A single 10 V pulse with a width of 100 μs is applied to the RC integrator in Figure 20-8,
(a) To what voltage will the capacitor charge?
(b) How long will it take the capacitor to discharge if the internal resistance of the pulse source is 50 Ω?
(c) Draw the output voltage waveform.

Question

Accepted Answer

(a) The circuit time constant is

[latex] au =RC= (100 \ k\Omega )(0.001 \ \mu F)= 100 \ \mu s[/latex]

Notice that the pulse width is exactly equal to the time constant. Thus, the capacitor will charge approximately 63% of the full input amplitude in one time constant, so the output will reach a maximum voltage of

[latex]V_{out}[/latex]= (0.63)10 V = 6.3 V

(b) The capacitor discharges back through the source when the pulse ends. You can neglect the 50 Ω source resistance in series with 100 kΩ. The total approximate discharge time, therefore, is

[latex]5 au = 5(100 \ \mu s)= 500 \ \mu s[/latex]

(c) The output charging and discharging curve is shown in Figure 20-9.

Question 20.1: A single 10 V pulse with a width of 100 μs is applied to the...

Related Answered Questions