# Chapter7 - The 555 timer¶

## Ex 7.1 - page : 233¶

In [2]:
from __future__ import division
th=4 #ms
VCC=10 #V
C=0.05 #micro F(choosen between 0.01<=C<=1)
R=th*10**-3/(1.1*C*10**-6)/1000 #kohm
C1=0.01 #micro F(assumed)
C2=0.01 #micro F(choosen between 0.01<=C<=1)
R2=th*10**-3/(10*C2*10**-6)/1000 #kohm
C3=10 #micro F
print "Design values are : "
print "Capacitance C = %0.2f micro F " %C
print "Resistance R = %0.1f kohm " %R
print "Capacitance C1 = %0.2f micro F " %C1
print "Capacitance C2 = %0.2f micro F " %C2
print "Resistance R2 = %0.2f kohm " %R2
print "Capacitance C3 = %0.2f micro F " %C3
#Answer of R2 is wrong in the book.

Design values are :
Capacitance C = 0.05 micro F
Resistance R = 72.7 kohm
Capacitance C1 = 0.01 micro F
Capacitance C2 = 0.01 micro F
Resistance R2 = 40.00 kohm
Capacitance C3 = 10.00 micro F


## Ex 7.2 - page : 236¶

In [3]:
from __future__ import division
ft=2 #kHz
C=0.01 #micro F
T=1/ft #ms
n=3 #for divide-by-3 circuit
th=(0.2+(n-1))*T #ms
R=th/(1.1*C) #kohm
print "Value of Resistance R = %0.2f kohm " %R

Value of Resistance R = 100.00 kohm


## Ex 7.3 - page 239¶

In [6]:
from __future__ import division
fo=2 #kHz
D=70 #%(duty cycle)
T=1/fo #ms
VCC=12 #V
tC=D*T/100 #ms
tD=T-tC #ms
C=0.05 #micro F(choosen between 0.01<=C<=1)
RB=tD*10**-3/(0.69*C*10**-6)/1000 #kohm
RA=tC*10**-3/(0.69*C*10**-6)/1000-RB #kohm
print "Design values are : "
print "Capacitance C = %0.2f micro F " %C
print "Resistance RA = %0.1f kohm " %RA
print "Resistance RB = %0.1f kohm " %RB
#Answer is not accurate in the textbook.

Design values are :
Capacitance C = 0.05 micro F
Resistance RA = 5.8 kohm
Resistance RB = 4.3 kohm


## Ex 7.4 : page 243¶

In [8]:
from __future__ import division
fo=2 #kHz
D=50 #%(duty cycle)
T=1/fo #ms
VCC=10 #V
tC=D*T/100 #ms
tD=T-tC #ms
C=0.1 #micro F(choosen between 0.01<=C<=1)
RB=tD*10**-3/(0.69*C*10**-6)/1000 #kohm
RA=T*10**-3*1.45/(C*10**-6)/1000-RB #kohm
print "Design values are : "
print "Capacitance C= %0.2f micro F " %C
print "Resistance RA = %0.1f kohm " % RA
print "Resistance RB = %0.1f kohm " %RB
# RA & RB should be equal for 50% duty cycle.

Design values are :
Capacitance C= 0.10 micro F
Resistance RA = 3.6 kohm
Resistance RB = 3.6 kohm


## Ex 7.5 - page : 244¶

In [11]:
from __future__ import division
fo=2 #kHz
D=40 #%(duty cycle)
T=1/fo #ms
VCC=10 #V
tC=D*T/100 #ms
tD=T-tC #ms
C=0.22 #micro F(choosen between 0.01<=C<=1)
RB=tD*10**-3/(0.69*C*10**-6)/1000 #kohm
RA=T*10**-3*1.45/(C*10**-6)/1000-RB #kohm
print "Design values are : "
print "Capacitance C = %0.2f micro F " %C
print "Resistance RA = %0.1f kohm " %RA
print "Resistance RB = %0.f kohm " %round(RB)

Design values are :
Capacitance C = 0.22 micro F
Resistance RA = 1.3 kohm
Resistance RB = 2 kohm


## Ex 7.6 - page : 245¶

In [12]:
from __future__ import division
fo=700 #Hz
D=50 #%(duty cycle)
T=1/fo*1000 #ms
VCC=10 #V
tC=D*T/100 #ms
tD=T-tC #ms
C=0.05 #micro F(choosen between 0.01<=C<=1)
RB=tD*10**-3/(0.69*C*10**-6)/1000 #kohm
RA=T*10**-3*1.45/(C*10**-6)/1000-RB #kohm
print "Design values are : "
print "Capacitance C = %0.2f micro F " %C
print "Resistance RA = %0.2f kohm " %round(RA)
print "Resistance RB = %0.2f kohm " %round(RB)

Design values are :
Capacitance C = 0.05 micro F
Resistance RA = 21.00 kohm
Resistance RB = 21.00 kohm


## Ex 7.7 - page : 246¶

In [14]:
from __future__ import division
fo=800 #Hz
D=60 #%(duty cycle)
T=1/fo*1000 #ms
VCC=10 #V
tC=D*T/100 #ms
tD=T-tC #ms
C=0.047 #micro F(choosen between 0.01<=C<=1)
RB=tD*10**-3/(0.69*C*10**-6)/1000 #kohm
RA=tC*10**-3*1.45/(C*10**-6)/1000-RB #kohm
print "Design values are : "
print "Capacitance C = %0.3f micro F " %C
print "Resistance RA = %0.2f kohm " %round(RA)
print "Resistance RB = %0.2f kohm " %round(RB)

Design values are :
Capacitance C = 0.047 micro F
Resistance RA = 8.00 kohm
Resistance RB = 15.00 kohm