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Having experience in shaping custom solar cells, a colleague made me question my basic understanding of photovoltaic operation recently. He pointed out that a so-called Short Circuit Current in a solar cell conventionally appears at zero voltage between the cell's terminals. Given that the current is a flow of charges where voltage is the driving source—it becomes unclear how to interpret the Short Circuit Current at no voltage.

Through the web search, hypotheses to understanding this controversy were suggested:

  1. If the photocurrent is a result of the absorption of photon energy by electrons, the passage of those electrons with sufficient energy away from bulk material is considered to be current, regardless of no applied potential hence Zero Voltage. Link: https://www.quora.com/Why-is-there-some-photocurrent-present-even-at-zero-potential

The question is, how the energy of electrons is not present here in the form of voltage if the voltage is directly derived as the energy (per charge)?

  1. If p–n junction contains an internal voltage even at thermodynamic equilibrium, the voltage at Short Circuit Current may be zero only externally to the solar cell, when internally the electrostatic potential difference hence the voltage is still present and does the job of separating charge carriers. Link: https://en.wikipedia.org/wiki/Fermi_level#The_Fermi_level_and_voltage

The questions are: if it's true that the internal p-n junction's electrostatic potential is the driving source for the charges to flow, and if so, what defines the cell's output voltage, like in between zero and Open Circuit Voltage?

  1. And the last, my hypothesis is that typical I-V curves represent idealized solar cells when in reality a fraction of resistance at shorting the cell's terminals makes the voltage different from zero. Thus, the practical line on the I-V curve should never reach the Y-axis.

The question is if it's true that practical voltage at Short Circuit Current is different from zero (for our cells this is true but we blamed the internal resistance of the power meter)? And if so, how it is possible for an idealized solar cell to have a different from zero current at zero voltage in the light of Ohm's Law (at U=0 the I must be 0)?

We likely miss causation, so point us in the right direction, please!

enter image description here


First, thanks to all the contributors who provided a great view on the question. Thanks to this, my understanding of the operation of the solar cells (and semiconductors along with general circuits too :) has been updated as follows:

To not forget, the photocurrent in a solar cell is not only a result of the absorption of photon energy in bulk material with the creation of photocarriers but also a separation of photocarriers by, importantly, the action of a pre-existing electric field associated with the p-n junction (https://en.wikipedia.org/wiki/Electromotive_force#Solar_cell). It is this electric field that pumps photocarriers and originates electromotive force as a consequence.

The built-in potential seems to be the function of majority carriers' concentrations on the edges of depletion region (https://ocw.tudelft.nl/wp-content/uploads/Solar-Cells-R4-CH4_Solar_cell_operational_principles.pdf). At the first "contact" of p and n semiconductors there would be no potential yet, but it establishes through uncompensated diffusion.

Under darkness, the established p-n junction is at equilibrium mode, when further diffusion of majority carriers is compensated with the rise in the electric field, preventing this action. Thus, the built-in potential is rather static, not allowing electrons to flow at established equilibrium, and cannot be measured by the voltmeter.

Under illumination at open circuit mode, the built-in potential pumps photocarriers creating a potential difference by, importantly, minority carriers hence the electromotive force. The potential difference of minority carriers enables electrons to flow, has all the qualities of electromotive force, and can be measured by the voltmeter.

Under illumination at short circuit mode, the built-in potential of majority carriers is unchanged, the minority carriers separation keeps occurring, devise functions as a current source regardless of zero voltage drop across its terminals.

It would be great if someone with sharp knowledge could confirm or refine this understanding further. Thanks!

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    \$\begingroup\$ I believe that the source of your confusion is that you try to consider a semiconductor as a metal. In semiconductors you can have a current without a 'causing' voltage. It's called diffusion current and it's at the root of the workings of all PN junctions, hence all semiconductor transistors, diodes, and solar cells. \$\endgroup\$ – Sredni Vashtar Mar 21 at 15:01
  • \$\begingroup\$ Current is the flow of charge no matter the driving source. You can create an ion beam in zero field, or build up a static charge on a helium balloon and let go. Both are electric current. \$\endgroup\$ – Matt Mar 21 at 16:40
  • \$\begingroup\$ Any electrical power source can be forced to have an output voltage arbitrarily close to zero if you short it out with a sufficiently low resistance wire (or superconductor if it comes to that). Just because the voltage is zero (or arbitrarily close to zero) does not mean that no current can flow. Likewise, if you see the current approaching a limit as the V and R approach zero, there is no reason to believe that something magical or discontinuous will happen at R = 0. For a solar cell, as you get closer and closer to a zero ohm short, the current gets closer and closer to Isc. No magic. \$\endgroup\$ – mkeith Mar 25 at 1:40
  • \$\begingroup\$ Thank you mkeith, I see your point clearly in the light of the equivalent circuit. The initial question can be rephrased as: "If we see no voltage across shorted terminals, does it mean there is current without a potential difference?" If we add the internal resistance, things become rational and the potential difference is always present. Nevertheless, the equivalent circuit doesn't answer the question where we can measure this potential difference hence the voltage at short. I suggest that the equivalent circuit doesn't reflect the nature of the phenomenon, which makes me study further. \$\endgroup\$ – Nicolay Mar 25 at 19:25
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Given that the current is a flow of charges where voltage is the driving source—it becomes unclear how to interpret the Short Circuit Current at no voltage.

Consider the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

10mA flows through this circuit. According to Ohm's law, there is 1V across R1 but 0V across R2. V = IR, so when R=0, so does V.

This is exactly equivalent to what happens when you short circuit a solar panel. The solar panel has a source of electromotive force (equivalent to V1 in the circuit above) and an internal resistance (equivalent to R1 in the circuit above). There is current flowing through the short (equivalent to R2 in the circuit above), and there is 0 voltage across the short (equivalent to the 0 voltage across R2).

The question is, how the energy of electrons is not present here in the form of voltage if the voltage is directly derived as the energy (per charge)?

The voltage across a component is not the energy per unit charge of the moving charges, it is the work required to move a charge through the component. No work is required to move a charge through a 0 \$\Omega\$ resistor, so the voltage across it is 0, even though the electrons may be moving quite fast and have energy.

The question is, is it true that practical voltage at Short Circuit Current is different from zero

No. The voltage across a short circuit (0 \$\Omega\$ resistor) is 0, even though current flows through the short circuit.

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  • \$\begingroup\$ Good answer sir! So, in your analogy, we measure short circuit current across R2 terminals. Also in the analogy, it would be possible to measure the electromotive force across V1 or R1 terminals. Can you provide a suggestion where it would be possible to measure the electromotive force in a real solar cell at short circuit condition? \$\endgroup\$ – Nicolay Mar 21 at 3:22
  • \$\begingroup\$ I'm not sure whether one can get a meaningful measurement of emf in a short circuit condition. 10V/10 ohms gives the same current as 1V/1 ohm. One often measures emf at open circuit, because when no current flows, there is no resistive voltage drop. However, the non-linear nature of semi-conductors in general, and solar panels in particular, means that the open circuit voltage one measures, is not necessarily useful for determining what happens in non-open-circuit conditions. Useful maybe, but not necessarily for accurate prediction. \$\endgroup\$ – Math Keeps Me Busy Mar 21 at 3:29
  • \$\begingroup\$ You're right, I agree with your comment. My question was asked to better understand the nature of the short circuit condition in a solar cell. Thanks for your contribution! \$\endgroup\$ – Nicolay Mar 21 at 3:34
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    \$\begingroup\$ For a solar cell, the current at 0.000V is pretty much the same as the current at 0.01V. It doesn't change much at very low voltages. So you don't need to use a superconductor to short circuit it. And it is possible to measure current using hall effect techniques also. So that is another option. Hall-effect DC current probe with a heavy gauge copper wire short. \$\endgroup\$ – mkeith Mar 21 at 3:36
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    \$\begingroup\$ Also note that a solar cell is a complex network of circuits - there are many local series and shunt resistances that influence the local voltage and current sources. These local voltages and currents are present when the cell is biased with illumination, regardless of the external operating condition (e.g. short-circuit, open-circuit). The voltages and currents of the internal network change as necessary to satisfy the external operating condition. \$\endgroup\$ – HypeInst Mar 21 at 4:27
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I believe you are still making the mistake of considering a semiconductor as a metal. In semiconductors you can have a current without a 'causing' voltage. It's called diffusion current and it's at the root of the workings of all PN junctions, hence all semiconductor transistors, diodes, and solar cells. All you need is a gradient in carrier concentration, the thermal agitation at any T>0K will do the work for you.

Even the built-in voltage that 'separates' the charges released by photon interaction within a semiconductor junction is a result of the different concentrations in the p-doped and n-doped materials. In a way you could see that voltage as well as 'caused' by the thermal (IR) photons that hit the junction at ambient temperature. At thermal equilibrium you get a built-in voltage with diffusion currents in differently doped materials going both ways through the junction - hence zero total current (internal or external, it's the same). Does it strike you as equally odd that inside a diode you can have a voltage without a net current?
The built-in voltage is not visible from the outside because to read it with a voltmeter you would need metal contacts that, at thermal equilibrium, will end up compensating it with an equal and opposite voltage. (If it were not so, we would have a perpetuum machine and the second law of thermodynamics would not be happy).

I don't want to start another row about "voltage causes current" vs "current causes voltage", so let me quote verbatim (only the emphasis is mine) a passage from Millman's "Electronic Devices and Circuits" (1967), p. 588:

If a forward bias is applied, the potential barrier is lowered, and the majority current increases rapidly. When this majority current equals the minority current, the total current is reduced to zero. The voltage at which zero resultant current is obtained is called the photovoltaic potential. Since, certainly, no current flows under open-circuited conditions, the photovoltaic emf is obtained across the open terminals of a PN junction.

An alternative (but of course equivalent) physical explanation of the photovoltaic effect is the following: In Sec. 6-1 we see that the height of the potential barrier at an open-circuited (nonilluminated) p-n junction adjusts itself so that the resultant current is zero, the electric field at the junction being in such a direction as to repel the majority carriers. If light falls on the surface, minority carriers are injected, and since these fall down the barrier, the minority current increases. Since under open-circuited conditions the total current must remain zero, the majority current (for example, the hole current in the p side) must increase the same amount as the minority current. This rise in majority current is possible only if the retarding field at the junction is reduced. Hence the barrier height is automatically lowered as a result of the radiation. Across the diode terminals there appears a voltage just equal to the amount by which the barrier potential is decreased. This potential is the photovoltaic emf and is of the order of magnitude of 0.5 V for a silicon and 0.1 V for a germanium cell. The photovoltaic voltage Vmax corresponds to an open-circuited diode, if I = 0 substituted into Eq. (19-10), we obtain

open circuit voltage
Note from the XXI century: \$I_s\$ is the photocurrent (we call that \$I_{ph}\$), \$I_0\$ is the saturation current (we call that \$I_s\$), \$\eta\$ is 1 for Ge, 2 for Si.

Since, except for very small light intensities, Is/Io » 1, then Vmax increases logarithmically with ls and hence with illumination. Such a logarithmic relationship is obtained experimentally, as indicated in Fig. 19-25a.

from Millman, a guy who really understood electronics

When you load the cell with a resistor R, you are in the fourth quadrant of the V-I plane, in photovoltaic mode. The current and voltage that are common to the cell/photodiode and the load resistor (there could be a difference in sign due to the different convention for passive or active two-ports) is the intersection of the photodiode and resistor characteristics like so:

exponential knee in the fourth quadrant, from Millman
Photovoltaic mode in the fourth quadrant. Source: Millman, as above

The voltage that falls across the load resistance due to the current supplied by the photodiode/solar cell is actually contrasting the built-in voltage. For \$R_{load} = \infty\$ you have the maximum open circuit voltage mentioned above. For \$R_{load} = R_{ideal}\$ you have the maximum power delivered to the load. And finally, for \$R_{load} = 0\$ you are no longer contrasting the built-in voltage and you get (approximately) the maximum current possible, which is the term Iph added to the Shockley equation.

$$ I = I_S \left[ exp \left( \frac{V}{ \eta V_{Th}} \right) - 1 \right] - I_{ph} $$

All of this falls neatly in place when you look at the equivalent circuit of a solar cell (here shown with shunt and series resistances that represents losses and parasitic effects, and with the addition of a variable load resistor):

Loaded equivalent circuit of a solar cell with losses
Equivalent circuit of a solar cell (with losses) and load resistor (Source)

Once you accept that the current source depends on the illumination, you see that no matter what the load resistance is, you cannot exceed the voltage drop of a diode as far as the voltage 'supplied' go. And the maximum current you can get is that for the load R = 0, which will shunt the diode voltage to zero as well.

So, to get to the point of your question: voltage and current are a consequence of the illumination. Don't be surprised if one can be zero while the other is not.


As a side note, with the convention for the current I in the above equivalent circuit, we are looking at the V-I characteristics with the I axis inverted. The device should then be described by this equation (again I am neglecting the losses associated with the series and shunt resistors):

$$ I = I_{ph} - I_S \left[ exp \left( \frac{V}{ \eta V_{Th}} \right) - 1 \right] $$

and the characteristics we've seen above will be shows in the first quadrant, like this

inverted VI - conventions... pfff

Now the load resistor characteristic is the well know line in the first quadrant.

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  • \$\begingroup\$ Dear Sredni Vashtar, thanks a lot for the great effort put into this answer. I see some differences between my understanding and your reference—I will study them ) \$\endgroup\$ – Nicolay Mar 25 at 18:48

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