0
\$\begingroup\$

I am trying to make a datalogger with an Arduino, that should measure the voltage and current comming out of a small solar panel. When I measure the voltage and current using a multimeter, I get a result of around 3.4 volts, and a current of around 122 microamperes. An acquaintance told mee that, since I want to measure both values at once, I could just connect a resistance in series with the solar panel, so that I can measure the voltage and then use ohm's law to find the current, since I know the resistance.

In this case I am using a 10 ohm resistance, but that means that when I measure the voltage and get a value of 3.4 volts I should be reading a current of 0.34 Amperes. However, my multimeter reads 124 microamperes. What am I doing wrong? How can I fix this?

enter image description here

Edit: As suggested, I tried measuring in parallel. When I connected the setup shown below, I measured 1,5 v and about 3 microamperes without the resistance, and as soon as I connected it, it dropped to 0. What did I do wrong? I tried to connect as suggested. enter image description here enter image description here

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Put the multimeter in parallel with the resistance. \$\endgroup\$
    – Solar Mike
    Oct 1 '19 at 20:23
  • \$\begingroup\$ I'm sorry, I am really new at this. What would that look like? I don't have an intuition yet on these things \$\endgroup\$ Oct 1 '19 at 20:26
2
\$\begingroup\$

To measure current with a series resistance, you would use your multimeter as a voltmeter and measure the voltage across the resistor.

Your drawings show a resistor in series with your meter, and the notes say that you are making measurements with voltmeter and ammeter. When you measure the voltage, you get the same result as measuring the voltage directly.

Connect the 10 ohm resistor to the solar panel so that the circuit is complete.

Measure the voltage across the resistor.

Now \$E=IR\$ so \$I=E/R\$ so divide the measured voltage by 10 to get the current.

It would greatly surprise me if yoy get anything like 120uA. That value that you have measured looks wrong to me. A solar cell that only delivers 120uA would be very weak - there's not much you can do with that little current.

I think you are probably measuring current wrong. Most multimeters require that you plug the red lead into a separate jack to measure current. Double check your connections, and see if you get a more reasonable amount of current.

To measure current, you would usually want to measure it while your solar cell is powering a load. What you are measuring is the short circuit current. That's the maximum current it can deliver, but it is at the minimum voltage.


Since you aren't sure what "parallel" and series are, try this:

Connect the resistor from the + to the - of the solar cell.

Connect the multimeter red lead to the solar cell +, and connect the black multimeter lead to the solar cell -.

Your multimeter is now in parallel with the resistor.

\$\endgroup\$
2
  • \$\begingroup\$ I tried connecting what you described on a breadboard, but I got 0 reading on everything. I added pictures to the question. Why could that be? \$\endgroup\$ Oct 1 '19 at 22:18
  • \$\begingroup\$ It looks like the red wire is plugged in to the wrong row. It seems to be in row 10, but the resistor and the wire to the multimeter are in row 11. \$\endgroup\$
    – JRE
    Oct 2 '19 at 11:27
0
\$\begingroup\$

enter image description here

Here is how to wire it for the test, and make sure your probes aren't touching, it's hard to tell from the picture

\$\endgroup\$
6
  • \$\begingroup\$ Thank you! I don't think that my probes were touching, since I was getting a reading until i put the resistance in place. Does that make sense? I do not understand why all reading would be 0 when I put the resistance in. \$\endgroup\$ Oct 1 '19 at 22:41
  • \$\begingroup\$ When you had the circuit wired as before (as in your drawing), you were actually looking at the "open circuit" voltage of the solar cell. What that means is there was no load (except for the very small one associated with the meter itself). Now you have a 10 ohm load on the solar cell which is vastly less resistance than the volt-meter alone. You may have to take it outside into the sun in order to see an appreciable current through the resistor, it depends on the characteristics of the cell. \$\endgroup\$
    – io_silver
    Oct 1 '19 at 22:44
  • \$\begingroup\$ I am always confused with the wiring when 3 cables touch. Should I just put the tip of all 3 cables together? \$\endgroup\$ Oct 1 '19 at 22:48
  • \$\begingroup\$ A larger value resistor will produce a larger voltage for a given amount of current through the circuit, so consider using a larger value (like 1K or bigger) just so see if you can get "something" \$\endgroup\$
    – io_silver
    Oct 1 '19 at 22:49
  • \$\begingroup\$ @Nick Heumann yes, just connect all three things together by any means. Your breadboard photo appears correct, although breadboards can sometimes result in sketchy connections. \$\endgroup\$
    – io_silver
    Oct 1 '19 at 22:50

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.