Determine the output voltage waveform for the RL differentiator in Figure 20-45.

Question

Accepted Answer

First, calculate the time constant.

[latex] au = \frac{L}{R}= \frac{20 \ m H}{10 \ k\Omega } = 2 \ \mu s[/latex]

On the rising edge, the inductor voltage immediately jumps to +25 V. Because the pulse width is [latex]5 \ \mu s[/latex], the inductor charges for only [latex]2.5 au[/latex], so you must use the formula for a decreasing exponential derived from Equation 13-8: [latex]v=V_{F}+ (V_{i}- V_{F})e^{-Rt/L}[/latex] with [latex]V_{F}[/latex] = 0 and [latex] au =L/R[/latex] .

[latex]v_{L}= V_{i}e^{-t/ au }= 25e^{-5\mu s/2\mu s}= 25e^{-2.5}= 25(0.082)= 2.05 \ V[/latex]

This result is the inductor voltage at the end of the [latex]5 \ \mu s[/latex] input pulse. On the falling edge, the output immediately jumps from +2.05 V down to —22.95 V (a 25 V negative-going transition). The complete output waveform is shown in Figure 20-46.

Question 20.10: Determine the output voltage waveform for the RL differentia...

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