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What is the formula which decides the capacitance of the Filtering Capacitor.

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msivabalaji

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What is the formula which decides the capacitance of the Filtering Capacitor in a simple power supply.
How to decide the input filtering capacitor for a regulator, and the output filtering capacitor for the regulator.
 
Here's the simplist formula I know:

[latex] C = \frac{10I}{V}[/latex]

I = maximum DC current drawn (mA)
V= maximum ripple voltage across the capacitor.
C = capacitance (µF)

It's valid for a full wave rectifier at 50 to 60Hz only.

If you're using a bridge rectifier, just double the capacitor.
 
Here's the simplist formula I know:

[latex] C = \frac{10I}{V}[/latex]

I = maximum DC current drawn (mA)
V= maximum ripple voltage across the capacitor.
C = capacitance (µF)

It's valid for a full wave rectifier at 50 to 60Hz only.

If you're using a bridge rectifier, just double the capacitor.

Hi Hero,

Isn't the general formula for a Bridge rectifier(at least) supposed to be

C=I ΔT/ΔV

Where C is value of the capacitor,
ΔV the maximum value of the allowed ripple voltage.
ΔI is the maximum current drawn.
ΔT is the 1/2 the period of the AC input waveform.

This was for the Filer Capacitor.

End quote answer. :)

For the Input capacitor, i think you are asking for a low pass or high or a low pass filer on the AC input,

For each of them use this formula for the 3db frequency.

F3db=1/(2ΠRC)

It is not usually needed since you take input from mains.
 
Hi Hero,

Isn't the general formula for a Bridge rectifier(at least) supposed to be

C=I ΔT/ΔV

Where C is value of the capacitor,
ΔV the maximum value of the allowed ripple voltage.
ΔI is the maximum current drawn.
ΔT is the 1/2 the period of the AC input waveform.

This was for the Filer Capacitor.

End quote answer. :)
Well my formula was derived from the one you posted with the frequency fixed at 50Hz, the current is in mA and capacitance in µF but in reality it's a little more complicated

Δt is actually shorter than the half mains cycle because the capacitor finishes discharging and starts charging before half the mains cycle has finished. In reality this doesn't matter because the formula always oversizes capacitors a bit which is fine.

Here's a formula which takes the time the capacitor is discharging for into account.
[latex]C = \frac{I}{V_R} \times \left(\frac{1}{4f} + \frac{arcsin{\left(\frac{V_{IN}-V_R}{V_{IN}} \right)}}{2\pi f} \right)\\
C=\text{Minimum capacitance required}\\
V_R= \text{Maximum ripple}\\
V_{IN}= \text{Input voltage, excluding diode losses}\\
I=\text{Current drawn by regulator}\\
f= \text{Frequency}\\
[/latex]

Note that it still isn't 100% accurate because I've not put the diode losses in or accounted for the effect of the transformer's series resistance and leakage inductance.

As you can see, it gets more complicated as more things are factored in, so much so that it isn't worth it.
 
The actual value of large filter capacitors have a tolerance of plus and minus 20%. Some are plus 100%, minus 50%. So accuracy is not possible.
 
It's valid for a full wave rectifier at 50 to 60Hz only.

If you're using a bridge rectifier, just double the capacitor.

To avoid confusion, it might be worthwhile explaining the distinction between a full wave rectifier and bridge rectifier with regards to your formula.
 
I made a mistake, a bridge rectifier is a full waver rectifier.

The formula is for a full wave rectifier I meant to say you'll need to double the capacitor value for a half wave rectifier.
 
Thanks for all your replies, But I got 3 different formulae from each of you, which one to follow. My application circuit is at the electronic speedometer in an automotive(both 2wheeler and 4wheeler). Which would be the apt formula to use?
Please reply...............

Thankyou
 
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