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?


2 Answers 2


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

  • \$\begingroup\$ 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? \$\endgroup\$
    – Jcb Rb
    Commented Apr 25, 2020 at 16:11
  • \$\begingroup\$ @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. \$\endgroup\$
    – wbeaty
    Commented Apr 27, 2020 at 9:40
  • \$\begingroup\$ @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.) \$\endgroup\$
    – wbeaty
    Commented Apr 27, 2020 at 9:48
  • \$\begingroup\$ 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? \$\endgroup\$
    – Jcb Rb
    Commented Apr 28, 2020 at 13:39
  • \$\begingroup\$ @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. \$\endgroup\$
    – wbeaty
    Commented Apr 28, 2020 at 22:13

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:


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.

  • \$\begingroup\$ 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? \$\endgroup\$
    – Jcb Rb
    Commented Apr 23, 2020 at 22:45
  • \$\begingroup\$ 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. \$\endgroup\$
    – Dave Tweed
    Commented Apr 23, 2020 at 22:48
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jcb Rb
    Commented Apr 24, 2020 at 0:04
  • \$\begingroup\$ 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.) \$\endgroup\$
    – wbeaty
    Commented Apr 24, 2020 at 3:27
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jcb Rb
    Commented Apr 24, 2020 at 17:16

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