A little simple question

124

02:16:43

Yesterday i had an exam and there was that circuit at the top (with an ideal diode), i couldn't figure out the output waveform, my problem of understanding that circuit simplifies to the bottom circuit, so here is my question, why the right side of the capacitor has 5V? i knew that if it was a resistor it would be 5V like the lower part of the circuit, but actually i don't know the reason for that either i just knew that it was like that :(

I'm trying to fill up the learning holes that i have, i'm not actually that noob

published 7 years ago

selman

7 years ago

We'll the original circuit looks like a clamper (DC offset adder). with the exception of the DC source, was it there in the exam ???

BillyT

7 years ago

The diode does not allow the capacitor to discharge to ground, so the charge across the capacitor Is basically open circuit to th negative and will balance out to what the diode will allow.

fancycontrollers

7 years ago

At start the cap is discharged and so it's short circuit, hence the 5V on the right side. Ones closing the top switch, this will become 0V and stay that after switching off again. Try putting a switch over the cap to discharge it and after doing so, open the switch again and you will find the 5V again.

hunar

7 years ago

(selman) yes it was there, luckely for me there was 8 questions and we only had to answer 5 😂 so I skipped this and answered another one asking (Oscilloscope measures voltage, how to measure current with it) that one was easy :D

hunar

7 years ago

(BillyT)(fancycontrollers) thanks :)

selman

7 years ago

So in that case the output waveform will be exactly the same as the voltage at the top of rail.

selman

7 years ago

Frankly i would choose this one, but how do you measure current with oscilloscope, what did you answer ???

hunar

7 years ago

It's the same as creating Ameter with galvanometer, you connect the terminals of the Oscilloscope to a small resistor and place it at a point were a Ameter is used, I'll quickly make a circuit now

hunar

7 years ago

Here it is http://everycircuit.com/circuit/6491559058210816

selman

7 years ago

Introducing a resistance to the circuit... But that may disturb the circuit giving you wrong results. I would go with hall effect sensors.

hunar

7 years ago

Yes but Ammeters also introduce a resistor to the circuit

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