Electronic – Confused with voltage doubler circuit


This is how the book explains the process:

enter image description here

Let me consider the first 90° of the input, the capacitor charges up to the peak voltage less the diode drop of D1, D1 is forward biased and D2 is reverse biased, that's ok, but what I don't understand is, right after 90° to 180°, why isn't D2 forward biased? Shouldn't the cathode of D2 be negative with respect to the anode because of the capacitor's polarity (C1)? Likewise D1 should be reverse biased because of the capacitor, so why isn't current flowing through D2?


Best Answer

Let us analyze this circuit but we will freeze the time. I assumed \$V_p = 10V\$.

For positive input from 0° to 90°, the situation is clear for you D2 is OFF. And will look like this:

enter image description here

Now let us see what will happen between 90° and 180° when \$V_p = 8.5V\$.

enter image description here

As you can see the D2 diode will conduct current because C2 was initially empty.

But this will only happen at startup (when C2 is empty). Not during normal(steady-state) operation.

And your book shows you the "steady-state" situation.

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