The curve given (pic attachment) shows that there are two parts of supercapacitor discharge: resistive and capacitive.
The explanation given for the resistive voltage drop was:
“A voltage drop at the initial discharge is an energy loss due to the equivalent series resistance (ESR) that is embedded in any kind of capacitors”

My thinking is that:
Why does the ESR cause such a drop? In basic capacitor theory, the capacitor discharges exponentially through a resistive load. So, can’t the ESR be seen as part of the “load” by the capacitor and thus, shouldn’t the discharge be a smooth exponential?

Instead the curve shows a resistive voltage drop initially, and an almost straight line drop after that. Doesn’t this contradict capacitor behavior? Or am I missing something about supercapacitors?

Attached Images

3.Supercapacitor Discharge.jpg (12.2 KB, 25 views)

 

THat's strange. I've never heard or seen anything like that before.

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comforting to know im not the only clueless one here.

 

Yeah, doesn't make sense to me either?

 

The ESR is inside the capacitor so it reduces the voltage output immediately when there is a load. Same as a battery.
The curve is a straight line because the load is constant current, not just a resistor.

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Uncle $crooge

 

Hmm... I think I understand your explanation of the ESR causing a voltage drop. Its more of a practical thing, but should not happen theoretically, correct?

What exactly does "constant current load" mean? And how does it cause straight-line discharge?

 

Oh yeah, that's right. I didn't notice that. THe load is constant current and not a resistor. So the reasoning of the ESR being combined with the load resistance does not apply.

But still, since it is constant current the voltage drop across the ESR should be constant too which would just offset the entire discharge curve, not make a two segment curve. If this was an *actual* measurement rather than a theoretically drawn graph, I would attribute it to some startup current pulse required by the constant current load.

Do you know what a capacitor-resistor discharge curve looks like? As the capacitor discharges, the capacitor voltage drops. Since this voltage drops, the discharge current across the resistor also drops. THe result is that the discharge rate of the capacitor continuously decreases as the capacitor gets and the discharge voltage and current of the capacitor reaches an assymptotic zero.

But in your graph the voltage decrease linearily. That means that as the voltage is decreasing, the "so-called resistive load" that is discharging the capacitor is increasing (the current load is increasing which means that the resistance is decreasing) to maintain the voltage drop at a constant rate and stop it from slowing down.

In reality, other devices and physical phenomena are used to produce current sources that behave the same way rather than combining a variable resistor with a self-adjustment mechanism controlled by a current sensor.

In short, a constant current load is simply a load that adjusts the voltage across it to maintain the same amount of current flowing through it. A current sources adjusts the voltage across it to do the same thing except it is supplying current. A constant voltage load would be something like a diode that "for the most part" keeps the same voltage across it regardless of the current flowing through it.

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Hmm... Well the place I got the curve from mentined nothing about a constant current load. But no that you explain it, I get it.

I also get what you're saying about the ESR. Why is it causing two segments when it should simply shift the curve down by a constant voltage?

Anyone?

 

I think the beginning of the curve should be vertical then the discharge slope should begin a short distance down on the graph due to the ESR of the capacitor.

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Uncle $crooge

 

By George, I think you got it!

THat does make sense. No current drawn at the start would mean no ESR induced voltage drop. As soon as the CC load turns on (assuming it turns on instantly and the current ramps up instantly with no trace inductance) a ESR induced voltage drop would nick the capacitor's initial voltage off the start (producing that offset) which would then continue decreasing. Maybe it's just a ramp because CC sources don't turn on instantly.

Makes a lot more sense than my hypothesis of the initial current pulse from the constant current load starting up inducing a capacitor voltage drop, anyways.

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