Diode Output Predictions.

1. Jun 29, 2009

Petrucciowns

This post is similar to the one that I made the other day. There are two problems that I am having trouble understanding. The link is below:
http://img510.imageshack.us/img510/4396/diodeproblem.jpg [Broken]

They are numbered 4-5 (Ignore 3) . Once again I know the predictions are correct , because I did them in class. (This was a while back and I'm taking notes now)

#4 If the AC source was on the positive cycle then the diode would be reverse bias. Wouldn't the diode have a .7 volt drop with the resistor having a 0 volt drop or is it true that the diode takes control of the source voltage?

#5 The same goes here since the sum of the voltages is -5 volts does the voltages of the diode become source, whereas the resistor becomes zero?

I have a feeling this is true, I just always thought a diode stays at a constant +7 or -7 volts.

Last edited by a moderator: May 4, 2017
2. Jun 29, 2009

cepheid

Staff Emeritus
Yes

A diode only has a 0.7 V drop (from anode to cathode) when it is conducting. This diode is not conducting. You can consider it to be like an open circuit (on the positive half-cycle).

I don't quite understand what that means.

3. Jun 29, 2009

cepheid

Staff Emeritus
Sorry, I sort of missed the point with my previous post. Here's the correction.

If you are trying to "predict" the voltages across the output ports (the two dangling wires), and what you have handwritten there are your predictions, then I should point out that the prediction for #4 is not correct. The output would not be a 10 V peak-to-peak sine wave, because, as you have stated, the diode would be forward-biased on the negative half cycle and would conduct. If the diode conducted, then the two output wires would be nearly shorted together (actually they would differ by a diode drop). If we take a diode drop to be 0.7 V, then the anode (ground in this case) would be 0.7 V above the cathode. Since the output voltage is measured from the cathode to the anode, it would be -0.7 V. So my prediction would look something like this:

http://img5.imageshack.us/img5/9799/rect.th.jpg [Broken]

Edit: The process of of using a converting AC to DC by eliminating one of the voltage polarities is known as rectification, and a diode is sometimes called a rectifier for that reason.

Last edited by a moderator: May 4, 2017
4. Jun 29, 2009

cepheid

Staff Emeritus
The voltage across the diode is the same as the voltage across the source, because the diode does not conduct, which implies that there is no current in the circuit, which implies that there are no voltage drops anywhere. If this is what you mean when you say that the diode "takes control" of the source voltage, then okay, sure. IMO that statement doesn't make too much sense, and a better way of saying it would be that the input voltage appears across the output.

Sorry, I'm not understanding this either. "Stays constant" in what sense? And where did the seven volts come from?

5. Jun 29, 2009

Petrucciowns

But, on the positive half cycle the diode is reverse bias so wouldn't it be correct then?

6. Jun 29, 2009

Petrucciowns

yes my wording is a little off that's what I meant.

I meant to put .7 volts and -.7 volts not 7 volts.

7. Jun 29, 2009

cepheid

Staff Emeritus
The point I was making is that it is no longer a sine wave. Think about what 10 V peak-to-peak means. Is it true for the output voltage?

8. Jun 29, 2009

cepheid

Staff Emeritus
Oh okay. What did you mean with the "stays constant" part, by the way?

Basically there is a forward voltage drop of 0.7 V across the terminals of a diode (in the direction of the forward current) when it is conducting. That's it.

9. Jun 29, 2009

Petrucciowns

I just meant that the voltage across the diode does not change with a varying voltage like the voltage of say a resistor does. It stays at a constant .7 , .-7 or source voltage. I guess it does with the later.

Why would it no longer be a sine wave? What if I translated it to peak voltage or RMS?

10. Jun 29, 2009

cepheid

Staff Emeritus
For an ideal diode this is true, because the I vs. V curve is vertical at V = 0.7 V (or whatever the turn on voltage is). A vertical I vs. V curve means that you have constant forward voltage no matter what the forward current is. However, for a real diode (i.e. a real component in real life), this is not the case. The I vs. V curve for a real diode is *exponential.* Granted, that means it is very steep, but it is not vertical. So, the forward voltage drop will change (albeit minutely) as the forward current increases (substantially). Still, it won't change by much, and none of this is relevant to your problem, in which you're considering an ideal diode.

As you've pointed out, an exponential I vs. V curve for a diode is very different from the I vs. V curve for a resistor (which is linear).

For the reason that I explained in my second post. During the negative half-cycle, the diode conducts, meaning that the output terminals are (nearly) shorted together (separated only by a diode drop). So the full source voltage does not develop across the output. Instead, during the negative half cycle, the voltage is clamped to a constant -0.7 V. I showed you a picture of that. Does that look like a sine wave to you? It is not. It has been rectified.