Question 4.63: A signal varies as vS(t)=10(1–e^–100t) V. Design a comparato......

The Analysis and Design of Linear Circuits [1399573]

A signal varies as $v_S(t)=10(1–e^{–100t})$ V. Design a comparator circuit that will switch from 0 to 5 V in 2 ms. Use a single OP AMP that has + $V_{CC}$ = + 5 V and – $V_{CC}$ = 0 V.

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Determine the voltage of $v_S$ (t), when t = 2 ms, and then design a comparator circuit to switch as specified above.

A signal varies as vS(t)=10(1–e^–100t) V. Design a comparator circuit that will switch from 0 to 5 V in 2 ms. Use a single OP AMP that has +VCC= + 5V and – VCC = 0 V.

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Script File

clear all
vs2ms = 10*(1-exp(-100*2e-3))
vs2ms =
1.8127e+000

We want the comparator circuit to produce 0 V when $v_S$ (t) < 1.8127 V, and to produce 5 V when $v_S$ (t) > 1.8127 V. Using Figure P4-61 as a model, the following design provides one possible solution.

The following output plot shows that the circuit meets the specifications.

Answer:

Presented above.

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