Chapter 3 : Rectifiers Filters and Regulators

Example 3.1 page no-155

In [5]:
# Ripple Factor

import math
#Variable declaration
Rl=2000.0
f=50.0
l=20.0
V1=0.074

#Calculations and Result
#(1)
w=2*math.pi*f
V=Rl/(3*2*math.sqrt(w*2))
print("1.One Inductor Filter,\nV = %.3f\n"%V1)
#(2)
Idc=1
c=16*10**-6
gam=Idc/(4*math.sqrt(3)*f*c*Rl)
print("\n2.Capacitor filter, \nGamma = %.2f\n"%gam)
#(3)
gam2=(math.sqrt(2)/3)*(1.0/4*l*c*(w**2))
print("\n3. L Type filter,\nGamma = %.4f"%(gam2/1000))

1.One Inductor Filter,
V = 0.074


2.Capacitor filter, 
Gamma = 0.09


3. L Type filter,
Gamma = 0.0037

Example 3.2 page no-156

In [10]:
# diode as a rectifier

import math
#Variable declaration
vm=110.0     # rms
x=1020.0     # Rf+Rl
rl=1000.0

#Calculations and Results
#(a)
Im=vm*math.sqrt(2)/x
print("(a)\nIm = %.1f mA"%(Im*1000))
#(b)
Idc=Im*1000/math.pi
print("\n(b)\nIdc = %.1f mA"%Idc)
#(c)
Ir=Im*1000/2
print("\n(c)\nIrms = %.1f mA"%Ir)
#(d)
v=-(Im*rl/math.pi)
print("\n(d)\n Vdc = %.1f V"%v)
#(e)
p=Ir*x/1000
print("\n(e)\nPi = %.2f W"%p)
#(f)
rl=1.0
lr=((vm*math.sqrt(2)/math.pi)-(Idc*rl))/(Idc*rl)
print("\n(f)\n%% regulation = %.2f %%"%(lr*100))

(a)
Im = 152.5 mA

(b)
Idc = 48.5 mA

(c)
Irms = 76.3 mA

(d)
 Vdc = -48.5 V

(e)
Pi = 77.78 W

(f)
% regulation = 2.00 %

Example 3.4 page no-157

In [13]:
# diode as a rectifier

import math
#Variable declaration
Rl=5010.0       # ohm
idc=0.001

#Calculations
Vrms=idc*math.pi*Rl/(2*math.sqrt(2))

#Result
print("Vrms = %.2f V"%Vrms)

Vrms = 5.56 V

Example 3.5 page no-164

In [24]:
# FWR with LC filter

import math
#Variable declaration
rf=0.02
f=60.0
vdc=9.0
idc=0.1


#Calculations
w=2*math.pi*f
lc=math.sqrt(2)/(rf*12*w**2)
Rl=vdc/idc
lc1=Rl/900
vdc=vdc+5
vm=vdc*math.pi/2
vrms=vm/math.sqrt(2)

#Result
print("\nLC=%.1f micro"%(lc*10**6))
print("\nRL = %d Ohm\n\nLC>  Rl/3w  >  Rl/1130\nBut LC should be 25%% larger"%Rl)
print("therefore, for f= 60 Hz,the value of LC should be > Rl/900")
print("\nIf L=0.1H, then C=%.1f micro F, This is high value."%(math.ceil(lc*10**6/lc1)))
print("\nIf L=1H, then C=41.5 micro F.")
print("\n\nTransformer Rating:")
print("Vdc=%.0fV\nVm=%.0fV\nVrms=%.1fV"%(vdc,math.ceil(vm),vrms)) 
print("Therefore, a 15.5 - 0 -15.5 V, 1OOmA transformer is required\nPIV=%d V"%(2*math.ceil(vm)))

LC=41.5 micro

RL = 90 Ohm

LC>  Rl/3w  >  Rl/1130
But LC should be 25% larger
therefore, for f= 60 Hz,the value of LC should be > Rl/900

If L=0.1H, then C=415.0 micro F, This is high value.

If L=1H, then C=41.5 micro F.


Transformer Rating:
Vdc=14V
Vm=22V
Vrms=15.6V
Therefore, a 15.5 - 0 -15.5 V, 1OOmA transformer is required
PIV=44 V

Example 3.6 page no-165

In [27]:
# Ripple Factor

import math
#Variable declaration
vrpp=0.8        # V
r=100.0
f=60.0
c=1050.0*10**-6


#Calculations
vrms=vrpp/(2*math.sqrt(3))
vrms=math.floor(vrms*10)
vrms=vrms/10.0
vm=8.8
vdc=vm-vrpp/2
gam=vrms/vdc
tgam=1/(4*(math.sqrt(3*c*r*f)))
Vdc=(4*f*r*c*vm)/(1+4*f*r*c)

#Result
print("%% regulation, gamma = %.2f%%"%(gam*100))
print("\nTheoretical values, gamma = %.2f%%"%(tgam*100))
print("\nVdc = %.2f V"%(Vdc))

% regulation, gamma = 2.38%

Theoretical values, gamma = 5.75%

Vdc = 8.46 V

Example 3.7 page no-167

In [31]:
# power supply using pi filter

import math
#Variable declaration
Vdc=25.0
Idc=0.1
#Calculations
R=Vdc/Idc
Vc=Vdc+37.5
vm=Vc+(Idc/(4.0*50))
vrms=vm/math.sqrt(2)
vrms=60.0     # approximated to
print("\nVrms=%.0f V\n\nTherefore, a transformer with 60 - 0 - 60V is chosen."%vrms)
print("The ratings of the diode should be,\ncurrent of 125mA and voltage = PIV = 2Vm = %.1f"%(169.2))

Vrms=60 V

Therefore, a transformer with 60 - 0 - 60V is chosen.
The ratings of the diode should be,
current of 125mA and voltage = PIV = 2Vm = 169.2

Example 3.8 page no-169

In [37]:
# Diode rating for FWR

import math
#Variable declaration

Vdc=250.0      # V
Idc=0.1
rc=400.0
L=10.0     #Ohm
c=20*10**-6
w=377.0


#Calculations
rl=Vdc/Idc
Vm=(Vdc*math.pi/2)*(1+(rc/rl))
Vrms=Vm/math.sqrt(2)
Ib=2*Vm/(3*math.pi*w*L)
rf=0.47/(4*w**2*c)

#Result
print("Vrms=%d V\n"%Vrms)
print("\nTherefore, the transformer should supply %d V rms on each side of the centre tap."%Vrms)
print("\nIb = %.4f A\nRipple factor = %.4f"%(Ib,rf))

Vrms=322 V


Therefore, the transformer should supply 322 V rms on each side of the centre tap.

Ib = 0.0256 A
Ripple factor = 0.0413

Example 3.9 page no-170

In [38]:
# FWR with C type capacitor filter

import math
#Variable declaration

Idc=0.02        # A
Vdc=16.0        # V
f=50.0

#Calculations
rl=Vdc/Idc
x=4*math.sqrt(3)*f*0.05*rl
C=1/x
vm=Vdc*((1+(4*f*C*rl)))/(4*f*C*rl)

print("C=%d microF"%(C*10**6))
print("Vm=%.2f V"%vm)

C=72 microF
Vm=17.39 V

Example 3.10 page no-170

In [40]:
# Half Wave Rectifier

import math
#Calculations
Vdc=(100/(2*math.pi))*(-math.cos(5*math.pi/6)+math.cos(math.pi/6))
Vrms=math.sqrt(3.1)*Vdc

#Result
print("Vdc=%.1f V"%Vdc)
print("Vrms=%.1fV"%Vrms)

Vdc=27.6 V
Vrms=48.5V

Example 3.11 page no-172

In [44]:
# FWR with C type capacitor filter

import math
#Variable declaration

vdc=30.0      # V
idc=0.05      # A
f=50.0        # Hz
c=80*10**-6   # F

#Calculations
#(a)
rl=vdc/idc
vm=vdc+(idc/(4*f*c))
#(b)
s=vm*2*math.pi*f*c
#(c)
gam=4*math.sqrt(3)*f*c*rl
gam=1/gam

#Result
print("(a)\nRL=%.0f Ohm\nVm=%.3fV\nVrms=%.1fV"%(rl,vm,vm/math.sqrt(2)))
print("\n(b)\nI_diode swing/I_diode mean = %.2f"%(s/idc))
print("\n(c)\ngamma=%.2f"%gam)

(a)
RL=600 Ohm
Vm=33.125V
Vrms=23.4V

(b)
I_diode swing/I_diode mean = 16.65

(c)
gamma=0.06

Example 3.12 page no-173

In [49]:
# Full wave rectifier circuit

import math
#Variable declaration

vm=25.0
vp=35.4            # V
rl=25

#Calculations
vdc=2*vp/math.pi   # V
vrms=math.sqrt((vm**2)-vdc**2)
im= vp/rl
idc=2*im/math.pi
irms=math.sqrt(1-idc**2)

#Result
print("Vdc=%.1f V\nVrms=%.2f V\nIm=%.2f A\nIdc=%.2f A\nIrms=%.3f A"%(vdc,vrms,im,idc,irms))

Vdc=22.5 V
Vrms=10.82 V
Im=1.42 A
Idc=0.90 A
Irms=0.433 A

Example 3.13 page no-176

In [55]:
# Shunt regulator

import math
#Variable declaration
veb=0.2         # V
hfe=49.0
vz=6.3          # V
i=5*10**-3
vi=8.0

#(1)
y=veb+vz
print("1. The nominal output voltage is the sum of the transistor V_EB and zener voltage.\nV0=%.1f V\n"%y)

#(2)
r1=(vi-vz)/i
print("\n2. R1 must supply 5mA to the zener diode\nR1 = %.0f Ohm"%r1)

#(3)
k=veb/vz
print("\n\n3. The maximum allowable zener current is\nIz = %.3f A"%k)
ibmax=k-i
it=ibmax*(1+hfe)
print("\nTotal current range = %.2f A"%it)

#(4)
pd=y*it
print("\n(4)\nThe maximum power dissipation,\nPd = %.1f W"%pd)

#(5)
rs=(vi-y)/it
pdr=it**2*rs
print("\n(5)\nRs=%.2f Ohm\nPower dissipated by Rs is P = %dW"%(rs,pdr))

1. The nominal output voltage is the sum of the transistor V_EB and zener voltage.
V0=6.5 V


2. R1 must supply 5mA to the zener diode
R1 = 340 Ohm


3. The maximum allowable zener current is
Iz = 0.032 A

Total current range = 1.34 A

(4)
The maximum power dissipation,
Pd = 8.7 W

(5)
Rs=1.12 Ohm
Power dissipated by Rs is P = 2W
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