to IC2

reducing IB1 and consequently IL down to the limited current value.

Q2

Q1

5. High value of IL will produce high VBE at Q2

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Shunt Voltage Regulators

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dcShunt voltage regulator


Vm

Vdc


Vdc

Rs

VBE

Vi

VZ

Vo

-+

+

-+

-

RL

+

-

Vo = VZ+VBE = Regulated voltage

VRs = Vi – VZ – VBE which is constant if Vi is constantIRs = (VRs/Rs) is also constantIC = IRs will all flow through BJT (when no load RL and BJT is hot)

if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold)

Therefore Shunt Regulators are suitable to supply circuits where RL is always connected.

BKBIET, Pilani


n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

2

1Zo

21

2oZ2R R

R1VV

RR

RVVVBecause Ii=0, a small power

Zener may be used to regulate the whole load current in RL

+

-

IC Shunt voltage regulator


Vm

Vdc


Vdc

RL

+

-

VRs = Vi – Vo which is constant if Vi is constantIRs = (VRs/Rs) is also constantIC = IRs will all flow through BJT (when no load RL and BJT is hot)

IC Shunt Voltage Regulator

if load RL is connected IRs will mostly drain by the load RL making IC small (when there is load BJT is cold)

Therefore Shunt Regulators are suitable to supply circuits where RL is always connected.

R1

Vi

R2

R3

VZ

Vo

VZ

VZ

Ii=0

Ii=0

Rs

BKBIET, Pilani


Voltage Regulator IC

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

IC Part Number Output Voltage Min input Voltage

7805 +5V +7.3V

7806 +6V +8.3V

7808 +8V +10.5V

7810 +10V +12.5V

7812 +12V 14.6V

7815 +15V +17.7V

7818 +18V +21V

7824 +24V +27.1V

Fixed Positive Voltage Regulator IC

C2 RL

1

Voltage regulator IC

2

3

7812

Vo=+12V+Vi


Vm

Vdc

Regulated power supply +12V

VdcC1 is a ripple filter capacitor (C1 > 100 F)

C2 is a high frequency filter (C2 < 0.1 F)

BKBIET, Pilani



Vm

Vdc

15V

7.3V

1


2

3

7805

Vo=+5V+Vi=15Vmax

IL = ?

Example:Given the IC regulator circuit shown, 7805 requires a minimum input voltage of 7.3V. (a) find the maximum load current IL that can be used. (b) if a maximum load current IL = 500mA is required, what is the size of the filter capacitor? Take ac supply frequency 50Hz.

V22.232

7.7

32

V)rms(VV7.7V3.7V15V r

mA77A077.0310050422.23fC422.2I

3fC4

I)rms(VvoltageRippleV22.2)a(

dc

dcr

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

C1 = 250mF

F650350422.2

A5.0

3f422.2

IC

3fC4

I)rms(VvoltageRippleV22.2)b(

dc

dcr

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n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

Fixed Negative Voltage Regulator IC

C2 RL

1


2

3

7912

Vo=-12V-Vi


-Vm

-Vdc

Regulated power supply -12V

-Vdc

IC Part Number Output Voltage Min input Voltage

7905 -5V -7.3V

7906 -6V -8.3V

7908 -8V -10.5V

7910 -10V -12.5V

7912 -12V -14.6V

7915 -15V -17.7V

7918 -18V -21V

7924 -24V -27.1V

C1 is a ripple filter capacitor (C1 > 100 F)

C2 is a high frequency filter (C2 < 0.1 F)

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Adjustable Voltage Regulator IC

The IC regulator LM317 can be used to produce any regulated output voltage between 1.2V to 37V. Typical IC values are: Vref = 1.25V and Iadj = 100mA

Adjustable Voltage regulator IC

1.2V < Vo< 37V-Vi

VIN VOUT

ADJ

LM317

R1

R2Iadj = 100 A

Vref = 1.25V

2adj1

2ref2

1

refref2adj1Rref2R1Ro RI

R

R1VR

R

VVRIIVVVV

21

2o RA100

R

R1V25.1V

n:1+1

Vac=220Vf=50Hz

Vmax

VmaxC1

I dc

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VIN VOUT

ADJ

LM317

R1

R2

n:1+1

Vac=220Vf=50Hz

Vmax

Vmax C1

I L

470F

C2

IL

RLC3100 A

1.25V

Analysis of the LM317 IC Regulator circuit

1. Unregulated power supply enters LM317 with a dc voltage of VIN and a ripple peak-to-peak voltage of DV=Vmax-Vmin depending upon the value of C1 and IL used.

5. C3 is unwanted high-frequency low capacitance bypass capacitor which is always necessary at the output of all regulator circuits.

3. Unwanted voltage spikes are suppressed by the two diodes connected from VOUT to ADJ and VIN terminals of the IC.

4. C2 is a ripple filter across R2 .

2. Note that LM317 data are Vref = 1.25V and Iadj = 100mA and will produce a regulated dc voltage VOUT determined by R1 and R2

BKBIET, Pilani


VIN VOUT

ADJ

LM317

R1

R2

V85.131.075.13k1A100100

k11V25.1RA100

R

R1V25.1V)b( 2

1

2o

n:1+1

Vac=220Vf=50Hz

Vmax

Vmax C1

I L

470F

C2

IL

RLC3

V577.032

2

32

V)rms(VV2VVVifcaseIn rminmax

mA942.933470504577.03fC4577.0I

3fC4

I)rms(VvoltageRippleV577.0)a(

dc

dcr

100 A

1.25V

Example:Given the LM317 IC regulator circuit shown, (a) find the maximum load current IL that can be used, if a ripple of peak-to-peak voltage of 2V is present at the input of the regulator. (b) what is the output voltage of the variable IC regulator if R1 = 0.1kW and R2= 1k W ? Take ac supply frequency 50Hz.

BKBIET, Pilani


Than

ksBKBIET, Pilani

by s k rai electronics engineering department bkbiet, ceeri road pilani (raj.)-333 031

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