# Chapter 14 Special Function ICs¶

## Example 14.1 Pg 415¶

In [1]:
from __future__ import division

# to determine the regulated voltage
R1 = 250 # #ohm
R2 = 2500 # # ohm
Vref = 2 # #V #reference voltage

#the output voltage of the adjustable voltage regulator is defined by
print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '

the output voltage of the adjustable voltage regulator is = 22.25  V


## Example 14.2 Pg 416¶

In [2]:
# to determine the current drawn from the dual power supply
V = 10 #  # V
P = 500 # # mW

# we assume that  each power supply provides half power supply to IC
P1 = (P/2)#

# the total power dissipation of the IC
# P1 = V*I #
I = P1/V #
print 'the total power dissipation of the IC is = %0.2f'%I,' mA '

the total power dissipation of the IC is = 25.00  mA


## Example 14.3 Pg 416¶

In [3]:
# to determine the output voltage
R1 = 100*10**3 # #ohm
R2 = 500*10**3 # # ohm
Vref = 1.25 # #V #reference voltage

#the output voltage of the adjustable voltage regulator is defined by
Vo = Vref*(R1+R2)/R1#
print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '

the output voltage of the adjustable voltage regulator is = 7.50  V


## Example 14.4 Pg 417¶

In [4]:
# determine the output voltage of the switching regulator circuit
d = 0.7 #  # duty cycle
Vin = 5 # # V # input voltage

# The output voltage of switching regulator circuit is given by
Vo = d*Vin #
print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '

The output voltage of switching regulator circuit is = 3.50  V


## Example 14.5 Pg 417¶

In [5]:
# determine the duty cycle of the switching regulator circuit
Vo = 4.8 # # V # output voltage
Vin = 5 # # V # input voltage

# The output voltage of switching regulator circuit is given by
# Vo = d*Vin #

# Duty cycle is given as
d =Vo/Vin #
print 'The output voltage of switching regulator circuit is = %0.2f'%d,'  '

The output voltage of switching regulator circuit is = 0.96


## Example 14.6 Pg 418¶

In [6]:
from __future__ import division
# determine the duty cycle of the switching regulator circuit
T =120 # #msec # total pulse time
# T = ton + toff #
ton = T/2 #

# The duty cycle of switching regulator circuit is given by
d = ton/T#
print 'The output voltage of switching regulator circuit is = %0.2f'%d,'  '

The output voltage of switching regulator circuit is = 0.50


## Example 14.7 Pg 418¶

In [7]:
from __future__ import division
# determine the duty cycle of the switching regulator circuit
ton = 12 # #msec # on time of pulse
# ton = 2*toff #  given
# T = ton + toff #
toff = ton/2 #
T = ton+toff #  # total time

# The duty cycle of switching regulator circuit is given by
d = ton/T#
print 'The output voltage of switching regulator circuit is = %0.2f'%d,'  '

The output voltage of switching regulator circuit is = 0.67


## Example 14.8 Pg 419¶

In [8]:
from __future__ import division
# determine the output voltage of the audio power amplifier IC LM380
Vcc = 12 # # V
Ic3 = 12*10**-6 #  # A  # collector current of the transistor Q3
Ic4 = 12*10**-6 #  # A  # collector current of the transistor Q4
R11 = 25*10**3 # # ohm
R12 = 25*10**3 #  # ohm

# the collector current of Q3 is defined as
# Ic3 = (Vcc-3*Veb)/(R11+R12)#
Veb = (Vcc-(R11+R12)*Ic3)/3 #
print 'The emitter bias voltage is = %0.2f'%Veb,' V '

# the output voltage of the IC LM380
Vo = (1/2)*Vcc+(1/2)*Veb#
print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '

The emitter bias voltage is = 3.80  V
The output voltage of the IC LM380 is = 7.90  V


## Example 14.9 Pg 420¶

In [9]:
from __future__ import division
# determine the output voltage of the audio power amplifier IC LM380
Vcc = 10 # # V
Ic3 = 0.01*10**-6 #  # A  # collector current of the transistor Q3
Ic4 = 0.01*10**-6 #  # A  # collector current of the transistor Q4
R11 = 25*10**3 # # ohm
R12 = 25*10**3 #  # ohm

# the collector current of Q3 is defined as
# Ic3 = (Vcc-3*Veb)/(R11+R12)#
Veb = (Vcc-(R11+R12)*Ic3)/3 #
print 'The emitter bias voltage is = %0.2f'%Veb,' V '

# the output voltage of the IC LM380
Vo = (1/2)*Vcc+(1/2)*Veb#
print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '

The emitter bias voltage is = 3.33  V
The output voltage of the IC LM380 is = 6.67  V


## Example 14.10 Pg 421¶

In [10]:
from __future__ import division
from numpy import inf
from math import sqrt, pi
# Design a video amplifier of IC 1550 circuit
Vcc = 12 # # V
Av = -10 #
Vagc = 0 # # at bandwidth of 20 MHz
hfe = 50 # # forward emitter parameter
rbb = 25 #  # ohm  # base resistor
Cs = 1*10**-12 #  # F  # source capacitor
Cl = 1*10**-12 #  # F  # load capacitor
Ie1 = 1*10**-3 # # A # emitter current of Q1
f = 1000*10**6 # # Hz
fT = 800*10**6 # # Hz
Vt = 52*10**-3 #
Vt1 = 0.026 #

# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3
# i.e Ic1=Ie1=Ie3
Ie3 = 1*10**-3 # # A # emitter current of Q3
Ic1 = 1*10**-3 #  # A  # collector current of the transistor Q1

# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite

re2 = inf #

# emitter resistor of Q3
re3 = (Vt/Ie1)#
print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm ( at temperature 25 degree celsius) '

# the trans conductance of transistor is
gm = (Ie1/Vt1)#
print 'The trans conductance of transistor is = %0.1f'%(gm*1000),' mA/V ' # Round Off Error

# the base emitter resistor rbe
rbe = (hfe/gm)#
print 'The base emitter resistor rbe is = %0.2f'%(rbe/1000),' K ohm ' # Round Off Error

# the emitter capacitor Ce

Ce = (gm/(2*pi*fT))#
print 'The emitter capacitor Ce = %0.2f'%(Ce*1e12),' pF ' # Round Off Error

# the voltage gain of video amplifier is
# Av = (Vo/Vin) #
# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl))))
# At Avgc = 0 i.e s=0 in the above Av equation
alpha3 = 1 #
s = 0 #
# Rl = -((alpha3*gm)/(rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av)))#

# After solving above equation for Rl We get Rl Equation as
Rl = 10/(37.8*10**-3)#
print 'The value of resistance RL is = %0.2f'%Rl,' ohm '

# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency
Rl = 675 #
# fa = 1/(2*pi*Rl*(Cs+Cl))#
# after putting value of Rl ,Cs and Cl we get
fa = 1/(2*3.14*264.55*1*10**-12)#
print 'The pole frequency fa is = %0.2f'%(fa*10**-3/1000),' M Hz '# Round Off Error

#fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))#
# after putting value of Ce rbb and rbe we get
fb = 1/(2*pi*6.05*10**-12*24.5)#
print 'The pole frequency fb is = %0.2f'%(fb*10**-3/1000),' M Hz '

fc = 1/(2*pi*Cs*re3)#
print 'The pole frequency fc is = %0.2f'%(fc*10**-3/1000),' M Hz '

print 'Hence fa is a dominant pole frequency '

The emitter resistor of Q3 is = 52.00  ohm ( at temperature 25 degree celsius)
The trans conductance of transistor is = 38.5  mA/V
The base emitter resistor rbe is = 1.30  K ohm
The emitter capacitor Ce = 7.65  pF
The value of resistance RL is = 264.55  ohm
The pole frequency fa is = 601.91  M Hz
The pole frequency fb is = 1073.74  M Hz
The pole frequency fc is = 3060.67  M Hz
Hence fa is a dominant pole frequency


## Example 14.11 Pg 423¶

In [11]:
from __future__ import division
from numpy import inf,pi
# Design a video amplifier of IC 1550 circuit
Vcc = 12 # # V
Av = -10 #
Vagc = 0 # # at bandwidth of 20 MHz
hfe = 50 # # forward emitter parameter
rbb = 25 #  # ohm  # base resistor
Cs = 1*10**-12 #  # F  # source capacitor
Cl = 1*10**-12 #  # F  # load capacitor
Ie1 = 1*10**-3 # # A # emitter current of Q1
f = 1000*10**6 # # Hz
Vt = 52*10**-3 #
Vt1 = 0.026 #

# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3
# i.e Ic1=Ie1=Ie3
Ie3 = 1*10**-3 # # A # emitter current of Q3
Ic1 = 1*10**-3 #  # A  # collector current of the transistor Q1

# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite
re2 = inf #

# emitter resistor of Q3
re3 = (Vt/Ie1)#
print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm '

# the trans conductance of transistor is
gm = (Ie1/Vt1)#
print 'The trans conductance of transistor is = %0.1f'%(gm*1e3),' mA/V '

# the base emitter resistor rbe
rbe = (hfe/gm)#
print 'The base emitter resistor rbe is = %0.1f'%(rbe/1e3),' kohm '

# the emitter capacitor Ce
Ce = (gm/(2*pi*f))#
print 'The emitter capacitor is = %0.2f'%(Ce*1e12),' pF '

# the voltage gain of video amplifier is
# Av = (Vo/Vin) #
# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl))))
# At Avgc = 0 i.e s=0 in the above Av equation
alpha3 = 1 #
s = 0 #
Av =-10 #
Rl = -((alpha3*gm)/((rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av))))#
Rl = (1/Rl)#
print 'The value of resistance RL is = %0.2f'%Rl,' ohm '

# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency
Rl = 265
fa = 1/(2*pi*Rl*(Cs))/1e6#
print 'The pole frequency fa is = %0.2f'%fa,'MHz '

fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))/1e6
print 'The pole frequency fb is = %0.2f'%fb,'MHz '

fc = 1/(2*pi*Cs*re3)/1e6
print 'The pole frequency fc is = %0.2f'%fc,'MHz '

print  'Hence fa is a dominant pole frequency '

The emitter resistor of Q3 is = 52.00  ohm
The trans conductance of transistor is = 38.5  mA/V
The base emitter resistor rbe is = 1.3  kohm
The emitter capacitor is = 6.12  pF
The value of resistance RL is = 265.00  ohm
The pole frequency fa is = 600.58 MHz
The pole frequency fb is = 1060.00 MHz
The pole frequency fc is = 3060.67 MHz
Hence fa is a dominant pole frequency


## Example 14.12 Pg 425¶

In [12]:
# Determine the output voltage of an isolation amplifier IC ISO100
Vin = 5.0 # # V
Rin = 10*10**3 #
Rf = 55*10**3 # # ohm # feedback resistance

# the input voltage of an amplifier 1
# Vin = Rin*Iin
Iin = Vin/Rin #
print 'The input current is = %0.2f'%(Iin*1e3),'mA '

# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction
# Iin = -Iout
# the output of an op-amp
# Vo = -Rf*Iout
Vo = Rf*Iin#
print 'The output of an op-amp is = %0.2f'%Vo,' V '

The input current is = 0.50 mA
The output of an op-amp is = 27.50  V


## Example 14.13 Pg 426¶

In [13]:
# Determine the output voltage of an isolation amplifier IC ISO100
Vin = 12.0 # # V
Rin = 1*10**3 #
Rf = 17*10**3 # # ohm # feedback resistance

# the input voltage of an amplifier 1
# Vin = Rin*Iin
Iin = Vin/Rin #
print 'The input current is = %0.f'%(Iin*1e3),'mA '

# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction
# Iin = -Iout
# the output of an op-amp
# Vo = -Rf*Iout
Vo = Rf*Iin#
print 'The output of an op-amp is = %0.f'%Vo,' V '

The input current is = 12 mA
The output of an op-amp is = 204  V