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Ohm's Law doesn't work - Weird?

Discussion in 'General Electronics Discussion' started by NuLED, Jun 11, 2013.

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  1. NuLED

    NuLED

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    Jan 7, 2012
    I have a tiny solar panel I took out from a cheap toy.

    With the DMM it gives 2.744 volts at the constant lighting I have.

    I took a resistor, and connected it with alligator clips, etc. I measured the TOTAL RESISTANCE across all the wires and the resistor. I got 979 ohms.

    So, E = IR, and I = E/R, predicting that I should get some 2.7 mA current.

    And yet measuring the actual current in series with the DMM, I only get 81 uA.

    I don't understand? I measured everything again with another DMM and I get the same results. The solar panel voltage is correctly measured, the total resistor + wires measured is corrected (i.e., the entire circuit resistance minus the solar cell).

    It is off by a tremendous amount, with predicted milliamps and yet actual is microamps.

    ???
     
  2. NuLED

    NuLED

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    Jan 7, 2012
    I checked it against a 4 x AA in series (5.12 volts) and it worked as predicted. (5.2 mA give or take).

    So, something special about the solar cell... What is it?
     
  3. duke37

    duke37

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    Jan 9, 2011
    The solar cell will have a resistance which must be taken into account.
    The batteries will have a low internal resistance.
     
  4. NuLED

    NuLED

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    Jan 7, 2012
    Thanks; any way to check internal resistance?

    I assume you can't just put the DMM in ohmmeter mode when the voltage source is active?

    Or does it have to be calculated indirectly by measuring current?
     
  5. NuLED

    NuLED

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    Jan 7, 2012
    (or measure resistance at zero light levels?)
     
  6. davenn

    davenn Moderator

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    Sep 5, 2009
    also was that 2.744V from the solar cell with or without load ?

    The voltage out of the cell will drop significantly with load applied
    you mauy find its load voltage is only ~ 1.5V
    that will also screw up your calcs

    Dave
     
  7. john monks

    john monks

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    Mar 9, 2012
    At zero light level the internal resistance of a typical solar cell is very high, maybe well in the millions of ohms. When light is applied to a solar cell the resistance, sort of speak, is much lower. Now there are two types of resistance:
    1. Alternating current resistance or impedance.
    2. Direct current resistance that you normally measure with a DMM.
    In solar cells I normally think in terms of impedance so I can ignore the effects of voltage being generated. And the impedance of a solar cell drop greatly with light coming into the cell. And typically DMM's do not measure impedance. Now if you are measuring the resistance of a solar cell you may get some inaccuracies because light interring into the cell will cause the DMM to give a false reading. And the DMM has its own voltage source that can cause the solar cell to go into an abnormal condition such as forward biasing the diodes that comprise a solar cell. To measure the impedance of a solar cell in operation can be tricky and may require some mathematical calculations.
    If you posted a schematic of your circuit maybe we can go into greater detail.
     
  8. NuLED

    NuLED

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    Jan 7, 2012
    Hi Dave, the 2.744 v was from the solar cell only with no load. Just DMM across solar cell.

    John, I don't know how to draw the schematic and upload but it's basically:

    SOLAR CELL --- RESISTOR (979 Ohms including all wires)

    (and DMM in series when measuring current).

    With the 4xAA battery pack, it all calculates out correctly.
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,384
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    Jan 21, 2010
    Firstly, Ohms Law always works because it's a Law.

    A Law is something which is observed to be true. Same as the Law of gravity, the Laws of thermodynamics, etc.

    If you're not getting the results you expect then one of several things may be wrong:

    1) you're making a mistake
    2) the resistance is non-linear
    3) there is a concept you don't understand
    4) there is something you're doing wrong
    5) your equipment is faulty
    6) you don't understand the equipment or its limitations
    7) there is something you've not taken account of
    8) there is an extra resistance hiding in there
    9) there is an extra voltage source hiding in there
    10) there is an extra current source hiding in there

    In your case the answer is (7) because of (8) leading to (6), (4), (3), and finally (1).

    What you need is to concentrate on fixing (6) first.

    Measure the voltage from the cell when under load, then measure the current through the resistor. Then compare that to what Ohms Law suggests.

    You might like to read the voltage across both the battery and the solar panel unloaded and loaded.

    Then try what you're doing now but using a 10 ohm resistor. Do you get unusual results from the batteries now? (be careful you don't burn yourself on the resistor).

    Note that your meter won't ever give you completely accurate results due to issues such as impedance, burden voltage, and absolute accuracy, but you should get close.
     
    Last edited: Jun 12, 2013
  10. NuLED

    NuLED

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    Jan 7, 2012
    Hey thanks! I didn't think about that:

    "Measure the voltage from the cell when under load"
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Great thread! And good advice, as usual. This highlights the subject of internal resistances of voltage sources, which are often not understood by newcomers to electronics.

    Every voltage source has an internal resistance. A PC power supply rated at 30A and 3.3V will have an extremely low internal resistance - perhaps less than 1/1000th of an ohm. This can be described as a very "stiff" or "firm" voltage source - the voltage will not vary significantly over a wide range of load currents. A typical dry cell's internal resistance will be on the order of an ohm, depending on its size. Other voltage sources can have all kinds of internal resistance figures.

    Yes it's possible to measure the internal resistance of a voltage source. You compare the voltage that you measure without load against the voltage you measure with load. For example if you measure 2.744V with just the multimeter connected (10 megohm load - negligible load current) and only 0.0793V with a 979 ohm load connected (I calculated that voltage from your 81 uA figure), the ESR (effective series resistance) of the solar panel, at that illumination level, is about 33 kilohms!

    I calculated that from R = V / I, where V is the voltage dropped across the internal resistance (2.744V - 0.0793V) and I is 81 uA. The exact figure is not really meaningful, since it's not really a linear resistance, but you can get an idea of why you got those unexpected readings if you imagine the solar cell as being a 2.75V source (approximately), in series with a 33 kilohm resistor. Any load resistor you connect across it will form a voltage divider with the internal resistance (ESR), and the terminal voltage you see will be less than the unloaded voltage.

    Those numbers aren't exactly right because I assumed that the multimeter (10 megohm load resistance) was measuring the actual unloaded voltage of the cell. With an ESR of 33 kilohms in the solar cell, even a 10 megohm load will affect the voltage slightly. So those numbers are just approximations, to give you an idea of the scale of the internal resistance of the cell.

    This is why small solar panels are really only adequate to power low-current devices like calculators with LCD displays!
     
    Last edited: Jun 12, 2013
  12. duke37

    duke37

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    Jan 9, 2011
    The maximum power outpuit is obtained when the source resistance equals the load resistance. When this is so, the source voltage will drop to half the open circuit voltage. You could use a potentiometer as a load and adjust it to give half voltage. Then the resistance of the potentiometer can be measured.

    I presume the resistance will vary with the light level.
     
  13. NuLED

    NuLED

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    Jan 7, 2012
    Wow these are excellent responses. I had no idea when I started asking this and thanks again as usual for great insight. In fact I only learned about ESR recently so this is surreptitious.

    Darn it, I thought wow, I can recharge an AA with just this tiny little solar cell!!! I guess I can't, or it will take years.
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    You're welcome! And I think you might mean serendipitous!

    Right. Solar cells are still pretty inefficient. You need a large area and bright sunlight if you want to get a significant amount of power from them.
     
  15. NuLED

    NuLED

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    Jan 7, 2012
    Damn I didn't get a good night's sleep and you are right!!! I should surreptitiously take a nap today! :-D
     
  16. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    LOL :)
     
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