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Explain me the output of a diode in series with a capacitor

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jetsam

Newbie level 3
clamping diodes

Can anyone please explain the output of a diode in series with a capacitor?

/ File deleted, look in next post by jetsam for schematic.

You have to be more specific.
Where is the input and where is the output in your schematic?
What is the input signal? Sine or squarewawe or other?
I would recommend you use a spice simulator to simulate the circuit, then you can watch the signals yourself.

voltge detector

It looks more like a peak voltage detector than a clamp.

This circuit is nonsense. No generator! No input! Unknown output!

sorry for the confusion. i'm a newbie so bear with me.

anyway, here is the corrected one. I just want to know the output waveform at the instant the switch closes. I know a capacitor acts as an open circuit in a DC ckt.

diode direction

The diode is reverse biased when the switch is closed and so with an ideal diode you get the full 10 V on the output. (If the diode were not ideal, you would get the 10 V minus the diode leakage current drawn through the resistor.) If the diode was the other direction, the output would be the diode forrward drop plus the voltage on the capacitor. The voltage would rise to 10 V with a time constant of RC and you would end up with all 10 V on the capacitor because the diode forward drop at low currents approaching zero is itself nearly zero.

Even if you turn around the diode I don't see the point of placing the diode in this RC circuit.
What kind of circuit are you trying to make?
The circuit shown makes no sense.

I didn't design it. I saw it in Electronics Engineer's Handbook. I just can't understand the ouput it showed. That's why I'm asking for a second opinion.

Btw, If you want to check it, the circuit is under the Passive Waveform Generators under section 16-13.

The handbook is kinda old. I saw it in the library. It's copyright is 1975

poor circuit

This circuit will follow the graph you show only when the capacitor has a voltage on it and only for a short time because the voltage will change when the capacitor is discharged. A better form of this circuit is to have a DC source where the capacitor is now.

clamping of ST

clamping of ST

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