# Capacitor bank circuit

Discussion in 'General Electronics Chat' started by AmitC, Feb 7, 2015.

1. ### AmitC Thread Starter New Member

Feb 7, 2015
2
0
Dear All,

Hi..
I have one requirement. I am using dc motor in one of my project. The controller is PLC based. I want to make one power retentive circuit which will provide power to PLC, Motor and H-Bridge circuit even after mains supply is off for at least 5 sec. Can you please help me to build this circuit. The details are-
Supply voltage : 24 vdc
current: upto 4 Amp
I am not getting how much capacitance or charge is required.

Oct 2, 2009
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3. ### wayneh Expert

Sep 9, 2010
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In case it wasn't clear, a battery is a far better solution than a capacitor. A battery is cheaper and smaller for a given amount of energy, and it gives up its energy at a nearly constant voltage. The capacitor drops voltages as it discharges.

Four amps for 5 seconds is 20A-sec or 20 coulombs. If you could live with a 2V drop while the capacitor discharges, that means you'd need a capacitor with capacity in Farads = coulombs per volt, C = Q/V = 20 coulombs/2V = 2F. Limiting the ∆V to one volt would require 4F.

 Oops, how did I screw it up so badly? I mean:
C = Q/V = 20 coulombs/2V = 10F. Limiting the ∆V to one volt would require 20F.

Last edited: Feb 7, 2015
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4. ### AmitC Thread Starter New Member

Feb 7, 2015
2
0

I think 20F is very high and not possible practically.

I tried with formula: v(t)=V0 * e^(-t/RC)
Here I used, v(t)= Voltage across capacitor (about 22vdc)
V0=Applied voltage (24vdc)
t=discharge time (5 sec.)
R=load resistance (I am not sure about it. I calculated the resistances of PLC, H-bridge ckt & Motor in series. near about 45k)
C= capacitance

This formula gives some capacitance value but it doesnt work...

Please tell me which formula I should use?