Capacitor discharge
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 9th June 2005, 09:00 PM #1 thkking   diyAudio Member   Join Date: Mar 2005 Location: wpb, fl Capacitor discharge I've been interested in designing large capacitors to use for lessening ripple voltage in car audio, and for that matter the practical uses of the current "stiffening capacitors" being manufactured. I thought I had a decent grasp on most for some time now, but as I was thinking about design I found myself stumped on a question which is probably fairly easy to answer. It relates to discharge of a capacitor in the car's circuit. When using the equation t=R*C, R is the load resistance when referring to discharge, and how would that be determined or estimated in a car? By adding the resistances incurred at a given current through every electrical component as it's wired to the circuit? Or is the internal resistance of the capacitor used in determining both the charging and discharging rates? Thanks for understanding my ignorance Grant
 9th June 2005, 10:37 PM #2 sss   diyAudio Member     Join Date: Jul 2003 Location: Israel , haifa the internal resistance of the cap is not important for that matter __________________ if you are not living on the edge you are taking too much space
 10th June 2005, 12:15 AM #3 thkking   diyAudio Member   Join Date: Mar 2005 Location: wpb, fl Ok, so your saying internal resistance is only a factor during the charging. But can you please expand an your answer, and incorporate how to best estimate the load resistance. Also, for that matter, wouldn't internal resistance be at least a factor in determining the total load resistance? Thanks
Bill Fitzpatrick
diyAudio Member

Join Date: Jun 2001
Location: Eugene, OR
Re: Capacitor discharge

Quote:
 Originally posted by thkking I've been interested in designing large capacitors
So, you're going into the capacitor design and manufacturing business?

 10th June 2005, 06:39 AM #5 thkking   diyAudio Member   Join Date: Mar 2005 Location: wpb, fl LOL, no but it would probably be pretty lucrative if I did . I'm purely doing this for myself. I'm mainly curious on various properties of different large capacitor types and their effects on a car's electrical in theory vs. real world use. Like inductive properties, capacitance, leakage, voltage/current phase shift, etc... But I'm also curious what effects things like charge vs discharge besides the obvious variables I can plug into an equation. Before I can do that, I need to figure out what would apply as the load resistance? Any idea? It seems it would be an easy answer, like measuring voltage at the load side of the cap and current due to load, then figuring out resistance, I=E/R. But isn't that oversimplifying things??
 10th June 2005, 07:05 AM #6 Bill Fitzpatrick   diyAudio Member     Join Date: Jun 2001 Location: Eugene, OR Do you suppose you could rephrase your questions? It's hard to figure out what you're after.
 10th June 2005, 08:25 AM #7 sss   diyAudio Member     Join Date: Jul 2003 Location: Israel , haifa when there is nothing connected to the cap it is being discharged through the internal resistance , if u put it in the car , connected near the power amp , then it is d=ischarged through the amp , the internal res is not important in this case because its too high compared to the amp __________________ if you are not living on the edge you are taking too much space
thkking
diyAudio Member

Join Date: Mar 2005
Location: wpb, fl
Quote:
 Originally posted by Bill Fitzpatrick Do you suppose you could rephrase your questions? It's hard to figure out what you're after.
When using the equation t=R*C, R is the load resistance when referring to discharge, and how would that be determined or estimated in a car? I'm trying to plug in numbers to solve for the RC time constant. How is R found, when a capacitor is wired in parallel to amp, battery, etc
Quote:
 Originally posted by sss when there is nothing connected to the cap it is being discharged through the internal resistance , if u put it in the car , connected near the power amp , then it is d=ischarged through the amp , the internal res is not important in this case because its too high compared to the amp
I do know how capacitors are discharged when wired in series to a resistor or not wired to anything. But when it's placed in a car's circuit, traditionally it's wired in parallel to all electronics and battery. So really it wouldn't release its energy just to the amp, but into the load, correct? So what would be used I determining the resistance for discharge.
Also, if the internal resistance of the cap. is the highest resistance as seen by the cap(which I don't see how it could be higher than the load resistance in a car), then wouldn't it be the influencing factor in discharge?

 10th June 2005, 02:54 PM #9 sss   diyAudio Member     Join Date: Jul 2003 Location: Israel , haifa i think u are confused here amp is the load !!all the electric circuits of your car are the load can u tell us what are u trying to do ? that equation is also worthless in that case because the load is changing __________________ if you are not living on the edge you are taking too much space
 10th June 2005, 05:46 PM #10 Bill Fitzpatrick   diyAudio Member     Join Date: Jun 2001 Location: Eugene, OR The cap only releases energy when the car's supply voltage drops below the cap voltage because that's the only condition under which cap current can flow. In this case current flows from the cap only until the cap voltage is equal to the supply voltage. In normal car operation (no high power audio) the cap will track the supply voltage with a lag of 100 microseconds, give or take. It can be said, then, that there is essentially no load on the cap. So, things like lights, air, etc. don't load the cap except for the extremely brief period of time it takes for the car's electrical system to recover after these items are first switched on. The charged cap shortens the recovery time but that's beside the point. So, now to audio. A huge bass note causes much current to flow into the amp. This causes the car's voltage to drop because it's trying to supply a lot of current through it's own internal resistance. Fortunately for us, we have a big cap attached to the amp's supply. The cap senses the reduced supply voltage and attempts to bring it back up to it's own voltage. This works only until the cap has discharged to the point where it's own voltage is equal to the supply voltage. At that point, it quits helping. That's OK though because the bass note is probably over by now anyway. The car's voltage goes back up and the cap recharges. All this occurs in about 1ms again, give or take. Maybe you knew all this. The point is that there is NO load on the cap except during the time that the car's supply voltage is less that that of the cap. If you were to look at the current through the cap with some loud bass content music playing, you would see that it is pulsating and these pulses are pretting much in sync with the voltage differential between the cap and the car's supply. I say pretty much in sync because the ESR of the cap causes a slight delay. So, regarding the time constants. The factors are ESR, cable resistance, capacitance and voltage differential between the cap and the supply line. BTW, contrary to a previous post, a cap does not discharge through it's ESR. Unattached, a cap discharges through it's leakage resistance.

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