I need help with figuring something out for my Science Fair. I need to test Solar Panels in different temperatures, and I need to test a solar panel's current and voltage inside a freezer. Is there any way I can take my experiment out of the freezer and see the voltage and currents produced by the solar panel when it was in the freezer? I can only use a multimeter (any type), my solar panel, and my freezer.

Proposal for experiment:

Warming Up To Solar Panels

I will investigate the effect of different temperatures on the efficiency of a solar panel. My project helps the environment because this may make solar panels a more efficient energy source, so it can replace non eco-friendly energy sources that are going to run out.

The experiment is testing a solar panel at different temperatures, using a metal plate, ventilation, a freezer, and a fan to adjust temperature. I will carry out the experiment by building a structure for the solar panel to rest on, which incorporates the factors mentioned above (fan, etc.). A lamp will be the source of light. I will be using a new 40 watt incandescent light bulb for each test. From research I know that this light bulb will produce 450 lumens. The metal plate will start at 30 - 40 ° Celsius, with ventilation around 25 - 30° Celsius, with the fan it will go down around 15 - 20° Celsius and I will set my freezer to -18 - -25 ° Celsius. I will be recording temperatures with a thermometer. The control variables are the solar panel and the lumens on the solar panel, the dependent variables are the voltage and amperes from the solar panel, and the independent variable is the temperature. I will repeat the temperatures 3 times, and take these 3 measurements for each temperature and average them out to reveal a trend.

The materials I will need are a solar panel, a voltage meter, a infrared thermometer, a heating bath, an ammeter, a fan, a metal plate, LEGO bricks, a lamp a freezer, a thermometer. I will need 3 weeks for completing the experiment, 3 weeks for research, as well as 2 weeks to complete the write-up.

(google doc version)

  • \$\begingroup\$ Please describe your test protocol in more detail. Do you have the light source inside of the freezer to illuminate the panel? How do you monitor the temperature of the panel? \$\endgroup\$ Jan 3, 2017 at 2:27
  • \$\begingroup\$ Inside a freezer is generally pretty dark - I wouldn't expect to get any power from a solar panel inside a freezer. Depending on your location, you may be able to get data at low temperatures simply by leaving the panel outdoors. \$\endgroup\$ Jan 3, 2017 at 2:30
  • \$\begingroup\$ @PeterBennett Sorry! I forgot to link my proposal, but I edited the thread with it now \$\endgroup\$
    – Joshua
    Jan 3, 2017 at 2:42
  • 2
    \$\begingroup\$ Is there something that prevents you from running the wires out of the freezer, making the measurements through the wires while keeping the entire setup (solar panel, light source) in the freezer? \$\endgroup\$ Jan 3, 2017 at 2:53
  • \$\begingroup\$ true... that could work. thanks for the answer! it was nice to find an answer, usually on other stack exchanges my questions get locked for being off topic :/ \$\endgroup\$
    – Joshua
    Jan 3, 2017 at 2:59

1 Answer 1


Assuming the light source is also in the freezer, and that you are not, then it is "easy enough" to bring out test points.

You need a current sense resistance to measure current - there are other means but that is easy and cheap.

In normal operation R_Isense should be sized so that the voltage across Rsense is small compared to panel output voltage. eg if you have a 6V panel voltage under normal operating conditions then a peak voltage across R_Isense should be say no greater than about 5% of Vpanel = 0.3V, and ideally no more than about 1% = 0.06V.

In this case if the load is a resistor for test purposes (which is a good idea) then both V and I can be determined from a single reading if the value of R is known. If the load is not resistive - eg an LED or light bulb or motor then a small sense resistor is needed.


Panel max power in test = Wmp Panel voltage during test = Vmp Panel max current in test = Imp

(1) For a resistive test load.

eg set load to load panel properly at max power.
R = V/I = Vmp/Imp = 6V/0.1A = 60 Ohm load resistor.
Now Vpanel = V load resistor
I panel = Iload resistor
But I = V/R so Ipanel = Vpanel/60 in this case.

To measure Vpanel externally you need run only 2 wires out of the test area to an external meter. From Vpanel you cam also calculate I panel as above.

(2) For a non linear test load - eg LED, bulb, motor etc.

You will need to use Rsense as above to measure I panel.
You will now need 3 wires from the test space.
Connect Vpanel+ to load+
Vpanel - to Rsense+
Rpanel - to "ground".
Rsense - to ground.
Run wires from - Load+ (= Vpanel+)
- Rsense+ (= Load-)
- Ground

Vpanel = Vpanel+ to ground voltage
Ipanel = (Vsense+ above ground) / Rsense.

Set Rsense to about
R = Vmp/Imp/k
or R = Wmp / Impp^2 /k

where k is between say 10 and 100
(Lower affects Vload less but produces lower Vsense readings.)

ie Rsense will be about 1% to 10% of the effective load resistance at full power.


Using a sense resistor - Example only

Vmp = 6V, Imp = 100 mA, Wmp = 600 mW. k= 100 say

R = Vmp/Imp/k = 6V/0.1A/100 = 0.6 Ohms. or R = Wmp / Impp^2 /k = 0.6W / (0.1A)^2 / 100 = 0.6 Ohms (ie same result).

For ease of implementation use 1 1 Ohm sense resistor. Then at eg 100 mA, Vsense = IR = 0.1A x 1 Ohm = 0.1V. ie Vsense in mV = I load in mA.


If Rload is inside the test chamber it will affect the temperature a little.
If the light source is inside the chamber it will affect the temperature - how much depends on intensity and efficiency of the source -
eg a 300W halogen bulb at about 2 feet is approximately as bright as sunlight overall but has far more % infrared than sunlight and will heat a test chamber substantially.
An LED source will provide less IR per light output BUT the narrow bands of light energy may cause lower net panel efficiency.


Panel operated as an ~ current source for output measurement.

A good way to measure panel output is to operate it with a load resistor much lower than the optimum resistor that gives max power. It then operates as an approximate current source with Ipanel ~= proportional to light energy.

This means that Vpanel will be much lower than Vmp and not a useful indication of power output, but it is directly proportional to panel current and a good indicator of panel output under various conditions.

"Shorting" the panel with an ammeter (which will usually have low resistance) is a good approximation to the above.

A multimeter on a 200 mA range will usually have a resistance in the 1 to 20 Ohm range (some but not all are nearer the low end of that range.)

A multimeter on eg a 10A DC range will usually have a resistance under 1 Ohm and is a good load for a panel acting in current-source mode. _______________________________________


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