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.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ Jan 6, 2017 at 19:03
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    \$\begingroup\$ Contrary to Hollywoodian beliefs, defibrillators cannot be used as makeshift tasers! \$\endgroup\$ Jan 7, 2017 at 13:04
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    \$\begingroup\$ Op was not asking about movies. That was edited in by someone else. \$\endgroup\$
    – Passerby
    Jan 7, 2017 at 20:39
  • \$\begingroup\$ Not only will it output a waveform, except the ancient ones but they have been decommissioned, they will also sense the heart rhythm and if the patient does not have one or it's even enough (despite dying), it will not fire. Unlike Hollywood movies, you can't revive someone with an AED, only reset the persons heart rythm if it's very irregular (again a dying condition). In hospitals they have more high powered stuff, adrenaline shots to the heart and what not, but in the feild, don't be surprised if the AED does not fire. Be prepared to do CPR for minutes until paramedics arrive. \$\endgroup\$
    – winny
    Jan 7, 2017 at 20:57
  • \$\begingroup\$ @Passerby The O.P. had actually asked this question in the context of a movie, oddly enough. \$\endgroup\$ Jan 8, 2017 at 19:15

2 Answers 2


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

enter image description here

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:

enter image description here

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.


Below, you see the lower circuit board of an AED from the late 90's. Along the top are two coordinated capacitor banks, with the large capacitors to the left forming one and the smaller capacitors to the right forming another. The bank on the left is charged by the DC/DC converter a little below the third and fourth caps from the left, black sides, white top. The right bank is charged by the Datel DC/DC converter on the right of the board. Each cap on the left is rated to 450 V with a capacitance of 740 µF, and we can assume the caps on the right are rated no higher than the larger ones on the left.

AED Lower Circuit Board

The energy capacity of a capacitor is \$\frac{CV^2}{2}\$

For seven capacitors rated to 450 V, each with capacitance 740 µF, the math goes like this:

$$E_{max} = 7 \cdot \frac{740 \textrm{µF} \cdot (450 \textrm{V})^2}{2}$$ $$E_{max} = 524.5 \textrm{J}$$

Current medical practice says you start at 200 Joules for the first shock, then increase if necessary. The device is only meant to output a specific biphasic waveform, lasting a couple of milliseconds. If you wanted to get a continuous discharge, you'd need to repeat this biphasic waveform. Assuming an upper bound on discharge time of 25 ms, and 100% capacitor efficiency, you could stretch the discharge to last \$\frac{E_{max}}{E_{discharge}} \cdot t_{discharge}\$ and adding in the numbers \$\frac{524.5 \textrm{J}}{200 \textrm{J}} \cdot 25 \textrm{ms}\$, or about 65.6 ms, as an absolute upper limit for what you could get out of this AED before you'd need to stop and let the caps recharge.

The standard 200 J shock, delivered in up to 25 ms, means a power output of 8 kW, roughly equivalent to the power requirements of 10 refrigerators. If you wanted to charge the capacitor as fast as you discharge it, you'd need some crazy battery pack capable of matching that power output. A realistic battery pack might output 12V at a max current of 20A before you start melting things. That's a power output of 240 W, or 3% of the 8 kW you're looking for. That means for every 65.6 ms discharge you'd need to wait 1/3% of that, or 33.3 times. That's 2.18 s, so one pulse every 2.25 s. Have fun with your continuous discharge defibrillator.

To answer your more direct questions, that means there is a "small upper bound on charge delivery time" of 65.6 ms, and an "operation time duty cycle" of 3%, definitely no ongoing shocking.


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