A heart pacemaker circuit is shown in Fig. 7.29. An SCR (silicon-controlled rectifier) is a solid-state device that has two distinct modes of operation. When the voltage across the SCR is increasing but less than 5 V, the SCR behaves like an open circuit, as shown in Fig. 7.30a. Once the voltage

Question

A heart pacemaker circuit is shown in Fig. 7.29. An SCR (silicon-controlled rectifier) is a solid-state device that has two distinct modes of operation. When the voltage across the SCR is increasing but less than 5 V, the SCR behaves like an open circuit, as shown in Fig. 7.30a. Once the voltage across the SCR reaches 5 V, the device functions like a current source, as shown in Fig. 7.30b. This behavior will continue as long as the SCR voltage remains above 0.2 V. At this voltage, the SCR shuts off and again becomes an open circuit.
Assume that at t = 0, [latex]v_{C}[/latex](t) is 0 V and the 1- μF capacitor begins to charge toward the 6-V source voltage. Find the resistor value such that [latex]v_{C}[/latex](t) will equal 5 V (the SCR firing voltage) at 1 s. At t = 1 s, the SCR fires and begins discharging the capacitor. Find the time required for [latex]v_{C}[/latex](t) to drop from 5 V to 0.2 V. Finally, plot [latex]v_{C}[/latex](t) for the three cycles.

Accepted Answer

For t < 1 s, the equivalent circuit for the pacemaker is shown in Fig. 7.31. As indicated earlier, the capacitor voltage has the form

[latex]v_{C}(t) = 6 - 6e^{-t/RC}[/latex] V

A voltage of 0.2 V occurs at

[latex]t_{1}[/latex] = 0.034RC

whereas a voltage of 5 V occurs at

[latex]t_{2}[/latex] = 1.792RC

We desire that [latex]t_{2} - t_{1}[/latex] = 1 s. Therefore,

[latex]t_{2} - t_{1}[/latex] = 1.758RC = 1 s

and

RC = 0.569 s and R = 569 kΩ

At t = 1 s the SCR fires and the pacemaker is modeled by the circuit in Fig. 7.32. The form of the discharge waveform is

v(t) = [latex]K_{1} + K_{2} e^{-(t-1)/RC}[/latex]

The term (t-1) appears in the exponential to shift the function 1 s, since during that time the capacitor was charging. Just after the SCR fires at t = [latex]1^{+}[/latex] s, [latex]v_{C}[/latex](t) is still 5 V, whereas at t = [latex]\infty[/latex], [latex] v_{C}[/latex](t) = 6 - IR. Therefore,

[latex]K_{1} + K_{2}[/latex] = 5 and [latex]K_{1}[/latex] = 6 - IR

Our solution, then, is of the form

[latex]v_{C}[/latex](t) = 6 - IR + (IR - 1)[latex]e^{-(t-1)/RC}[/latex]

Let T be the time beyond 1 s necessary for v(t) to drop to 0.2 V. We write

[latex]v_{C}(T + 1) = 6 - IR + (IR - 1)e^{-T/RC}[/latex] = 0.2

Substituting for I, R, and C, we find

T = 0.11 s

The output waveform is shown in Fig. 7.33.

Question 7.14: A heart pacemaker circuit is shown in Fig. 7.29. An SCR (sil...

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