# Voltage across diode

If the diodes are ideal, there is no reverse leakage current, & no forward voltage drop. If the ac source is an ideal current source, the voltage across the diode pair is always zero, since 1 of the 2 diodes is forward biased.

If the ac source is an ideal voltage source, you get a paradox. A zero resistance perfect voltage source is loaded by a perfect diode with zero voltage drop, resulting in the current ramping up towards infinity.

Claude
I want to ask about the bold part. In the case, what is the voltage across the diode pair, zero or voltage source (Vin) when it is in positive half?

meBigGuy
Gold Member
I=E/R and division by zero is undefined. That's the cop-out that I'd use.

Basically, you are asking what happens when you short an ideal voltage source. No need for a diode.

How about an ideal current source in series with an ideal voltage source?

The answer is : sorry, does not compute! Qualifies as a paradox.

Thank you.
Basically, you are asking what happens when you short an ideal voltage source. No need for a diode.
Yup.
Consider a circuit including an ideal voltage source, V and an ideal wire is connected across two ends of the source.
Let's call the voltage between two ends of the ideal wire v.
$$v = \lim_{R \rightarrow 0} i.R = \lim_{R \rightarrow 0} \frac{V}{R} . R = V$$

(by apply L'Hospital's Rule for 0/0)
How about an ideal current source in series with an ideal voltage source?

The answer is : sorry, does not compute! Qualifies as a paradox.
Can you tell me why this is a paradox?

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meBigGuy
Gold Member
Can you tell me why this is a paradox?
No idea how I came to that conclusion. I should go back and edit it out. LOL

Applying limits is the only valid approach for dealing with infinity, but I can't help you with whether this is a valid application of L'Hospital's Rule.

meBigGuy
Gold Member
Actually, thinking about it a bit, I think you nailed it. In the limit it makes sense.

A current source connected to a voltage source is no paradoxat all. One will deliver power to the other. Connect a 1 volt source to a 1 amp source in a loop. The relative polarity determines which which one delivers and which one receives power.
Claude

meBigGuy
Gold Member
didn't I just say that I was wrong? Which doesn't seem to be something you are capable of.