# Capacitor discharge

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#### thkking

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

#### sss

the internal resistance of the cap is not important for that matter

#### thkking

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

thkking said:
I've been interested in designing large capacitors

So, you're going into the capacitor design and manufacturing business?

#### thkking

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??

#### Bill Fitzpatrick

Do you suppose you could rephrase your questions? It's hard to figure out what you're after.

#### 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

#### thkking

Bill Fitzpatrick said:
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
sss said:
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?

#### sss

i think u are confused here
can u tell us what are u trying to do ? that equation is also worthless in that case because the load is changing

#### Bill Fitzpatrick

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.

#### thkking

Yes, I understand everything you said but, I'm not looking for a definition of "how a capacitor works", no offense.

Let me expand on the question. But first, I will reiterate that I understand when and how a capacitor discharges into a circuit, I understand power factors, how parallel resistance is in effect the leakage, etc, etc, etc...
A point to expand upon before I explain.
Originally posted by Bill Fitzpatrick 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
I know this, but this voltage fluxing is constantly occurring while a car is running - you mentioned this, I know, since most car's have electronics installed which are constantly drawing various currents depending on their operations. Amplifiers are one very good example, computer operated electronics, engine electronics, etc... not to mention cable losses causing resistive losses- I know you mentioned this. So I'm looking for a real world answer.
**This is very important (assuming that voltage/current is constantly fluxing), because this directly relates to the load resistance.
Now in your first paragraph you basically summarized what I am trying to use for my question, but also add the fact that car stereo's are constantly on, stock or A/M, people has A/M electronics in their car, etc.
So, my question is, how would one go about finding the load resistance, taking a real world approach at it. Given that the circuit voltage is going to be fluxing, causing the capacitor's energy to discharge/charge. Now I understand that I would have to take basically an instantaneous measurement. But to what extent should I go in determining the load resistance? Or is it as simple as determining the current at any given instant (during capacitor discharge during a voltage drop) and the related voltage, then calculating resistance based on OHM's law? Or is there another method of finding this, because this seems too easy...
SO: sss, yes, I know the load is constantly changing, but I'm questioning how to find the load resistance at any given point in time during a capacitor discharge.....lol I hope thats expanded enough to get an answer.

#### Lligior

I started a post finished it and realized it probably wasn't relevant. So I'm taking another stab at it.

The "resistance" of the cap should be put in series with the cap, ie added to the load. But this number is always given at a certain frequency. Given the fact that the power flows out of the cap based on the switching frequency of the amps, the demands of the sound being produced, and the charge state of the cap this number would be useless. So it is effectively impossible to calculate and would be difficult to computer model.

disregard what follows.

For one cap alone (not connected to any other electronics)

The number you want is called "DC leakage current" it can be found on the datasheet. It is dependant on the capacitance and voltage present in the capacitor. Real electronics have datasheets than will give this information, I'm not sure about car supercaps.

for example:
dc leakage current given as .06CV + 10uA after 2mins
cap is 10000uF@14.4V

this is almost zero

If you want this in resistance terms:
R=V/I=14.4/.000011=1.3M
note however that as the voltage goes down the aparent value of this "resistor" goes up

#### inhale

Bill Fitzpatrick said:
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.

thats not true, capacitors are ment to hold charge. you could severly burn your self. now im not sure, but you could discharge a CRT monitor tube with a screwdriver, so to discharge a capicator, connect it to a grounded wire? shouldnt that be correct?

#### sss

thkking said:
SO: sss, yes, I know the load is constantly changing, but I'm questioning how to find the load resistance at any given point in time during a capacitor discharge.....lol I hope thats expanded enough to get an answer.

ok lol , there is a thing called wattmeter , charge your cap to lets say 12V ,disconnect the car battery , connect the wattmeter and the cap to the cars electrical circuit . now u will see the power taken by your cars electrical circuit , u can play music in your car and see how its changing , its gonna last a few secunds or less , depends on the cap lol . also connect a voltmeter to the cap to messure the voltage on it , in the end .... R=(U^2/P)

now can u plz tell us what exactly are u trying to do?
thats car audio , dont try to make it a rocket science

#### johnnyx

Rather than trying to deal with load resistance, the following is probably more useful;-

Voltage change = Current X Time / Capacitance.

Although it still isn't clear what you are trying to do, in that, once you find the effective resistance at any given point in time, what are you going to do with it?

#### Bill Fitzpatrick

thkking said:
but I'm questioning how to find the load resistance at any given point in time during a capacitor discharge.....lol I hope thats expanded enough to get an answer.

I'm not sure I have an answer. An insturmentation engineer would.

The first thing that popped into my mind was to use a current probe and a circuit ( digital? analog? four quadrant divider chip?) to divide the voltage by the current.

You haven't said whether you want a permanently installed system to monitor activity and perform some action or and external system to make some measurements and then be disconnected.

#### thkking

Lligior said:

The "resistance" of the cap should be put in series with the cap, ie added to the load. But this number is always given at a certain frequency. Given the fact that the power flows out of the cap based on the switching frequency of the amps, the demands of the sound being produced, and the charge state of the cap this number would be useless. So it is effectively impossible to calculate and would be difficult to computer model.
Good point, thanks.

sss said:

now can u plz tell us what exactly are u trying to do?
thats car audio , dont try to make it a rocket science
LOL
I was preparing myself to study material composition of capacitors and how the various materials affect the characteristics of capacitors, in particular in a replicated 12V car environment. Why? Because I think it's interesting how capacitors interact with rectified A/C, and I'm interested in the findings. + I'm interested in applying my findings to an actual car's circuit. Whether or not this question applies to my purpose, I just was having trouble identifying the answer.

johnnyx said:

Voltage change = Current X Time / Capacitance.

Bill Fitzpatrick said:

I'm not sure I have an answer. An insturmentation engineer would.

The first thing that popped into my mind was to use a current probe and a circuit ( digital? analog? four quadrant divider chip?) to divide the voltage by the current.

You haven't said whether you want a permanently installed system to monitor activity and perform some action or and external system to make some measurements and then be disconnected.
External & portable: Laptop w/software , voltage/current probes, o-scope. Only the computer is mine

Thanks though everyone, you've got my thoughts more focused now. I think I would have to use the load resistance as an unknown and solve for the known. Also, Bill, I was thinking about using a 4 quad analog multiplier test circuit, but I want to have a realistic model and be able to compare my results to results in the actual simulated environment. I think I've kind of answered my own question and I've benefited from all your inputs.
Thanks
Grant

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