# Ch-18 : Rectifiers and Power Supplies¶

## Page No. 548 Example 18.1.¶

In :
from math import pi
im=325./(100+1000)  # in A
x1=im*10**3  # in mA
print "(a) Peak value of current,  Im = Vm / rf+RL = %0.2f mA"%x1
idc=295.45/pi # in mA
print "    Average current,  Id.c. = Im / pi = %0.2f mA"%idc
irms=295.45/2  # in mA
print "    RMS value of current,  Irms = Im / 2 = %0.2f mA"%irms
pdc=((94.046*10**-3)**2)*1000  # in W
print "(b) D.C. power output,  Pd.c. = (Id.c.)**2 * RL = %0.2f W"%pdc
pac=((147.725*10**-3)**2)*1100 # in W
print "(c) AC input power,  Pac = (Irms)**2 * (rf+RL) = %0.2f "%pac
eta=(8.845/24)*100  # in percentage
print "(d) Efficiency of rectification,  eta = Pdc / Pac = %0.2f %%"%eta

(a) Peak value of current,  Im = Vm / rf+RL = 295.45 mA
Average current,  Id.c. = Im / pi = 94.04 mA
RMS value of current,  Irms = Im / 2 = 147.72 mA
(b) D.C. power output,  Pd.c. = (Id.c.)**2 * RL = 8.84 W
(c) AC input power,  Pac = (Irms)**2 * (rf+RL) = 24.00
(d) Efficiency of rectification,  eta = Pdc / Pac = 36.85 %


## Page No. 548 Example 18.2.¶

In :
from math import pi
icd=(24./500)*10**3  # in mA
print "Average value of load current,  Id.c.= Vdc / RL = %0.2f mA"%icd
im=pi*48 # in mA
print "Maximum value of load current,  Im= pi * Idc = %0.2f mA"%im
print "Therefore, maximum ac voltage required at the input,"
vm=550*150.8*10**-3 # in V
print "      Vm = Im * (rf+RL) = %0.2f V"%vm

Average value of load current,  Id.c.= Vdc / RL = 48.00 mA
Maximum value of load current,  Im= pi * Idc = 150.80 mA
Therefore, maximum ac voltage required at the input,
Vm = Im * (rf+RL) = 82.94 V


## Page No. 549 Example 18.3.¶

In :
from math import sqrt,pi
x1=230./5 # in V
vm=sqrt(2) * 46  # in V
vdc=65./pi  # in V
im=65./300  # in A
pm=0.217**2 * 300  # in W
idc=20.7/300 # in A
pdc=(0.069**2)*300 # in W
print "(a) The transformer secondary voltage = %0.2f V"%x1
print "    Maximum value of secondary voltage,  Vm = %0.2f V"%vm
print "    Therefore, d.c. output voltage,  Vd.c. = Vm / pi = %0.2f V"%vdc
print "(b) PIV of a diode  = Vm = 65 V"
print "(c) Maximum value of load current,  Im= Vm / RL = %0.2f A"%im
print "    Therefore, maximum value of power delivered to the load,"
print "           Pm = Im**2 * RL = %0.2f W"%pm
print "(d) The average value of load current,  Id.c.(A) = Vdc / RL = %0.2f A"%idc
print "    Therefore, average value of power delivered to the load,"
print "        Pd.c. = (Idc)**2 * RL = %0.2f W"%pdc

(a) The transformer secondary voltage = 46.00 V
Maximum value of secondary voltage,  Vm = 65.05 V
Therefore, d.c. output voltage,  Vd.c. = Vm / pi = 20.69 V
(b) PIV of a diode  = Vm = 65 V
(c) Maximum value of load current,  Im= Vm / RL = 0.22 A
Therefore, maximum value of power delivered to the load,
Pm = Im**2 * RL = 14.13 W
(d) The average value of load current,  Id.c.(A) = Vdc / RL = 0.07 A
Therefore, average value of power delivered to the load,
Pd.c. = (Idc)**2 * RL = 1.43 W


## Page No. 550 Example 18.4.¶

In :
from math import sqrt,pi
x1=230./5  # in V
vrms=46./2 # in V
vdc=(2.*23*sqrt(2))/pi # in V
idc=(20.7/1000)*10**3 # in mA
pdc=((20.7*10**-3)**2)*900  # in W
piv=2*23*sqrt(2)  # in V
vrrms=sqrt(23**2 - 20.7**2)  # in V
f=2*60  # in Hz
print "The voltage across the two ends of secondary(in V) = 230 / 5 = %0.2f V"%x1
print "Voltage from center tapping to one end, Vrms = %0.2f V"%vrms
print "(a) d.c. voltage across the load,  Vdc = 2Vm / pi = %0.2f V"% vdc
print "(b) d.c. current flowing through the load,"
print "      Idc = Vdc / (rs+rf+RL) = %0.2f mA"%idc
print "(c) d.c. power delivered to the load,"
print "      Pdc = (Idc)**2 * RL = %0.2f W"%pdc
print "(d) PIV across each diode = 2Vm = %0.2f W"%piv
print "(e) Ripple voltage,  Vr,rms = sqrt(Vrms**2 - Vdc**2) = %0.2f V"%vrrms
print "    Frequency of ripple voltage = %0.2f Hz"%f

The voltage across the two ends of secondary(in V) = 230 / 5 = 46.00 V
Voltage from center tapping to one end, Vrms = 23.00 V
(a) d.c. voltage across the load,  Vdc = 2Vm / pi = 20.71 V
(b) d.c. current flowing through the load,
Idc = Vdc / (rs+rf+RL) = 20.70 mA
(c) d.c. power delivered to the load,
Pdc = (Idc)**2 * RL = 0.39 W
(d) PIV across each diode = 2Vm = 65.05 W
(e) Ripple voltage,  Vr,rms = sqrt(Vrms**2 - Vdc**2) = 10.03 V
Frequency of ripple voltage = 120.00 Hz


## Page No. 552 Example 18.5.¶

In :
from math import sqrt,pi
imax=0.8*400  # in mA
print "Therefore,  Imax = %0.2f mA"%imax
print "The maximum value of the secondary voltage,"
vm=sqrt(2)*100  # in V
print "      Vm = %0.2f V"%vm
print "Therefore, the value of load resistor that gives the largest d.c. power output"
RL=141.4/(320*10**-3)
print "      RL = Vm / Imax = %0.2f ohm"%RL
vdc=(2*141.4)/pi
print "(b) D.C.(load) voltage,  Vdc(V) = (2*141.4)/pi = %0.2f V"%vdc
idc=90./442
print "    D.C. load current,  Idc = Vdc / RL = %0.2f A"%idc
print "(c) PIV of each diode  = 2Vm = 282.8 V"

Therefore,  Imax = 320.00 mA
The maximum value of the secondary voltage,
Vm = 141.42 V
Therefore, the value of load resistor that gives the largest d.c. power output
RL = Vm / Imax = 441.88 ohm
(b) D.C.(load) voltage,  Vdc(V) = (2*141.4)/pi = 90.02 V
D.C. load current,  Idc = Vdc / RL = 0.20 A
(c) PIV of each diode  = 2Vm = 282.8 V


## Page No. 553 Example 18.6.¶

In :
from math import sqrt,pi
print "D.C. power delivered to the load,"
print "            Pdc = Vdc**2 / RL"
vdc=sqrt(50.*200)
print "Therefore,  Vdc = sqrt(Pdc*RL) = %0.2f V"%vdc
print "The ripple factor,  gamma = Vac / Vdc"
print "i.e.                 0.01 = Vac / 100"
print "Therefore, the ac ripple voltage across the load, Vac = 1 V"

D.C. power delivered to the load,
Pdc = Vdc**2 / RL
Therefore,  Vdc = sqrt(Pdc*RL) = 100.00 V
The ripple factor,  gamma = Vac / Vdc
i.e.                 0.01 = Vac / 100
Therefore, the ac ripple voltage across the load, Vac = 1 V


## Page No. 554 Example 18.7.¶

In :
from math import sqrt,pi
Vrms=230./4  # in V
vm=sqrt(2)*57.5  # in V
vdc=(2*81.3)/pi  # in V
pdc=52.**2/1000  # in W
print "(a) The rms value of the transformer secondary voltage,"
print "      Vrms = %0.2f V"%Vrms
print "    The maximum value of the secondary voltage"
print "      Vm = %0.2f V"%vm,
print "Therefore, d.c. output voltage,  Vdc = 2Vm / pi = %0.2f V"%vdc
print "(b) D.C. power delivered to the load,"
print "      Pd.c. = (Vdc)**2 / RL = %0.2f W"%pdc
print "(c) PIV across each diode = Vm = 81.3 V"
print "(d) Output frequency = 2 x 50 = 100 Hz"

(a) The rms value of the transformer secondary voltage,
Vrms = 57.50 V
The maximum value of the secondary voltage
Vm = 81.32 V Therefore, d.c. output voltage,  Vdc = 2Vm / pi = 51.76 V
(b) D.C. power delivered to the load,
Pd.c. = (Vdc)**2 / RL = 2.70 W
(c) PIV across each diode = Vm = 81.3 V
(d) Output frequency = 2 x 50 = 100 Hz


## Page No. 556 Example 18.8.¶

In :
L=0.0625/0.04 # in H
print "We know that the ripple factor for inductor filter is gamma = RL / 3*sqrt(2)*omega*L"
print "Therefore,    L = %0.2f Henry"%L

We know that the ripple factor for inductor filter is gamma = RL / 3*sqrt(2)*omega*L
Therefore,    L = 1.56 Henry


## Page No. 558 Example 18.9.¶

In :
print "We know that the ripple factor for capacitor filter is"
print "    gamma = 1 / 4*sqrt(3)*f*C*RL"
c=(0.722)/0.01  # in pF
print "Therefore,    C = %0.2f pF"%c

We know that the ripple factor for capacitor filter is
gamma = 1 / 4*sqrt(3)*f*C*RL
Therefore,    C = 72.20 pF


## Page No. 560 Example 18.10¶

In :
rl=10./(200*10**-3)  # in ohm
lc=1.194/0.02
print "The effective load resistance  RL = %0.2f ohm"%rl
print "We know that the ripple factor,  gamma = 1.194 / LC"
print "i.e.      LC = %0.2f "%lc
print "Critical value of L(mH) = RL / 3*omega = 50 / 3*2*pi*f = 53mH"
print "Taking L = 60 mH (about 20% higher), C will be about 1000 uF"

The effective load resistance  RL = 50.00 ohm
We know that the ripple factor,  gamma = 1.194 / LC
i.e.      LC = 59.70
Critical value of L(mH) = RL / 3*omega = 50 / 3*2*pi*f = 53mH
Taking L = 60 mH (about 20% higher), C will be about 1000 uF


## Page No. 561 Example 18.11¶

In :
from math import sqrt,pi
rl=(10./(200*10**-3))  # in ohm
c2=11.4/0.02
c=sqrt(570.)  # in uF
print "    RL = %0.2f ohm"%rl
print "    0.02 = 5700 / L*C1*C2*50 = 114 / L*C1*C2"
print "If we assume L = 10 mH and  C1 = C2 = C, we have"
print "    0.02 = 114 / L*C**2 = 11.4 / C**2"
print "    C**2 = %0.2f "%c2
print "therefore, C = %0.2f uF"%c

    RL = 50.00 ohm
0.02 = 5700 / L*C1*C2*50 = 114 / L*C1*C2
If we assume L = 10 mH and  C1 = C2 = C, we have
0.02 = 114 / L*C**2 = 11.4 / C**2
C**2 = 570.00
therefore, C = 23.87 uF


## Page No. 562 Example 18.12.¶

In :
pz=10.*40*10**-3  # in W
print "    Pz = Vz * Iz_max = %0.2f W"%pz
print "Hence a 0.5Z 10 zener can be selected"
print "Value of load resistance, RL"
rlmin=10./(50*10**-3)  # in ohm
print "  RL_min = Vo / IL_max = %0.2f ohm"%rlmin
rlmax=10./(30*10**-3)  # in ohm
print "  RL_max = Vo / IL_min = %0.2f ohm"%rlmax
print "Value of input resistance, R"
rmax=(30.-10)/((30+40)*10**-3)  # in ohm
print "  Rmax = Vin(max)-Vo / ILmin+IZmax = %0.2f ohm"%rmax
rmin=(20.-10)/((50+20)*10**-3)  # in ohm
print rmin,"  Rmax(ohm) = Vin(min)-Vo / ILmax+IZmin ="
r=(286+143.)/2
print "  R = Rmax+Rmin / 2 = %0.2f ohm"%r  # answer in textbook is wrong

    Pz = Vz * Iz_max = 0.40 W
Hence a 0.5Z 10 zener can be selected
RL_min = Vo / IL_max = 200.00 ohm
RL_max = Vo / IL_min = 333.33 ohm
Value of input resistance, R
Rmax = Vin(max)-Vo / ILmin+IZmax = 285.71 ohm
142.857142857   Rmax(ohm) = Vin(min)-Vo / ILmax+IZmin =
R = Rmax+Rmin / 2 = 214.50 ohm


## Page No. 563 Example 18.13.¶

In :
rl=5./(20*10**-3)  # in ohm
print "    RL = Vo / IL = %0.2f ohm"%rl
r=(8.-5)/((5.+20)*10**-3)  # in ohm
print "Hence, the series resistance  R(ohm) = Vi(min)-Vo / IZ(min)+IL = %0.2f ohm"%r
print "The various values are given in the Zener regulator shown in Fig. 18.19"

    RL = Vo / IL = 250.00 ohm
Hence, the series resistance  R(ohm) = Vi(min)-Vo / IZ(min)+IL = 120.00 ohm
The various values are given in the Zener regulator shown in Fig. 18.19


## Page No. 564 Example 18.14.¶

In :
print "(i) Let IZ = IZ(min) and IL = 0"
print "    The total current  I = IL + IZ = 10 mA"
r=10./(10*10**-3)  # in ohm
print "    Therefore,  R = %0.2f ohm"%r
print "(ii) For IZ = IZ(max) = 100 mA and IL = 20 mA"
i=20+100.  # in mA
print "     I = IL + IZ = %0.2f mA"%i
r=10./(120*10**-3)
print "     Therefore,  R = %0.2f ohm"%r
print "(iii) The range of R varies from 83.33 ohm to 1000 ohm"

(i) Let IZ = IZ(min) and IL = 0
The total current  I = IL + IZ = 10 mA
Therefore,  R = 1000.00 ohm
(ii) For IZ = IZ(max) = 100 mA and IL = 20 mA
I = IL + IZ = 120.00 mA
Therefore,  R = 83.33 ohm
(iii) The range of R varies from 83.33 ohm to 1000 ohm


## Page No. 565 Example 18.15.¶

In :
rl=5./(10*10**-3)  # in ohm
print "Here, load resistance is RL = Vo / IL = %0.2f ohm"%rl
iz=400./5  # in mA
print "Maximum Zener Current Iz_max = %0.2f mA"%iz
print "The minimum input voltage required will be when Iz = 0. Under this condition,"
print "    I = IL = 10 mA"
print "Minimum input voltage  Vi_min = Vo + IR"
vi=10.-2  # in V
print "Hence,    Vi_min = %0.2f V"%vi
print "or    8 = 5 + (10*10**-3)R"
rmax=3./(10*10**-3)  # in ohm
print "Therefore,  Rmax = %0.2f ohm"%rmax
print "Now, maximum input voltage, Vi_max = 5 + [(80+10)10**-3]R"
rmin=7./(90*10**-3)  # in ohm
print "    Rmin = %0.2f ohm"%rmin
print "The value of R is chosen between 77.77 ohm and 300 ohm"

Here, load resistance is RL = Vo / IL = 500.00 ohm
Maximum Zener Current Iz_max = 80.00 mA
The minimum input voltage required will be when Iz = 0. Under this condition,
I = IL = 10 mA
Minimum input voltage  Vi_min = Vo + IR
Hence,    Vi_min = 8.00 V
or    8 = 5 + (10*10**-3)R
Therefore,  Rmax = 300.00 ohm
Now, maximum input voltage, Vi_max = 5 + [(80+10)10**-3]R
Rmin = 77.78 ohm
The value of R is chosen between 77.77 ohm and 300 ohm


## Page No. 566 Example 18.16.¶

In :
il=(24./1200)*10**3  # in mA
print "The load current,  IL( = Vo / RL = %0.2f mA"%il
iz=600./24  # in mA
print "Max. Zener current,  Iz_max = %0.2f mA"%iz
rmax=(32.-24)/((20+25)*10**-3)  # in ohm
print "  Rmax(ohm) = Vi-Vo / IL_min+IZ_max = %0.2f ohm"%rmax

The load current,  IL( = Vo / RL = 20.00 mA
Max. Zener current,  Iz_max = 25.00 mA
Rmax(ohm) = Vi-Vo / IL_min+IZ_max = 177.78 ohm


## Page No. 567 Example 18.17.¶

In :
vi=15.+3   # in V
print "Refer to fig.18.24. We know that"
print "    Vi_min(V) = Vo + 3V = %0.2f V"%vi
vi=18.+1  # in V
print "Assuming the ripple voltage Vr = 2V(max), the input voltage is"
print "  Vi = Vi(min) + Vr/2 = %0.2f V"%vi
vz=19./2  # in V
print "Then    Vz = Vi /2 = %0.2f V           (use the zener diode 1N758 for 10V)"%vz
print "Therefore,  Vz = 10 V"
print "            Iz = 20 mA"
r1=(19.-10)/(20*10**-3)  # in ohm
print "    R1 = Vi-Vz / Iz = %0.2f ohm"%r1
print "Let    I2 = IB(max) = 50 uA"
r2=((15.-10)/(50*10**-6))*10**-3  # in k-ohm
print "    R2 = Vo-Vz / I2 = %0.2f kohm"%r2
r3=(10./(50*10**-6))*10**-3  # in k-ohm
print "    R3 = Vz / I2 = %0.2f kohm"%r3
print "Select    C1 = 50 uF"
print "Specification of transistor Q1"
vce=19.+1  # in V
print "    VCE_max = Vi_max(V) = Vi + Vr/2 = %0.2f V"%vce
print "    IE = IL = 50 mA"
p=((19.-15)*50)  # in mW
print "    P = VCE*IL = (Vi-Vo) * IL = %0.2f mW"%p
print "Use the transistor 2N718 for Q1"

Refer to fig.18.24. We know that
Vi_min(V) = Vo + 3V = 18.00 V
Assuming the ripple voltage Vr = 2V(max), the input voltage is
Vi = Vi(min) + Vr/2 = 19.00 V
Then    Vz = Vi /2 = 9.50 V           (use the zener diode 1N758 for 10V)
Therefore,  Vz = 10 V
Iz = 20 mA
R1 = Vi-Vz / Iz = 450.00 ohm
Let    I2 = IB(max) = 50 uA
R2 = Vo-Vz / I2 = 100.00 kohm
R3 = Vz / I2 = 200.00 kohm
Select    C1 = 50 uF
Specification of transistor Q1
VCE_max = Vi_max(V) = Vi + Vr/2 = 20.00 V
IE = IL = 50 mA
P = VCE*IL = (Vi-Vo) * IL = 200.00 mW
Use the transistor 2N718 for Q1


## Page No. 568 Example 18.18.¶

In :
rlmin=20./(50*10**-3)  # in ohm
print "Selection of Zener diode"
print "  RLmin = Vo / ILmax = %0.2f ohm"%rlmin
vz=20./2  # in V
print "  Vz = Vo / 2 = %0.2f V"%vz
print "The current flowing through the Zener,"
iz=10.+10  # in mA
print "  Iz = IE2 + IR1 = %0.2f mA"%iz
pz=10.*20*10**-3  # in W
print "  Pz = Vz*Iz = %0.2f W"%pz  # > 0.5 W
print "Selection of transistor Q1"
ie1=10.+10+50  # in mA
print "  IE1 = IR1 + IR2 + IL = %0.2f mA"%ie1
print "  Vi(max) - Vo = 30 -20 = 10 V"
print "For transistor SL100, the rating are"
print "    IC(max) = 500 mA"
print "   VCE(max) = 50 V"
print "        hre = 50 - 280"
print "Hence, SL100 can be chosen for Q1"
print ""
print "Selection of transistor Q2"
vce2=20.6-10  # in V
print "  Therefore,  VCE2_max = (Vo + VBE1) - Vz = %0.2f V"%vce2
print "For transistor BC107, the rating are"
print "    VCEO(max) = 45 V"
print "      IC(max) = 200 mA"
print "          hFE = 125 - 300"
print "Hence, transistor BC107 is selected for Q2"
print "Selection of resistor R1, R2 and R3"
vr1=20-10.  # in V
print "  VR1 = Vo - Vz = %0.2f V"%vr1
r1=10./(10)  # in k-ohm
print "  R1 = VR1 / IR1 = %0.2f kohm"%r1
vr2=20.-10.6  # in V
print "  VR2 = Vo - VR3 = %0.2f V"%vr2
r2=9.4/(10*10**-3)  # in ohm
print "  R2 = VR2 / IR2 = %0.2f ohm"%r2
vr3=10+0.6  # in V
print "  VR3 = Vz + VBE2(sat) = %0.2f V"%vr3
r3=10.6/(10*10**-3)  # in ohm
print "  R3 = VR3 / IR3 = %0.2f ohm"%r3
print "Selection of resistor R4"
vb1=20+0.6  # in V
print "  VB1 = VC2(V) = Vo + VBE1 = %0.2f V"%vb1
ib1=70./50  # in mA
print "  IB1 = IC1 / beta = %0.2f mA"%ib1
ir4=11.4  # in mA
print "  IR4 = IB1 + IC2 = %0.2f mA"%ir4
r4max=(30-20.6)/(11.4*10**-3)  # in ohm
print "  R4_max = VR4(max) / IR4 = Vi(max)-VB1 / IR4 = %0.2f ohm"%r4max
r4min=(22-20.6)/(11.4*10**-3)  # in ohm
print "  R4_min = VR4(min) / IR4 = Vi(min)-VB1 / IR4 = %0.2f ohm"%r4min
r4=(825.+123)/2  # in ohm
print "  R4 = R4(max)+R4(min) / 2 = %0.2f ohm"%r4

Selection of Zener diode
RLmin = Vo / ILmax = 400.00 ohm
Vz = Vo / 2 = 10.00 V
The current flowing through the Zener,
Iz = IE2 + IR1 = 20.00 mA
Pz = Vz*Iz = 0.20 W
Selection of transistor Q1
IE1 = IR1 + IR2 + IL = 70.00 mA
Vi(max) - Vo = 30 -20 = 10 V
For transistor SL100, the rating are
IC(max) = 500 mA
VCE(max) = 50 V
hre = 50 - 280
Hence, SL100 can be chosen for Q1

Selection of transistor Q2
Therefore,  VCE2_max = (Vo + VBE1) - Vz = 10.60 V
For transistor BC107, the rating are
VCEO(max) = 45 V
IC(max) = 200 mA
hFE = 125 - 300
Hence, transistor BC107 is selected for Q2
Selection of resistor R1, R2 and R3
VR1 = Vo - Vz = 10.00 V
R1 = VR1 / IR1 = 1.00 kohm
VR2 = Vo - VR3 = 9.40 V
R2 = VR2 / IR2 = 940.00 ohm
VR3 = Vz + VBE2(sat) = 10.60 V
R3 = VR3 / IR3 = 1060.00 ohm
Selection of resistor R4
VB1 = VC2(V) = Vo + VBE1 = 20.60 V
IB1 = IC1 / beta = 1.40 mA
IR4 = IB1 + IC2 = 11.40 mA
R4_max = VR4(max) / IR4 = Vi(max)-VB1 / IR4 = 824.56 ohm
R4_min = VR4(min) / IR4 = Vi(min)-VB1 / IR4 = 122.81 ohm
R4 = R4(max)+R4(min) / 2 = 474.00 ohm


## Page No. 569 Example 18.19.¶

In :
from math import sqrt,pi
print "A bridge rectifier or full wave rectifier is used to get the pulsating d.c. output."
rl=9./(100*10**-3)  # in ohm
print "  RL = Vdc / TL = %0.2f ohm"%rl
print "A capacitor filter is used to remove the ripple and get a smooth output."
print "    Ripple factor gamma = 1 / 4*sqrt(3)*f*C*RL"
print "Assume the ripple factor to be 0.03"
c=(1./(4*sqrt(3)*50*0.03*90))*10**6  # in uF
print "  C = %0.2f uF"%c  # = 1000 uF
print "The short circuit resistance Rsc connected with the series pass transistor is"
rsc=0.7/(150*10**-3)  # in ohm
print "  Rsc = VBE / Ilim_it = %0.2f ohm"%rsc
print "Assume 7.6 V Zener diode in series with 1.5 k-ohm"
print "The designed circuit is shown in fig.18.32."

A bridge rectifier or full wave rectifier is used to get the pulsating d.c. output.
RL = Vdc / TL = 90.00 ohm
A capacitor filter is used to remove the ripple and get a smooth output.
Ripple factor gamma = 1 / 4*sqrt(3)*f*C*RL
Assume the ripple factor to be 0.03
C = 1069.17 uF
The short circuit resistance Rsc connected with the series pass transistor is
Rsc = VBE / Ilim_it = 4.67 ohm
Assume 7.6 V Zener diode in series with 1.5 k-ohm
The designed circuit is shown in fig.18.32.