In Figure 12-39, calculate the voltage across the capacitor every 0.1 ms for one complete period of the input.Then sketch the capacitor waveform. Assume the Thevenin resistance of the generator is negligible.

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Accepted Answer

[latex] au = RC= (15 \ k\Omega )(0.0056 \ \mu F)= 0.084 \ ms[/latex]

The period of the square wave is 1 ms. which is approximately 12 [latex] au [/latex]. This means that 6 [latex] au [/latex] will elapse after each change of the pulse, allowing the capacitor to fully charge and fully discharge.

For the increasing exponential,

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

At 0.1 ms : v = 2.5 V( 1 - [latex]e^{-0.1ms/0.084ms}[/latex]) = 1 .74 V

At 0.2 ms : v = 2.5 V( 1 - [latex]e^{-0.2ms/0.084ms}[/latex]) = 2.27 V

At 0.3 ms : v = 2.5 V( 1 - [latex]e^{-0.3ms/0.084ms}[/latex]) = 2.43 V

At 0.4 ms : v = 2.5 V( 1 - [latex]e^{-0.4ms/0.084ms}[/latex]) = 2.48 V

At 0.5 ms : v = 2.5 V( 1 - [latex]e^{-0.5ms/0.084ms}[/latex]) = 2.49 V

For the decreasing exponential.

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

In the equation, time is shown from the point when the change occurs (subtracting 0.5 ms from the actual time). For example, at 0.6 ms, t = 0.6 ms — 0.5 ms = 0.1 ms.

At 0.6 ms: v = 2.5 V([latex]e^{-0.1ms/0.084ms}[/latex]) = 0.76 V

At 0.7 ms: v = 2.5 V([latex]e^{-0.2ms/0.084ms}[/latex]) = 0.23 V

At 0.8 ms: v = 2.5 V([latex]e^{-0.3ms/0.084ms}[/latex]) = 0.07 V

At 0.9 ms: v = 2.5 V([latex]e^{-0.4ms/0.084ms}[/latex]) = 0.02 V

At 1.0 ms: v = 2.5 V([latex]e^{-0.5ms/0.084ms}[/latex]) = 0.01 V

Figure 12-40 is a plot of these results.

Question 12.16: In Figure 12-39, calculate the voltage across the capacitor ...

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