How to measure verry smal current (amperes) with multi meter?


I'm trying to measure how much current can be generated from tomatoes.
Please have a look at http://maaikedevos.blogspot.com and my question will make a lot more sence.

What option should i use ( 200u, 2000u, 20m, 200m) if i use 20 on the voltage and i want to use the formula W=VxA ?
It seems to me that the measuring in itselfs alters the numbers. The brightness of the LED changes.
What is happening here ? How can i measure amperes ?

On the picture:

1/ LED hooked up to a battery. Multimeter placed in series to measure the current. (the LED seems to light up but this is reflection)

2/ I turn the multimeter to 200m, the LED lights up very brightly. On the display: 03,8

3/ Multimeter is turned to 20m, the LED lights up. On the display: 3,29

4/ Multimeter is turned to 2000u, the LED lights up. On the display: 1197

5/ Multimeter is turned to 200u, the LED lights up but not verry strong. On the display 1  .

Thanks to any one who can explain this to me in clear language. This are my first staps in elctricity-land.


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You are VERY lucky you haven't destroyed your LED by running it with no current limiting resistor....

Forget using LEDS to measure it.

The meter's own resistance is varied by switching in different current ranges,at least in some meters.

The method you should use is to find a suitable potentiometer - I'll guess at ~500 Ohms. Connect it as a variable resistor. Measure first of all, the output voltage of your tomato WITH NO LOAD. Connect your pot, now adjust the pot until the meter reading is exactly HALF the first measurement.

Measure the resistance of the pot, in the position you left it. This is the equivalent to the internal resistance of your source.

The LOWER the internal resistance of the source, the more power you can extract.

With a load EQUAL to the internal resistance, you are extracting the MAXIMUM you can from that source.
dr zeezucht (author)  steveastrouk4 years ago
Hey thanks for the swift reply's .

It seems that my multimeter is the kind that changes his internal recistance when switching in an other current range. (seems strange to me i must admit)

I don't have a potentiometer. Is there an other way to measure the intern recistance in my tomato ?

If i could do that i could use the formula: I=V/R

If not using a LED as a LOAD what would you suggest ? The nice thing about the LED is that you can see if it works a little bit yet. ;-)
Most of the electrical devices work, or they don't. I going to connect a lot of tomatoes in the end, and i need to check the current now and than.

The LED is a lousy device, because you can't adjust it, and its non-linear anyway, so you can't predict much either.....

Try a piece of heater wire from an old toaster. Measure the resistance of it - I can't see you needing more than 100 Ohms to try it. connect one end of your tomato to one end of the wire, with one terminal of your meter (volts range) connect the other end of the tomato to the other terminal of your meter, and touch THAT up and down the heater until you find the sweet spot I mentioned earlier, where the meter reads half the O/C voltage of your tomato.

Now you can scale off the reading by saying
resistance of tomato = length to half volts x resistance of whole resistor/ length of whole resistor. Now you have the maximum power point of the tomato.

You realise of course that the generation is only occuring between the acid soaked metal parts - sprinkling salt on your battery may get you more output too perhaps.

I think what Steve is saying is that you can model your tomato battery as a voltage source in series with an internal resistance.  A model is often called a "Thevenin equivalent" source.

You can see some diagrams of what this circuit looks like in this Wiki article:

The game here is to pick some different resistors that are close to the internal resistance of your tomato battery, and use them as loads, and plot I and Vload as you load the tomato battery.  Your loading graph will also include the points (Voc, 0) and (0, Isc) where:

Where Voc is the "open circuit voltage"
(measured when Rload = infinity)

and Isc=Voc/Rin is the "short circuit" current 
(measured when Rload = 0)

The slope of your plot of I vs Vload should be (-1/Rin), since:

Vosc = I*Rin + Vload 

I = (Vosc-Vload)/Rin = Iosc - Vload/Rin
I was avoiding short circuiting her tomato....
My method, as outlined is gentler than the method you propose....