calculating what capacitor to use.

what is the formula for calculating which capacitor to use with your circuit?

i totally forget what it is, and some how im failing at finding it via google search.

It really depends upon what the function of the capacitor is susposed to do.
We need more details.
Is a filter for a power supply?
A rf bypass?
For a resonate circuit?
Etc.

i just need it to hold a 5V charge for ~4ms.

i have a micro switching between several leds at a time, and im trying to make them brighter.

it switches so fast that the leds dont reach full brightness, so they look dimmer.

If it's an LED matrix the caps may make matters worse. Increasing the current or lowering the duty cycle might help.

if they are all on at the same time, put them in parallel, and cut the current limiting resistor to 1/3 of the previous value

Berserk87 said:

i just need it to hold a 5V charge for ~4ms.

Click to expand...

I = cΔV/Δt
Δt = .004
If you can stand a 5% droop in the 5v, then Δv = 0.25
So you only need to spec' I to find C.

Not politically correct, but the way I remember it is "Civil Engineers are queer", i. e. C*ΔV=Q, or the the charge stored in the capacitor Q Coulombs = the capacitance C Farads times the change ΔV in Volts as the capacitor is either charged or discharged.

You also need to know that an Ampere of current is a Coulomb of charge per second, I=Q/Δt.

Substituting and rearranging gives:

Q=I*Δt and Q= C*ΔV, so I*Δt=C*ΔV

or C = I*Δt/ΔV

So to only allow a 1V drop (20% if 5V) while delivering 100mA for 4ms, you would need 0.1*0.004/1 = 4e-4 = 400e-6 = 400uF.

the above calculations assume that the output resistance of whatever you are charging the capacitor with is low enough to bring the voltage back up in a reasonable amount of time.

another way of thinking about it is if it is discharging (supplying current) for 100% of the time, and charging for 1/3 of the time, the current supplied by whatever is charging has to be 3 times the LED current.

It is not a good idea to hook a microcontroller directly up to a capacitive load like this.