Introduction to Electric Circuits

190 Transient analysis

so that Vc + VR- 0. The voltage across the resistor will therefore have to

change immediately to -Vc at that instant. The voltage across the capacitor will

now begin to decay to zero in accordance with Equation (8.19) and will take a

time equivalent to five time constants to do so. In the same time the voltage

across the resistor will rise to zero. The waveforms are thus as shown in

Fig. 8.28, which is drawn for a pulse width T - 2z.

V ......

0.865V

0

-0.865V

.... [~ .~s

Vo

V VR = Vo

Figure 8.28

Example 8.9

The differentiator circuit shown in Fig. 8.26 has R = 2 Mfl and C = 1.25 pF.
Obtain the waveform of the output voltage Vo for each of the following input
conditions:
1 a single pulse of amplitude 10 V and width 15 ~s;
2 a train of pulses, each of amplitude 10 V and width 5 I~S, separated by 15 ~s.

Solution

1 When the pulse is applied to the circuit, Vc + VR = 10 V and all of this

appears across the resistor because the capacitor voltage cannot change

instantaneously. The output voltage Vo therefore immediately becomes

equal to Vin (= 10 V). The time constant of the circuit is

z- CR - 1.25 • 10 -12 x 2 • 10 6 - 2.5 ~s

The pulse width is 15 ~zs and since this is >5z then the capacitor voltage will

have reached V (= 10 V) and the output voltage Vo (=VR) will have reached

zero before the pulse is removed. The output voltage will remain at zero

until the trailing edge of the pulse arrives at the input terminals. During this

time the capacitor voltage vc = 10 V. When the pulse is removed the input

voltage is zero and so Vc + VR = 0. The capacitor voltage cannot suddenly