by s k rai electronics engineering department bkbiet, ceeri road pilani (raj.)-333 031
TRANSCRIPT
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
EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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
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EE2603-01Electronic Circuit Analysis
+
-
-+-
+
+-
-
+
-
+
-
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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EE2603-01Electronic Circuit Analysis
Shunt Voltage Regulators
n:1+1
Vac=220Vf=50Hz
Vmax
VmaxC1
I dcShunt voltage regulator
Unregulated power supply
Vm
Vdc
Regulated power supply
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
EE2603-01Electronic Circuit Analysis
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
Unregulated power supply
Vm
Vdc
Regulated power supply
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
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EE2603-01Electronic Circuit Analysis
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
Unregulated power supply
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
EE2603-01Electronic Circuit Analysis
Unregulated power supply
Vm
Vdc
15V
7.3V
1
Voltage regulator IC
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
BKBIET, Pilani
EE2603-01Electronic Circuit Analysis
n:1+1
Vac=220Vf=50Hz
Vmax
VmaxC1
I dc
Fixed Negative Voltage Regulator IC
C2 RL
1
Voltage regulator IC
2
3
7912
Vo=-12V-Vi
Unregulated power supply
-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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EE2603-01Electronic Circuit Analysis
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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EE2603-01Electronic Circuit Analysis
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
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EE2603-01Electronic Circuit Analysis
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.
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EE2603-01Electronic Circuit Analysis
Than
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