# Circuits with diode

1. Feb 23, 2013

### Clandry

I have attached the problem along with the answers.

I have a question on both of the questions.

For the first question, if the diode is ideal, does that mean it's automatically "forward biased?"
The problem statement says "assume the diode is ideal (i.e. has a 0V forward bias voltage), but I don't think that's necessarily stating it's forward biased, is it?
A forward biased diode means that current is flowing from left to right (in this figure). The voltage to the left of the diode must be greater than the voltage to the right of the diode.
My question is, the solutions assumed it's an ideal diode, but I don't understand how they could assume that without figuring if the voltage to the left of the diode is greater than the right.

For the second question, when a capacitor just starts charging, it pretty much acts like a short right? And an inductor would act like a open.

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2. Feb 23, 2013

### ehild

No. It means that the voltage across it is zero when it is "open". You can substitute an open diode with zero resistance. (A closed one means infinite resistance)

No, it does not mean that.

It is the best if you start from the beginning, when the capacitor is uncharged, and there is zero voltage on it. The right-hand side of the diode is positive with respect to the left hand side. If it is an ideal diode, its resistance is infinite when reverse biased. All current flows into the capacitor. So the capacitor is charged, its voltage increases till 2 V, what happens then with the diode? .What is the steady state voltage across the capacitor?

Yes.

ehild

3. Feb 23, 2013

### Clandry

You mean when the voltage across it is zero when it is a "short/closed" right?
An open diode should have infinite resistance and a closed diode should have 0 resistance (no voltage drop), I think.

How come the capacitor's voltage increases till 2V?

4. Feb 23, 2013

### Staff: Mentor

Both of you are right. Unfortunately. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon9.gif [Broken]

There is a huge ambiguity associated with the terms "open" and "closed", and I think the vernacular should be strenuously avoided, for reasons evidenced here.

ehild was using "open" in the sense that an open gate allows free passage, and a closed gate blocks passage. This usage appears at odds with the electronics conventions "open circuit" and "closed circuit".

Confusion reigns!!

Last edited by a moderator: May 6, 2017
5. Feb 23, 2013

### ehild

Oh, I learnt it just the opposite way. Call it as you learnt: forward biased means a closed switch and reverse biased means an open switch.

If the voltage across the capacitor gets a bit greater then the left hand side of the diode would be more positive than the right hand side. What happens to the diode then?

ehild

6. Feb 23, 2013

### Staff: Mentor

A silicon diode conducts when it is forward biased by about 0.6 volts. (It even conducts when the voltage is less than this, but it doesn't conduct well.) A germanium diode conducts when it is forward biased by about 0.3 volts, but if you allow it much current the voltage will rise to 0.5 or more and it can get warm or hot so might need a heatsink.

An ideal diode conducts when it is forward biased by 0.00 volts. Even when you allow it a heavy current, the voltage across it remains at 0.00 volts and it doesn't even get warm because it has zero power losses. Questions where an ideal diode is specified do so in the interests of simplicity.

It's a shame that no manufacturer produces these ideal diodes, they I'm certain they would sell like hot cakes.

7. Feb 23, 2013

### Clandry

Thanks for the replies.
That means the diode will act as a short.
In steady state, the capacitor would behave as a gap. I still don't quite understand how you determined the left hand side of the diode is greater than the right. I guess my question comes down to: why can't the diode be reverse biased at steady state?
I know the capacitor becomes a gap, but I don't quite see why that would mean the left hand side of the diode's voltage is greater than the right.

8. Feb 23, 2013

### ehild

If the diode is reverse biased you can eliminate the right-hand side of the network (diode and 2 V source). The 6V source charges the capacitor through the resistor. What will be the steady-state voltage of the capacitor?

ehild

9. Feb 23, 2013

### Clandry

It would be VA=5V across the capacitor since no current will be flowing through this circuit at steady state.

Is this the reason why the left hand side of the diode's voltage is greater than the right? Thus, creating a forward biased diode.

10. Feb 23, 2013

### ehild

Yes. But the diode gets forward biased as soon as the capacitor voltage excesses 2 V.

ehild