How Do Defibrillators Work?
Defibrillators are devices that send a short, high voltage pulse of current pulse through a person’s heart. It helps restore a heart’s natural rhythm when it goes into a state of arrhythmia or tremor (fibrillation)
There are many things that can be discussed regarding the defibrillator; however, due to the scope of this exercise, a few components of the inner circuitry will by looked at, attempting to demystify a life-saving device.
In its most basic form, a relies on a large capacitor, inductor, and transformer to produce the desired effect – that is, a short spike of current within a high impedance load (the human body).
A circuit diagram shown below can clear things up: A transformer takes the input voltage and steps it up (also known as a step-up transformer) and bring its potential up to 1000-5000 volts.
The voltage charges a large capacitor in the center of the circuit, which, when the defibrillator paddles are on the body, a switch is depressed to disengage it from the source and engage the capacitor which discharges very rapidly. We will treat the source as a DC signal as opposed to an AC signal (transformers need an AC signal).
Figure 1: A simple defibrillator circuit
The current travels from the capacitor through an inductor. The inductor’s job is to slow down the current a bit. Inductors don’t like current spikes at all and slows the inrush current down. We will pretend like the inductor is not part of the circuit for this exercise.
The capacitor takes some time to charge, conventionally five time-constants, as positive charges accumulate on one plate, and negative charges on the other. To understand how this is happening, we’re going to look at a few equations, and by the end, we will be able to solve for current and voltage
Inner workings and equations:
Let’s assume the transformer steps the voltage up to 5000V, and we pick a decently sized 50 microfarad capacitor. When the switch is closed, the capacitor starts to charge. The capacitor reaches a steady state, meaning that the current in the circuit stays the same as time goes by. This can be modeled with the equation:
∮ 𝐸⃑ ∙ 𝑑𝑙 = 0 This is a modified version of the integral:
− ∫ 𝐸⃑ ∙ 𝑑𝑙
And this is because we are adhering to Kirchhoff’s circuit laws, which state that the sum of all currents in a system must equal to 0, or:
∑ 𝐼𝑡𝑜𝑡𝑎𝑙 = 0 Similarly, the sum of all voltages in a system must equal 0, or:
∑ 𝑉𝑡𝑜𝑡𝑎𝑙 = 0
It is also worth noting that a wire used to conduct electricity does not have an electric field. The electromotive forces created in the battery is what is causing the electrons within the
conductor to move. The wire just stays a wire and does nothing more.
The capacitor charges at a steady rate modeled by the equation, and its total charge is:
𝑄 = 𝐶 ∆𝑉
= 50𝜇𝐹 × 5000𝑉 = (50𝐸 − 6)(5000𝑉) = 1/4 𝐶
Using the equation V = IR, we can determine different things about our circuit, whether it is power dissipation, or resistance. Taking the units for current (C/s) we can determine how much charge flows in the patient, depending on how quickly the capacitor discharges. We can determine the charging rate of the capacitor and the discharge rate of the capacitor with:
𝑉𝑐 = 𝑉(1 − 𝑒−𝜏𝑅𝐶) 𝑡𝑖𝑚𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜏 = 𝑅𝐶
= (10𝑘Ω)(50𝐸−6𝐹)
= 0.5𝑠
𝑎𝑡 5𝜏 = 2.5𝑠 𝑐ℎ𝑎𝑟𝑔𝑒 𝑡𝑖𝑚𝑒
Since the capacitor acts as a battery, after being disconnected from the source, the current can be modeled by I=V/R:
𝐼 =𝑉 𝑅
= 5000𝑉 10𝑘Ω
= 0.5𝐴
To see how the current diminished over time, you can multiply the exponential right next to that equation, and solve for time to see how much current is flowing at that time (for example, after 5 time constants):
𝐼 =𝑉
𝑅(1 − 𝑒−𝑡𝑅𝐶)
=5000𝑉
10𝑘Ω (1 − 𝑒
−5
(10000)(50𝐸−3))
= 22𝐸 − 6 𝐴
Half an amp of electricity is passing through the body! That is a large amount, however, it is for a very brief amount of time. Now we can solve for time:
𝐼 =∆𝑄 0.25𝐶∆𝑡
0.50𝐴= 500𝑚𝑠
Note: This is an exercise using hypothetical values. The values do not represent actual values used in a defibrillator. The time it takes for the capacitor to discharge is most likely less than a few milliseconds, and the current inrush could very well be greater than half an amp. The capacitor might also be much bigger. This is just a demonstration of how the equations work with each other.
Figure 2: Voltage discharge and current discharge in a capacitor:
While we’re at it, we can determine how much energy is stored in the capacitor, and subsequently released into the patient:
𝐸 = 1 2𝐶𝑉2
= 1
2(50𝐸−6𝐹)50002
= 625𝐽
We can also find exactly how much power was dissipated through the patient’s torso by the product of our voltage and out current:
𝑃𝑚𝑎𝑥 = 𝐼𝑚𝑎𝑥𝑉𝑚𝑎𝑥
= (0.5𝐴) ∗ (5000𝑉) = 2500𝑊 𝑚𝑎𝑥!
Figure 3: Current when our capacitor is charging from the source:
Figure 4: Current when our capacitor is discharging through the patient:
To summarize everything up: A source charges a large capacitor, rated at some large voltage (in this case around 5000V). The capacitor charges, and as soon as contact is made with the patient with the paddles, it goes through a current-smoothing inductor, and the current displaced runs through the heart, defibrillating it! It’s a relatively simple, yet life-saving mechanism!
References
https://www.explainthatstuff.com/defibrillators.html
https://www.researchgate.net/figure/The-defibrillator-circuit-showing-the-various- components-the-step-up-transformer-and_fig3_275304884
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html https://www.electronics-tutorials.ws/rc/rc_2.html
