# calculating what capacitor to use.

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#### Berserk87

##### New Member
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.

#### k7elp60

##### Active Member
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.

#### Berserk87

##### New Member
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.

#### blueroomelectronics

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

#### FlinxSL

##### New Member
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

#### Willbe

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

#### MikeMl

##### Well-Known Member
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.

#### FlinxSL

##### New Member
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.

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