## Example 5-1, Page 146¶

In :
Vsmin=20             #Source voltage minimum(V)
Vsmax=40             #Source voltage maximum(V)
Vbd=10               #Breakdown voltage(V)
R=0.82               #Resistance(KOhm)

Vr1=Vsmin-Vbd        #voltage across resistor(V)
Is1=Vr1/R            #Minimum current(mA)
Vr2=Vsmax-Vbd        #voltage across resistor(V)
Is2=Vr2/R            #Maximum current(mA)

print 'Ideally, zener diode acts as a battery(of breakdown voltage = 10V) shown in figure 5-4b'
print 'Minimum current Is1=',round(Is1,2),'mA'
print 'Maximum current Is1=',round(Is2,2),'mA'

Ideally, zener diode acts as a battery(of breakdown voltage = 10V) shown in figure 5-4b
Minimum current Is1= 12.2 mA
Maximum current Is1= 36.59 mA


## Example 5-2, Page 149¶

In :
Vs=18                 #supply voltage(V)
Rs=0.27               #source resistance(KOhm)
Vz=10                 #Zener voltage(V)

VTH=(RL/(Rs+RL))*Vs    #Thevenin voltage(V)

print 'Thevenin voltage VTH = ',round(VTH,2),'V'
print 'Thevenin voltage is greater than zener voltage, zener diode is operating in breakdown region.'

Thevenin voltage VTH =  14.17 V
Thevenin voltage is greater than zener voltage, zener diode is operating in breakdown region.


## Example 5-3, Page 149¶

In :
Vs=18                 #supply voltage(V)
Rs=0.27               #source resistance(KOhm)
Vbd=10                #Zener voltage(V)

Vr=Vs-Vbd            #voltage across resistor(V)
Is=Vr/Rs              #Current(mA)
IL=Vbd/RL            #Current(mA)
Iz=Is-IL             #Zener current(mA)

print 'Load current IL = ',IL,'mA'
print 'Zener current Iz = ',round(Iz,2),'mA'

Load current IL =  10 mA
Zener current Iz =  19.63 mA


## Example 5-7, Page 153¶

In :
Iz=20                    #zener current(mA)
Rz=8.5                   #zener resistance(Ohm)
Vbd=10                   #Zener voltage(V)

print 'Change in load voltage DVL =',DVL,'V'
print 'Load voltage with second approx., VL =',VL,'V'

Change in load voltage DVL = 0.17 V
Load voltage with second approx., VL = 10.17 V


## Example 5-8, Page 154¶

In :
Rs=270                   #Source resistance (Ohm)
Rz=8.5                   #zener resistance(Ohm)
VRin=2                   #Zener voltage(V)


Load ripple voltage VRout= 62.96 mV


## Example 5-10, Page 157¶

In :
Vil=22                 #input voltage range low(V)
Vih=30                 #input voltage range high(V)
Vz=12                  #regulated output voltage(V)

RSmax=Rl*(float(Vil)/float(Vz)-1)  #Maximum series resistance

print 'Maximum series resistance RSmax =',round(RSmax,2),'V'

Maximum series resistance RSmax = 116.67 V


## Example 5-11, Page 157¶

In :
Vil=15                     #input voltage range low(V)
Vih=20                     #input voltage range high(V)
Vz=6.8                     #regulated output voltage(V)

RSmax=(Vil-float(Vz))/Ih*1000  #Maximum series resistance

print 'Maximum series resistance RSmax =',RSmax,'V'

Maximum series resistance RSmax = 410.0 V


## Example 5-12, Page 168¶

In :
Vi=50                    #voatage supply(V)
Rs=2.2                   #series resistance(KOhm)
Vf=2                     #forward approx. voltage

Is=(Vi-Vf)/Rs

print 'LED current Is =',round(Is,2),'mA'

LED current Is = 21.82 mA


## Example 5-13, Page 168¶

In :
Vs=9                     #voatage supply(V)
Rs=470.0                 #series resistance(Ohm)
Vf=2                     #forward approx. voltage

Is=(Vs-Vf)/Rs

print 'LED current Is =',round((Is*1000),2),'mA'

LED current Is = 14.89 mA


## Example 5-14, Page 169¶

In :
import math

Vac=20                   #AC voatage supply(V)
Rs=680.0                 #series resistance(KOhm)

Vacp=1.414*Vac            #peak source voltage(V)
Is1=(Vacp/Rs)*1000        #approx. peak current(mA)
Is2=Is1/math.pi              #average of half-wave current through LED(mA)
P=(Vac)**2/Rs             #Power dissipation(W)

print 'approx. peak LED current Is1 =',round(Is1,2),'mA'
print 'average of half-wave current through LED Is2 =',round(Is2,2),'mA'
print 'Power dissipation P =',round(P,2),'W'

approx. peak LED current Is1 = 41.59 mA
average of half-wave current through LED Is2 = 13.24 mA
Power dissipation P = 0.59 W


## Example 5-15, Page 170¶

In :
import math

Vs=120                   #AC voatage supply(V)
f=60                     #frequency(Hz)
C=0.68                   #series resistance(KOhm)

Xc=1/(2*math.pi*f*C)*1000    #capacitive reactance(KOhm)
Vacp=Vs*1.414
Is1=(Vacp/Xc)             #approx. peak current(mA)
Is2=Is1/math.pi              #average current through LED(mA)

print 'Capacitance reactance Xc = ',round(Xc,2),'KOhm'
print 'approx. peak LED current Is1 =',round(Is1,2),'mA'
print 'average current through LED Is2 =',round(Is2,2),'mA'

Capacitance reactance Xc =  3.9 KOhm
approx. peak LED current Is1 = 43.5 mA
average current through LED Is2 = 13.85 mA

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