# CHAPTER 20 : SILICON CONTROLLED RECTIFIERS¶

## Example 20.2 : Page number 559¶

In :
#Variable declaration
I=50.0;                             #Surge current, A
t=12.0;                             #Time for which surge current lasts, ms
circuit_fusing_rating_max=90;       #Maximum circuit fusing rating, A²s

#Calculation
circuit_fusing_rating=I**2*(t*10**-3);              #Circuit fusing rating, A²s

#Result
if(circuit_fusing_rating<circuit_fusing_rating_max):
print("The device will not be destroyed.");
else:
print("The device will be destroyed.");

The device will not be destroyed.


## Example 20.3 : Page number 559¶

In :
#Variable declaration
I2_t_rating=50.0;                   #circuit fuse rating, A²s
Is=100.0;                           #Surge current, A

#Calculation
t_max=(I2_t_rating/Is**2)*1000;            #Maximum allowable duration, ms

#Result
print("The maximum allowable duration =%dms"%t_max);

The maximum allowable duration =5ms


## Example 20.4 : Page number 559¶

In :
#Variable declaration
R=220.0;                    #Gate resistor, Ω
I_G=7.0;                    #Gate current, mA
V_GK=0.7;                   #Junction voltage, V

#Calculation
V_in=V_GK+(I_G/1000)*R;                    #Input voltage, V (Kirchhoff's voltage law)

#Result
print("The required input voltage=%.2fV."%V_in);

The required input voltage=2.24V.


## Example 20.5 : Page number 564¶

In :
from math import asin
from math import cos
from math import pi

#Variable declaration
I_G=1.0;                #Gate current, mA
V_m=200.0;              #Peak value of input sinusoidal voltage, V
v=100.0;              #Forward breakdown voltage of SCR, V

#Calculation
#(i)
#v=Vm*sin(theta)
#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage
theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad
theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees

alpha=round(theta,0);                    #Firing angle, degrees

#(ii)
phi=180-alpha;                  #Conduction angle, degrees

#(iii)
V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180));           #Average voltage, V
I_avg=V_avg/R_L;                                    #Average current, A

#Result
print("(i) The firing agle=%d°"%alpha);
print("(ii) The conduction angle=%.0f°"%phi);
print("(iii) The average current=%.4fA "%I_avg);

#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.

(i) The firing agle=30°
(ii) The conduction angle=150°
(iii) The average current=0.5940A


## Example 20.6 : Page number 564¶

In :
from math import asin
from math import cos
from math import pi
from math import floor

#Variable declaration
I_G=1.0;                #Gate current, mA
V_m=400.0;              #Peak value of input sinusoidal voltage, V
v=150.0;                #Forward breakdown voltage of SCR, V

#Calculation
#(i)
#v=Vm*sin(theta)
#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage
theta=asin(v/V_m);            #angle for input voltage = forward breakdown voltage, rad
theta=theta*180/pi;             #angle for input voltage = forward breakdown voltage, degrees

alpha=theta;                    #Firing angle, degrees

#(ii)
V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10;           #Average voltage, V

#(iii)
I_av=V_av/R_L;                                    #Average current, A

#(iv)
P_out=V_av*I_av;                                #Output power, W

#Result
print("(i) The firing agle=%d°"%alpha);
print("(ii) The average output voltage=%.1f V"%V_av);
print("(iii) The average current=%.3fA "%I_av);
print("(iv) The output power=%.2f W"%P_out);

(i) The firing agle=22°
(ii) The average output voltage=122.6 V
(iii) The average current=0.613A
(iv) The output power=75.15 W


## Example 20.7 : Page number 564-565¶

In :
from math import asin

#Variable declaration
v=180.0;                    #Forward breakdown voltage, V
V_m=240.0;                  #Peak value of input voltage, V
w=314.0;                    #Angular frequency of input ,rad/s

#Calculation
#v=Vm*sin(w*t)
#So, t=asin(v/Vm)/w
t=(asin(v/V_m)/w)*1000;                        #Time for which SCR remains off, ms

#Result
print("The SCR remains off for %.1f ms."%t);

The SCR remains off for 2.7 ms.


## Example 20.8 : Page number 565¶

In :
from math import pi
from math import cos
from math import floor

#Variable declaration
alpha=30.0;                 #Firing angle, °

#Calculation
I_av=I_dc;                                                 #Average current(= d.c current), A

#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL

#Result

Peak-load current=3.36 A.


## Example 20.9: Page number 565¶

In :
from math import sqrt
from math import pi
from math import sin

#Variable declaration
alpha=60.0;                 #Firing angle, °
P=100.0;                    #Power rating of tungsten lamp, W
V=110.0;                    #Voltage rating of tungsten lamp, V
V_ac=110.0;                 #a.c supply voltage, V

#Calculation
V_m=V_ac*sqrt(2);               #Peak value of input voltage, V

#Since, E_rms²=(1/2*pi) ∫ V_m²sin²(theta) d(theta), limits: alpha to pi
# E_rms²=Vm²*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),
# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),
E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)));            #r.m.s voltage developed in the lamp, V

I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A

#Result
print("The r.m.s voltage developed in the lamp=%d V."%E_rms);
print("The r.m.s current developed in the lamp=%.2f A."%I_rms);

The r.m.s voltage developed in the lamp=70 V.
The r.m.s current developed in the lamp=0.58 A.


## Example 20.10 : Page number 567¶

In :
from math import cos
from math import pi

#Variable declaration
V_m=200.0;                  #Peak a.c voltage, V
alpha=60;                   #firing angle, °

#Calculation
#(i)
V_av=(V_m/pi)*(1+cos(alpha*pi/180));                #D.C output voltage, V

#(ii)

#Result
print("(i)  d.c output voltage=%.1f V."%V_av);

(i)  d.c output voltage=95.5 V.


## Example 20.11: Page number 567¶

In :
from math import sqrt
from math import pi
from math import sin

#Variable declaration
alpha=60.0;                 #Firing angle, °
P=100.0;                    #Power rating of tungsten lamp, W
V=110.0;                    #Voltage rating of tungsten lamp, V
V_ac=110.0;                 #a.c supply voltage, V

#Calculation
V_m=round(V_ac*sqrt(2));               #Peak value of input voltage, V

#Since, E_rms²=(1/2*pi) ∫ V_m²sin²(theta) d(theta), limits: alpha to pi
# E_rms²=Vm²*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),
# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),
E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi));            #r.m.s voltage developed in the lamp, V

I_rms=E_rms/RL;                 #r.m.s current developed in the lamp, A

#Result
print("The r.m.s voltage developed in the lamp=%.1f V."%E_rms);
print("The r.m.s current developed in the lamp=%.2f A."%I_rms);

The r.m.s voltage developed in the lamp=98.9 V.
The r.m.s current developed in the lamp=0.82 A.


## Example 20.12 : Page number 572¶

In :
#Variable declaration
VCC=15;                     #Suuply voltage, V
V_T=0.7;                    #Gate trigger voltage, V
I_T=7.0;                    #Gate trigger current, mA
I_H=6.0;                    #Holding current. mA
R_Vin=1;                    #Resistance at Vin, kΩ
R_VCC=100;                  #Resistance at Vcc, Ω

#Calculation
#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor
V_out=VCC;                      #Output voltage, when SCR is off, V

#(ii)
V_in=V_T+I_T*R_Vin;             #Input voltage required to trigger the SCR, V

#(iii)
#Since, I_H=(Vcc-VT)/R_Vin;
VCC_SCR_open=(I_H/1000)*R_VCC+V_T;                 #Decreased value of supply voltage at which SCR opens, V

#Result
print("(i) The output voltage when SCR is off=%dV."%V_out);
print("(ii) The input voltage required to trigger the SCR=%.1f V."%V_in);
print("(iii) The decreased supply voltage at which SCR opens=%.1f V."%VCC_SCR_open);

(i) The output voltage when SCR is off=15V.
(ii) The input voltage required to trigger the SCR=7.7 V.
(iii) The decreased supply voltage at which SCR opens=1.3 V.


## Example 20.13 : Page number 572-573¶

In :
#Variable declaration
Vz=5.6;                 #zener voltage, V
V_T=0.7;                #Trigger voltage of SCR, V

#Calculation
VCC=Vz+V_T;                 #Required supply voltage to turn on the crowbar, V

#Result
print("The required supply voltage to turn on the crowbar=%.1fV."%VCC);

The required supply voltage to turn on the crowbar=6.3V.


## Example 20.14 : Page number 573¶

In :
#Variable declaration
Vz=12;                  #Zener breakdown voltage, V
V_T=1.5;                #Trigger voltage, V
tolerance_z=10.0;         #Tolerance of zener diode, %

#Calculation
Vz_max=Vz*(1+tolerance_z/100);          #Maximum value of zener breakdown, V
Vz_min=Vz*(1-tolerance_z/100);          #Minimum value of zener breakdown, V
V_crowbar=Vz_max+V_T;                   #Maximum value of supply voltage for crowbarring, V

#Result
print("The maximum value of supply voltage for crowbarring=%.1fV"%V_crowbar);

The maximum value of supply voltage for crowbarring=14.7V


## Example 20.15 : Page number 573¶

In :
#Variable declaration
VCC=25.0;               #Supply voltage, V

#Calculation
#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,
V_out=0;                #Output voltage, V

print("Output voltage when bright light is on=%dV"%V_out);

#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,
V_out=VCC;                #Output voltage, V
print("Output voltage when bright light is off=%dV"%V_out);

Output voltage when bright light is on=0V
Output voltage when bright light is off=25V

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