# Chapter 8 - 555 Timer¶

## Example 8.1 Page No.315¶

In [1]:
# Given data

R = 100*10**3
T = 100*10**-3

# Solution

C = T/(1.1*R)

print "The value of C =",round(C*10**6,1),"uF"

The value of C = 0.9 uF


## Example 8.2 Page No.322¶

In [2]:
# Given Data

Ra = 6.8*10**3
Rb = 3.3*10**3
C = 0.1*10**-6

# Solution
# Solution of part a

tHIGH = round(0.69*(Ra + Rb)*C,5)*10**3

# Solution pf part b

tLOW = round(0.69*(Rb)*C,5)*10**3

# Solution of part c

f = int(1.45/((Ra + 2*Rb)*C))*10**-3

# Solution for part d

D = round(Rb/(Ra + 2*Rb),2)

print "The value of tHIGH        =",tHIGH,"ms"
print "The value of tLOW         =",tLOW,"ms"
print "The value of frequency f  =",f,"kHz"
print "The value of Duty cycle D =",D,"or",D*100,"%"

The value of tHIGH        = 0.7 ms
The value of tLOW         = 0.23 ms
The value of frequency f  = 1.082 kHz
The value of Duty cycle D = 0.25 or 25.0 %

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