Electronic – Vdc in ripple factor

capacitorfilterrectifierripplevoltage

V_p(rect) is the rectified unfiltered peak voltage

V_r(pp) is the peak-to-peak voltage after filtering by a capacitor

Here is a picture:

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V_r(pp) is given by the expression:

enter image description here

Which I have no problem with its derivation.

Now, as you can see with the first picture, the book shows that V_dc = V_p(rect) – 1/2 V_r(pp)

Why is that?

When he derives an expression for V_dc , this is what he does:

enter image description here

In short, why is V_dc the difference between the peak rectified unfiltered voltage and the peak-to-peak filtered voltage?

Here are also the circuit diagrams:

enter image description here

Best Answer

In short, why is V_dc the difference between the peak rectified unfiltered voltage and the peak-to-peak filtered voltage?

That's not what $$ V_{dc} = V_{p(rect)} - \frac {1}{2} V_{r(pp)} $$

is saying. It is saying that "Vdc is the peak rectified voltage and the peak minus half the ripple". This is clear from the graph.

In fact, and it is visible in the graph as drawn, this is an approximation. The area between the blue line and the average when above the blue line is greater than when below.

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