Diode

Discussion in 'Homework Help' started by hupsenfg, Oct 6, 2017.

1. hupsenfgNew Member

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Is my answer correct?

2. throbscottleWell-Known Member

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

3. ci139Active Member

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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

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5. throbscottleWell-Known Member

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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.

6. KeepItSimpleStupidWell-Known MemberMost Helpful Member

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The voltage source sucks. it's not
). 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?

Last edited: Oct 7, 2017
7. jpanhaltWell-Known MemberMost Helpful Member

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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.

• Informative x 1
8. KeepItSimpleStupidWell-Known MemberMost Helpful Member

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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.

9. throbscottleWell-Known Member

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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!)

10. KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Check out the "Ideal diode controllers" from Linear Technology.

11. throbscottleWell-Known Member

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Huh. Clever clogs...

12. ci139Active Member

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← the circuit is not correct , but the waveform is

Last edited: Oct 9, 2017
13. throbscottleWell-Known Member

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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?

14. RatchitWell-Known Member

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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