# Problems with solar cell efficiency

1. Dec 21, 2009

### sc3005

As the solar cell gets further away from the light source, does it get more efficient or less efficient?
At the moment I am able to work out how much power enters the solar cell.
However, it is pretty weird because as I found out from my experiment, the efficiency of the solar cell gets better as the cell moves away from the light source.

2. Dec 21, 2009

### sc3005

It seems as there is more lux then the solar cell would be less efficient

3. Dec 22, 2009

Can you try describe the situation a bit more fully, to help people understand the problem?

For instance, are you making practical measurements, or is this simulated by computer?

What is the load connected to the solar cell? Is it a resistor?

Having said that, one way in which efficiency may fall at high input powers is that the load voltage may get big enough to cause significant forward conduction in the photocell.

4. Dec 22, 2009

### sc3005

Oh I'm making practical measurements.
The solar cell is connected to a voltmeter and an ammeter to record the current and voltage and to find out power output.
To measure power input, I used a 100W bulb and shone the light at the solar cell from different distances. I recorded the lux readings from distances of 0 to 50cm with 5 cm intervals.
Since there was the efficacy reading on the bulb, which was 12.5, I used it to get my power input, since I've already had the lux reading and could calculate lumens.

5. Dec 22, 2009

### JaWiB

The maximum power point will change with the light intensity (e.g., as shown here) though I don't know how much that would affect your measurements. It might be best to plot an I-V curve at each distance to find the maximum power point.

6. Dec 22, 2009

### sc3005

so you mean that the maximum power it can reach will change with light intensity, efficiency will not change?

7. Dec 22, 2009

Hello again,

You mention an ammeter and a voltmeter, but what kind of load is being used? The load is the device that dissipates the photocell output. For instance, if you are using a fixed resistance, this can only be optimal at one light intensity: efficiency will be degraded at other intensities.

On the other hand, perhaps are you using a variable resistance and attempting to set this for maximum power at each intensity. As has already been mentioned, it might help to plot curves - The V/I characteristics would be useful in themselves, and if you wanted to be sure of getting Pmax at each level, you might also plot power against load resistance, that is V*I versus V/I - look for the peak value.

As to how the efficiency varies, probably we shouldn't spell it out, but you might try thinking about how the voltage and current available from the cell vary with intensity.

Possibly it would be accurate enough to say that the current generated into a short-circuit is a constant factor times the input power, i.e. so many Amps per Watt of optical input.

Remember that the available voltage is limited by forward diode conduction - look at the curves JaWiB has provided. How does the optimum load voltage change with intensity?

At quite high intensities, there may be significant complicating factors, such as heating of the cell increasing forward conduction, or voltage drop in internal series resistance.

8. Dec 22, 2009

### sc3005

Oh no, I didn't use a load cuz I didn't think it was needed.
The current and voltage definitely varies depending on the light intensity.
Because as the distance increases between the solar cell and the light source, the current and voltage decreases.

9. Dec 23, 2009

If you were not using a load, how could you get current and voltage values?

Were you making current and voltage measurements separately? That is, measurements of the current directly into an ammeter (short-circuit), and then voltage measurements with only a voltmeter connected (open-circuit)? Multiplying these figures together would give an over-estimate of the output power - can you think why?

It would be very helpful if you could post a schematic (circuit diagram) of your setup.

10. Dec 23, 2009

### sc3005

i'm sorry i don't know how to post a picture of my setup.
But i can help to describe the setup.
Ok there is a light source that shines on the solar cell.
The solar cell is then connect in parallel to the voltmeter (multimeter), and series to the ammeter (another multimeter). That's it

11. Dec 23, 2009

From what you have just said, I think you may have a setup which will not give a valid measurement of the photodiode output. With only the voltmeter and the ammeter in circuit, the cell probably cannot deliver much power.

There are two possibilities:

1) Ammeter in series with cell, then voltmeter in parallel. Here the effective load will be the (large) input resistance of the voltmeter. This will most likely be above the optimum load, except perhaps at very low illumination.

2) Voltmeter directly in parallel with cell, then ammeter across voltmeter. Here the effective load will be the (very low) input resistance of the ammeter. This will most likely be much less than the optimum load.

Perhaps this is something you should (if possible) discuss with a teacher or lecturer.

12. Dec 23, 2009

### sc3005

It looks like I'm taking the first possibility for my experiment at the moment.
I'm not sure what you mean by load, I checked wiki and it says it has something to do with power output?

I was supposed to carry out two experiments, one was this light intensity one and the other was altering temperature.
For the temperature experiment, the results went according to plan.
The efficiency decreased with a temperature increase.
This experiment was carried out using the first possibility you suggested.

13. Dec 23, 2009

### JaWiB

The load is where the power is dissipated. Could be a device that consumes power, or it could just be a resistor that turns the power into heat. Or in your case, you might also use a variable resistor (potentiometer)

Keep in mind that in the ideal case the ammeter has zero resistance (behaves as a short circuit) whereas the voltmeter has infinite resistance (behaves as an open circuit). That means that if you connect both to your cell with no load you either get virtually no current and are measuring the open circuit voltage of the cell, or you get virtually no voltage and are measuring the short circuit current of the cell. Either way, I would expect that your measured power would be close to zero.

14. Dec 23, 2009

### sc3005

Thanks you so much guys, I understand what's wrong with my experiment now.
It seems that I cannot get my resistor till after the holidays so in the mean time I've got to work on the theory of my extended essay first...

15. Dec 24, 2009

### sc3005

Actually I have one more question to ask, in theory does the efficiency of the solar cell increase with light intensity? Or the maximum power point changes only?

16. Dec 25, 2009

c3005,
I'm not sure what theory would be expected at your level of study. Do you have a suitable textbook to refer to? That said, here are some hints which may help you see what to expect, in a qualitative way.

1) Over quite a wide range of optical input powers, the current which the cell would deliver into a short-circuit is likely to be be proportional to the optical input power. The short-circuit output current can be approximated as so many Amps per Watt of input.

2) If you look closely at the I-V curves for a solar cell (look at the graph from JaWib), you will see that the maximum output voltage gets somewhat larger as input power increases. The voltage at the maximum power point also increases, which would predict better efficiency at higher inputs, provided that the current at maximum power remained in similar proportion to the short-circuit value (possibly a reasonable assumption, but note I'm not giving you any justification for it).

3) Unfortunately, increasing the input power will also tend to increase the cell temperature, because much of the optical input will be lost as heat. As the cell gets hotter, the voltage at the maximum power will reduce, lowering the output power and hence the efficiency.

4) At high power inputs, the cell current may be large enough to cause a significant voltage drop in internal series resistance within the cell. This also lowers efficiency.

17. Dec 26, 2009

### sc3005

first of all i'm at high school level
second of all, i'm asking in a situation where the temperature does not increase. so let's say the temperature does not increase, will there be a change in efficiency?
for my high school essay, i'm conducting 2 experiments.
1 on temperature against efficiency and 1 on light intensity against efficiency. i have actually conducted the temperature one.

18. Dec 26, 2009

Didn't you read item 2) in my last post?
The voltage available from a (single photo-diode) cell is limited by its forward conduction characteristic. This can be approximated by the diode conduction equation. I = IS(evD/nVT - 1)
The bottom line is that for moderate intensities and at constant temperature, the available voltage should increase by roughly 20mV per doubling of input power. The Amps / Watt ratio should be nearly constant, giving a modest increase of efficiency with intensity.

Note however that the above assumes optimal loading at each intensity. It seems likely that your experiment was made with a very high load resistance, so the cell power output could hardly increase as the intensity was increased - hence falling efficiency.

Finally a suggestion - take your existing data for cell output voltage and intensity. Plot voltage (y axis) versus log (intensity). Do you get a straight line, at least over part of the range? What is its slope?

19. Dec 26, 2009

### sc3005

It seems that i've got to get a a variable resistor for this experiment then.
At the current moment i'm experiencing falling efficiency.