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I understand that a solar cell is, essentially, a P-N junction. A P-N junction, roughly speaking, will do the following: The free electrons on the N-side will flow over to fill holes on the P-side. This produces a voltage gradient where the P-side is slightly negative and the N-side is slightly positive.

Now, when light strikes the P-N junction, it excites free electrons all over the place. These electrons, if they are close to the junction, are drawn towards the positively charged N-side. This creates current and this is how a solar cell produces power.

The way it has been described to me is that, at first, the N-side is slightly positive and the P-side is slightly negative (it's acting like a normal diode). But after light hits, this reverses. The N-side becomes very negative and the P-side very positive. By attaching a conduit between these sides, you can draw power from that current.

Here is my question

After light strikes the P-N junction, and all those electrons are drawn to the positively charged N-side....wouldn't the N-side quickly lose it's positive charge and return to being neutrally charged? This would destroy the drift voltage... so any free electrons would have no reason to build up on the N side and there wouldn't be any voltage.

What am I missing here?

Jcb Rb
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2 Answers2

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When light strikes the silicon of a solar cell, it produces electron-hole pairs "all over the place", as you put it. But the only ones that matter are the ones that are created within the depletion region of the PN junction. They are the ones that get separated by the E field within the junction to produce the terminal voltage that you can measure (\$V_{oc}\$). The rest just recombine on their own almost right away.

As you say, this is effectively a current source that is in parallel with the diode junction:

schematic

simulate this circuit – Schematic created using CircuitLab

The point is, this current flows all the time. If you don't draw it off through an external circuit, it simply flows through the diode junction itself.

There is no reversal of voltage across the junction. The normal zero-current voltage is merely reduced by the photocurrent until a new equilibrium is reached. It is this difference between the quiescent voltage and the reduced voltage that we can read using an external voltmeter, and the anode is positive with respect to the cathode.

Dave Tweed
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  • Thanks for the response! Let me add to my questions: So you have a P-N Junction. Let's say this particular P-N junction, in the dark, the drift voltage is 0.7 V. There are no wires connected to it. Now, you shine a light on it. Does that drift voltage increase? Decrease? Does it decrease and then re-stabilize back at 0.7 V? – Jcb Rb Apr 23 '20 at 22:45
  • Was I not clear enough? It decreases to the point where the forward current through the junction is equal to the (reverse) photocurrent, establishing a new equilibrium. – Dave Tweed Apr 23 '20 at 22:48
  • You were clear. I just have a thick head. So you're saying that, when you DO connect a wire with a resistor...the Voltage across that wire's resistor...as in, the voltage put out by the solar cell, is the difference between the DARK drift current and the LIGHT drift current. Again, I have a thick head, so I'm probably asking redundant questions here. I apologize. – Jcb Rb Apr 24 '20 at 00:04
  • The real answer involves the two other junction-potentials, out where the metal leads make contact. If there's 0.7V at the PN junction in the dark, then there will be around 0.3 to 0.4V at each of the "ohmic" non-rectifying contacts. If the two metal terminals are shorted together externally, there's no current, since these "ohmic" potential-steps will exactly oppose the 0.7V appearing at the PN junction. (Thermocouples do something similar, with always some millivolts across the copper-iron contact, but exactly equal millivolts across the other iron-copper contact.) – wbeaty Apr 24 '20 at 03:27
  • Bill Beaty from Amasci! I found your website a few years ago and it is awesome! Also, thanks for this comment. This explanation makes sense. – Jcb Rb Apr 24 '20 at 17:16
  • @JcbRb Thanks! The trick with this topic is to google "Volta Potential." These voltages were discovered by Alessandro Volta by using a foil-leaf electroscopes amplified by a few-kilovolts power supply (instrument called a Quadrant Electrometer, with a tiny mirror to bounce a sunbeam on the distant laboratory wall.) If we can easily measure the "floating" voltage on any piece of metal, then we'll find that neutral metals will charge up when touched to other types of metal, or to semiconductors, or to wet objects. Pairs of metals produce mV, wet objects give up to 6V! (water-gold contact) – wbeaty Apr 27 '20 at 10:04
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Light DOES destroy the static (or built-in) potential of the PV cell. That's how PV cells work! (At least, at the macro level, that's how the internal voltage-drops behave.)

During darkness, we might find 1.0V static potential between the n-type and p-type parts of the solar cell. This potential is really there, it's the energy-hill appearing in the depletion zone. But it can only be detected by non-contact electrometers (where no metal probes make contact, so no unwanted metal/silicon junctions are formed.)

Besides the junction-potential, we'd also find another potential where the metal terminal connects to the n-type, and a third where the other terminal connects to the p-type. These two "ohmic" or "non-rectifying" built-in potentials are much like static potential of any thermocouple ...though quite a bit higher volts. Roughly 0.5V each. They sum to the same value as the PN junction potential. That way, if the solar-cell contacts are shorted, there will be zero current, even though in darkness the junction exhibits an energy-hill of one entire volt!

And so, when light strikes the junction, carriers flood the depletion zone, and the staticf junction-potential is "shorted out." But this doesn't affect the two 0.5V steps located where metal contacts touch silicon.

As a result, the potential measured across the terminals has an upper limit: roughly 1.0V. PV output-voltage doesn't rise continuously as light-intensity rises. That one-volt limit was the same value as static built-in potential of the PN junction during darkness. That potential is now missing, so it will now appear at the output terminals of the PV cell.

Very strange, eh? Solar cells are actually driven by the sum of the two potential-steps found at their ohmic contacts. It almost seems like perpetual motion! But the electromotive force at the metal junctions can only pump some charges when photons are injecting energy into the PN junction, and therefore "promoting" valence electrons into the conduction band, without them having to be pushed there by an external power supply. The depletion zone is shorted out by the sudden new population of carriers there. The diode becomes forward-biased. Yet the energy needed to do this ...is exactly the energy being injected by the ohmic contacts, as carriers "fall down" the contact-potentials found at the metal-semiconductor bonds.

"Contact potentials" are weird stuff!

And now you can get an idea about Peltier module function. A thermoelectric module works because p-type and n-type blocks are welded onto little copper straps, with no diodes being formed. Yet the potential-hills are really there. They all sum to zero around the circuit ...unless half of the metal-semiconductor contacts are heated up, and the other half cooled down. As with solar cells, the heated contacts become "shorted out" because of injected energy at the micro-level: phonon absorption, which knocks the carriers uphill over the built-in junction potentials. Heh, a Peltier thermogenerator is a bunch of PHONO-voltaic cells in series! (As in: lattice thermal energy, junctions "illuminated" by phonons not photons.)

Confusing regarding measuring barrier potential of a pn junction using a voltmeter

wbeaty
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  • Ok so I'm working through this answer and trying to grasp it. When you say the Voltages at the contacts are 0.5 V and 0.5 V respectively which sum to the P-N voltage of 1.0 V....is this because of Kirchhoff's Voltage Law? – Jcb Rb Apr 25 '20 at 16:11
  • @JcbRb nope, it's just "Contact-" or "Volta-Potential," which normally obeys energy-conservation, where they sum to zero around any closed circuit. If p-type against n-type produces 0.6V difference across that junction, then p-type against copper will produce roughly 0.3V, and n-type against copper will produce another 0.3V in the opposite direction. The 0.6V of the PN junction is perfectly canceled by the 0.3V produced by each copper-silicon contact in series. But heat or cool any one junction, and its voltage changes (so a current will appear.) That, or shine light on the PN junction. – wbeaty Apr 27 '20 at 09:40
  • @JcbRb in other words, ALL junctions are perpetual power supplies! But even though these voltage-sources are perfectly real, we can't get at them, or use them directly. Yet thermocouples, Peltier modules, PV cells, LEDs, all rely upon them indirectly. When charge-carriers cross any junction, they must go up or down a "potential hill," even if the junction isn't a diode. Touch iron against copper, and both metals become charged opposite, like a self-charging capacitor. (We'd need an electrostatic "field-mill" or similar instrument, if we wanted to actually measure these voltages.) – wbeaty Apr 27 '20 at 09:48
  • Let me see if I understand what you're saying. At baseline, in the dark, the PN junction will reach an equilibrium where the PN Junction is 1.0V. The copper terminal on the N-type side will be -0.5V. The copper terminal on the P-type side will also be -0.5V. The total voltage of the circuit is 0 because they all balance out. But when light shines on the depletion region of the PN Junction, that voltage will drop...maybe to 0.4V...or something. Copper terminals don't change. SO now, there's a total circuit voltage of -0.6V....so we get a current. Is this correct? – Jcb Rb Apr 28 '20 at 13:39
  • @JcbRb Exactly right. As light acts to reduce the PN junction's potential, that same change in potential will appear across the copper terminals. As a result, the max output voltage possible on the copper terminals is the same as the potential found across the PN junction in darkness. In other words, the light only "shorts out" the PN junction, but doesn't cause its voltage to go past zero, becoming opposite to the usual built-in polarity. – wbeaty Apr 28 '20 at 22:13
  • Correct me if I'm wrong. Those voltages (1.0V, -0.5V, -0.5V) are created by electrons moving around in conduction/valence bands a la band theory until equilibrium is reached. When light hits PN junction, that screws up the equilibrium. Electrons are swept into N-type, wouldn't that momentarily change the -0.5V to something like -0.3V? And wouldn't those electrons then want to diffuse into the copper terminal? Those electrons would then push the free electrons in the copper until they pop out at the terminal in the P-type. Am I getting this right? – Jcb Rb Apr 29 '20 at 19:22
  • @JcbRb Yep, although it's better to view it as electrostatics/capacitors, like water-filled tanks. (Injecting some water molecules into ONE tiny spot inside a closed tank will pressurize the ENTIRE tank.) Because of the band-structure, equilibrium is reached after the two different conductors pump some charge and create a voltage across their junction. Cu against Ag does this, also Cu against Si. Next, light is basically pumping some charges backwards, against the PN junction's "self-charging" effect. That collapses the junction's energy hill (voltage,) but not the hills at the Cu/Si junctions – wbeaty Apr 30 '20 at 01:35
  • You keep saying the hills at the Cu/Si junctions don't collapse. But I don't understand how this could possibly be true. I get that, if you create a closed loop, all the charges will flow around until they balance each other out, hence the total baseline voltage of the circuit is 0. I do NOT understand why, if you shine light on one junction, the other two don't change. If electrons flow OUT of P and IN to N....then the P-N voltage is lower. But now, MORE electrons in N means N/Cu gradient will become lower. Additionally, LESS electrons in P means P/Cu gradient becomes lower. – Jcb Rb Apr 30 '20 at 13:45
  • @JcbRb There is no closed loop. This is electrostatics physics, where capacitors or batteries can produce a voltage across their floating terminals, with no closed circuit needed. If we take hunks of copper and n-type silicon, briefly bump them together, the hunks will spontaneously transfer electrons, develop equal and opposite charges, but only until an equilibrium is reached, and 0.5V appears between them. This "equilibrium-seeking" does not involve any closed circuit. It's roughly the same way that a battery will transfer internal charges, and produce its output voltage, wo/any circuit. – wbeaty Apr 30 '20 at 19:32
  • @JcbRb WITHOUT any closed loop, electrons flow right across each junction, same as batteries creating constant voltage. (Each junction is like a 2-terminal battery, and tries to always maintain the same voltage between its terminals.) Look at "batteries in series" rule. We have three in series: one 1.0V battery pointing forwards, and two 0.5V batteries pointing backwards. The voltages add up, and across the ends of the series we see zero volts (=1.0v-0.5v-0.5v). Next, short out the 1.0V battery. What voltage now appears across the ends of the chain? No longer zero, now it's 1.0V. – wbeaty Apr 30 '20 at 19:34
  • Thank you for patiently explaining all this. You are awesome and I have a thick head! I guess what I'm confused about is why would the voltages all cancel each other out at baseline? Why should the Cu/N or the Cu/P junctions be affected by the P/N junction gradient in the dark (they perfectly balance out the voltage at the PN junction), but then suddenly not care once light hits the junction? – Jcb Rb Apr 30 '20 at 22:33
  • They way you're describing this seems to refute everything I thought I understood about solar cells, that's why I'm having so much trouble. My understanding was originally that light would hit the PN junction. This would send electrons from the depletion region sliding downhill to the N-side. These electrons would then be highly motivated to push their way through a copper wire and rejoin with holes on the P-side. Along the way, they could do some work, like lighting up an LED. But you're telling me that the power is coming from the built in gradient at the Cu/Silicon terminals! – Jcb Rb Apr 30 '20 at 22:38
  • @JcbRb No, the power comes from the shorted-out PN junction, and that "shorting" is caused by energy injected during creation of electron-hole pairs (via the absorbed light.) In circuits it's never wise to follow individual electrons. It usually leads to wrong ideas. Circuits are like circular rivers, where if we pour in a bucket of water here, the entire water-level changes everywhere, all at once. Pour in a bucket of red-dyed water, and the red water stays right there. Yet miles away, the water level increases. Electrons in conductors flow at feet per year, but the watts move at light speed – wbeaty May 01 '20 at 06:33
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/107442/discussion-between-wbeaty-and-jcb-rb). – wbeaty May 01 '20 at 06:37