Having trouble understanding capacitors - Page 1
Forum section: electronics / beginners

author: diynoise
author_rank: Contributor
timestamp: January 13, 2018, 11:00:06 am
content: Hi!I'm reading about capacitors. I seem to understand that you can charge one by hooking it up to a voltage source. Let's say it's a mean 700F ultracap, so we have time to see what is happening.I seem to understand that if we hook that up to a bench psu, we are going to have to wait a couple of minutes to have it completely charged.But everything I'm reading on capacitors says that the instantaneous current for a capacitor is the capacitance times the time derivative of the voltage. The power supply is supplying a steady voltage, so the current for the capacitor should be 0, except for a very short spike when we turn on the power supply. How can the capacitor charge if the current is 0?I'm obviously missing or misunderstanding something very fundamental here.

author: paulca
author_rank: Super Contributor
timestamp: January 13, 2018, 11:13:25 am
content: If you connect that capacitor across the bench supply and your supply has no current limit functionality.... fireworks.The capacitor will appear almost as a short circuit and will pull very large current to charge. As the charge rises the resistance will rise and the current will fall away.The same will happen discharging.Be careful with such large caps.A way to look at a capacitor is like a VERY, VERY fast battery. They behave very similarly. Unlike a battery a capacitor can dump it's entire energy very, very quickly.These are worth a watch:

author: diynoise
author_rank: Contributor
timestamp: January 13, 2018, 11:17:29 am
content: Quote from: paulca on January 13, 2018, 11:13:25 amI get this, and I'm not actually charging any capacitor right now. I'm just trying to understand how the capacitor can pull very large currents if the formula for its current is proportional to dv/dt, which should be 0. Maybe my question isn't very clear?

author: paulca
author_rank: Super Contributor
timestamp: January 13, 2018, 11:24:42 am
content: I'm not great at maths, but the equation is:i = C * dv / dtIf you change the 700F capacitor's voltage by 1 volt per second, thats:i = C * 1/1i = 700 amps.Which doesn't sound right either, but it wouldn't surprise me. That's a big capacitor.Something like a 1uF capacitor discharged by sticking a small resistor across it's terminals and discharging it from 16V in 1 ms.i = 0.000001 * 16 / 0.001i = 16mASeems more reasonable.

author: diynoise
author_rank: Contributor
timestamp: January 13, 2018, 11:28:42 am
content: Just to make sure I understand, you are saying the v in dv/dt is referring to the voltage between the plates of the capacitor, not in the supply voltage in the circuit?

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: January 13, 2018, 11:31:51 am
content: Perhaps looking at the problem in a different light would be more useful.Assuming the capacitor has a low ESR and can thus charge/discharge large amounts of current, and that the power supply operates in constant current mode.We can then re-arrange the equation to give (by integrating both sides, sorry for horrendous formatting):V(t) = 1/C * integral[ i(t) dt ]Assuming constant current 'k', the integral computes to (and assuming zero initial conditions):V(t) = k*t/C

author: paulca
author_rank: Super Contributor
timestamp: January 13, 2018, 11:33:28 am
content: Its the change in voltage across the capacitor over time.If the capacitor is at 16V and you drop that voltage to 15V in 1 second it will generate X current.This is what capacitors are used for. They resist change in voltage by sourcing/sinking current. It's why, for example you put a capacitor across a power rail to smooth noise out. if the supply drops a fraction of a volt the capacitor will discharge towards that new voltage delivering current to prop the circuit up momentarily. The reverse is true, if the voltage rises slightly the capacitor will adsorb current momentarily. If the changes are fast enough the voltage on the circuit the capacitor is on will smooth out.

author: diynoise
author_rank: Contributor
timestamp: January 13, 2018, 11:38:25 am
content: All right, I think everything is clear now.Quote from: paulca on January 13, 2018, 11:33:28 amI already understood this conceptually, so I guess the main problem was a misunderstanding of the equation.Thanks everyone for the help!

author: MrAl
author_rank: Super Contributor
timestamp: January 13, 2018, 03:06:32 pm
content: Hello,Questions like this get deep into the nuances of theory vs the practical.In theory, using an ideal voltage source, you can charge a capacitor up in zero time but it requires an infinite current. What that amounts to is what is called an "impulse". An impulse is allowed in theory, but in practice it is hard to generate and harder to use.In some cases a voltage impulse is used, and in theory that should be an infinitely high amplitude voltage pulse that lasts for zero seconds, but in practice a 1000v pulse of finite but short duration may be used if it is compatible with the equipment. That gives the user some idea what the impulse response is.In theory, we use a source that is simply called an impulse, and strangely enough that also has a magnitude associated with it.The definition of an impulse is based on an area and im not sure if you want to get into this right now but in short it is a pulse signal that has an area of 1 as the pulse width approaches zero, which means the amplitude approaches infinity.Charging a cap with a power supply though means the power supply has to be current limited, and that current should not exceed the rating of the capacitor. When capacitors blow out they actually can blow up like dynamite, so much caution has to be applied. Look on the data sheet to find out the recommendations from the manufacturer.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: January 13, 2018, 03:15:16 pm
content: Quote from: diynoise on January 13, 2018, 11:28:42 amI think you are getting confused because you are confusing the "ideal voltage source" in those equations with the "practical" voltage source you have on a lab bench.Even the best bench powersupply in the world will not be able to maintain it's set voltage while charging a capacitor. This is because this would require an infinite amount of current because it would need to charge it in an infnitly small amount of time in order to not have it's voltage drop.A real powersupply will have a voltage drop - either this be in the leads of whatever you are using to connect said supply to the capacitor, or in the supply itself. It will then follow that dv/dt rating you spoke of while charging.The thing is just that with small capacitors this happens so fast you can't see it happening. Because it isn't visible, you never realized it was happening. But now you found an extreme example (by going to a super cap of 700 F instead of having a ''normal" capacitor of maybe a few uF) and it "broke" your simplified model and this is what left you confused.If you were to hook up a oscilloscope across a capacitor, you will see this slope rise as we have the source charge the capacitor.