Capacitor, Reverse bias diode and resistor circuit

Thread Starter

13th Warrior

Joined Oct 30, 2008
I have a circuit with capacitor in series with a parallel reverse bias diode and resistor circuit. Vo is across the parallel circuit. Vi is a square wave. Vγ is 0V. Assuming Vc is 0v initially, I thought the Vc would charge up to Vi then discharge through the resistor. But answer to the book has Vo as 2xVi. Does it add Vi and Vo since the resistor is shorted through the diode?


Joined Aug 10, 2008

I'll get the ball rolling, here, and then people will correct me and you will get your question answered, by the real experts, I'm just a hobbyist so do not take what I share here for final say, but wait until all the replies come in then youll get the real answer.

You probable have a clamping circuit.

if the resistor is high enough then the output will be double the input signal, but offset by around 0.7 volts above or below zero depending on the diode direction.

Basically whats happening is if the diode is pointing with its cathode at the capacitor and it's anode to ground, then

when the signal goes negative with respect to ground the capacitor plate on the diode resistor side charges through the diode after a 0.7v. offset below 0v. ref. so the capacitor now has the full input voltage seen at the output, on the next cycle when the signal input goes positive with respect to ground

then the positive signal plus the already pos. volt. on the cap. plate add together to produce the double of signal voltage on the output.

The diode allows the capacitor to charge up, then the next cycle it is reversed biased so the signal and cap. + volt. add across the resistor.

Now only take this as getting the ball rolling, and wait for the real replies to come in....

Hope this helps...
Last edited:


Joined Aug 10, 2008
Your welcome.

I couldn't open your attrachment, so my guess was just that, not knowing what the schematic looked like.

Thankyou for showing the courtesy to reply back...

Have a great day...


Joined Feb 17, 2009
The schismatic look like this:

And this circuits is very simple "peek voltage detector " or "dc restoration".

And if we assume that time constant is large then input signal frequency then we can start analysis.

If Vin is equal -20V then capacitor is start charging very quickly through diode to Vc=Vin-Vd=19.3V and Vo is voltage on diode Vo=-0.7V.

So know if Vin change to +20V. The voltage on capacitor cannot change suddenly (charged capacitor act very similar as a voltage source, empty capacitor act like short-circuit)
So know Vo = Vin+Vc=39.3V. And capacitor is slowly discharged through resistor until Vin change to -20V and capacitor will be recharged by the diode.