A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated quivering of the heart muscle fibers—by delivering an electric shock to the heart. The Lown defibrillator was developed by Dr. Bernard Lown in 1962. Its key feature, shown in Fig. 7.52a, is its voltage

Question

A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated quivering of the heart muscle fibers—by delivering an electric shock to the heart. The Lown defibrillator was developed by Dr. Bernard Lown in 1962. Its key feature, shown in Fig. 7.52a, is its voltage waveform. A simplified circuit diagram that is capable of producing the Lown waveform is shown in Fig. 7.52b. Let us find the necessary values for the inductor and capacitor.

Accepted Answer

Since the Lown waveform is oscillatory in nature, we know that the circuit is underdamped ([latex]\zeta [/latex] < 1) and the voltage applied to the patient is of the form

[latex]v_{o}(t) = K_{1} e^{-\zeta \omega_{o}t} sin[\omega t][/latex]

where

[latex] \zeta \omega_{o} = \frac{R}{2L}[/latex]
[latex] \omega = \omega_{o} \sqrt{1 - \zeta^{2}}[/latex]

and

[latex]\omega_{o} = \frac{1}{ \sqrt{LC}}[/latex]

for the series RLC circuit. From Fig. 7.52a, we see that the period of the sine function is

T = 10 ms

Thus, we have one expression involving [latex]\omega_{o}[/latex] and [latex] \zeta [/latex]

[latex]\omega = \frac{2\pi}{T} = \omega_{o} \sqrt{1 - \zeta^{2}}= 200 \pi [/latex] rad/s

A second expression can be obtained by solving for the ratio of [latex]v_{o}[/latex](t) at t = T/4 to that at t = 3T/4. At these two instants of time, the sine function is equal to +1 and -1, and respectively.
Using values from Fig. 7.52a, we can write

[latex]\frac{v_{o}(t/4)}{-v_{o}(3T/4)} = \frac{K_{1} e^{-\zeta \omega_{o}(T/4)}}{K_{1} e^{-\zeta \omega_{o}(3T/4)}} = e^{\zeta \omega_{o}(T/2)} \approx \frac{3000}{250}[/latex] = 12

or

[latex]\zeta \omega_{o}[/latex] = 497.0

Given R = 50 Ω, the necessary inductor value is

L = 50.3 mH

Using our expression for [latex]\omega[/latex],

[latex]\omega^{2} = (200\pi)^{2} = \omega_{0}^{2}-(\zeta \omega_{o})^{2}[/latex]

or

= [latex](200\pi)^{2} = \frac{1}{LC} - (497.0)^{2}[/latex]

Solving for the capacitor value, we find

C = 31.0 μF

Let us verify our design using the circuit shown in Fig. 7.53a. The output voltage plot shown in Fig. 7.53b matches the Lown waveform in Fig. 7.52a; thus, we can consider the design to be a success.
While this solution is viable, it is not the only one. Like many design problems, there are often various ways to satisfy the design specifications.

Question 7.21: A defibrillator is a device that is used to stop heart fibri...

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