Scientists develop mega-thin solar cells that could be shockingly easy to produce: ‘As rapid as printing a newspaper’::These cells could be laminated onto various kinds of surfaces, such as the sails of a boat to provide power while at sea.
Scientists develop mega-thin solar cells that could be shockingly easy to produce: ‘As rapid as printing a newspaper’::These cells could be laminated onto various kinds of surfaces, such as the sails of a boat to provide power while at sea.
The electrons are being replaced, otherwise the system would become charged over time. They go the long way round the P side of the junction and bond with a hole again in the depleted region.
Solar cells do deteriorate over time but it’s not due to use, or not directly. The structure of various parts gets damaged through lattice migration due to heat/thermal cycling, UV radiation and higher can cause excitation to reactive states that damage crap, dopants can migrate around over time (like how carbon can leech from steel) and reduce the conductive efficiency etc.
I think this might be what you’re confused by? there are a finite number of available charge carriers in the depletion region and damage to the region uses them up, but it’s not because they’re used up it’s because of structural and chemical changes caused by damage that occurs due to the environment.
Are they being replaced? I dont remember these circuits being closed loop. I also dont remember if photons do anything other than create a wave excitation to move electrons… where is the replenishment coming from?
I understand phonons deteriorate lattice structures. I dont know if that is a seperate issue though.
So on the arse ends of the junction are little wires (one transparent). If you leave the panel in an open circuit the carriers will separate till the charge build up overcomes the depletion zone field and no more charge separation happens during excitation.
In this configuration the cell is essentially a capacitor.
If you close the circuit the P side of the electric field will propagate through the circuit and the load, pushing charge carriers through the load, out the other end, and into the P side of the junction where they combine with the separated holes.
Electrons have to come from somewhere. They arent coming from sunlight. Im not understanding how even if its closed loop it will carry charge back so there is no defecit in charge. This is why i keep saying it makes sense as long as there is an assumption of infinite electrons. If you take that away, where does the potential come from…
The electrons are all already in the material. They are never created or destroyed just paired and unpaired with holes.
This video might help? from approx 5 minutes. https://www.youtube.com/watch?v=WfP5YdJn-c4
The energy comes from the sun, it excites electrons and holes, causing the cell to hold a small charge, that charge is the potential energy that drives the circuit. It is depleted by electrons flowing back into the P side from the circuit, they cannot go from the N side because of the field across the depletion zone. Recombinant electrons from the circuit can then be excited again, excess electrons in the N side flow out of the silicon into the load so that electrons can move from the load into the P side.
This all happens at once. In a very long time, eventually the very same electron that was originally excited by a photon will recombine in the depletion layer. There isn’t any loss here.
I am getting frustrated because i am pretty sure the qty of electrons is conserved so at some point equilibrium needs to be reached. I read this paper on diffusion lengths and i got confused so i started watching this older lecture about short and open circuit recombinations. My first impression is that the recombination rate (or qty of electrons coming back into the cell) depends on some thickness of material at the very least.
I was definitely wrong about how the whole loops work but that was me being dumb. Again, this is very out of my brain territory. I used to love this stuff though.
I’m not sure I understand the confusion. The material thickness does matter in that thinner materials will resist current flow due to a lack of charge carriers meaning each must travel faster, which means that more charge separation is required for a given recombination rate or current.
If the charge required is too high (more than the potential difference across the depletion zone) then the build up will prevent charge separation after excitation. They will recombine in the depletion zone making a bunch of heat and burning out the cell.
Maybe that’s the thing? super thin films under too much irradiation or load cooking themselves due to inability to move excited charges away fast enough?