# Homework Help: Photodiode and negative resistance

1. Aug 5, 2014

### ndnbolla

Our professor has asked us why would a photo diode when shined with light, produce no negative resistance.

So far, the only thing I can come up with is because it is in photo conductive orientation (as opposed to photo voltaic) increasing Dark Current thereby eliminating the negative resistance effects.

Can any one clarify this for me?

2. Aug 5, 2014

### Staff: Mentor

Can you clarify what is meant by negative resistance in this context?

3. Aug 5, 2014

### ndnbolla

From my notes and his lecture, negative resistance = power

4. Aug 5, 2014

### Staff: Mentor

No, negative resistance is not power. For one thing, they have different units, so they could not be equal.

Ohms and Watts are not equal units.

5. Aug 5, 2014

### ndnbolla

Shining light into a depleted region decreases its potential creating voltage

Current >>> Electron Hole Pairs

Indeed they are not equal, there is or are step(s) missing.

6. Aug 5, 2014

### Staff: Mentor

I would ignore the animation at the top right corner of the page -- I have no idea what it is trying to illustrate. The rest of the article seems good, though.

7. Aug 5, 2014

### ndnbolla

If the voltage in Figure 2 (from wiki link) became reversed biased, would that make the slope positive meaning "positive resistance"?

8. Aug 5, 2014

### Staff: Mentor

I certainly don't see any negative slopes in the V-I graph of photodiodes, either in Photovoltaic Mode or in Photoconducting Mode:

http://en.wikipedia.org/wiki/Photodiode

.

9. Aug 5, 2014

### ndnbolla

now even more confused, let me think a bit

10. Aug 5, 2014

### Eimacman

An interesting problem. I can only think that if a photo-diode exhibits a negative resistance it would have be due possibly to quantum tunneling caused by photon excitement of the PN junction, that would cause decreased current drop across the junction at higher voltages when exhibiting negative resistance. The photo electric effect could cause negative resistance but I am not sure of the mechanism involved.

11. Aug 5, 2014

### Staff: Mentor

What's a "decreased current drop across the junction"?

12. Aug 5, 2014

### Staff: Mentor

When the lines go up and to the right on the V-I diagram, that is positive resistance.

13. Aug 7, 2014

### Eimacman

Greetings berkeman:

Being that this conversation is about photo diodes, the junction of a semiconductor device determines the characteristics of such a device just as the amounts of 'dopant' and elements used to make the conducting P or N substrates. In a liner device if forward voltage is increased current drop will remain constant, positive junction resistance. If the junction is thick, maybe over a couple of thousand angstroms or more, maybe a couple of hundred atoms thick and capable of quantum tunneling, the junction when it reaches avalanche voltage will exhibit negative resistance in that the current drop across junction will decrease weather this is caused by quantum tunneling or the photo voltaic effect. It is the junction of the device that makes it a semiconductor. Even if that junction is formed by a crude quartz crystal being jabbed by an iron whisker.

As for the V-I diagram the voltage applied is measured by a volt meter at the source, neglecting resistance of the wires, and the current, junction current drop, is measured across the device by an ampere meter, there would be a resistive load in series with the device to prevent destruction of the device, the resistance of which and the voltage drop would be determined in advance mathematically.

I hope that this clears this up in that I had to make this post quickly.

Eimacman.