Electronic – Can an external defibrillator deliver a continuous charge

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Can an external defibrillator, used medically to restore normal heart rhythm, deliver a continuous charge? I've been trying to find references to how a defibrillator works, whether it sources with batteries or capacitors, and so on, but the information seems to be evading me.

Edit: Considering that defibrillators deliver a complicated AC waveform, I can see how this question could be unclear. What I'm asking is whether defibrillators are capable only of producing one-off "shocks," or whether they can "shock" continuously, like inserting a finger into a light socket. For a mathematically better defined problem, I'm asking whether typical external defibrillators used by medical staff have an operation time duty cycle (how many milliseconds in every x milliseconds the device can be run, as opposed to a PWM duty cycle) and a small upper bound on charge delivery time. In other words, could I, for instance, deliver a 2-minute charge?

Context: There's a scene in the TV show Dexter wherein the titular character electrocutes two people simultaneously with external defibrillators. He doesn't cause arrhythmias, as far as I can tell. He actually electrocutes them with extended shocks. My first thought upon seeing this was, "that's not how defibrillators work." I wanted to do some additional research just to be sure. I asked it here after googling around and not finding anything of note that actually describes modern defibrillator function.

Best Answer

An external defibrillator will deliver a biphasic waveform usually. This would look like something like this:

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The waveform can also be monophasic or triphasic. So we are not talking about continuous charge. Current is equal to charge per second.

A defibrillator circuit is fairly simple. Generally speaking, you would only need a battery, capacitor, inductor, and a set of paddles. The most reduced version of the circuit would look like this:

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When the switch is connected to 'A', the capacitor is charging. When it is connected to 'B', the capacitor is discharging through the inductor and the paddles. I am assuming the inductor is there to control/limit the amount of current surging through the paddles. Current is what will kill you, not voltage.