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 This topic has 7 replies, 3 voices, and was last updated 6 years, 10 months ago by BenLee3000.

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January 1, 2013 at 3:32 am #20164firemankurtParticipant
I have a 12v 4A camera system and I want it to switch from AC adapter to 12v Battery in a power failure.
I am planning to use a relay held on by AC power to switch the load to battery when AC goes away.I want to use a capacitor to keep power flowing while relay flops over.How do I figure out the Capacitor size?I thought I would try keep enough capacity for 1/2 second of 12v power at 4A. (relay should switch much faster than 1/2 second).Thanks for any pointers on this.January 1, 2013 at 4:08 am #21119Windell OskayKeymasterCapacitors are not like batteries. They do not maintain a fixed voltage when discharged, but instead decay exponentially. Look at how an “RC” circuit discharges for an example. So, what you really need to know is how far down that curve you can go. Is 11 V an acceptable endpoint? Or is 9 V? Or do you need 11.5 V? Once you figure that out, you should be able to find how much capacitance you need to let you stay at or above that level for 1/2 second.
January 1, 2013 at 12:37 pm #21120firemankurtParticipantOkay, considering that I did not phase the original question correctly.
How about:What is the formula to figure capacitor size when a 2v drop over 500milliseconds is desired on disconnect of a circuit where 4A is being drawn from an original 12v power source?January 1, 2013 at 5:35 pm #21121Windell OskayKeymasterYou can approximate this as an RC circuit. http://en.wikipedia.org/wiki/RC_circuit
If 12 V draws 4 A, then you could pretend that it’s a 3 ohm resistive load.Then V(t) = V_0 exp(t/RC), or V(t) / V_0 = exp(t/RC)Taking the natural log of both sides,ln(V(t)/V_0) = t/RCFor your case, we are solving for t=0.5 s. V(t = 0) = 12 V, and V(t = 0.5 s) = 10 V.Then, ln(10/12) = 0.5 s/RCApproximately given by .182 = 0.5 s/RC, or RC = 0.5 s/ .182 = 2.74Putting back in R = 3 ohms, that gives C = 2.74/3, or about 0.9 F.So, you’ll need roughly a 1 F capacitor rated for at least 12 V, or in practice 1.5 X higher just in case the actual voltage goes higher. It’s actually common to find capacitors in this range (15 F, 1624 V rating), manufactured for use in car stereos, in the range $20$100.January 2, 2013 at 3:59 pm #21122firemankurtParticipantThanks a bunch Widell.
The“If 12 V draws 4 A, then you could pretend that it’s a 3 ohm resistive load.”part was tripping me up when I was trying to use formulas found on internet.Most of this electronic circuitry stuff I can bumble through but obviously much of it still eludes me.June 30, 2014 at 1:11 pm #21123BenLee3000ParticipantGreat answer Widell. I follow all but how you got V(t=5) = 10.
June 30, 2014 at 1:18 pm #21124Windell OskayKeymaster@BenLee3000: @firemankurt wrote “What is the formula to figure capacitor size when a 2v drop over 500milliseconds is desired on disconnect of a circuit where 4A is being drawn from an original 12v power source?”
If the initial voltage was 12 V at time t = 0, and it drops by 2 V in 500 ms, then V(t = 500 ms) = 12 V – 2 V = 10 V.June 30, 2014 at 3:07 pm #21125BenLee3000ParticipantOh wow! Thanks Windell! Look at me over complicating things! Here I was ready to solve a multi variable differential equation! Jk! Look for the simple answer first! Lesson learned! Thanks again!

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