I made some homemade vinegar batteries. The voltage of each cell was measured around 0.45V. I have 18 of them in series and I measured around 6.5V for the whole chain.

Since the vinegar batteries have very low current output, I am using a solar engine to harvest the energy. I used this exact design: http://www.solarbotics.net/library/circuits/se_t1_1381.html

The 1381 that I bought was designed to trip at 4.2V. However, it never seems to produce any output. The output has been steadily staying at around 0.6V. When I measured the voltage across the capacitor, I can see that it was rising when I first connected the circuit. However, it stopped rising at around 2.2V.

Why wouldn't the voltage capacitor get to the same level as the batteries? Does it have something to do with the internal resistance of the battery? In that case, how do I figure out the internal resistance and the trip voltage level that I would need?


Notes on battery internal resistance for an individual cell

The vinegar battery actually behaved quite differently when it was freshly made and after about 3 days.
For a fresh battery cell (when the electrolyte was just put into the vinegar), when I loaded it with a 10K ohm resistaer. the voltage and current kept dropping as I measured them. The voltage and current drawn was really low. The resulte below isn't really that meaningful since the measurements were still changing, but that give you an idea of the magnitude.

V_oc = 0.46V
R_load = 9.8k ohm
I_load = 32.7 uA
V_load = 0.336V

This gives a internal resistance of more than 5.75k ohm!

After a few day (I did not touch it or have it powering anything), the open circuit voltage actually dropped to 0.192V. I loaded it again with 10K ohm. This time, the current drawan was really stable and the voltage dropped was stable as well.

V_oc = 0.192V
V_load = 0.183V
I_load = 17.0 uA
R_load = 9.8k Ohm

This gives a internal resistance of 529ohm.

Not sure why that's happening. Am I doing the internal resistance test correctly? Note that the 6.5V overall voltage I mentioned above is the voltage right now (I just made it). Perhaps it will drop tomorrow.

  • \$\begingroup\$ Can you measure the output resistance of your batteries? You've measured the output with just your meter loading the cell. Now add maybe 10 or 100 kOhm in parallel with the meter and read that voltage. The drop between those two conditions will let you calculate the cell resistance. The resistance of 18 cells in series will be (more or less) 18x the resistance of a single cell. \$\endgroup\$
    – The Photon
    May 27, 2013 at 20:23
  • \$\begingroup\$ I added the note about how I tested the battery's internal resistance to the question. Thanks! \$\endgroup\$
    – tuzzer
    May 27, 2013 at 20:39
  • 1
    \$\begingroup\$ When you say the voltag changes during the measurement, It sounds to me like your batteries are being exhausted as you measure them, or at least the amount of reactive material available at the electrodes is limited so the chemical reaction can't keep up with the amount of charge you're drawing to do the measurement. \$\endgroup\$
    – The Photon
    May 27, 2013 at 20:43

1 Answer 1


you have to understand one the thing for eg,if you connect a capacitor across a 9V PP3 battery or some other battery (not fruit or vegetable battery) the capacitor would get immediately charged to the battery voltage (but it would take a certain time because of the presence of internal resistance would reduce the flow of charge Q; C=Q/V) and the current through the capacitor would get immediately reduced.

Click here and simulate the circuit by giving different R values like a)0.1 ohm b)1 ohm c)100 ohm d)1K and observe the volatge and current of capacitor for different values of resistor.


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