Determine how much the capacitor in Figure 20-10 will charge when the single pulse is applied to the input.

Question

Accepted Answer

Calculate the time constant.

[latex] au =RC= (2.2 \ k\Omega )(1 \ \mu F)= 2.2 \ ms [/latex]

Because the pulse width is 5 ms, the capacitor charges for 2.27 time constants(5 ms/2.2 ms = 2.27). Use the exponential formula from Chapter 12 (Eq. 12-19 :[latex] v=V_{F}(1-e^{-t/RC})[/latex]) to find the voltage to which the capacitor will charge. With [latex]V_{F}[/latex] = 25 V and t = 5 ms, the calculation is as follows:

[latex]v=V_{F}(1-e^{-t/RC})[/latex]

[latex]\ \ =(25 \ V)(1-e^{-5ms/2.2ms})= (25 \ V)(1-e^{-2.27})[/latex]

[latex]\ \ =(25 \ V)(1-0.103)= (25 \ V)(0.897)= 22.4 \ V[/latex]

These calculations show that the capacitor charges to 22.4 V during the 5 ms duration of the input pulse. It will discharge back to zero when the pulse goes back to zero.

Question 20.2: Determine how much the capacitor in Figure 20-10 will charge...

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