A 10 V pulse with a width of 1 ms is applied to the RL integrator in Figure 20-38.Determine the voltage level that the output will reach during the pulse. If the source has an internal resistance of 30 Ω, how long will it take the output to decay to zero ? Draw the output voltage waveform.
Question 20.8: A 10 V pulse with a width of 1 ms is applied to the RL integ...
The inductor charges through the 30 Ω source resistance plus the 470 Ω external resistor. The lime constant is
\tau = \frac{L}{R_{tot}}= \frac{500 \ mH}{470 \ \Omega + 30 \ \Omega } = \frac{500 \ mH}{500 \ \Omega } = 1 \ msNotice that in this case the pulse width is exactly equal to \tau. Thus, the output V_{R} will
reach approximately 63% of the full input amplitude in 1\tau. Therefore, the output voltage gets to 6.3 V at the end of the pulse. After the pulse is gone, the inductor discharges back through the 30 Ω source resistance and the 470 Ω resistor. The output voltage takes 5\tau to completely decay to zero.
The output voltage is shown in Figure 20-39.
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