Chapter 18 : Power Circuits and Systems

Example 18.1a, Page No 780

In [1]:
import math
#initialisation of variables

Vo=25.0     #in V
ro=10.0     #in ohm

print('select a silicon reference diode')
print('two IN7555 diodes are provided')
Rz = 12.0        #in ohm
Vo=25.0        #output voltage in V
Vr = 7.5 + 7.5   #because two diodes are used
Iz = 20.0    #in mA
Ie2=10.0     #in mA
Ic2 = Ie2
Icmax=30.0   #in mA
Vcemax=45.0  #in V
hFE2=220.0 
hfe2=200.0 
hie2=800.0    #in ohm
Id=10.0       #in mA
Il = 1000.0   #in mA
Vi = 50.0     #in V
dVi = 10.0    #change in input voltage
dIl = 1.0     #change in load current

#Calculations

#For D1 and D2 operate
Iz = Id + Id
Rd = (Vo-Vr)/Id
print("The value of Rd is %.2f K " %Rd)
Ib2 = (1000.0*Ic2)/hFE2
print("The value of Ib2 is %.2f microA " %Ib2)
#Since we require I1>Ib2 , we select
I1=10*(10**-3)#in A
Vbe = 0.7#in V
V2 = Vbe + Vr
print("The value of V2 is %.2f v " %V2)
R1 = (Vo-V2)/I1
R2 = V2/I1
print("The value of R1 is %.2f ohm " %R1)
print("The value of R2 is %.2f ohm " %R2)
#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required
print('We are selecting Texas Instruments 2N1722 silicon power transistor')
Ic1 = 1#in A
hFE1=125.0
hfe1=100.0
hie1=20.0


#Results
Ib1 =(1000*I1 + Il + Id)/hFE1
print("The current through resistor R3 is %.2f mA " %(Ib1+Ic2))
I=Ib1 + Ic2

R3 = (Vi - (Vbe + Vo))/I
print("The value of R3 is %.2f K " %R3)

select a silicon reference diode
two IN7555 diodes are provided
The value of Rd is 1.00 K 
The value of Ib2 is 45.45 microA 
The value of V2 is 15.70 v 
The value of R1 is 930.00 ohm 
The value of R2 is 1570.00 ohm 
We are selecting Texas Instruments 2N1722 silicon power transistor
The current through resistor R3 is 18.16 mA 
The value of R3 is 1.34 K

Example 18.1b, Page No 780

In [2]:
import math

#initialisation of variables
Vo=25.0    #in V
ro=10.0    #in ohm

Rz = 12.0 #in ohm
Vo=25.0   #output voltage in V
Vr = 7.5 + 7.5#because two diodes are used
Iz = 20.0 #in mA
Ie2=10.0 #in mA
Ic2 = Ie2
Icmax=30.0 #in mA
Vcemax=45.0 #in V
hFE2=220.0 
hfe2=200.00 
hie2=800.0 #in ohm
Id=10.0 #in mA
Il = 1000.0 #in mA
Vi = 50.0 #in V
dVi = 10.0  #change in input voltage
dIl = 1.0  #change in load current

#Calculations

#For D1 and D2 operate
Iz = Id + Id
Rd = (Vo-Vr)/Id
Ib2 = (1000*Ic2)/hFE2

#Since we require I1>Ib2 , we select
I1=10*(10**-3)#in A
Vbe = 0.7#in V

V2 = Vbe + Vr

R1 = (Vo-V2)/I1
R2 = V2/I1

#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required
Ic1 = 1#in A
hFE1=125.0
hfe1=100.0
hie1=20.0 

Ib1 =(1000.0*I1 + Il + Id)/hFE1
#The current through resistor R3
I=Ib1 + Ic2

R3 = (Vi - (Vbe + Vo))/I
Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) 
Sv = (10**-3)/(Gm*R3)

#Results

print("The value of Sv is %.2f " %Sv)

The value of Sv is 0.02

Example 18.1c Page No 780

In [3]:
import math

#initialisation of variables
Vo=25.0 #in V
ro=10.0  #in ohm
Rz = 12.0  #in ohm
Vo=25.0  #output voltage in V
Vr = 7.5 + 7.5#because two diodes are used
Iz = 20.0 #in mA
Ie2=10.0 #in mA
Ic2 = Ie2
Icmax=30.0 #in mA
Vcemax=45.0 #in V
hFE2=220.0
hfe2=200.0
hie2=800.0  #in ohm
Id=10.0  #in mA
Il = 1000.0 #in mA
Vi = 50.0 #in V
dVi = 10.0 #change in input voltage
dIl = 1.0  #change in load current

#Calculations
#For D1 and D2 operate
Iz = Id + Id
Rd = (Vo-Vr)/Id

Ib2 = (1000.0*Ic2)/hFE2

#Since we require I1>Ib2 , we select
I1=10*(10**-3)#in A
Vbe = 0.7#in V

V2 = Vbe + Vr

R1 = (Vo-V2)/I1
R2 = V2/I1

#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required
Ic1 = 1#in A
hFE1=125.0
hfe1=100.0
hie1=20.0

Ib1 =(1000*I1 + Il + Id)/hFE1
#The current through resistor R3 is
I=Ib1 + Ic2

R3 = (Vi - (Vbe + Vo))/I

Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) 

#Results
print("The value of Gm is %.2f " %Gm)
Ro = (ro + (((1000*R3) + hie1)/(1+hfe1)))/(1 + (Gm*((1000*R3) + ro)))
print("The output impedence is = %.2f K " %Ro)

The value of Gm is 0.03 
The output impedence is = 0.51 K

Example 18.1d, Page No 780

In [4]:
import math
#initialisation of variables
Vo=25.0 #in V
ro=10.0 #in ohm

Rz = 12.0  #in ohm
Vo=25.0  #output voltage in V
Vr = 7.5 + 7.5#because two diodes are used
Iz = 20.0  #in mA
Ie2=10.0  #in mA
Ic2 = Ie2
Icmax=30.0 #in mA
Vcemax=45.0 #in V
hFE2=220.0 
hfe2=200.0 
hie2=800.0 #in ohm
Id=10.0 #in mA
Il = 1000.0 #in mA
Vi = 50.0 #in V
dVi = 10.0  #change in input voltage
dIl = 1.0 #change in load current

#Calculations

#For D1 and D2 operate
Iz = Id + Id
Rd = (Vo-Vr)/Id

Ib2 = (1000.0*Ic2)/hFE2

#Since we require I1>Ib2 , we select
I1=10.0*(10**-3)#in A
Vbe = 0.7#in V

V2 = Vbe + Vr

R1 = (Vo-V2)/I1
R2 = V2/I1

#We are selecting Texas Instruments 2N1722 silicon power transistor, so following parameters are required
Ic1 = 1#in A
hFE1=125.0
hfe1=100.0
hie1=20.0

Ib1 =(1000.0*I1 + Il + Id)/hFE1
#The current through resistor R3 is
I=Ib1 + Ic2

R3 = (Vi - (Vbe + Vo))/I

Gm = hfe2*(R2/(R2+R1))*(1/((R1*R2/(R1+R2))+hie2+(1+hfe2)*Rz)) 
Sv = (10**-3)/(Gm*R3)

Ro = (ro + (((1000*R3) + hie1)/(1+hfe1)))/(1 + (Gm*((1000*R3) + ro)))

dVo = (Sv*dVi)+(Ro*dIl)

#Results
print("Change in output voltage = %.2f v " %dVo)

Change in output voltage = 0.74 v

Example 18.2a Page No 792

In [5]:
import math
from scipy import integrate
#initialisation of variables
Vs=230.0  #in V
Rl=200.0  #in ohm
#Trigger is adjusted so that conduction starts after 60degree of start of cycle
#Instantaneous Current il = (230*2^0.5*sin(a))/200

#Calculations
# to find rms value
xo = math.pi/3.0 #lower limit of integration
x1 = math.pi     #upper limit of integration
def f(x):
    return((230.0*(2.0**0.5)*math.sin(x))/200.0)**2.0


X = integrate.quad(f,xo,x1)
Irms = (X[0]/(2*math.pi))**0.5

#Results
print("The value of Irms is %.2f A " %Irms)

The value of Irms is 0.73 A

Example 18.2b Page No 792

In [6]:
import math
from scipy import integrate
#initialisation of variables
Vs=230.0 #in V
Rl=200.0 #in ohm
#Trigger is adjusted so that conduction starts after 60degree of start of cycle
#Instantaneous Current il = (230*2^0.5*sin(a))/200
#It is noted that between 0 to pi/3 SCR voltage equals line voltage and between pi/3 to pi it is zer and for the rest it is equal to line voltage
#Vl = 230*2^0.5*sin(x)
#To find instantaneous power

x0=math.pi/3#lower limit of integral
x1=math.pi#upper limit of integral
def f(x):
    return((230*230*2*(math.sin(x)**2))/200)

X = integrate.quad(f,xo,x1)
P = X[0]/(2*3.14)
print("The value of P is %.2f W " %P)

The value of P is 106.45 W

Example 18.2c Page No 792

In [7]:
import math

#initialisation of variables

Vs=230.0 #in V
Rl=200.0 #in ohm
#Trigger is adjusted so that conduction starts after 60degree of start of cycle
#Instantaneous Current il = (230*2**0.5*sin(a))/200

#To find Vrms

#Calculations
#It is noted that between 0 to pi/3 SCR voltage equals line voltage and between pi/3 to pi it is zer and for the rest it is equal to line voltage
xo=0      #lower limit of first integral
x1=math.pi/3.0  #upper limit of first integral
x2=math.pi      #lower limit of second integral
x3=2.0*(math.pi)    #upper limit of second integral

def f(x):
    return((230*(2**0.5)*math.sin(x))**2)
	
def g(x):
    return((230*(2**0.5)*math.sin(x))**2)

X1 = integrate.quad(f,xo,x1)
X2 = integrate.quad(g,x2,x3)
Vrms = ((X1[0]+X2[0])/(2*math.pi))**0.5

#Results
print("The value of Vrms is %.2f v " %Vrms)

The value of Vrms is 177.82 v

Example 18.3, Page No 794

In [8]:
import math
#initialisation of variables

#Caption:SCR Relaxation Oscillator Phase control Circuit
#Given Data
C=0.1#in microF
V=60.0 #in V
Vb=32.0 #in V
Vh=10.0 #holding voltage in V
Ih=100.0 #in microA
c=45.0 #conductance angle in degree
cd = 360.0 - c#angle in which capacitor will get charged
td = (cd/360.0)*(1.0/60)#in ms

#Calculations
#if the anode voltage is positive,the SCR will fire when vc=32V
vc=32#in V
#let time constant = t = R*C
#vc-Vh = (V-Vh)(1-exp(-td/t))
t = -td/math.log(1-((vc-Vh)/(V-Vh)))

#Results

print("The value of time constant = is %.2f sec " %t)
R = t/C#Resistance in K
print("The value of R is %.2f K " %(R*1000))

The value of time constant = is 0.03 sec 
The value of R is 251.52 K