Hi there,

I have a question about discharging capacitors, solving the differential equation for a simple RC circuit, we all know the formula Q(t)=Q0*exp(-t/RC), differentiating and substituting Q=CV, we get a formula for the current; i(t)=(V/R)*exp(-t/RC), the peak current is then just V/R as Ohms law states, this to me is fine, but! there must be some limiting factors like the size of the capacitor? I thought it was strange that the current is not some function of capacitance as if we have 12 volts and a R = 0.1ohm we can get 120A, i would normally associate this with large capacitors but from the equations i have presented, it is possibel to get it from a small capacitor too, which to me, does not seem possible! Can someone clarify the limitations on the peak current (maybe ESL or ESR), maybe also tell me why people who build capacitor banks all the time always quote the energy stored in the capacitor when there is no direct dependance on the energy according to the derived equations?

Any help is appreciated!

 

You have to recall that the charge contained in the capacitor is quite finite. As soon as the discharge begins, the charge starts to equalize and the current drops accordingly as the voltage difference approaches zero. The initial current will be limited by internal resistance in the capacitor and in the shorting device, but will have a high instantaneous value.

 

Thanks for the reply, but how does the capacitance of the capacitor and hence its ability to store charge (energy) affect the peak current at the time when a switch is closed and the capacitor current follows the exponential decay curve? Energy is the ability to do work, in this case to move electrons in the shorting wire so the capacitance must have some effect?

 

You are confusing an ideal capacitor with a real one. Every real capacitor has ESR (equivalent series resistance) and also ESL (equivalent series inductance) which limit the peak current when the capacitor is shorted out. In an ideal cap, the peak current is NOT dependent on amount of capacitance.

 

You are confusing an ideal capacitor with a real one. Every real capacitor has ESR (equivalent series resistance) and also ESL (equivalent series inductance) which limit the peak current when the capacitor is shorted out. In an ideal cap, the peak current is NOT dependent on amount of capacitance.

Thats what I was thinking, Im designing a test rig at work for supplying a mock stator winding with a high inrush current (to simulate transient fault currents and observe the electromechanical effect in the impregnation epoxy) and I am needing a capacitor bank, but need to size the capacitors and I have a current in my mind we need to work at, and possibly vary it by chraging the capacitor up to various levels of voltage and them discharge them through the coil BUT I had no idea how to go about sizing the capacitors to keep the rigs cost down? I have never built one before and need advice on their construction and constraints!

 

If you have a 100 nF capacitor at 12 V and try to get 120 A out of it you will discharge the capacitor as follows:

q = CV = 100n * 12 = 1.2 uC
q = It, or t = q/I = 1.2uC/120 = 10 ns

This ignores what beenthere said, that the voltage is dropping and hence the current will be reducing. This is just a rough calculation to give an idea of the situation.

If you try to have a current of 120 A build up on the order of 1 ns then series inductance on the order of 1 nH will resist with a voltage of:
V = L di/dt = 1 nH * 120 A / 1 ns = 120 V

The calculation again isn't giving a very relevant number, it's just the idea. The point is that tiny inductance simply wouldn't let it happen.

Here's a plot of a 100 nF capacitor charged to 12 V discharging with 1 pH, 100 pH, and 1 nH of series inductance through a series resistance of 0.1 ohms. Notice the peak is dropping significantly.



As for energy it is relevant. By having higher C you have stored more energy and hence can sustain the current for longer.

Here's the same situation but with 10 uF instead of 100 nF (same scale).



Aside from these considerations another one is how much current the capacitor can actually handle. Try to force too much for too long through a capacitor and it will overheat and could even explode.
Paralleling capacitors lets you reduce the series inductance and resistance, raising the peak current, along with reducing the current demand from each individual cap. You also get more stored energy because the capacitance is adding together. Even more benefit is that you're buying many cheap(er) caps instead of a few very expensive ones.

 

If you have a 100 nF capacitor at 12 V and try to get 120 A out of it you will discharge the capacitor as follows:

q = CV = 100n * 12 = 1.2 uC
q = It, or t = q/I = 1.2uC/120 = 10 ns

This ignores what beenthere said, that the voltage is dropping and hence the current will be reducing. This is just a rough calculation to give an idea of the situation.

If you try to have a current of 120 A build up on the order of 1 ns then series inductance on the order of 1 nH will resist with a voltage of:
V = L di/dt = 1 nH * 120 A / 1 ns = 120 V

The calculation again isn't giving a very relevant number, it's just the idea. The point is that tiny inductance simply wouldn't let it happen.

Here's a plot of a 100 nF capacitor charged to 12 V discharging with 1 pH, 100 pH, and 1 nH of series inductance through a series resistance of 0.1 ohms. Notice the peak is dropping significantly.

View attachment 21129

As for energy it is relevant. By having higher C you have stored more energy and hence can sustain the current for longer.

Here's the same situation but with 10 uF instead of 100 nF (same scale).

View attachment 21128

Aside from these considerations another one is how much current the capacitor can actually handle. Try to force too much for too long through a capacitor and it will overheat and could even explode.
Paralleling capacitors lets you reduce the series inductance and resistance, raising the peak current, along with reducing the current demand from each individual cap. You also get more stored energy because the capacitance is adding together. Even more benefit is that you're buying many cheap(er) caps instead of a few very expensive ones.

Thanks Ghar, It was what I expected but I just thought id ask around to see if anyone could tell me otherwise! do you have any tips for building a capacitor bank and optimising its design for maximum current output apart from paralleling? aswell as selecting the right capacitor values? Any help would be greatly appreciated.

 

I've never built one so I'm sure people can give better advice than I can but really your goals are:
1) minimize series impedance. This means choosing capacitors that have low losses (low ESR and ESL, and generally designed for fast discharge applications) and keeping your layout nice and tight with small loop areas. Twisted wires are better than not twisted. A loop of wire (or circuit trace) the size of your wrist is already about 100 nH.
2) maximize stored energy. This means choosing as many of the highest value capacitors as makes sense in your test. Figure out how much current you want for how long and at what voltages. Keep in mind capacitor tolerances can be as bad as 20% so you must add margin. Check datasheets.
3) keep the current in each capacitor within safe limits.

Paralleling capacitors really helps both of those.

I'd ask the manufacturers if they recommend their caps to be used for such a thing and always be wary of all the deratings you need to the rated voltage and current and everything else.

 

This kind of raises a red flag, as we do get questions about capacitor banks that involve dangerous things like rail guns. What is the application?

 

This kind of raises a red flag, as we do get questions about capacitor banks that involve dangerous things like rail guns. What is the application?

He mentions it in his other post:

Im designing a test rig at work for supplying a mock stator winding with a high inrush current (to simulate transient fault currents and observe the electromechanical effect in the impregnation epoxy) and I am needing a capacitor bank

Of course that doesn't stop anyone with other intentions from reading this thread but I'll let you moderators figure that one out

 

Thanks, I missed that post.

 

Im thinking of using AVX BestCap Ultra Low ESR High Power Pulse Supercapacitors as the charge storage, has anyone used these or this type of capacitor before?

 

Help us out here. What are the expected magnitudes of voltage and current involved? Those supercaps are amazing for a number of things, but price and voltage are not in that number. They are usually rated for 2.7 and 5.5 volts.

If high voltage is not important, but some sustained level of current is, perhaps a big DC arc welder would be more the thing?

 

Help us out here. What are the expected magnitudes of voltage and current involved? Those supercaps are amazing for a number of things, but price and voltage are not in that number. They are usually rated for 2.7 and 5.5 volts.

If high voltage is not important, but some sustained level of current is, perhaps a big DC arc welder would be more the thing?

High voltage is not important, a short current pulse is required, 5.5V would be suffice, currents would need to be ideally around the 100's of Amps, say upto 500 maximum for a short burst. The load has a very small impedance, around 0.04 ohms.

 

High voltage is not important, a short current pulse is required, 5.5V would be suffice, currents would need to be ideally around the 100's of Amps, say upto 500 maximum for a short burst. The load has a very small impedance, around 0.04 ohms.

You also said you are testing stator windings. What is the inductance of the winding? The inductance is kind of eqal to the ESL of the caps but probably a lot higher, so it could prevent you from achieving high current.

If the inductance is significant, you might need a bigger capacity to overcome the initial charging of the inductance and still having enough voltage on the caps to get the current after the inductance is charged.

 

You also said you are testing stator windings. What is the inductance of the winding? The inductance is kind of eqal to the ESL of the caps but probably a lot higher, so it could prevent you from achieving high current.

If the inductance is significant, you might need a bigger capacity to overcome the initial charging of the inductance and still having enough voltage on the caps to get the current after the inductance is charged.

The stator windings I am using are only mock stator windings, it will be parallel wires of 1m length, these will be dipped in epoxy, that our stator windings are and the current pulsed into the lengths of wire, the purpose is to observe and analyse the electromechanical effect on the breakdown of the epoxy due to the force experienced by the wires when a transient fault current occurs. The inductance of the 'winding' would be approx 220mH.