CHAPTER 21 : POWER ELECTRONICS

Example 21.3: Page number 585

In [1]:
#Variable declaration
V_GT=2;                 #Gate triggering voltage, V
V_F=0.7;               #Forward voltage for diode D1

#Calculation
#(i)Triggering only by a positive gate voltage,
#A diode is connected at the gatewith the n-side connected to thegate of the device,
V_A=V_F+V_GT;                   #Required voltage to trigger the device, V

print("The  required voltage to trigger the device only by positive voltage=%.1fV."%V_A);

#(ii)
print("In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.");

The  required voltage to trigger the device only by positive voltage=2.7V.
In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.

Example 21.4 : Page number 585-586

In [3]:
#Variable declaration
R=50.0;                   #Resitor, Ω
V=50.0;                   #Supply voltage, V
V_drop=1.0;               #Drop across the triac in conduction, V

#Calculation
#(i) Ideal triac
#Since the triac is ideal, voltage drop across it is zero,
I=V/R;                  #Current through the 50 Ω resistor, A

print("(i) The cuurent through the 50 Ω resistor when the triac is ideal=%dA."%I);

#(ii) Triac has a drop of 1V
I=(V-V_drop)/R;                     #Current through the 50 Ω resistor, A

print("(ii) The current through the 50 Ω resistor when the triac has a drop of 1V=%.2fA."%I);

(i) The cuurent through the 50 Ω resistor when the triac is ideal=1A.
(ii) The current through the 50 Ω resistor when the triac has a drop of 1V=0.98A.

Example 21.5 : Page number 588-589

In [4]:
#Variable declaration
V_GT=2;             #Gate triggering voltage, V
V_BO=20;            #Breakover voltage,V

#Calculation
print("The triggering level is raised by using a diac.");
V_A=V_BO+V_GT;                  #Gate trigger signal, V

#Result
print("In order to turn on the triac, the gate trigger signal=%dV."%V_A);

The triggering level is raised by using a diac.
In order to turn on the triac, the gate trigger signal=22V.

Example 21.6 : Page number 589

In [5]:
#Variable declaration
V_BO=30;            #Breakover voltage of diac, V
V_GT=1;             #Trigger voltage of the triac, V
I_T=10;             #Trigger current, mA


#Calculation
V_A=V_BO+V_GT;              #Voltage required for triggering the triac, V

#Result
print("The minimum capacitor voltage that will trigger the triac=%d V."%V_A);

The minimum capacitor voltage that will trigger the triac=31 V.

Example 21.7 : Page number 593

In [6]:
#Variable declaration
eta=0.6;                        #Intrinsic stand-off ratio for UJT
R_BB=10;                        #Inter-base resistance, kΩ

#Calculation
#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),
#eta=RB1/RBB.
R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, kΩ
R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, kΩ 

#Result
print("Resistance of the bar between B1 and emitter junction=%d kΩ."%R_B1);
print("Resistance of the bar between B2 and emitter junction=%d kΩ."%R_B2);

Resistance of the bar between B1 and emitter junction=6 kΩ.
Resistance of the bar between B2 and emitter junction=4 kΩ.

Example 21.8 : Page number 593

In [7]:
#Variable declaration
V_BB=10;                #Interbase voltage, V
eta=0.65;                #Intrinsic stand-off ratio for UJT
V_D=0.7;                #Voltage drop in the pn junction, V

#Calculation
V_stand_off=eta*V_BB;               #Stand off voltage, V
V_P=V_stand_off+V_D;                #Peak-point voltage, V

#Result
print("Stand off voltage=%.1f V."%V_stand_off);
print("Peak-point voltage=%.1f V."%V_P);

Stand off voltage=6.5 V.
Peak-point voltage=7.2 V.

Example 21.9 : Page number 593

In [8]:
#Variable declaration
V_BB=25;                #Interbase voltage, V
eta_max=0.86;           #Maximum intrinsic stand-off ratio for UJT
eta_min=0.74;           #Minimum intrinsic stand-off ratio for UJT
V_D=0.7;                #Voltage drop in the pn junction, V

#Calculation
V_P_max=eta_max*V_BB+V_D;           #Maximum peak-point, V
V_P_min=eta_min*V_BB+V_D;           #Minimum peak-point, V

#Result
print("Maximum peak-point voltage=%.1fV"%V_P_max);
print("Minimum peak-point voltage=%.1fV"%V_P_min);

Maximum peak-point voltage=22.2V
Minimum peak-point voltage=19.2V

Example 21.10 : Page number 593-594

In [11]:
#Variable declaration
eta=0.65;                        #Intrinsic stand-off ratio for UJT
R_BB=7.0;                          #Inter-base resistance, kΩ
R1=100.0;                          #Resistor R1, Ω
R2=400.0;                          #Resistor R2, Ω
V_S=12.0;                          #Source voltage, V
V_D=0.7;                         #Voltage drop in the pn junction, V

#Calculation
#(i)
#Since, eta=RB1/RBB,
R_B1=eta*R_BB;                  #Resistance of the bar between B1 and emitter junction, kΩ
R_B2=R_BB-R_B1;                 #Resistance of the bar between B2 and emitter junction, kΩ 

print("(i)  Resistance of the bar between B1 and emitter junction=%.2f kΩ."%R_B1);
print("     Resistance of the bar between B2 and emitter junction=%.2f kΩ."%R_B2);

#(ii)
V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000));          #Voltage  across B2-B1, V (voltage divider rule)
V_P=eta*V_B2_B1+V_D;                        #Peak-point voltage, V

print("(ii) The voltage across the base B2-B1=%.1fV."%V_B2_B1);
print("     Peak-point voltage=%.2fV"%V_P);

(i)  Resistance of the bar between B1 and emitter junction=4.55 kΩ.
     Resistance of the bar between B2 and emitter junction=2.45 kΩ.
(ii) The voltage across the base B2-B1=11.2V.
     Peak-point voltage=7.98V

Example 21.11 : Page number 596

In [12]:
from math import log

#Variable declaration
RE_initial=5;                   #Initial value of emitter resistor, kΩ
RE_adjusted=10;                 #Adjusted value of emitter resistor, kΩ
C=0.2;                          #Capacitance, μF
eta=0.54;                         #intrinsic stand-off ratio

#Calculation
#(i)
t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms
f=(1/t)*1000;                                             #frequency, Hz

print("Frequency for 5kΩ setting=%dHz."%f);

#(i)
t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms
f=(1/t)*1000;                                             #frequency, Hz

print("Frequency for 10kΩ setting=%dHz."%f);

Frequency for 5kΩ setting=1282Hz.
Frequency for 10kΩ setting=645Hz.

Example 21.12 : Page number 596-597

In [13]:
from math import log

#Variable declaration
V_S=12;                         #Supply voltage, V
R_BB=5;                         #Interbase resistance, kΩ
R_1=50;                         #Resistor R1, kΩ
R_2=0.1;                        #Resistor R2, kΩ
C=0.1;                          #Capacitance, μF
eta=0.6;                          #intrinsic stand-off ratio
V_D=0.7;                        #Voltage drop across pn junction, V

#Calculation
#(i)
#Since, η=R_B1/R_BB,
R_B1=eta*R_BB;                            #Resitance between base B1 and emitter junction, kΩ

#Since, R_BB=R_B1+R_B2
R_B2=R_BB-R_B1;                         #Resitance between base B2 and emitter junction, kΩ

#(ii)
V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2);           #Voltage drop across R_B1 and R_2 resistors, V
V_P=V_D+V_RB1_R2;                           #Peak-point voltage, V

#(iii)
t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2);                #Time period, ms
f=(1/t)*1000;                                                    #frequency, Hz


#Result
print("(i)   R_B1=%dkΩ and R_B2=%dkΩ"%(R_B1,R_B2));
print("(ii)  The peak-point voltage to turn on the UJT=%.0fV."%V_P);
print("(iii) Frequency of oscillations=%dHz."%f);

(i)   R_B1=3kΩ and R_B2=2kΩ
(ii)  The peak-point voltage to turn on the UJT=8V.
(iii) Frequency of oscillations=218Hz.
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