T_Vert

Will a cap discharge over time even w/ no load? I am going to use capacitors to prevent brown out during crank. i was going to put a diode between the cap and the ACC hot and the head unit to prevent voltage bleed but will that also allow the cap to retain it's initial charge? Can anyone tell me what the charge time would be to get to full capacity on a .25F capacitor? My concern is that after the car sits for a few days and I go to crank there will not be enough time between ign. on and crank to charge the cap enough to prevent the brown out.

Thanks
Dave

bobeast

Yes capacitors discharge over time. think of there being a resistor across the terminals except that resistor is made of air. Granted of extremely high resistance, but nonetheless there is some leakage. There is also internal leakage to contend with.

Adding a diode will discharge the cap even faster, since there is always a small amount of leakage past the anode. A relay would be a better device to minimize discharge, than would a diode.

T_Vert

Great thanks. I Just need to calculate recharge time of the cap then.

Dave

bobeast

It depends on the size of the resistor through which you charge it, the supply voltage, wire gauge, the initial charge state of the cap, etc.

try this;

http://mustcalculate.com/electronics...=13&c=.25&r=50

This shows that using a 50 Ohm 5 watt resistor with a charge voltage of 12 volts, will get it charged to ~99% (from zero) in roughly 32 seconds.

You can halve the time by halving the resistance but you'l need to double the resistor wattage to 10 Watts. Also note that a capacitor will never reach 100% of charge voltage.

T_Vert

Thanks. I'm confused on the resistor. Would that just be the draw on the cap? I ended up using a 470 uf cap w/ a 1A diode and I get 7 or 8 seconds before it shuts off. It seems I can get all the way down to 7v before it actually kills the radio which is fine with me.

Dave

bobeast

The resistor is just an exercise to see how a specific resistance value affects charge times. You could just as easily use a light bulb.

Without some resistance in the circuit, the cap will act like a short circuit. It would charge the cap almost instantly but would fry the wires and likely the battery. A higher resistance would charge the cap more slowly, but would draw less current. The trick is to find a value that gives you the best compromise.

A diode acts similarly to a one direction resistor. That is to say it presents a very high resistance in one direction (discharge) and a very low resistance in the opposite direction (charge). Same principle. The advantage of the diode is that it allows for a relatively quick charge cycle and a relatively slow discharge cycle. The important thing to note is that a diode will always provide some resistance and so in the absence of input voltage, will completely discharge the cap, even with no other load. If you want to avoid bleeding off any residual charge on the cap when the radio is off, you can consider using an inline relay powered from the accessory circuit. Your cap will still eventually discharge but the chance of it having to go through an entire charge cycle during the next ignition cycle is greatly diminished.

In a real world situation, the resistor represents the sum of the load resistance (your radio in this case) and the resistance of the inline diode.