# Photovoltaics are heat sinks?

1. Apr 6, 2009

### Smacal1072

Hey all,

First off, thks to everyone here on physicsforums, this is such a great resource for anything physics-related.

I was thinking today about the photoelectric effect, and a scenario popped into my head which seemed counter-intuitive. Please correct me if I'm wrong, it's been a while since I've been at the University:

Photovoltaic cells utilize the photoelectric effect to produce a voltage whenever an incident photon's energy is higher than the band gap of the silicon used for the cell.

All matter above absolute zero emits thermal radiation, and the energy of the radiant photons is distributed by Planck's law. Planck's law implies that at all (even very low temperatures) high energy photons are radiated. Let's assume I throw a photovoltaic cell in an isolated physical system that is above absolute zero. Assuming that the voltage produced is being utilized outside of the system, how is this not an ideal heat sink?

2. Apr 6, 2009

### Staff: Mentor

It is not ideal because it only converts a fraction of the incident energy into electricity. But yes, if, for example, you don't connect a solar panel to anything that will dissipate the electrical energy, it will get hotter than if you do.

Also, I'm not sure if there is such a thing as a heat sink that operates by absorbing radiant heat. I can't fathom an application for such a thing.

3. Apr 6, 2009

### Smacal1072

I'm sorry, I appreciate your response but I think I've neglected to mention that in this scenario the solar cell is connected to a load outside of the system (or even inside the system). Maybe it drives a tiny motor, which raises a tiny weight. Or it's charging an ideal capacitor, with infinite capacitance. Or it's electrolyzing water. Anything but converting it back to heat.

And I also have to apologize: heat sink is the wrong word (it's been a while since thermo 101). I was trying to express that the cell is converting a portion of the heat energy in the system to a scalar potential (by way of radiant radiation). So, it's reducing the overall temperature of the system. In a roundabout way it is a thermoelectric transducer.

My point of confusion is this: How can this system be physical? It's breaking a rule of thermodynamics (it's reducing the entropy of the system by being a heat engine without a cold reservoir).

4. Apr 8, 2009

### Bob S

This is an interesting question. Suppose we have an isolated system, meaning an ideal evacuated isolated black box at say 293 kelvin with a photovoltaic cell inside, and the band gap were very small, so the photocell could detect far infra red (including part of Planck spectrum). We further assume that the photocell is at equilibrium with the box. Now, attach wires to the photocell, and draw current through the wall of the black box. Does the temperature inside the black box drop?

5. Apr 8, 2009

### Smacal1072

Thks Bob that summarizes the problem much better.

Assuming this situation is possible, the only thing I can imagine is that the system becomes more and more negatively ionized as we use the photocell's current, since electrons are ejected from the cell by the photoelectric effect. Maybe once it becomes ionized enough it reaches an equilibrium?

But even if this is the case, it isn't really a satisfactory answer - we've just made another problem because we've spontaneously ionized the system.

6. Apr 8, 2009

### Staff: Mentor

No, you did say that. I was showing the difference between the two scenarios to confirm/emphasize your point.
You started another thread asking the same question. The answer is simply that you are violating your system boundary. You are analyzing the situation incorrectly. Draw the boundary around the entire system: the sun, the PV cell, and the electrical load, and you will see that the system does not spontaneously decrease its own entropy.

7. Apr 8, 2009

### Bob S

My system assumed that all the current (or electrons) on one wire returned on the other wire, so no net charge left or entered the system. No net charge either left or returned to the photocell. But energy was being extracted from the system and dissipated on a resistor outside. So where did the energy come from?.

8. Apr 8, 2009

### Staff: Mentor

Charge is simply a carrier for energy, like water in a dam, steam in a steam engine, etc. It transfers energy when you move it around. The energy to move it in this case came from the sun.

9. Apr 8, 2009

### Staff: Mentor

A photocell at thermal equilibrium with its surroundings will not get power from the surroundings. Remember, a photocell is a semiconductor junction, and when a semiconductor's band-gap is on the order of the temperature times k then the photocell short circuits due to thermal leakage current. Roughly it works out that a photocell won't really get power from a blackbody unless it is more than 3 times hotter than the photocell. So in a photocell you always have energy going down the temperature gradient, consistent with the 2nd law of thermo.

10. Apr 8, 2009

### Staff: Mentor

Oops, missed that part...

11. Apr 8, 2009

### Smacal1072

That would explain it - thks DaleSpam

Russ - You're right I did start another thread on this topic...I apologize but I was so curious! I was afraid this thread appeared resolved. I do appreciate your replies

12. Apr 15, 2009

### Blenton

So would this mean that cooling a solar cell down to very low temperatures would increase its efficiency?

13. Apr 15, 2009

### Smacal1072

Sounds like thats the case

This makes sense, but I just thought of this today: The band gap of Si is about 1.11 eV at 300K. It maxes out at 1.2 eV as Si is cooled to 0 Kelvin, and decreases as it's heated. A photocell begins to short circuit when kT approaches the band gap energy right (T here is ~ 10,000K)? So long as a photocell is below this temperature, short circuiting does not arise?

Last edited: Apr 15, 2009
14. Apr 15, 2009

### Staff: Mentor

Yes! You should be able to find temperature vs efficiency curves for commercially available solar cells via google.

15. Apr 15, 2009

### Staff: Mentor

It is not a sharp cut off like that. Remember that the Boltzmann distribution has some pretty large tails. Also, don't forget that there are a lot of holes, so even a relatively minor fraction of them can easily swamp the small photoelectric current.