0
$\begingroup$

I have trouble understanding the $I(U)$-plot of a solar panel: enter image description here

So without any light it should probably just be the characteristic curve of a diode, right? Which means that at some point, when the voltage is high enough, we have a current.

And what about the lighted part? Why is it negative? Is it because the diode is still in reverse direction and the photo current $I_{ph}$ is "coming out" of the panel? And if we increase the voltage, at some point ($V_{OC}$ we have an opposite current of the diode that evens out the photo current? So technically speaking it is (assuming the lighting is constant) a constant current $I_{ph}$ which overlaps with the exponential current $I_D\propto\exp(U)$ from the diode?

And how does one "change the voltage" on a solar panel?

$\endgroup$
5
$\begingroup$

I will try to answer your questions one at a time:

  1. So without any light it should probably just be the characteristic curve of a diode, right?

Correct. In principle, a simple solar panel is just a big pn-junction.

  1. When the voltage is high enough, we have a current.

Right. In the sense that the current follows the simple equation $$I(V) = I_S (e^{V/V_t} -1)$$ for a perfect pn-junction.

  1. Why is it negative? Is it because the diode is still in reverse direction and the photo current Iph is "coming out" of the panel?

I guess the best way to understand the direction of the current is to look at the band diagram of a pn-junction.  A p–n junction in thermal equilibrium with zero-bias voltage applied. (from Wikipedia) If you apply a positive bias, electrons (holes) start to flow from the p (n) region to the n (p) region. In contrast to that, if you have a solar panel, you use your pn-junction in the other direction. As the free carriers in your device, generated by the absorbed photons, follow the electric field in the device (see picture), the electrons (holes) flow from the n (p) region to the p (n) region (if they are close enough to the junction).

  1. And if we increase the voltage, at some point (VOC we have an opposite current of the diode that evens out the photocurrent? So technically speaking it is (assuming the lighting is constant) a constant current Iph which overlaps with the exponential current ID∝exp(U) from the diode? The point of the solar panel is that you control the voltage by the load you apply to it. I am not sure what you mean with applying voltage to the diode, as there is no external voltage source. You are right, that in principle you could apply a voltage to you solar panel and you could control the current, as it follows the current law described above (hence exponentially) but this is not what you do. What you do is to impose a current which defines your $$I(V=0) = I_{ph}$$ and then the current-voltage characteristics of your solar panel (pn-junction) define which voltage corresponds to which current (that is why the curve shown in your figure is always the same and is just shifted in y-direction)

  2. And how does one "change the voltage" on a solar panel? As described in 4., the voltage is controled by the load you apply. For example, the red line indicates, that the maximal voltage is 0.45 V and between 0 V and 0.35 V the current is nearly constant, no matter of voltage. That is why you should consider the solar panel more like a current source than a voltage source.

$\endgroup$
  • 1
    $\begingroup$ Of course, one can also bias a solar panal (like a diode) and measure the current at the same time. Years ago I was using a curve tracer to measure the characteristics of a diode, which like many was in a black plastic package. I was getting really weird inconsistent results, and finally figured out that stuff happened when I moved my hand to adjust the curve trace display. This cast a shadow on the diode, dropping the measured current. Yes, the black plastic packaged diode was responding to the fluorescent lights in the lab. The rest of the characterization was done in the dark $\endgroup$ – Jon Custer Dec 19 '18 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.