Introduction to Electric Circuits

188 Transient analysis

Example 8.8

The integrator circuit shown in Fig. 8.20 has R = 10 kf~ and C = 0.01 txF.
Sketch the waveform of the output voltage Vo for each of the following input
conditions:

1 the input signal is a single pulse of amplitude 10 V and width T = 600 Ixs;

2 the input is a train of pulses of amplitude 10 V, width 100 Ixs and separation

500 ~s.

Solution

The time constant of the circuit is given by

-r = CR = 0.01 x 10 -6 x 10 x 103 = 100 x 10-6s

1 The pulse width is greater than five time constants so that the capacitor

voltage Vc will reach its steady state value of 10 V before the pulse is

removed. It will remain at this value until the pulse is removed at

t - 600 p~s, after which it will decay to zero as shown in Fig. 8.24.

IOV

0 500 600 1100 "{(~s)

Figure 8.24

2 The pulse width is 100 Ixs which is less than five time constants (in fact it is

one time constant) so that the capacitor will only be partially charged

before a pulse is removed. After 100 ~xs,

Vo - Vc = V(1 - exp (-t/CR) - 10{1 - exp [-100/(0.01 x 10 x 10s)]}

= 6.32 V

The pulse separation is 500 Ixs which is 5~" so the capacitor will be fully

discharged just as the next pulse in the train arrives at the input terminals.

The waveform of Vo is given in Fig. 8.25.

I

l~ I ...-- .....

0 100 600 700 1200 1300 "{(l~S)

Figure 8.25