# Confusion regarding the I-V characteristics of a solar cell

We know that I-V characteristics of a solar cell are described through the following equation :

This equation can be translated graphically to the following (ideal) circuit :

Now according to the following equation and circuit, the solar cell current consists of two-component currents, one is for the diode current Id and the other due to the light generated current IL. Now I have the following questions :

1. I think that I am confused about the understanding of the operation of the solar cells. According to my understanding, the solar cell depends on the light generated current that is generated due to photons energy. This current is simply due to the minority carriers generated in each semiconductor region ( p and n) and this current is supported by the built-in voltage and exists between the p and n regions. Also, the solar cell depends on the diode current ID that simply represents the portion of the IL lost by recombination in the bulk region before reaching the depletion region. This discussion was according to the formula above I=ID+IL and also my discussion agrees with the fact that the solar cell is a source not a passive. So do I understand right?
2. V in the above equation represents the built-in voltage which is around 0.6 v for silicon as we know. Now this means that we need to apply an 0.6v across the solar cell to make it work? If we do so then the sollar cell will not become a source because now we have inserted some sort of electrical energy to it although it should solely depend on light energy?
3. In the above equation there are two quantities Io and IL, the reverse saturation current and the light-generated current. The reverse situation current is the current that flows in the opposite direction due to the minority carriers but this is the same as IL as I have discussed above, then what are the differences between them?
• The direction arrow for $I$ on your diagram is not consistent with your equation for $I$. Commented Dec 23, 2020 at 21:42
• Aren't solar cells used without an outside source? Commented Dec 23, 2020 at 21:56
• @ThePhoton Yest thats right . All we need to do is to change the sign of the current components I= IL-Id Commented Dec 23, 2020 at 22:05
• @TheForceAwakens Yes, I was just confused. It's just a drop of 0.6 V across the cell. Commented Dec 23, 2020 at 22:06

...we need to apply an 0.6v across the solar cell to make it work?

No. The circuit diagram that you embedded in your question really is a good model. The sunlight forces a constant current flow, and when current is forced through a diode in the forward direction, there must be a voltage "drop" (i.e., the 0.6V). If the external circuit is able to accept that much current at a significantly lower voltage, then substantial current will flow through the external circuit, and not so much through the diode.

The edge cases would be a short circuit (virtually all of the current flows through the short, and the voltage is practically zero because of the low resistance), or an open circuit (All of the current flows through the diode because there is no other available path, and the terminal voltage is the full "diode drop.")

• Many thanks , but what is the difference between Io and IL ? Commented Dec 23, 2020 at 22:04
• @AAA Your diagram is a bit fuzzy but I'll assume that $I_o$ is the "diode" current in your diagram, and $I_L$ is the source current. In that case, it's just a matter of applying KCL: $I_L$ must equal the sum of $I_o$ and the current delivered to the load. Commented Dec 23, 2020 at 22:08
• $I_o$ is one factor in the equation for diode current, it isn't the whole diode current. Commented Dec 24, 2020 at 2:31

the solar cell depends on the light generated current that is generated due to photons energy. This current is simply due to the minority carriers generated in each semiconductor region ( p and n) and this current is supported by the built-in voltage and exists between the p and n regions.

The photocurrent is mainly generated in the depletion region, not in the p and n regions.

If a photon excites an electron-hole pair in the depletion region, the two carriers will be quickly separated by the strong electric field in that region, preventing them recombining. If there was a pair generated in the p or n region, the minority carrier would quickly recombine with one of the majority carriers present and not contribute to the terminal current.

Also, the solar cell depends on the diode current ID that simply represents the portion of the IL lost by recombination in the bulk region before reaching the depletion region.

The diode current is the same current that would be present in any PN junction with no light exposure.

It comes from carriers that are thermally excited across the barrier formed by the depletion region. It increases if the junction is forward biased (decreasing the width of the depletion region) and decreases if the junction is reverse biased (increasing the width of the depletion region).

It is a separate process from the current produced by photocarriers, but you are right that it works in opposition to the photocurrent and prevents the output voltage of the device becoming large even if there is no load.

my discussion agrees with the fact that the solar cell is a source not a passive.

The diode current would be present even if you operated the cell as a passive device (i.e. didn't expose it to light and applied an external voltage to it).

V in the above equation represents the built-in voltage which is around 0.6 v for silicon as we know.

V in your equation is the voltage across the terminals, whether it's applied by an external source or generated by the photocurrent output of the cell flowing through a load.

The built-in voltage is an internal potential difference across the PN junction. It's normally treated as a characteristic parameter of the device, not a value that changes depending how you operate the device.

The reverse situation current is the current that flows in the opposite direction due to the minority carriers but this is the same as IL

What you call "reverse situation current" is more often called "reverse saturation current".

It is the current that flows when the junction is strongly reverse biased, and no light is incident.

It has no real connection to $$\I_L\$$ (the photocurrent).

• Sorry it is the spell checker :D I will edit the question . Commented Dec 23, 2020 at 22:16