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According to what I have read online, if a capacitor's ESR increases, the capacitor temperature will also increase and this has me left a little confused.

Consider the simple single resistor circuit below:

Enter image description here

If my resistor value was 100 ohms, I would be dissipating 0.25 watts of energy. If I was to replace the 100 ohms resistor with a 200 ohms resistor I would be dissipation 0.125 watts of energy. We can clearly see here that a higher resistance value produces fewer watts and fewer watts equals less heat.

So the question is why is the same principle not occurring in a capacitor when the ESR of the capacitor increases? I would think that more ESR means more resistance, more resistance equals less current, less current equals fewer watts and fewer watts equal less heat. So more capacitor ESR equals a cooler capacitor.

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  • \$\begingroup\$ ESR is modeled in series with the capacitance, not in parallel with the applied voltage. Also, try thinking of a power supply which presents a sawtooth current waveform to the capacitor. \$\endgroup\$ – Adam Lawrence Jul 17 '15 at 18:27
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The voltage across the ESR in a capacitor will be very low compared to the voltage across the capacitative part. As such the current will not vary significantly if the ESR varies. This is true because it wouldn't be a very good capacitor if the resistive component was significant in determining the overall impedance.

This means that it is equivalent to a resistor being fed from a constant current generator. In that case the power dissipated in the resistor will be proportional to the value of the resistance - P = I^2*R where I is constant.

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  • \$\begingroup\$ I agree. So if T555 wants to convert this to his simple model he should use a current source instead of a voltage soure. \$\endgroup\$ – Wouter van Ooijen Jul 18 '15 at 13:42
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It all depends on your voltage source internal resistance. If you'll read Maximum power transfer theorem then you will know that a capacitor's internal resistance (ESR) (as will any other resistive load) will dissipate the most power when the ESR is equal to voltage source's internal resistance. This is presented in the picture below (red curve). When RS (source resistance) is equal to RL (ESR) then power transfer reaches maximum (Pmax point at RL/RS = 1 -> RL = RS).

Power transfer

I am assuming that voltage source that you are modeling is perfect (i.e. it presents negligible internal resistance). Pmax point occurs at RS = RL, as I've mentioned before, and since you are experimenting with ESRs being far from negligible RS you are, in fact, far away on the right site of this chart where the function value (which is in fact proportional to the power being dissipated at your load) is decreasing with increasing RS. That's why you are observing a decrease in power with an increase in ESR.

Try to experiment with voltage sources that have a similar resistance to the ESR. If you set your source resistance to, let's say 100 ohm and your ESR to 90 ohm and then to 100 ohm, you'll notice that the ESR will dissipate more power when equal to 100 ohm.

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