# CHAPTER 6: SEMICONDUCTOR DIODE¶

## Example 6.2, Page number 81¶

In :
#Variable Declaration
Vf =20;                #Peak Input Voltage in V
rf=10;                  #Forward Resistance in ohms
V0=0.7;                  #Potential Barrier Voltage of the diodes in V

#Calculation
#(1)
If_peak=(Vf-V0)/(rf+RL);                  #Peak current through the diode in A
If_peak=If_peak*1000;                    #Peak current through the diode in mA
#(2)
V_out_peak =If_peak * RL/1000 ;               #Peak output voltage in V

#For an Ideal diode
If_peak_ideal=Vf/RL;                   #Peak current through the ideal diode in A
If_peak_ideal=If_peak_ideal*1000;      #Peak current through the ideal diode in mA

V_out_peak_ideal=If_peak_ideal * RL/1000;     # Peak output voltage in case of the ideal diode in V

#Result
print '(i) Peak current through the diode = %.1f mA '%If_peak;
print '(ii) Peak output voltage = %.1f V'%V_out_peak;
print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;
print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;

(i) Peak current through the diode = 37.8 mA
(ii) Peak output voltage = 18.9 V
(iii) Peak current through the ideal diode = 40 mA
(iv) Peak output voltage in case of the ideal diode = 20 V


## Example 6.3, Page number 82¶

In :
#Variable Declaration
V =10.0;                #Battery voltage in V
R1=50.0;                  #Resistor 1's resistance in ohms
R2=5.0;                 #Resistor 2's resistance in ohms

#Calculation
#Using Thevenin's Theorem to find current in the diode
E0=(R2/(R1+R2))*V;          #Thevenin's Voltage in V
R0=(R1*R2)/(R1+R2);         #Thevenin's Resistance in ohms

I0=E0/R0;                   #Current through the diode in A
I0=I0*1000;                 #Current through the diode in mA

#Result
print 'Current through the diode = %d mA '%Io;

Current through the diode = 200 mA


## Example 6.4, Page number 82-83¶

In :
#Variable Declaration
V =10.0;                #Battery voltage in V
R0=48.0;                  #Resistance of the resistor in ohms
Rd=1.0;                 #Forward resistance of the diodes in ohms
Vd=0.7;                 #Potential barrier of the diodes in V
#Calculation
V_net=V-Vd-Vd;        #Net voltage in the circuit in V
R_net=R0+Rd+Rd        #Net resistance of the circuit in ohms
I_net=V_net/R_net;    #Net current in the circuit in A
I_net=I_net*1000;     #Net current in mA

#Result
print 'Net current in the circuit = %d mA '%I_net;

Net current in the circuit = 172 mA


## Example 6.5, Page number 83¶

In :
#Variable Declaration
E1=24;           #Voltage of first source in V
E2=4;            #Voltage of second source in V
V0=0.7;          #Potential barrier of diodes in V
R=2000;          #Resistance of the given resistor in ohms
Rd=0;            #Forward resistance of the diodes in ohms

#Calculation
I=(E1-E2-V0)/(R+Rd);  #Current in the circuit in A
I=I*1000;             #Current in the circuit in mA

#Result
print 'Current in the circuit = %.2f mA '%I;

Current in the circuit = 9.65 mA


## Example 6.6, Page number 83-84¶

In :
#Variable Declaration
V=20;           #Voltage of source in V
V0=0.3;         #Potential barrier of Germanium diode in V
V0_Si=0.7;      #Potetial barrier of Silicon diode in V

#Calculation
#As only Ge diode is turned on due to less potential barrier,
VA=V-V0;        #Voltage VA acroos resistor of 3k ohms

#Result
print 'Voltage VA = %.1f mA '%VA;

Voltage VA = 19.7 mA


## Example 6.7, Page number 84¶

In :
#Variable Declaration
V=10;           #Voltage of source in V
V0=0.7;      #Potetial barrier of Silicon diode in V
# Resistance of all resistors in ohms
R1=2000;
R2=2000;
R3=2000;

#Calculation
Id=(V-V0)/(R2+2*R3);     #Current through the diodes in A
VQ=2*Id*R3;      #Voltage VQ across the grounded 2k ohm resistor in V
Id=Id*1000;      #Current through the diodes in mA

#Result
print 'Voltage VQ = %.1f V '%VQ;
print 'Current through the diodes, Id = %.2f mA '%Id;

Voltage VQ = 6.2 V
Current through the diodes, Id = 1.55 mA


## Example 6.8, Page number 84¶

In :
#Variable Declaration
V=15;           #Voltage of source in V
V0=0.7;      #Potetial barrier of Silicon diode in V
R=500       # Resistance of all resistors in ohms

#Calculation
I1=(V-V0)/R;      #total current in the circuit in A
Id1=I1/2;         #current in first diode in A
Id1=Id1*1000;     #current in first diode in mA
Id2=Id1           #current in second diode in mA

#Result
print ('Current in first diode = %.1f mA'%Id1);
print ('Current in second diode = %.1f mA'%Id2);

Current in first diode = 14.3 mA
Current in second diode = 14.3 mA


## Example 6.9, Page number 85¶

In :
#Variable Declaration
E=20;           #Voltage of source in V
V0_d1=0.7;      #Potetial barrier of first Silicon diode in V
V0_d2=0.7;      #Potetial barrier of second Silicon diode in V
R1=5600;       # Resistance of first resistor in ohms
R2=3300;       #  Resistance of second resistor in ohms

#Calculation
I2=V0_d2/R2;                  #Current I2 through resistor R2 in A
I2=round((I2*1000),3);                   #Current I2 through resistor R2 in mA
I1=(E-V0_d1-V0_d2)/R1;        #Current I1 through resistor R1 in  A
I1=round((I1*1000),2);                   #Current I1 through resistor R1 in  mA
I3=I1-I2;                     #Current I3 through diode D2 in mA

#Result
print 'Current I1= %.2f mA'%I1;
print 'Current I1= %.3f mA'%I2;
print 'Current I1= %.3f mA'%I3;

Current I1= 3.32 mA
Current I1= 0.212 mA
Current I1= 3.108 mA


## Example 6.10, Page number 85-86¶

In :
#Variable Declaration
E=10.0;           #Voltage of source in V
V0=0.7;      #Potetial barrier of Silicon diode in V
R1=2000;       # Resistance of first resistor in ohms
R2=8000;       #  Resistance of second resistor in ohms
R3=4000;       #Resistance of third resistor in ohms
R4=6000;       #Resistance of fourth resistor in ohms

#Calculation
#Assuming the given diode to be reverse bised and calculating voltage across it's terminals
V1=(E/(R1+R2))*R2;      #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V
V2=(E/(R3+R4))*R4;      #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V

#Result
if((V1-V2)>=V0):
print 'Our assumption was wrong and, the diode is forward biased';
else:
print 'The diode is reverse biased';

Our assumption was wrong and, the diode is forward biased


## Example 6.11, Page number 86¶

In :
#Variable declaration
V=2;                              #Supply voltage in V
V0=0.7;                           #Potential barrier voltage of the diode in V
R1=4000.0;                        #Resistance of first resistor in Ω
R2=1000.0;                        ##Resistance of second resistor in Ω

#Calculation
#Assuming the diode to be in ON state
I1=((V-V0)/R1)*1000;                  #Current through resistor R1, in mA
I2=(V0/R2)*1000;                    #Current through resistor R2, in mA
ID=I1-I2;                           #Diode current, in  mA

if(ID<0):
#Since the diode current is negative, the diode must be OFF
ID=0;                     #True value of diode current, mA

#As the diode is in OFF state it can be replaced by an open ciruit equivalent
VD=V*R2/(R1 +R2);                #Voltage across the diode, in V

#Result
print 'ID =%d mA'%ID;
print 'VD =%.1f V'%VD;

ID =0 mA
VD =0.4 V


## Example 6.12, Page number 89-90¶

In :
#Variable declaration
AC_Input_Power=100.0;          #Input AC Power in watts
AC_Output_Power=40.0;          #Output AC Power in watts
Accepted_Power=50.0;           #Power accepted by the half-wave rectifier in watt

#Calculation
R_eff=(AC_Output_Power/AC_Input_Power)*100;             #Rectification efficiency of the half-wave rectifier
Unused_power=AC_Input_Power-Accepted_Power;       #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt
Power_dissipated=Accepted_Power-AC_Output_Power;  #Power dissipated by the diode watt

#Result
print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;

print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);

The rectification efficiency of the half-wave rectifier= 40%
Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts


## Example 6.13, Page number 90¶

In :
from math import pi
from math import sqrt
#Variable declaration
Vrms=230.0;                #AC supply RMS voltage in V
Turns_Ratio=10/1;        #turn ratio of the transformer

#Calculation
Vpm=sqrt(2)*Vrms;         #Maximum primary voltage in V
Vsm=Vpm/Turns_Ratio;     #Maximum secondary voltage in V
#Case 1
Vdc=Vsm/(round(pi,2));              #Output D.C voltage, which is the average voltage in V
Vdc=round(Vdc,2);
#Case 2
PIV=Vsm;                 #Peak Inverse Voltage in V

#Result
print 'The output d.c voltage= %.2f V'%Vdc;
print 'The peak inverse voltage= %.2f V'%PIV;

The output d.c voltage= 10.36 V
The peak inverse voltage= 32.53 V


## Example 6.14, Page number 90-91¶

In :
from math import pi
#Variable declaration
rf=20.0;           #Internal resistance of the crystal diode in ohms
Vm=50.0;           #Maximum applied voltage in V

#Calculation
# 1
Im=Vm/(rf+RL);         #Maximum current in A
Im=Im*1000;            #Maximum current in
Im=round(Im,0);
Idc=Im/pi;             #Average voltage in mA
Idc=round(Idc,1);
Irms=Im/2;             #RMS value of the current in mA
Irms=round(Irms,1)

# 2
AC_Input_Power=pow(Irms/1000,2)*(rf+RL);     #Input a.c power in watt

DC_Output_Power=pow(Idc/1000,2)*RL;          #Output d.c power in watt

# 3
DC_Output_Voltage=(Idc/1000)*RL;               #Output d.c voltage in V

# 4
Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100;   # Efficiency of rectification of the half-wave rectifier

#Result
print ' i:';
print '   Im   = %d mA'%Im;
print '   Idc  = %.1f mA'%Idc;
print '   Irms = %.1f mA'%Irms;
print ' ii: ';
print '   a.c input power= %.3f watt'%AC_Input_Power;
print '   d.c output power= %.3f watt'%DC_Output_Power;
print ' iii: ';
print '   d.c output voltage = %.2f volts'%DC_Output_Voltage;
print ' iv: '
print '   Efficiency of rectification = %.1f%%'%Rectifier_efficiency;


 i:
Im   = 61 mA
Idc  = 19.4 mA
Irms = 30.5 mA
ii:
a.c input power= 0.763 watt
d.c output power= 0.301 watt
iii:
d.c output voltage = 15.52 volts
iv:
Efficiency of rectification = 39.5%


## Example 6.15, Page number 91¶

In :
from math import pi
#Variable declaration
Vdc=50.0;           #Output d.c voltage in V
rf=25;            #Diode resistance in ohm

#Calculation
Vm=(pi*(rf+RL)*Vdc)/RL;   #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge  of Vdc from the half-wave rectifier, in V
Vm=round(Vm,0);
#Result
print 'The a.c voltage required should have maximum value of = %d V' %Vm;

The a.c voltage required should have maximum value of = 162 V


## Example 6.16, Page number 95¶

In :
from math import sqrt
from math import pi
#Variable declaration
rf=20;              #Internal resistance of the diodes in ohm
Vrms=50;            #RMS value of transformer's secondary voltage from centre tap to each end of secondary

#Calculation
Vm=Vrms*sqrt(2);      #Maximum a.c voltage in V
Im=Vm/(rf+RL);        #Maximum load current in A
Im=Im*1000;           #Maximum load current in mA

# 1:

# 2:
Irms=Im/sqrt(2);       #RMS value of load current in A

#Result
print 'i:';
print'   The mean load current= %d mA'%Idc;
print 'ii:';
print '  The r.m.s value of the load current = %d mA'%Irms;

i:
The mean load current= 45 mA
ii:
The r.m.s value of the load current = 50 mA


## Example 6.17, Page number 95-96¶

In :
from math import sqrt
#Variable declaration
rf=0;                            #Internal resistance of the diodes in ohm
Turns_ratio=5/1;                 #Primary to secondary turns ratio of transformer
P_Vrms=230;                      #R.M.S value of voltage in primary winding in V
S_Vrms=P_Vrms/Turns_ratio;       #R.M.S value of voltage in secondary winding in V
S_Vm=S_Vrms*sqrt(2);        #Maximum voltage across secondary winding in V
Vm=S_Vm/2;                  #Maximum voltage across half seconfdary winding in V

#Calculation
# 1:
Idc=2*Vm/(pi*RL);            #Average current in A
Vdc=Idc*RL;                  #d.c output voltage in V

# 2:
PIV=S_Vm;                    #Peak Invers Voltage(= Maximum secondary voltage) in V

# 3:
Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL);         #a.c input power in watt
Pdc=(pow(Idc,2)*RL);                        #d.c output power in watt
R_eff=(Pdc/Pac)*100;      #Rectification efficiency
R_eff=round(R_eff,1);

#Result
print 'i:';
print '   The d.c output voltage= %.1f V'%Vdc;
print 'ii:';
print '   The peak inverse voltage= %d V'%PIV;
print 'iii:';
print '   Rectification efficiency= %.1f%%'%R_eff;

i:
The d.c output voltage= 20.7 V
ii:
The peak inverse voltage= 65 V
iii:
Rectification efficiency= 81.1%


NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%.

## Example 6.18, Page number 96-97¶

In :
from math import sqrt
#Variable declaration
fin=50;                       #frequency of input ac source in Hz
Turns_ratio=4/1;              #Transformers turns ratio, primary to secondary.
P_Vrms=230.0;                 #R.M.S value of voltage in primary winding in V
S_Vrms=P_Vrms/Turns_ratio     #R.M.S value of voltage in secondary winding in V
Vm=S_Vrms*sqrt(2);            #Maximum voltage across secondary winding in V

#Calculation
# 1:
Idc=2*Vm/(pi*RL);           # Average current in A
Vdc=Idc*RL;                 #Output d.c voltage in V
Vdc=round(Vdc,0);
# 2:
PIV= Vm;                    #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V

# 3:
fout=2*fin;                 #Output frequency in Hz

#Result
print 'i:';
print '   The d.c output voltage = %d V' %Vdc;
print 'ii:';
print '   The peak inverse voltage = %.1f V'%PIV;
print 'iii:';
print '   The output frequency = %d Hz'%fout;

i:
The d.c output voltage = 52 V
ii:
The peak inverse voltage = 81.3 V
iii:
The output frequency = 100 Hz


## Example 6.19, Page number 97¶

In :
from math import pi
from math import sqrt

#Variable declaration
Turns_ratio=5/1;                 #Primary to secondary turns ratio of the transformer
Vin=230.0;                       #R.M.S value of input voltage in V
fin=50;                          #Input frequency in Hz

#Calculation
Vs_rms=Vin/Turns_ratio;          #R.M.S value of the voltage in secondary winding, in v
Vs_max=Vs_rms*sqrt(2);           #Maximum voltage across secondary, in V

# (i)
#Case i: Centre-tap circuit
Vm=Vs_max/2;                     #Maximum voltage across half secondary winding, in V
Vdc=2*Vm*RL/(pi*RL);             #DC output voltage, in V
print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;

#Case ii:
Vm=Vs_max;                      #Maximum voltage across secondary, in V
Vdc=2*Vm*RL/(pi*RL);            #DC output voltage, in V
print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc;

# ii:
#Case i: Centre-tap circuit
Turns_ratio=5/1;                      #Turns ratio of the transformer
Vs_rms=Vin/Turns_ratio;               #R.M.S value of the secondary voltage in V
Vs_max=Vs_rms*sqrt(2);                #Maximum voltage across the secondary in V
Vm=Vs_max/2;                          #Maximum voltage across half of the secondary in V
PIV=2*Vm;                             #Peak Inverse Voltage in V
print 'PIV in case of centre-tap circuit = %d V'%PIV;

#Case ii: Bridge circuit
Turns_ratio=10/1;                             #Turns ratio of the transformer
Vs_rms=Vin/Turns_ratio;                       #R.M.S value of the secondary voltage in V
Vs_max=Vs_rms*sqrt(2);                        #Maximum voltage across the secondary in V
PIV=Vm;                                        #Peak Inverse Voltage in V
print 'PIV in case of bridge circuit = %.1f V'%PIV;

The d.c output voltage for the centre-tap circuit = 20.7 V
The d.c output voltage for the bridge circuit = 41.4 V
PIV in case of centre-tap circuit = 65 V
PIV in case of bridge circuit = 32.5 V


## Example 6.20, Page number 98¶

In :
from math import pi
from math import sqrt
#Variable declaration
rf=1;                    #forward resistance of diodes of the rectifier in ohm
Vrms=240.0;                #a.c supply voltage in V
Vm=Vrms*sqrt(2);         #Maximum a.c voltage in V

#Calculation
# 1:
Rt=2*rf+RL;             #Total circuit resistance at any instance in ohm
Im=Vm/Rt;               #Maximum load current in A
Idc=2*Im/pi;            #Mean load current in A

# 2:
Irms=Im/2;              #R.M.S value of current  in A
P=pow(Irms,2)*rf;       #Power dissipated in each diode in watt

#Result
print 'i:';
print '   Mean load current = %.2f A'%Idc;
print 'ii:';
print '   Power dissipated in each diode= %.3f W'%P;

i:
Mean load current = 0.45 A
ii:
Power dissipated in each diode= 0.124 W


NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W.

## Example 6.21, Page number 98-99¶

In :
from math import sqrt,pi
#Variable declaration
V0=0.7;                   #Potential barrier voltage of diodes in V
Vrms=12;                  #R.M.S value of input a.c voltage in V
Vs_pk=Vrms*sqrt(2);       #Peak secondary voltage in V

#Calculation
# 1:
Vout_pk=Vs_pk-(2*V0);      #Peak output voltage in  V
Vav=2*Vout_pk/pi;          #Average output voltage in V
Vav=round(Vav,2);

# 2:
Iav=Vav/RL;                #Average output current in A
Iav=Iav*pow(10,6);         #Average output current in μA

#Result
print 'i:';
print '   Average output voltage=%.2f V'%Vav;
print 'ii:';
print '   Average output current=%.1f μA'%Iav;

i:
Average output voltage=9.91 V
ii:
Average output current=825.8 μA


## Example 6.22, Page number 102¶

In :
#Variable declaration
Vdc_A=10;                      #Supply voltage of A in V
Vdc_B=25;                      #Supply voltage of B in V
Vac_rms_a=0.5;                 #Ripples in power supply A in V
Vac_rms_b=0.001;               #Ripples in power supply B in V

#Calculation
#For power supply A
ripple_factor_A=Vac_rms_a/Vdc_A;      #Ripple factor of power supply A

#For power supply B
ripple_factor_B=Vac_rms_b/Vdc_B;      #Ripple factor of power supply B

#Result
if(ripple_factor_A<ripple_factor_B):
print 'Power supply A is better';
else :
print 'Power supply B is better';

Power supply B is better


## Example 6.23, Page number 105-106¶

In :
from math import sqrt
#Variable declaration
C=50*pow(10,-6);            #Capacitance of the capacitor used in filter circuit in F
V0=0.7;                     #Potential barrier voltage of the diodes of the rectifier in  V
Vrms=115.0;                   #R.M.S value of input a.c voltage in V
fin=60;                     #Frequency of input a.c voltage in Hz
Turns_ratio=10/1;           #Primary to secondary, turns ratio of the transformer

#Calculation
Vp_prim=Vrms*sqrt(2);               #Peak primary voltage in V
Vp_sec=Vp_prim/Turns_ratio;         #Peak secondary voltage in V
Vp_in= Vp_sec - 2*V0;               #Peak full wave rectified voltage at the filter input in V
f=2*fin;                            #Output frequency in Hz
Vdc=Vp_in*(1-(1/(2*f*RL*C)));       #Output d.c voltage in V

#Result
print 'The output d.c voltage  is = %.1f V'%Vdc;

The output d.c voltage  is = 14.3 V


## Example 6.24, Page number 106¶

In :
from math import pi
#Variable declaration
R=25;                        #d.c resistance of the choke in ohm
Vm=25.7;                     #Maximum value of the pulsating output from the rectifier in V

#Calculation
V_dc=2*Vm/pi;                #d.c component of the pulsating output in V
V_dc=round(V_dc,1);
V_dc_out=(V_dc*RL)/(R+RL);   #Output d.c voltage in V
V_dc_out=round(V_dc_out,1);

#Result
print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;

 The output d.c voltage accross the load resistance is = 15.9 V


## Example 6.25, Page number 113-114¶

In :
#Variable declaration
Ei=120.0;                   #Input Voltage in V
Vz=50.0;                    #Zener Voltage in V
R=5000.0;                   #Resistance of the series resistor in ohm

#Calculation
V=Ei*RL/(R+RL);             #Voltage across the open circuit if the zener diode is removed
if(V>Vz):
#Zener diode is in ON state
# i:
Output_voltage=Vz;             #Voltage across load resistance, in V
#ii:
Voltage_R=Ei-Vz;               #Voltage across the series resistance R, in V
#iii:
IL=Vz/RL;                      #Load current through RL in A
IL=IL*1000;                    #Load current through RL in mA
I=Voltage_R/R;                 #Current through the series resistance in A
I=I*1000;                      #Current through the series resistance in mA
Iz=I-IL;                       #Applying Kirchhoff's first law, Zener current in mA

#Result
print 'i)   The output voltage across the load resistance RL = %d V'%Output_voltage;
print 'ii)  The voltage drop across the series resistance  R = %d V'%Voltage_R;
print 'iii) The current through the zener diode = %d mA'%Iz;


i)   The output voltage across the load resistance RL = 50 V
ii)  The voltage drop across the series resistance  R = 70 V
iii) The current through the zener diode = 9 mA


## Example 6.26, Page number 114-115¶

In :
#Variable declaration
Max_V=120.0;                  #Maximum input voltage in V
Min_V=80.0;                   #Minimum input voltage in V
R=5000.0;                     #Series resistance in ohm
Vz=50.0;                      #Zener voltage in V

#Calculation
#Case i: Maximum zener current
#Zener current will be maximum when the input voltage is maximum
V_R_max=Max_V-Vz;                #Voltage across series resistance R, in  V
I_max=V_R_max/R;                 #Current through series resistance R, in A
I_max=I_max*1000;                #Current through series resistance R, in mA
Iz_max=I_max-IL_max;                     #Applying Kirchhoff's first law, Zener current in mA;

#Case ii: Minimum zener current
#The zener will conduct minimum current when the input voltage is minimum
V_R_min=Min_V-Vz;                #Voltage across series resistance R, in V
I_min=V_R_min/R;                 #Current through series resistance R, in A
I_min=I_min*1000;                #Current through series resistance R, in mA
Iz_min=I_min-IL_min;                     #Applying Kirchhoff's first law, Zener current in mA

#Result
print 'Case i: ';
print 'Maximum zener current = %d mA'%Iz_max;
print 'Case ii: ';
print 'Minimum zener current = %d mA'%Iz_min;

Case i:
Maximum zener current = 9 mA
Case ii:
Minimum zener current = 1 mA


## Example 6.27, Page number 115¶

In :
#Variable declaration
Ei=12;                      #Input voltage in V
Vz=7.2;                     #Zener voltage in V
E0=Vz;                      #Voltage to be maintained across the load in  V
IL_max=0.1;                 #Maximum load current in A
IL_min=0.012;                  #Minimum load current in A
Iz_min=0.01;                  #Minimum zener current in A

#Calculation
#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL
R=(Ei-E0)/(Iz_min+IL_max);        #The value of series resistance R to maintain a voltage=E0 across load, in ohm

#Result
print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f Ω'%R;

The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 Ω


NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 Ω

## Example 6.28, Page number 115¶

In :
#Variable declaration
Ei_min=22;                #Minimum input voltage in V
Ei_max=28;                #Maximum input voltage in V
Vz=18;                   #Zener voltage in V
E0=Vz;                    #Constant voltage maintained across the load resistance in V
Iz_min=0.2;               #Minimum zener current in A
Iz_max=2;                 #Maximum zener current in A

#Calculation
IL=Vz/RL;                      #Constant value of load current in A
#When the input voltage is minimum, the zener current will be minimum
R=(Ei_min-E0)/(Iz_min+IL)      #The value of series resistance so that the voltage E0 across RL remains constant

print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f Ω.'%R;

The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 Ω.


## Example 6.29, Page number 116¶

In :
#Variable declaration
Vz=10                                #Zener voltage in V
Ei_min=13;                           #Minimum input voltage in V
Ei_max=16;                           #Maximum input voltage in V
Iz_min=0.015;                        #Minimum zener current in A
IL_min=0.01;                         #Minimum load current in A
IL_max=0.085;                        #Maximum load curremt in A
E0=Vz;                               #Constant voltage to be maintained in V

#Calculation
#The zener current will be minimum when the input voltage will be minimum  and at that time the load current will be maximum
R=(Ei_min-E0)/(Iz_min+IL_max);       #The value of series resistance R to maintain a constant voltage across load

#Result
print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d Ω'%R;

The value of series resistance to maintain a constant voltage across the load resistance is = 30 Ω


## Example 6.30, Page number 116¶

In :
#Variable declaration
Iz=0.2;               #Current rating of each zener in A
Vz=15;                #Voltage rating of each zener in V
Ei=45;                #Input voltage in V

#Calculation
# i: Regulated output voltage across the two zener diodes
E0=2*Vz;                     # V

# ii: Value of series resistance
R=(Ei-E0)/Iz;                  # Ω

#Result
print 'i) The regulated output voltage = %d V'%E0;
print 'ii) The value of the series resistance = %d Ω'%R;

i) The regulated output voltage = 30 V
ii) The value of the series resistance = 75 Ω


## Example 6.31, Page number 116-117¶

In :
#Variable declaration
Vz=10;                   #Voltage rating of each zener in V
Iz=1;                    #Current rating of each zener in A
Ei=45;                   #Input unregulated voltage in V

#Calculation
#Regulated output voltage across the three zener diodes
E0=3*Vz;               # V

#Value of series resistance to obtain a 30V regulated output voltage
R=(Ei-E0)/Iz;          # Ω

#Result
print 'Value of series resistance to obtain a 30V regulated output voltage = %d Ω'%R;

Value of series resistance to obtain a 30V regulated output voltage = 15 Ω


## Example 6.32, Page number 117¶

In :
#variable declaration
R=200.0;                     #Series resistance in Ω
Iz=0.025;                  #Zener current rating in A
E0=30.0;                     #Output regulated voltage in V

#Calculation
#Minimum input voltage will be required when Iz=0 A, and at  this condition
IL=E0/RL;                   #Load current during Iz=0, in A
I=IL;                       #According to Kirchhoff's law, total current, in A
Ei_min=E0+(I*R);            #Minimum input voltage in V

#The maximum input voltage required will be when Iz=0.025 A, and at that condition
I=IL+Iz;                    #According to Kirchhoff's law, total current, in A
Ei_max=E0+(I*R);            #maximum input voltage in V

#Result
print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max);

The required range of input voltage is from 33 V to 38 V


## Example 6.33, Page number 117-118¶

In :
#Variable declaration
Ei=16;                   #Unregulated input voltage in V
E0=12;                   #Output regulated voltage in V
IL_min=0;                #Minimum load current in A
IL_max=0.2;              #Maximum load current in A
Iz_min=0;                #Minimum zener current in A
Iz_max=0.2;              #Maximum zener current in A

#Calculation
#As the regulated voltage required across the load is 12V
Vz=E0;                   #Voltage rating of zener diode in V
V_R=Ei-E0;               #Constant Voltage that should remain across series resistance
#The minimum zener current will occur when the curent in the load in maximum
R=V_R/(Iz_min+IL_max);   #Series resistance in Ω

Max_power_rating=Vz*Iz_max;     #Maximum power rating of zener diode in W

#Result
print 'The regulator is designed using a Seris resistance of %d Ω and a zener diode of zener voltage %d V'%(R,Vz);
print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;

The regulator is designed using a Seris resistance of 20 Ω and a zener diode of zener voltage 12 V
The maximum power rating of the zener diode is = 2.4 W


## Example 6.34, Page number 118¶

In :
#Variable declaration
V=12;                       #Source voltage in V
R=1000;                     #Series resistance in Ω
Vz=6;                       #Voltage rating of zener in V

#Calculation
#Case i: zener is working properly
#The output voltage across the load will be equal to the zener voltage.
V0=Vz;                        # V

#Result
print 'Case i: Output voltage when zener is working properly is %d V'%V0;

#Case ii: zener is shorted
#As the zener is shorted, the potential difference across the load will be zero
V0=0;                        #V

#Result
print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;

#Case iii: zener is open circuited
#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule
V0=V*RL/(R+RL);                      #V

#Result
print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;

Case i: Output voltage when zener is working properly is 6 V
Case ii: Output voltage when zener is short circuited is 0 V
Case iii: Output voltage when zener is open circuited is 10 V