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Diode

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hupsenfg

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upload_2017-10-7_11-17-18.png


upload_2017-10-7_11-17-4.png


Is my answer correct?
 
Think about what the current through the diode is doing, What happens when it's not conducting?
 
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
 
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.
 
The voltage source sucks. it's not [latex]12sin(\omega t)[/latex]). 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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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.
 
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.
 
The voltage source sucks. it's not [latex]12sin(\omega t)[/latex]). 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.

Good link John - worth reading through (well, for me anyhow!)
 
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?
 
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.
hupsenfg.JPG

Ratch
 
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