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In a few questions ( Does the ISS need more heating or more cooling?, Why did Salyut 7 freeze over, while ISS requires massive cooling system? ) it is implied that the station inside heat budget is pretty much result of power draw of various systems - systems running = station needs cooling; systems inactive = station freezing, needs heating.

Still, taken as a whole, station and its solar cells, the system receives pretty much the same amount of solar irradiation regardless of whether the systems inside are running or not. And the same heat that would otherwise result in overheating the station (if not for the cooling system) should land somewhere if the station inside is not heated. That somewhere being most likely the solar panels. Likely reducing their electrical efficiency and as result leading to collecting even more heat.

Is that so? How does drawing power from the solar panels influence their temperature? Does it necessitate special procedures for reactivating panels that have been "off" for too long (turn "edge towards the Sun" to dissipate heat, for the efficiency to rise enough that facing the Sun they will not keep heating up, enough irradiation drawn away as electricity, instead of accumulating as heat with too low electrical efficiency to keep the temperature stable)?

Or am I wrong, and the panels even entirely switched off and facing the Sun "perfectly" still radiate more heat than they acquire, never exceeding temperatures where their efficiency drops too much?

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  • $\begingroup$ I can't answer all of your questions, but an efficient solar cell only converts 20% of sun light into electricity. 80% end up as heat in the cells either way. 25% higher heat intake should not make a major difference with respect to lifetime and degradation. $\endgroup$ – asdfex Jul 26 '16 at 17:58
  • $\begingroup$ Some of the unused incident energy would be reflected as well as be converted to heat. Drawing power would definitely increase their temperature because of the current flow. The exact increase would depend on the current/voltage/resistance in the circuit. As far as their orientation and relation to temperature and efficiency I'm not sure. I have a friend on the Johnson Space Center ISS operations team. I'll see if she can help. (no promises) $\endgroup$ – Andrew W. Jul 26 '16 at 21:19
  • $\begingroup$ @AndrewW. Drawing power actually decreases their temperature. See my answer below. $\endgroup$ – Tristan Jul 26 '16 at 22:11
  • $\begingroup$ Hum, you're right, the temperature would reduce temperature. However I think the reason is a bit more nuanced. The reason is that a current would reduce the number of electron hole pair recombinations, which in turn reduces waste heat. That's also assuming that the temperature increase from resistance is less than the cooling effect. $\endgroup$ – Andrew W. Jul 26 '16 at 23:57
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The ISS solar arrays are divided into dozens of individual segments that can be switched on and off electronically to meet the station's power demand. Solar cells will reach an equilibrium temperature that balances the solar flux with thermal radiance.

Drawing power from the solar cells actually cools them off. This is because the solar input (which is more or less constant) is divided between power delivered and waste heat. Delivering power results in less waste heat, which means the cells run cooler. You can see this effect in infrared views of the space station such as this.

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  • $\begingroup$ So if the electron hole pairs are not collected separately as current, they recombine in the cell. Since it's pretty much always non-radiative recombination in silicon, the energy is released back to the crystal as phonons? Amazing! Can you help point out in the video where we can see that cells delivering current are cooler than cells not delivering current - an annotated screenshot or at least time code? Thanks! $\endgroup$ – uhoh Jul 27 '16 at 1:05
  • $\begingroup$ I've asked this question about that. $\endgroup$ – uhoh Jul 27 '16 at 2:35
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    $\begingroup$ @uhoh Take a look at the solar arrays starting at around 0:22. You will see alternating dark and light bands, with more bands going light as the video progresses. The dark bands are cooler because they are switched on. As the batteries become charged up, the power need drops, so more array sections are switched off. $\endgroup$ – Tristan Jul 27 '16 at 15:30
  • $\begingroup$ @uhoh One other point -- unless an array segment is damaged to create an open circuit, current is always flowing. Power demand is met by selectively shunting portions of the arrays, resulting in the "switched off" portions delivering the same current but at near zero voltage. $\endgroup$ – Tristan Jul 27 '16 at 15:36
  • $\begingroup$ Hmmm.... I see what you mean. Between 00:33 and 00:36 especially. Because the angle of the camera view keeps moving, it could also look like a directional effect. Thermal imagers don't actually measure temperature - the IR brightness is also affected by material texturing and emissivity, and those can be directional also. But this is really interesting! $\endgroup$ – uhoh Jul 27 '16 at 15:39

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