# Chapter15-Applications and Design of Integrated Circuits¶

## Ex2-pg935¶

In [1]:
import math

##Example 15.2
C=20.*10**-6;
Req=1000.;
fC=1./(C*Req);
print"%s %.2f %s"%('\nclock frequency = ',fC,' KHz\n')
clock frequency =  50.00  KHz

## Ex3-pg936¶

In [2]:
import math

##Example 15.3
fC=10000.;
f3dB=1000.;
##x=C2/C1
x=2.*math.pi*f3dB/fC;
print"%s %.2f %s"%('\ncapacitances C2/C1= \n',x,'')
capacitances C2/C1=
0.63

## Ex4-pg940¶

In [3]:
import math

##Example 15.4
C=0.1*10**-6;
R=1000.;
fo=1/(2.*math.pi*R*C*math.sqrt(3.));
print"%s %.2f %s"%('\nthe oscillation frequency =  ',fo,'Hz\n')
##minimum amplifier gain=8
R=1.;##KOhm
R2=8.*R;
print"%s %.2f %s"%('\nR2= ',R2,' KOhm\n')
the oscillation frequency =   918.88 Hz

R2=  8.00  KOhm

## Ex6-pg953¶

In [4]:
import math

##Example 15.6
R1=10000.;
R2=90000.;
Vh=10.;
Vl=-10.;
Vth=R1*Vh/(R1+R2);
print"%s %.2f %s"%('\nupper crossover voltage= ',Vth,' V\n')
Vtl=R1*Vl/(R1+R2);
print"%s %.2f %s"%('\nlower crossover voltage= ',Vtl,' V\n')
x=Vth-Vtl;
print"%s %.2f %s"%('\nhysteresis width =  ',x,'V\n')
upper crossover voltage=  1.00  V

lower crossover voltage=  -1.00  V

hysteresis width =   2.00 V

## Ex7-pg958¶

In [5]:
import math

##Example 15.7
Vs=2.;
Vh=15.;
Vl=-15.;
##hysteresis width=x
x=60.*0.001;##(V)
##Vth-Vtl=(R1/(R1+R2))*(Vh-Vl)
##R2/R=y
y=(Vh-Vl)/x-1.;
print"%s %.2f %s"%('\nR2/R1= \n',y,'')
Vref=(1.+1./y)*Vs;
print"%s %.2f %s"%('\nreference voltage= ',Vref,' V\n')
R2/R1=
499.00

reference voltage=  2.00  V

## Ex10-pg969¶

In [6]:
import math

##Example 15.10
C=15.*10**-9;
T=100.*10**-6;##(s) time
R=T/(1.1*C);
R=R*0.001;##Kohm
print"%s %.2f %s"%('\nResistance R= ',R,' KOhm\n')
Resistance R=  6.06  KOhm

## Ex13-pg974¶

In [7]:
import math

##Example 15.13
Pl=20.;##power delivered to the load
Ps=20.;##(W)
Vp=math.sqrt(2.*Rl*Pl);
print"%s %.2f %s"%('\npeak output voltage= ',Vp,' V\n')
Ip=Vp/Rl;
print"%s %.2f %s"%('\npeak load current = ',Ip,'A\n')
Vs=math.pi*Rl*Ps/Vp;
print"%s %.2f %s"%('\nrequired supply voltage= ',Vs,' V\n')
peak output voltage=  20.00  V

peak load current =  2.00 A

required supply voltage=  31.42  V

## Ex14-pg979¶

In [8]:
import math

##Example 15.14
Vonl=5;
Vofl=4.96;
I1=0.005;
I2=1.5;
dVo=Vonl-Vofl;
dIo=I1-I2;
Rvf=-dVo/dIo;
print"%s %.2f %s"%('\noutput resistance= ',Rvf,' Ohm\n')
LR=100.*(Vonl-Vofl)/Vonl;
print"%s %.2f %s"%('\nload regulation =\n',LR,'')
output resistance=  0.03  Ohm

0.80

## Ex15-pg982¶

In [9]:
import math

##Example 15.15
Aol=1000.;
Vref=5.;
Vo=10.;
Io=0.1*0.001;
Vt=0.026;
Rof=2.*Vt*Vo/(Io*Vref*Aol);
print"%s %.2f %s"%('\noutput resistance= ',Rof,' mOhm\n')
##dVo/Vo=V  and dIo/Io=I
##V=-I*2*Vt/(Vref*Aol)
##V/I=x
x=-2.*Vt/(Vref*Aol);
print"%s %.2e %s"%('\npercent change=\n',x,'')
output resistance=  1.04  mOhm

percent change=
-1.04e-05

## Ex16-pg984¶

In [10]:
import math

##Example 15.16
Vz=6.3;
Vbe=0.6;
Veb=0.6;
Vo=8.;
R1=3.9;
R2=3.4;
R3=0.576;
Ic3=(Vz-3.*Vbe)/(R1+R2+R3);
print"%s %.2f %s"%('\nbias current = ',Ic3,' mA\n')
Vb7=Ic3*R1+2.*Vbe;
print"%s %.2f %s"%('\ntemperature compensated reference voltage= ',Vb7,' V\n')
R13=2.23;
R12=R13*Vo/Vb7-R13;
print"%s %.2f %s"%('\nR12= ',R12,' KOhm\n')
bias current =  0.57  mA

temperature compensated reference voltage=  3.43  V

R12=  2.97  KOhm