Diode

throbscottle

Well-Known Member
Think about what the current through the diode is doing, What happens when it's not conducting?

ci139

Active Member
i donno . . . i have questions
1. where is the reference/signal ground ??? at the center of sine supply ??? at-schematic-cathode of the sine supply ??? at "Vo" (as it's not V.something_else -- but is Vo ??? a "Zero" voltage ???) ??? e.c.
2. is the sine supply going ±12V or +6 ±6V as there is no marking "±" in front of 12V but there is "+" ???
3. assuming the "•" e.g. the "sine"-cathode is the signal ground (assumes from typical (diode-)clipping schematic) -- the reverse conducting ideal diode´s resistance → ∞ (infinity) and the forward conducting → 0 (Zero) . . . -- so we have to throw dice about the sine supply or provide 2 answeres

throbscottle

Well-Known Member
If the sine signal is centred at 0v I think there would be no signficant signal at Vo, so it seems unlikely. I had assumed it was single sided.

KeepItSimpleStupid

Well-Known Member
The voltage source sucks. it's not $12sin(\omega t)$). We might think of it as 12 V RMS which changes the problem. That's the only 12 that makes sense/ But, in the lab you probably font have a meter that would read 12 for that source -- Not enough frequency response.

Where is the reference? Vo relative to what?

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jpanhalt

Well-Known Member
This explains such clippers: https://electronicspost.com/write-short-notes-on-clipping-circuit-and-clamping-circuit/

For those concerned, the answer to this question is not actually given in that link, but there is a description of how to get there. Agreed that aspects of the question are ambiguous, but I would assume 12VAC mean peak to peak. Other interpretations change the answer quantitatively, but not the approach to it, which should be the intent of the question.

KeepItSimpleStupid

Well-Known Member
12 ac 12 *sqrt(2) * sin (wt) is like 120 VAC which is like ~ 155*sin(wt); There's no reference either.

I'd vote for 12 VAC.

throbscottle

Well-Known Member
The voltage source sucks. it's not $12sin(\omega t)$). We might think of it as 12 V RMS which changes the problem. That's the only 12 that makes sense/ But, in the lab you probably font have a meter that would read 12 for that source -- Not enough frequency response.

Where is the reference? Vo relative to what?
Since the question states it is an ideal diode, I don't think this would ever be in a lab! It probably assumes that the battery - terminal is the reference point. Paper exercise only.

KeepItSimpleStupid

Well-Known Member
Since the question states it is an ideal diode
Check out the "Ideal diode controllers" from Linear Technology.

throbscottle

Well-Known Member
Check out the "Ideal diode controllers" from Linear Technology.
Huh. Clever clogs...

ci139

Active Member
← the circuit is not correct , but the waveform is

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throbscottle

Well-Known Member
I've just realised, the battery is shown with it's + terminal at the bottom. Or is that just me getting it wrong my whole life?

Ratchit

Well-Known Member
No, it is not correct.

It should be OK to give you the correct answer since a month has passed since you asked for help.

Assuming the 12 volts is a RMS value, an ideal diode, and the voltage referenced from the positive side of the battery, which connects with the AC source, the output voltage Vo is clipped at -6 volts as shown on the plot below.

Ratch