What is the formula or how would I go about sizing a bank of capacitors to prevent/lessen voltage dip on a power supply bus during a load fault?

I have a DC power supply that feeds 2 different circuits. On one circuit (circuit 1) I want to include a bank of capacitors across the input to help maintain the power supply bus voltage during a load fault on this circuit.

Basically, I don't want any load faults on circuit 1 to effect circuit 2 in any way if possible.

To give some numbers, lets say power supply voltage is 24V and possible load current on circuit 1 is 100 Amps.

 

Is the fault current momentary, or likely to be sustained for long periods before you realise a fault has occurred?
If you have a 100A fault current that lasts for several minutes or even seconds then no amount of capacitance is going to help you! (Except maybe infinite capacitance, but you don't see those for sale much anymore...

 

The fault will last until the fuse or circuit breaker trips.
Lets assume 1 second.

 

The fault will last until the fuse or circuit breaker trips.
Lets assume 1 second.

Calculate how much charge is used for the fault. If you do not know how to do this look it up.

 

Calculate how much charge is used for the fault. If you do not know how to do this look it up.

That's why I posted .. I wasn't sure what forumla to use for that..

 

Charge Q = C x V

Hence if the voltage drops by dV, the drop in charge dQ is

dQ = C x dV

Charge Q is also I x t

If you know the current I and the time t, the charge Q = I x t

 

The capacitor formula is:

I(t) = C dv/dt

It is an expression for the current as a function of time and always applies, though it's a differential (calculus) formula. However, it is still valid when I(t) is a constant and you can then use a difference equation of:

I = C (delta V) / (delta T)

It's a good approximation for a small change in voltage. If you solve it for C you get:

C = I * (delta T) / (delta V)

You can read this as when either I or (delta T) increases you need more cap, also if (delta V) decreases you also need more cap.


To put some numbers on this say your load is 100 amps, you need to support this with no more then a 2 volt change for up to 1 second. Then:

C = 100 amps * (1 second) / (2 volts) = 50 Farads.

That be one big cap!

 

Perfect..
Thanks guys