Capacitor bank circuit

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

  1. AmitC

    Thread Starter New Member

    Feb 7, 2015
    Dear All,

    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.
    Awaiting for your reply....
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
  3. wayneh


    Sep 9, 2010
    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.

    [edit] 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
    planeguy67 likes this.
  4. AmitC

    Thread Starter New Member

    Feb 7, 2015
    Thanks for your reply..

    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?