Chapter 5: Choppers

example 5.1, Page No. 249

In [6]:
# chopper parameters

import math
#Variable declaration
Vi = 200.0               # input voltage
Vo = 150.0               # output voltage
R = 10.0                 # resistance

#Calculations
#(a)
alfa = Vo/Vi
#(b)
Iavg = Vo/R
Irms = math.sqrt(alfa)*Vi/R
#(c)
Irms_t = Irms
#(d)
Iavg_i = Iavg
#(e)
Reff = R/alfa

#Result
print("(a) Alfa = %.2f "%alfa)
print("(b) Average load current  = %.0f A\n    RMS load current = %.2f A"%(Iavg,Irms))
print("(c) RMS thyristor current = %.2f A"%Irms_t)
print("(d) Average input current = %.0f A"%Iavg_i)
print("(e) Effective input resistance = %.3f ohm"%Reff)

(a) Alfa = 0.75 
(b) Average load current  = 15 A
    RMS load current = 17.32 A
(c) RMS thyristor current = 17.32 A
(d) Average input current = 15 A
(e) Effective input resistance = 13.333 ohm
In [14]:
# periods of conduction and blocking in each cycle

import math
#Variable declaration
Vi = 230.0                    # input voltage
Vav = 150.0                   # output voltage
f = 1000.0                    # frequency

# Calculation
T = 1/f
Ton = Vav*T/Vi
Toff = T -Ton

#Result
print("Conduction Period = %.3f*10^-3 s\nBlocking period   = %.3f*10^-3 s"%(Ton*1000,Toff*1000))

Conduction Period = 0.652*10^-3 s
Blocking period   = 0.348*10^-3 s

example 5.3, Page No. 249

In [16]:
# periods of conduction and blocking

import math
#Variable declaration
Ra = 0.06                  # armature resistance
Rf = 0.03                  # field resistance
I = 15.0                   # Average circuit current
f = 500.0                  # chopper frequency
Be = 100                   # back emf of motor
Vi = 200                   # input voltage

#Calculations
Vav = I*(Ra+Rf)+Be
T = 1/f
Ton = Vav*T/Vi
Toff = T -Ton

#Result
print("Conduction Period = %.4f*10^-3 s\nBlocking period   = %.4f*10^-3 s"%(Ton*1000,Toff*1000))

Conduction Period = 1.0135*10^-3 s
Blocking period   = 0.9865*10^-3 s

example 5.4, Page No.250

In [22]:
# Duty Cycle

import math
#Variable declaration
N = 800.0            # rated speed in rpm
I = 20               # rated current
Ra = 0.5             # armature resistanc
V = 240.0            # supply voltage
N1 = 600.0           # motor speed

#Calculations
#(a)
Be_800 = V- I*Ra
Be_600 = Be_800*N1/N
Vav = Be_600 + I*Ra
D1 = Vav/V
#(b)
Ia = I/2.0
Vav2 = Be_600+Ia*Ra
D2 = Vav2/V

#Result
print("(a) Since torque is constant,Ia is also constant, i.e. %d A\n    Duty cycle = %.4f "%(I,D1))
print("(b) Since torque is reduced to half and flux is constant,armature current is reduced to half, i.e. %d A\n    Duty cycle = %.4f "%(Ia,D2))

(a) Since torque is constant,Ia is also constant, i.e. 20 A
    Duty cycle = 0.7604 
(b) Since torque is reduced to half and flux is constant,armature current is reduced to half, i.e. 10 A
    Duty cycle = 0.7396

example 5.5, Page No.256

In [48]:
# chopper parameters

import math
#Variable declaration
V = 200.0               # input voltage
R = 8.0                 # load resistance
Vt = 2.0                # voltage drop across thyristor
f = 800.0               # chopper frequency
d = 0.4                 # duty cycle
alfa = 0.5              # duty cycle for case 5

#Calculations
#(a)
Vav = d*(V-Vt)
#(b)
Vl = math.sqrt(d)*(V-Vt)
#(c)
Po = (Vl**2)/R
Pi = d*V*(V-Vt)/R
eff = Po/Pi
#(d)
Ri = R/d
#(e)
Vl2 = 2*(V-Vt)/math.pi
sqrt_2 = math.floor(math.sqrt(2)*1000)/1000
V1 = Vl2/sqrt_2
V1 = math.floor(V1*1000)/1000

#Result
print("(a) Vav = %.1f V\n(b) Vl = %.3f V\n(c) Chopper efficiency = %.0f%%\n(d) Input resistance = %.0f ohm"%(Vav,Vl,eff*100,Ri))
print("(e)\nThe fundamental component (n=1) is,\nVl = %.2fsin(%d*pi*t)\nThe rms value of fundamental component is, V1 = %.3f V"%(Vl2,2*f,V1))

(a) Vav = 79.2 V
(b) Vl = 125.226 V
(c) Chopper efficiency = 99%
(d) Input resistance = 20 ohm
(e)
The fundamental component (n=1) is,
Vl = 126.05sin(1600*pi*t)
The rms value of fundamental component is, V1 = 89.144 V

example 5.6, Page No. 257

In [50]:
# chopping frequency

import math
#Variable declartion
V = 400                 # input voltage
R = 0                   # resistance
L = 0.05                # inductance
alfa = 0.25             # duty cycle of chopper
I = 10                  # load current

#Calculation
Vav = V*alfa
Ton = L*I/(V-Vav)
T = Ton/alfa
f = 1/T

#Result
print("Chopper frequency = %d pulses/s"%(f))

Chopper frequency = 150 pulses/s

example 5.7, Page No. 257

In [65]:
# chopper parameters

import math
#variable declaration
R = 4.0                 # resistance
L = 6*10**-3            # inductance
V = 200.0               # voltage
alfa = 0.5              # duty cycle
f = 1000.0              # frequency
E = 0                   # back emf 
#calculations
T = 1/f
#(a)
x = R*T/L
Imax = ((V/R)*((1-math.e**(-alfa*x))/(1-math.e**(-x))))-(E/R)
Imax = math.ceil(Imax*100)/100
Imin = ((V/R)*(((math.e**(alfa*x))-1)/((math.e**(x))-1)))-(E/R)
Imin = math.floor(Imin*100)/100
#(b)
ripple = V/(f*R*L)
#(c)
Iavg = (Imax+Imin)/2
#(d)
Il = math.sqrt((Imin**2)+(((Imax-Imin)**2)/3)+(Imin*(Imax-Imin)))
Il = math.floor(Il*100)/100
#(e)
Iavg_i =0.5*Iavg
#(f)
Irms_i = Il*math.sqrt(0.5)

#Result
print("(a) Imax = %.2f A\tImin = %.2f A"%(Imax,Imin))
print("(b) Maximum Ripple = %.2f A\n(c) Average load current = %.0f A"%(ripple,Iavg))
print("(d) rms load current = %.2f A\n(e) Average input current = %.1f A\n(f) RMS input current = %.3f A"%(Il,Iavg_i,Irms_i))

(a) Imax = 29.13 A	Imin = 20.87 A
(b) Maximum Ripple = 8.33 A
(c) Average load current = 25 A
(d) rms load current = 25.11 A
(e) Average input current = 12.5 A
(f) RMS input current = 17.755 A

example 5.8, Page No.258

In [66]:
# Load inductance

import math
#Variable declaration
V = 300                    # input voltage
R = 5                      # load resistance
f = 250.0                  # chopping frequency
r = 0.20                   # 20% ripple
I = 30                     # Average load current

#Calculations
x = r*I
L = V/(4*f*x)

#Result
print("L = %.0f *10^-3 H"%(L*1000))

L = 50 *10^-3 H

example 5.9, Page No.258

In [88]:
# Current at the instannce of turn off of thryster and 5 msec after turn off

import math
#Variable declaration
R = 0.5                # Armature resistance
L = 16*10**-3          # Arature inductance
V = 200.0              # DC input voltage
Be = 100.0             # Back emf
Imin = 10.0            # minimum load current
Toff = 2*10**-3        # thyristor turn off time
delay = 5*10**-3       # current at the to be calculated 5 msec after turn off

#Calculation
#(a)
x = R/L
i = (((V-Be)/R)*(1-math.e**(-x*Toff)))+Imin*(math.e**(-x*Toff))
i = math.floor(i*100)/100
#(b)
i2 = i*(math.e**(-x*delay))

#Result
print("(a) Current at the instance of turn off of thyristor = %.2f A"%i)
print("(b) After the turn off of thyristor,the current freewheels 5 ms.\n\ti = %.2f A"%i2)
# answers in the book are wrong:12.12+9.39=21.41

(a) Current at the instance of turn off of thyristor = 21.51 A
(b) After the turn off of thyristor,the current freewheels 5 ms.
	i = 18.40 A

example 5.10, Page No. 258

In [90]:
# Speed of the motor

import math
#Variable declaration
V = 220.0              # input voltage
Ra = 1.0               # armature resistance
N = 1000.0             # no load speed of the motor
alfa = 0.6             # duty cycle
I = 20                 # motor current
E = 220.0              # back emf at no load

# Calculations
Vi = V*alfa
Be = Vi-Ra*I
N1 = Be*N/V

#Result
print("Speed of the motor = %.1f rpm"%N1)

Speed of the motor = 509.1 rpm

example 5.11, Page No.259

In [84]:
# Average load voltage of the chopper

import math
#Variable declaration
Vi = 230.0                     # input voltage
Ton = 25                       # on time in ms
Toff = 10                      # off time in ms

#Calculations
V = Vi*Ton/(Ton+Toff)

#Result
print("Average load voltage = %.3f V"%V)

Average load voltage = 164.286 V

example 5.12, Page No. 259

In [89]:
# max, min and average load current and load voltage

import math
#Variable declaration
V = 100.0               # applied voltage
Be = 0                  # back emf
L = 1.0*10**-3          # inductance
R = 0.5                 # Resistance
T = 3*10**-3            # Ton + Toff
Ton = 1*10**-3          # Ton

#Calculations
alfa = Ton/T
x = T*R/L
#(a)
Imax = (V/R)*((1- math.e**(-alfa*x))/(1-math.e**(-x)))
#(b)
Imin = (V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1))
#(c)
Vavg = alfa*V

#Result
print(" Imax = %.2f A\n Imin = %.1f A\n Average load current = %.1f A\n Average output voltage = %.2f V"%(Imax,Imin,(Imax+Imin)/2,Vavg))

 Imax = 101.30 A
 Imin = 37.3 A
 Average load current = 69.3 A
 Average output voltage = 33.33 V

example 5.13, Page No.259

In [10]:
# In One quadrant chopper, output voltage and current

import math
#Variable declaration
V = 100.0               # applied voltage
Be = 12.0               # back emf
L = 0.8*10**-3          # inductance
R = 0.2                 # Resistance
T = 2.4*10**-3          # Ton + Toff
Ton = 1*10**-3          # Ton

#Calculations
alfa = Ton/T
x = T*R/L
#(a)
Imax = (V/R)*((1- math.e**(-alfa*x))/(1-math.e**(-x)))
#(b)
Imin = (V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1))
#(c)
Vavg = alfa*V

#Result
print("(a) Imax = %.2f A\n(b) Imin = %.1f A\n(c) Average output voltage = %.2f V"%(Imax,Imin,Vavg))

(a) Imax = 245.13 A
(b) Imin = 172.7 A
(c) Average output voltage = 41.67 V

example 5.14, Page No. 260

In [12]:
# Series inductance

import math
#Variable declaration
f = 400.0              # chopping frequency
V = 500.0              # input voltage
I = 10                 # maximum current swing

#calculations
L = V/(4*f*I)

#Result
print("Series Inductance, L = %.2f*10^-3 H "%(L*1000))

Series Inductance, L = 31.25*10^-3 H

example 5.15, Pager No.260

In [16]:
# Motor speed and current swing

import math
#Variable declaration
V = 800.0          # input voltage
P = 300.0          # motor power rating in HP
eff = 0.9          # motor efficiency
R = 0.1            # total armature and field resistance
L = 100*10**-3     # inductance
alfa = 0.2         # duty cycle
N = 900.0          # motor rated speed 
f = 400.0          # chopping frequency

#Calculations
Po = P*735.5/1000
Pi = Po/eff
I = Pi*1000/V
Be = V-(I*R)
Be_duty = (alfa*V)-(I*R)
N2 = N*Be_duty/Be
di = (V- (alfa*V))*alfa/(L*f)

#Result
print("Motor speed = %.2f rpm\nCurrent swing = %.1f A"%(N2,di))

Motor speed = 151.32 rpm
Current swing = 3.2 A

example 5.16, Page No. 261

In [17]:
print("Theoretical example")

Theoretical example

example 5.17, Page No. 262

In [18]:
print("Theoretical example")

Theoretical example

example 5.18, Page No. 264

In [59]:
# step down chopper parameters

import math
#Variable declaration
V = 200.0               # input voltage
R = 2.0                 # resistance
L = 10*10**-3           # inductance
E = 20.0                # back emf
T = 1000*10**-6         # chopping cycle
Ton =300*10**-6         # Ton time of thyrister

#Calculations
#(a)
f = 1/T
x = (R*T)/L
alfa_min =(math.log(1+((E/V)*(math.e**(x)-1))))/x
alfa = Ton/T
#(b)
Imax = ((V/R)*((1-math.e**(-alfa*x))/(1-math.e**(-x))))-(E/R)
Imax = math.floor(Imax*10)/10
Imin = ((V/R)*((math.e**(alfa*x)-1)/(math.e**(x)-1)))-(E/R)
Imin = math.floor(Imin*100)/100
#(c)
Il = (alfa*V-E)/R
#(d)
x1 = alfa*V
x2 = (V*math.sin(math.pi*2*alfa))/math.pi
x3 = (V/math.pi)*(1-math.cos(alfa*2*math.pi))
x4 =  (V*math.sin(math.pi*2*2*alfa))/(math.pi*2)
x5 =  (V/(2*math.pi))*(1-math.cos(alfa*2*2*math.pi))
x6 =  (V*math.sin(math.pi*3*2*alfa))/(math.pi*3)
x7 =  (V/(3*math.pi))*(1-math.cos(alfa*2*3*math.pi))
d = f*2*math.pi
#(e)
Iavg = (alfa*(V-E)/R)-(L*(Imax-Imin)/(R*T))

#Result
print("(a) Minimum required value of alfa = %.4f\n    Actual value of alfa = %.1f"%(alfa_min,alfa))
print("    Since actual value is more than minimum required for continuous current,the load current is continuous.")
print("\n(b) Imax = %.1f A\tImin = %.2f A"%(Imax,Imin))
print("\n(c) Average load current = %.0f A"%Il)
print("\n(d) Vl = %d + %.2fcos(%.1ft) + %.2fsin(%.1ft)%.2fcos(%.1ft) + %.2fsin(%.1ft)%.2fcos(%.1ft) + %.2fsin(%.1ft)+...."%(x1,x2,d,x3,d,x4,2*d,x5,2*d,x6,3*d,x7,3*d))
print("\n(e) Average Input current = %.2f A"%Iavg)
# answer for Imin is slightly greater than the answer given in book and hence answer for avg input current

(a) Minimum required value of alfa = 0.1095
    Actual value of alfa = 0.3
    Since actual value is more than minimum required for continuous current,the load current is continuous.

(b) Imax = 22.1 A	Imin = 17.92 A

(c) Average load current = 20 A

(d) Vl = 60 + 60.55cos(6283.2t) + 83.33sin(6283.2t)-18.71cos(12566.4t) + 57.58sin(12566.4t)-12.47cos(18849.6t) + 4.05sin(18849.6t)+....

(e) Average Input current = 6.10 A

example 5.19, Page No.266

In [74]:
# load current values

import math
#variable declaration
V = 200.0               # input voltage
R = 4.0                 # resistance
L = 10*10**-3           # inductance
T = 1000*10**-6         # chopping cycle
Ton =300*10**-6         # Ton time of thyrister

#Calculations
f = 1/T
alfa = Ton/T
x1 = alfa*V
x2 = (V*math.sin(math.pi*2*alfa))/math.pi
x2 = math.ceil(x2*100)/100
x3 = (V/math.pi)*(1-math.cos(alfa*2*math.pi))
x3 = math.floor(x3*100)/100
x4 =  (V*math.sin(math.pi*2*2*alfa))/(math.pi*2)
x4 = math.floor(x4*100)/100
x5 =  (V/(2*math.pi))*(1-math.cos(alfa*2*2*math.pi))
x5 = math.floor(x5*100)/100
x6 =  (V*math.sin(math.pi*3*2*alfa))/(math.pi*3)
x6 = math.ceil(x6*100)/100
x7 =  (V/(3*math.pi))*(1-math.cos(alfa*2*3*math.pi))
x7 = math.floor(x7*100)/100

sqrt_2 = math.floor(math.sqrt(2)*1000)/1000
V1 = math.sqrt((x2**2+x3**2))/sqrt_2
V2 = math.sqrt((x4**2+x5**2))/sqrt_2
V3 = math.sqrt((x6**2+x7**2))/sqrt_2
Z1_mag = math.sqrt(R**2+(2*math.pi*f*L)**2)
Z1_angle = math.tan((2*math.pi*f*L)/R)
I1 = V1/Z1_mag

Z2_mag = math.sqrt(R**2+(2*2*math.pi*f*L)**2)
Z2_angle = math.tan((2*2*math.pi*f*L)/R)
I2 = V2/Z2_mag

Z3_mag = math.sqrt(R**2+(3*2*math.pi*f*L)**2)
Z3_angle = math.tan((3*2*math.pi*f*L)/R)
I3 = V3/Z3_mag

Iavg = alfa*V/4
Irms = math.sqrt(Iavg**2+I1**2+I2**2+I3**2)

print("RMS value of first harmonic component of load current  = %.3f A"%I1)
print("RMS value of second harmonic component of load current = %.4f A"%I2)
print("RMS value of third harmonic component of load current  = %.4f A"%I3)
print("RMS value of load current = %.4f A"%Irms)

RMS value of first harmonic component of load current  = 1.157 A
RMS value of second harmonic component of load current = 0.3406 A
RMS value of third harmonic component of load current  = 0.0492 A
RMS value of load current = 15.0485 A

example 5.20, Page No.273

In [79]:
# L and C of the auxiliary commutation circuit

import math
#Variable declaration
V = 200               # input voltage
I = 50                # average current
Il = 60               # peak load current
Toff = 15*10**-6      # thyristor turn off current
l = 1.5               # max current limit of 150%


#calculation
C = Toff*Il/V           # C should be greater than this value
C1 = 5 *10**-6          # assumed
Ic = Il*l-Il
L = (V**2*(C1))/(Ic**2)

#Result
print("C > %.1f*10^-6, Therefore assume C = %.1f*10^6 F"%(C*10**6,C1*10**6))
print("L = %.2f*10^-6 H"%(L*10**6))

C > 4.5*10^-6, Therefore assume C = 5.0*10^6 F
L = 222.22*10^-6 H

example 5.21, Page No.277

In [81]:
# period of conduction

import math
#Variable declaration
V = 200.0                    # input voltage 
Vav = 250.0                  # output voltage
Toff = 0.6*10**-3            # blocking period

#calculation
Ton = (Vav*Toff/V) - Toff

#Result
print("Ton = %.2f*10^-3 second"%(Ton*10**3))

Ton = 0.15*10^-3 second

example 5.22, Page No.277

In [83]:
# period of conduction

import math
#Variable declaration
V = 150.0                    # input voltage 
Vav = 250.0                  # output voltage
Toff = 1*10**-3            # blocking period

#calculation
Ton = (Vav*Toff/V) - Toff

#Result
print("Ton = %.4f*10^-3 second"%(Ton*10**3))

Ton = 0.6667*10^-3 second