# Capacitor Discharge Rate

Discussion in 'General Electronics Chat' started by GTeclips, Jul 30, 2013.

1. ### GTeclips Thread Starter Member

Feb 18, 2012
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0
Hello all. A while back, I was experimenting around (or at least trying to) with electromagnets and the relationship between, as electromagnet implies, electricity and magnets.

For a main experiment, I was trying to create an electromagnetic pulse. Now, in-spite of the the menacing name EMP, I genuinely didn't want to use it for an destructive purposes (I mention this simply for the curious). For this, I needed to make a small capacitor bank.

Now, I was pretty foolish and not too knowledgeable, much like now. I ended up getting some dinky capacitors that held 1000μf. Not only that, but the immensely slow rate they discharged at almost pained me to observe.

Now that I've learned a little bit more, I realize I should have done some research. The one thing I'm still stumped on is the term for how fast a capacitor discharges. I've tried internet searches, but none of the information I obtained made much sense to me. Do all capacitors have a logical pattern to their discharge that can be followed in a formula? Does some device in the capacitor control the discharge rate?

Could someone please offer some information. Thank you.

2. ### MrChips Moderator

Oct 2, 2009
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4,271
All capacitors have a logical pattern to their discharge.

The voltage across the capacitor C as it discharges through a resistor R is given by the equation:

$image=http://forum.allaboutcircuits.com/mimetex.cgi?%20v=v_%7Binit%7D%20%28e%5E%7B%5Cfrac%7B-t%7D%7BRC%7D%7D%29&hash=b0bb75d0ccc4a5bf53b607d00b808b56$

A simple formula to remember is the Time-Constant = R x C.

Remember to apply the units correctly,

Time-Constant (seconds) = R (ohms) x C (farads)

The time-constant is the time it would take the capacitor to discharge to 37% of the initial voltage.

So to answer your question, the external resistor R placed across the capacitor is what controls the discharge rate.

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3. ### wayneh Expert

Sep 9, 2010
13,607
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In some cases - high frequency switching power supplies - the ESR equivalent series resistance of a capacitor becomes relevant. But generally the small internal resistance of a capacitor is negligible and it's all about RC, where R is external.

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4. ### GTeclips Thread Starter Member

Feb 18, 2012
96
0
Thank you guys.

My conclusion is that the type/make of the capacitors is irrelevant. Let's say I go back to those 1000μf capacitors. If I want them to discharge at a faster rate, I should increase the conductivity of the wire I am going to use, Correct?

Edit: I apologize if I misinterpreted any of your information. I'm not very educated in the area of electronics (but I'm working at it), so I don't understand much simple terminology.

Last edited: Jul 30, 2013
5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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When you say wire, are you referring to the wire you use to connect the capacitor? I think you are on the right track, but looking at the wrong boxcar. The resistance you need to look at is the connected load/device.

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6. ### wayneh Expert

Sep 9, 2010
13,607
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A cap can discharge far faster than you can observe without equipment. As noted, whatever you are doing depends on the load, not the capacitor itself.

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7. ### GTeclips Thread Starter Member

Feb 18, 2012
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The reason I said wire is because my original circuit had had no load, just a coil connected to a cap bank. However, due to the below quote, I see now why it was so sluggish.

Yes, I realize now the tools I used to observe it are what slowed the discharge.

Thanks for all the help guys! I think I understand the solution now.

8. ### GTeclips Thread Starter Member

Feb 18, 2012
96
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Sorry to bring up this past thread. I should have asked this earlier.

For the above project, what type of capacitor would you recommend (voltage, capacitance)? I'd prefer not to tamper around with higher voltages.

9. ### wayneh Expert

Sep 9, 2010
13,607
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I don't understand what your project is. Do you just want to touch the capacitor to the leads of a coil and rapidly discharge the cap through that coil? If so, I believe you would want a low ESR capacitor to obtain the fastest possible discharge. The voltage rating should be ~50% more than whatever voltage you plan to use. Capacity is up to you. IMHO, more is better!

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10. ### BobTPH Active Member

Jun 5, 2013
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If you connect a charged capacitor to a coil, you have created a tank circuit, a circuit that is tuned to a particular frequency. The capacitor will discharge, creating a current in the coil, then, it will re-charge in the opposite direction, this continuing until the resistances involved sap all the energy out of the system. What you will get is a sine wave that declines in amplitude.

Bob

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11. ### GTeclips Thread Starter Member

Feb 18, 2012
96
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Yes, this is what I was planning t do. Thank you for the recommendation.
Edit: How can I calculate the amount of voltage the cap will discharge?

But, if the capacitor bank is in series circuit, how could they recharge themselves form the opposite direction?
Edit: I guess it doesn't really matter what type of circuit it is come to think of it.

Last edited: Aug 1, 2013
12. ### wayneh Expert

Sep 9, 2010
13,607
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The highest voltage will be the instant before you attach the coil, and that voltage will be set by you.

And yes, an LC circuit is always an RLC circuit. It's just a matter of how small you can make R. They ring at the tuned frequency determined by LC and decay due to the damping determined by RC. Look up "damped sine wave" or "RLC circuit" and you'll find lots of descriptions. I can't imagine why, but this was one of my favorite Physics topics!

13. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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And, polarized caps (i.e. electrolytic, tantalum) will rapidly self-destruct in this configuration.

14. ### BobTPH Active Member

Jun 5, 2013
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Because when current in an inductor is decreasing it develops a voltage in the opposite direction of the current.

Bob

15. ### Wendy Moderator

Mar 24, 2008
21,012
2,744
I was wondering when someone would mention LC circuits. It is a field unto itself, and the basis for most tuned circuits (filters).

The larger a coil is in Henrys, the slower it will charge. It also follows 1/RL function, the inverse of what a cap does. Building a magnetic field around itself is a form of storage. A cap has an inverse function, current in a cap can build slowly.

What you really need is a true power supply and a timer with a circuit to pulse the coil, and it will still have a limited duration for charging. The way to get around this is voltage, unfortunately.

16. ### wayneh Expert

Sep 9, 2010
13,607
4,404
The OP might want to view this "completed" project. It's a device I breadboarded for testing inductors, in support of trying to fix a TV.