Correct setup to allow capacitors to supply power

Question

I am looking to fit a 400W, 12V power supply to my car for using the lights and such without draining the battery.

My plan is to have a rotary switch to disconnect the battery and connect the mains->12V power supply in quick succession.

I think I have pretty much solved it apart from 1 final step - using capacitors to maintain my clock and stereo settings in the moment between the two power sources being switched over.

I have bought 4 x 1mF capacitors rated for 50V (so as not to risk damaging them with the voltage) and plan on using them in parallel to provide the momentary power. As I understand it, this should provide 48mA for about 1 second when charged to 12V (and as I will be twisting the rotary switch straight around, this shouldn't be an issue... as far as I understand it).

I would have the capacitor bank connected with the negatives to my car's ground, and the positives directly to the feed to the car (as in, after the switch) so that they are permanently connected to my car.

So now you know my plan, my question is this: would I need to put a resistor in series with the capacitors to limit the current, or would they only provide what current is needed by the clock and stereo in the same way as the battery currently does? I don't want to go firing a massive current down the line if it could damage my electronics!

Thanks in advance guys, and sorry for the long post!

Answer 1

Your car battery is certainly providing enough power to charge the capacitors, but how about the 400W 12V supply? So, switching from the car battery to the 400W 12V supply very slowly (as in switch being long amount of time in intermediate state where nothing is connected) could drain the capacitors fully and then put a high load on the 400W 12V supply, thus possibly triggering some kind of overcurrent protection. Good supplies limit the voltage to limit the current, but bad supplies may just turn off, which you don't want to happen.

Since you are planning to have only 48mA, a 5 Ohm resistor would lose 0.048 A * 5 Ohm = 0.24 V. It would also limit the maximum current to 12 V / 5 Ohm = 2.4 A, and therefore the maximum power to 12 V * 2.4 A = 28.8 W, which your 400W 12V supply can certainly provide.

So, go ahead and put a resistor between the capacitors and the connection from the switch to the car's electrical system. The minimum resistor would be 0.36 Ohm, as 12 V * 12 V / 0.36 Ohm = 400 W, but I would consider higher values between 0.5 Ohm and 5 Ohm.

Also do note that the capacitors may already have some internal resistance. I don't however believe it would exceed the required 0.36 Ohm.

Also, your understanding of providing 48 mA for one second is incorrect. The formula is It = Q = CU, where U is the acceptable voltage loss. So if U is 12 volts (you accept it to fully drain), the current is 0.004 F * 12 V / 1 s = 0.048 A = 48 mA. However, you don't accept a 12 V voltage loss. You accept only discharge from about 12 volts to about 11 volts, otherwise there may not be enough voltage for the radio. So, with 1 V accepted voltage loss, it would provide only 4 mA!

You need to measure your radio to see if the provided 4 mA is enough to maintain the presets. However, most switches would probably occur in less than a second, so perhaps your current budget could be higher.

Answer 2

use a switch that temporarily connects to both as you switch.

Answer 3

You may be better off considering a second battery and a “split-charge” system : these are used for many apllications such as caravans, vehicles with electric winches etc.