3
$\begingroup$

please explain the VI characteristics of a solar cell. The characteristics is given in my book without any explanation. How can the Voltage decrease on increasing current shouldn't it be opposite.

Solar Cell I-V characteristics
Solar Cell I-V characteristics
(Image from Electrical 4 U - Characteristics of a Solar Cell and Parameters of a Solar Cell)

$\endgroup$
0

4 Answers 4

3
$\begingroup$

If I invert your graph and change the direction of the current axis I get the graph in the middle.

enter image description here

That is an $IV$ characteristic which looks very like that of a forward biassed diode and in fact the solar cell is a diode which is designed to act as a current source when illuminated with light, the current delivered by the current source being proportional to the solar radiance (intensity) falling on it.
If I flip the centre graph back to the orientation of your graph you will see that the only difference is the direction of positive current and an offset of $I_{\rm sc}$.

The simple equivalent circuit os shown on the left with a current source $i_{\rm ph}$ and a forward biassed diode with a current $i_{\rm D}$ passing though it with $i=i_{\rm ph} -i_{\rm D}$ and the current $i_{\rm D}$ is very small.

This model can be improved to better represent a real solar cell by the introduction of two resistor $R_{\rm p} \approx 1000 \,\Omega$ and $R_{\rm s} \approx 0.5\,\Omega$ as shown in the right hand diagram.

enter image description here

$\endgroup$
2
$\begingroup$

VI characteristic graph of diode with and without illumination (breakdown not shown):

When light falls on the depletion region of the diode, electron-hole pairs are generated that try to reach their parent nuclei. The electrons move towards the n-side and the holes to the p-side. This leads to a photovoltage that tries to make current flow in reverse direction.

Let $V_p$ be the photovoltage and $V$ be the external applied voltage.

In the graph one can see (under illuminated conditions):

  • If $V=0$, there's a negative current (i.e., a reverse current) in the circuit due to the photovoltage.

  • In forward-biased, if $|V_p|>V$ then the current is negative and if $V>|V_p|$ then the current is positive.

  • In reverse-bias ($V<0$), the photovoltage is largely responsible for the negative current; the current does not increase much on increasing the applied voltage (until reaching breakdown).

OP's graph just seems concerned with the magnitude of current in case of solar cell (quadrant 4 in my diagram).

The graph is made using the diode equation which says current in the diode $i = i_0(e^{kV}-1) - i_p$, where $i_0$ is the saturation current without illumination, $i_p$ is the photocurrent, $V$ is the applied voltage and $k$ is a constant (more on that here).

$\endgroup$
0
$\begingroup$

A simple (but sufficient) model is to consider the sunlight as a stream of photons, each with their own energy depending on the colors in the sunlight. Each photon can give its energy to a single electron.

If you try to get out a small current from the solar cell, there are more than enough photons, but you can't change the energy of the photons. That limits the voltage.

If you try to get a high current out, you need a high voltage, but that gives a problem. Red photons only carry a limited amount of energy. Purple photons carry only a little bit more. So as you try to achieve a higher voltage, you quickly lose usable colors of photons. That's the sharp drop in current.

$\endgroup$
6
  • 1
    $\begingroup$ No, that is wrong. You will get the same curve with blue light. $\endgroup$
    – user137289
    Feb 4, 2020 at 13:31
  • 1
    $\begingroup$ @Pieter: Exactly the same? Or just the same shape? I didn't want to go into bandgaps, that's why I started by explicitly stating that it's a simplified model. $\endgroup$
    – MSalters
    Feb 4, 2020 at 13:46
  • $\begingroup$ Current depends on intensity of course, but one would get exactly the same value of the open-circuit voltage. $\endgroup$
    – user137289
    Feb 4, 2020 at 13:48
  • $\begingroup$ @Pieter: How would you even get a V-I graph of a open circuit? I'm not understanding what you're getting at. Or do you mean that the open-voltage of a solar cell (i.e. at I=0A) doesn't depend on the colors of the photons? $\endgroup$
    – MSalters
    Feb 4, 2020 at 13:54
  • $\begingroup$ The value $V_{oc}$ in the graph denotes the open-circuit voltage. $\endgroup$
    – user137289
    Feb 4, 2020 at 13:59
-1
$\begingroup$

The characteristic is measured under illumination, when the solar cell is a source of electrical power, connected to a load. When the resistance of the load varies, one can get different points on the curve.

Compare with a battery: voltage decreases when a load is connected.

$\endgroup$
3
  • $\begingroup$ Suppose i increase the resistance then the current should decrease keeping the voltage constant then how does voltage decrease?(as shown in the graph) Are we also taking internal resistance non zero so that (EMF=V - IR ) like a battery $\endgroup$ Feb 4, 2020 at 10:20
  • $\begingroup$ First of all, a solar cell is not linear, the internal resistance is not constant. At low voltages, it behaves like a current source (infinite internal resistance). The first sentence of your comment I do not understand, it seems to be contradictory about the voltage. $\endgroup$
    – user137289
    Feb 4, 2020 at 10:34
  • $\begingroup$ I know it is not a linear source just search ing for analogy there (i was only asking about the fundamental cause of the decrement in current as potential increase for the solar cell) $\endgroup$ Feb 4, 2020 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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