The One Upper
you guys flame too much.
Mostly I hear of people using a cap to stop their lights from dimming when they drive with their sub(s) on pretty loud. That would make sense because the only charge from the cap you would need was that to make up for the power draw from your amps/subs when they hit. The cap takes some blow from your car's electrical system for that extra half second or so when the amp(s) need that extra power. You only need as much exra charge to power your headlights/whatever for the difference in power consumption there is from dimmed lights to fully functional lights.
Main point I'm trying to make: Less than half a second for cap to charge - half-second extra amp draw = full power everywhere else in your car.
Makes sense to me... I could be totally wrong though
Nice post wexdafid
Why a Capacitor is NOT useless (corrections need to be made to the FAQ-Why a CAP is useless!!!! thread)
I am writing these because of all the incorrect post that have been posted to the Why a CAP is useless!!! Thread. The thread appears to have been brought to and end by Urainium with some people agreeing with what he wrote. While the post by Urainium is very insightful with lots of equations and calculations it is also incorrect.
The problem with Urainiums example is that he has made a very bad error based on an extremely poor assumption that any engineer with his education background should have realized. While the formulas and equations themselves are correct he poorly assumed the resistance of the wire between the cap and the battery being to high. He as also incorrectly assumed the amplifiers resistance. These misassumptions threw off his RC calculations therefore ruining any supporting evidence.
Your wire from the battery to the cap has a total resistance of 1 ohm Urainium
If the resistance is 1 ohm your amplifier would not even work because there would be too much voltage drop across the wire. An amplifier is capable of drawing anywhere from say 1 amp to 100s of amps. If the wire were 1 ohm then by I=V/R you could only have a maximum total current draw of 12 amps. 12 amps X 12 volts means we could only have at max a 144 watt amplifier. This is just not going to happen. The actually resistance in the wire between your battery and amplifier is many times smaller than 1 ohm. This is why your capacitor example is flawed and I will post corrected charge and recharge times for you later.
According to Introductory Circuit Analysis Seventh Edition by Boylestad, if we were using a 10 AWG# wire the resistance per 1000 ft. would be .9989 ohms. Lets say we are using 15ft. of wire from battery to Cap. That comes to .015 Ohms for 15 ft. of wire. Now I did not calculate the wire from the negative of the battery to the cap because as we all know it is grounded as soon as possible directly to the car. So for this example we will use 15ft. of wire.
Lets not get the capacitor charge/discharge calculations just yet. We have one more major error to fix.
When we discharge, the R is 4 ohm (for the speaker load). Urainium
If we assume that we have 1 amplifier with 1 speaker attached then by I=V/R the most we could apply to this speaker (assuming a 100% efficient amplifier) is 3 amps. Well we all know that our amplifiers put out well more than three amps to our speakers so I will give us an estimated resistance of the amplifier to the battery by using known values and ohms law. Lets say that we have an amplifier that outputs 500 Watts to the speakers and the amplifier is 70% efficient.
Output Watt / % efficiency = Total Watts consumed by amps and speakers
500 Watts / .70 = 714.28 Watts (lets keep it simple and say 714)
Now lets find the resistance of the amplifier
Watts / Voltage = Amps
714 Watts / 12 Volts = 59.5 Amps
V / I = R
12 Volts / 59.5 Amps = .20 Ohms
Now we can do the RC circuits.
Charging the Cap
Lets say we are using a 1 Farad cap. 5(time constants) * R * C = Time to charge in seconds. This comes to .075 seconds to fully charge a cap. Now I can tell you there is a problem here in that the calculated current draw at first is well more than what the battery could supply. So this .075s is not the actually amount of time but it is MUCH faster than the time you would get using Urainiums incorrect model.
Im not going to go into the rest of the charging and discharging of the capacitor at this point since I have already busted the model that for some reason this board has accepted to be correct. It will take a little longer for the cap to charge from 12 to 14 (about .045 sec) than to discharge from 14 to 12 (about .03 sec if the cap is the sole source, will probably last about .06 seconds when working together with the charging system). But at this point you need to take into account that music is transient and will be changing its current draw continuously so it would be interesting to see a chart of capacitor charging vs its discharging taking into account the ever changing current demand on the system
Why a capacitor is not always useless
I can speak from experience that some lower quality amplifiers will greatly benefit from the addition of a properly installed capacitor. I have also found that higher quality amps really dont gain any improved sound quality from a capacitor.
By no means is a capacitor worthless and it does not increase the load on the charging system anymore than if it wasnt there. A capacitor is not going to act as a band-aid for a weak charging system in anyway but it will act to stabilize voltage in any system. Its sole purpose is to oppose changing voltage.
If you think that a capacitor is totally useless then I invite you to take out the large stiffing caps that are inside your amps. Removing these caps will reduce performance and sound quality of your amplifiers. Adding more capacitance may help in some situations but never hurt. Before deciding if a cap will not help you should try one on a trail basis from a friend and see if it helps any. If it doesnt then you know you dont need one.
Bachelor of Science in Electrical Engineering from Purdue
What's worse, is that in order for a 1 Farad cap to discharge, first the alternator output must have maxed out, and the voltage must have dropped around 1.5 volts. But I thought a cap was supposed to prevent that (voltage drop)!!!!!????? Yep, you got the point.