Hi all,
I can't get how to compute the value of C to have let's say a 10% Voltage Ripple.
This is the example on what i'm studying:

So V1 is a 4 V peak to peak sinusoidal wave at 1kHz.
I have to add a capacitor in parallalel to R1 to get a 10% Ripple Voltage
I wanted to use the formula:
\( V_{rpp} = \frac {I_{load}}{fC} \)

From simulation (without capacitor) I got that \(V_{o max} = 1.351V\) and \(I_{R1} = 600 uA\)
I was wondering if the Iload in the formula is this one or not. And ripple voltage should be 10% of 1.351? Or 10% of Vdc?

 

Hi all,
I can't get how to compute the value of C to have let's say a 10% Voltage Ripple.
This is the example on what i'm studying:

So V1 is a 4 V peak to peak sinusoidal wave at 1kHz.
I have to add a capacitor in parallalel to R1 to get a 10% Ripple Voltage
I wanted to use the formula:
\( V_{rpp} = \frac {I_{load}}{fC} \)

From simulation (without capacitor) I got that \(V_{o max} = 1.351V\) and \(I_{R1} = 600 uA\)
I was wondering if the Iload in the formula is this one or not. And ripple voltage should be 10% of 1.351? Or 10% of Vdc?

Ripple is usually expressed as a percentage of the average DC voltage.

 

Hi,
The ripple is normally measured from the Vpk appearing on the smoothing cap and the point where the next half cycle voltage starts to recharge the cap.
E

 

Below is a graph of a full wave rectified sin wave. The dotted line shows the voltage at the output without a capacitor and the red line shows with a capacitor. As you can tell the capacitor is charged each time the AC voltage reaches highes(or lows in full wave rectification) on half wave the voltage will remain at 0 50% of the time.

V-Ripple is the ammount of voltage fluctuation on the capacitor(red line) from peak to lowest point.

 

Ripple is usually expressed as a percentage of the average DC voltage.

Oops. It should be expressed based on the peak DC voltage. With small ripple, the peak and average are nearly the same, so using either gives you nearly the same percentage. But it makes more sense to calculate the minimum voltage as a percentage of the peak.