teknofix wrote:
I am about to install an 8 Farad Capacitor parallel to Battery to improve my Audio system performance |
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This is yor first mistake. If you have not addresed the power source in your car (two guesses, and the first seven don't count), you are wasting your time and energy, thinking this is going to "fix" or "help" anything.
teknofix wrote:
I have been reading forums about how to install an Farad Capacitor, everybody is talking about charging the capacitor thro' the 12v bulb an installing as close to the Amplifier. That's all fine. |
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First, it's a capacitor. JUST capacitor. Not a farad capacitor. Farad is the capacitance value.
teknofix wrote:
The purpose of the Farad Capacitor is, smooth, bump-up and hold the stored voltage, they are rated up to 24 Volts. |
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Wrong. The purpose of a capacitor when installed in a system with an inadequate charging source (have you figured it out yet?) is to pad the profit margin of the dealer selling it to unsuspecting, uninformed and undereducated consumers.
teknofix wrote:
Ok. lets say Battery voltage is : +12V before entering the Farad Capacitor, and it is bumped up to +16 Volts by the Capacitor, so there is a 4 Volt difference between Battery and the Capacitor, nobody mentions how to isolate the voltage difference between the batttery and the Capacitor, another words preventing bumped-up voltage by Capacitor going back to battery but intsted only going to amp. |
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Wrong again. The capacitor cannot output more voltage than it is connected to. If you are connecting it to a 12V source, that is the voltage you will read across it's terminals.
teknofix wrote:
I think, This can be done by simply placing a high wattage rating Diode between the battery and Capacitor on the + wire , hence Diode conducts only one way, Voltage stored on the Capacitor will only go to the amplifier, and it will be blocked to go back to battery, but the battery will supply to the Capacitor.Anybody agree or disagree with me or any suggestions how to isolate the mentioned voltage difference? |
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I disagree. First, diodes are rated by current @ a given voltage, not "wattage". Second, knowing this now, any diode of sufficient current rating to pull amplifier current demands through, will be HUGE, and will probably require you to mount it securely to the chassis of the car, (as the heatsink) in
literally a ½" hole. I might also mention to you the expense of said diode! (Expect $100 or more...!)
teknofix wrote:
What happens if I place Diode between Batt. and Cap. as shown below : ?
Batt = + 13.8 V ---------------I>I-------------I I-------------> +V to Amp
Diode Farad Cap |
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Nothing. The cap doesn't go in series as you have indicated in your diagram. It's a parallel connected device.
teknofix wrote:
I know there will be 0.6V drop across the Diode, but what Will be the final Voltage
at the Cap + ( using 8 Farad Cap ) ? A = 13.2V B = 14.4 V C = 14.6 V D = None ? |
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First, it's .7V drop for silicon diodes, and about .3V for germanium diodes. (Not really sure who told you it was .6V...) So, in answer to your question, AND based on the illustrated voltage of 13.8V, your cap will be charged to 13.1 volts, using the most common diode material - silicon. It will be around 13.5 volts using the less common, and more expensive germanium. Again, using your diagram, what will be the voltage at your amplifier? Zero. Your amp will not even turn on.
teknofix wrote:
I will let you know, after I experiment to find correct answer (awaiting ordered 8 Farad Cap) |
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You already HAVE the correct answer: You are $200 poorer (give or take), will see little to absolutely zero benefit for that $200 loss, AND have still not addressed the POWER SOURCE in your car. (Have you guessed it yet?)
It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."