# Chapter09 : Voltage Regulators¶

## Exa 9.1 : page 380¶

In [1]:
#given data
Iq=5 #in mA
Vo=18 #in volts
Vreg=15 #in volts
R1=Vreg/(10*Iq*10**-3) #in Ohms  Iq must be in Amperes here
R2=(Vo-Vreg)/(11*Iq*10**-3) #in Ohms  Iq must be in Amperes here
print "the values of R1 and R2 are : %0.2f  & %0.1f ohms"%(R1,R2)

the values of R1 and R2 are : 300.00  & 54.5 ohms


## Exa 9.2 : page 381¶

In [4]:
#given data
Vreg=8 #in volts as IC 7808 is given
IL=100 #in mA
IR=100 #in mA
Iq=0 #in mA
RL=50 #in ohms
# let find the value of resistor to deliver required current
R=Vreg/(IR*10**-3) # in ohms  so current must be in amperes
print "Required Resistance = %0.2f ohms  "%R
Vo = (IR*10**-3)*R+(IL*10**-3)*RL
print "Vo = %0.2f Volts"%Vo
#considering 2 volt dropout
Vdropout=2 #in volts
VI=Vo+Vdropout
print "Input voltage, VI = %0.2f Volts "%VI

Required Resistance = 80.00 ohms
Vo = 13.00 Volts
Input voltage, VI = 15.00 Volts


## Exa 9.3 : page 382¶

In [13]:
from __future__ import division
#given data
RL1=100 #in ohms
RL2=8 #in ohms
RL3=1 #in ohms
VEBon=0.7 #in volts
Beta=25
R=5 #in ohms
#device used 7808 so V=8 volts
V=8
# part(i) for a laod of 100 ohms
IL1=V/RL1 #in amperes
VacR1=IL1*R
print "Part(i) : "
print "VacR = %0.2f Volts "%VacR1
print "  Which is less than the given VEBon. Hence Transistor remains OFF."
#so Io=IL and Ic=0
Io1=IL1 #in amperes
Ic1=0
print "Ic and Io for the 100 ohms load are %0.2f mA & %0.2f Amperes "%(Io1*1E3,Ic1)
# part(ii) for a laod of 8 ohms
IL2=V/RL2 #in amperes
VacR2=IL2*R
print "Part(ii) : "
print "The voltage across R will be = %0.2f Volts "%VacR2
print "Which is greater than the given VEBon.Hence Transistor will be ON."
#expression for Io
Io2=((IL2+(Beta*VEBon)/R))/(Beta+1) #in amperes
Ic2=IL2-Io2
print "Ic and Io for the 8 ohms load are %0.f & %0.f Amperes  "%(Io2*1e3,Ic2*1e3)

# part(iii) for a laod of 1 ohms
IL3=V/RL3 #in amperes
VacR3=IL3*R
print "Part(iii) : "
print "VacR = %0.2f Volt "%VacR3
print "  Which is greater than the given VEBon.Hence Transistor will be ON."
#expression for Io
Io3=((IL3+(Beta*VEBon)/R))/(Beta+1) #in amperes
Ic3=IL3-Io3
print "Ic and Io for the 1 ohms load are %0.f & %0.3f Amperes "%(Io3*1e3,Ic3)

Part(i) :
VacR = 0.40 Volts
Which is less than the given VEBon. Hence Transistor remains OFF.
Ic and Io for the 100 ohms load are 80.00 mA & 0.00 Amperes
Part(ii) :
The voltage across R will be = 5.00 Volts
Which is greater than the given VEBon.Hence Transistor will be ON.
Ic and Io for the 8 ohms load are 173 & 827 Amperes
Part(iii) :
VacR = 40.00 Volt
Which is greater than the given VEBon.Hence Transistor will be ON.
Ic and Io for the 1 ohms load are 442 & 7.558 Amperes


## Exa 9.4 : page 384¶

In [17]:
#given data
#for the given IC LM317:
Vref=1.25 #in volts
R1=240 #in ohms
#(i) For Vo=2 volts
Vo=2 #in volts
print "for Output 2 volts the requires value of resistance R2 = %0.f ohms "%(R2)
#(i) For Vo=12 volts
Vo1=12 #in volts
print "for Output 12 volts the requires value of resistance R2 = %0.2f kohms"%(R21/1e3)
print "Hence use 3kohm potentiometer to set R2."

for Output 2 volts the requires value of resistance R2 = 141 ohms
for Output 12 volts the requires value of resistance R2 = 2.03 kohms
Hence use 3kohm potentiometer to set R2.


## Exa 9.5 : page 385¶

In [19]:
#given data
#for the given IC LM317:
Vref=1.25 #in volts
R1=240 #in ohms
R2min=0 #in ohms
R2max=3000 #in ohms
print "Minimum output voltage = %0.2f Volts "%(Vomin)
print "Maximum output voltage = %0.3f Volts "%Vomax

Minimum output voltage = 1.25 Volts
Maximum output voltage = 17.175 Volts


## Exa 9.6 : page 389¶

In [24]:
#given data
#IC 723
Vsense=0.7#in volts
Vo=5 #in volts
Im=50 #in mA
Id=1 #in mA
Vr=7 #in volts
R1=(Vr-Vo)/(Id*10**-3)
R2=Vo/(Id*10**-3)
R3=(R1*R2)/(R1+R2)
Rcl=Vsense/(Im*10**-3)
print "various resistance values for the circuit is as follows : "
print "R1 = %0.2f kohm"%(R1/1e3)
print "R2 = %0.2f kohm"%(R2/1e3)
print "R3 = %0.2f kohm"%(R3/1e3)
print "Rcl = %0.2f ohm"%Rcl

various resistance values for the circuit is as follows :
R1 = 2.00 kohm
R2 = 5.00 kohm
R3 = 1.43 kohm
Rcl = 14.00 ohm


## Exa 9.7 : page 390¶

In [25]:
#given data
# IC 723
Id=1 #in mA
Vsense=0.7 #in volts
Vo=15 #in volts
Im=50 #in mA
Vr=7 #in volts
R1=(Vo-Vr)/(Id*10**-3)
R2=Vr/(Id*10**-3)
R3=(R1*R2)/(R1+R2)
Rcl=Vsense/(Im*10**-3)
print "various resistance values for the circuit is as follows : "
print "R1 = %0.2f kohm"%(R1/1e3)
print "R2 = %0.2f kohm"%(R2/1e3)
print "R3 = %0.2f kohm"%(R3/1e3)
print "Rcl = %0.2f ohm"%Rcl

various resistance values for the circuit is as follows :
R1 = 8.00 kohm
R2 = 7.00 kohm
R3 = 3.73 kohm
Rcl = 14.00 ohm


## Exa 9.8 : page 390¶

In [29]:
#given data
Im=100 #in mA
Vr=7 #in volts
R2=10 #in kohm
Vsense=0.7 #in volts
#using equation Vo=((R1+R2)/R2)*Vr
#for Vo=10 volts assuming R2=10kohm
Vo=10 #in volts
R1=((Vo*R2)/Vr)-R2#it gives 3R2=7R1
print "Part(i) : "
print "Value of resistance R1 = %0.2F kohms "%R1
#now let output voltage is 15 volts
Vo=15 #in volts
R1=((Vo*R2)/Vr)-R2#it gives 3R2=7R1
print "Part(ii) : "
print "Value of resistance R1 = %0.2f kohms"%R1
R3=(R1*R2)/(R1+R2)
print "value of resistance R3 = %0.2f kohms"%R3
Rcl=Vsense/(Im*10**-3) #in ohms
print "Value of Rcl = %0.2f ohms" %Rcl

Part(i) :
Value of resistance R1 = 4.29 kohms
Part(ii) :
Value of resistance R1 = 11.43 kohms
value of resistance R3 = 5.33 kohms
Value of Rcl = 7.00 ohms