I am using TEC1-12706 to generate electricity from hand heat. When I measure the voltage with multimeter , the reading is about 72mV and without connecting any load I also measured the current(about 64 micro watts). Is the multiplication of voltage and ampere actual watt output or I need to measure the voltage and current with a load connected to it?

Note: The peltier has internal resistance of 2-2.2 ohm depending on the temperature. enter image description here

  • \$\begingroup\$ Attaching a load will make the voltage drop, you need to measure both with the load attached and also take into account the burden voltage of your multimeter \$\endgroup\$
    – PlasmaHH
    Aug 31 '17 at 8:03

You might think you're measuring the open circuit voltage, and the short circuit current. Unfortunately, DMMs make much better voltmeters than they do ammeters. Depending on the selected range, your ammeter might have an input resistance comparable to the Peltier. Measure the short circuit current on several different ranges to see this effect.

It may be better to construct a load well below the Peltier resistance, say 100mohms, and measure the voltage across that load, at least then you know what your non-zero load resistance is. Better yet, make several low value loads, and extrapolate the readings down to zero ohms, to estimate the true s/c current.

Once you have the true o/c voltage and s/c current, your best power output will be at 25% of their product, into a load equal to the Peltier internal resistance.

  • \$\begingroup\$ Will I have to add internal resistance with load resistance while using the formula P=V^2/R ? \$\endgroup\$ Aug 31 '17 at 15:03
  • \$\begingroup\$ That would give you the total power delivered. If you want to know the power in the load, you would have to use \$R_{load}\$ and \$V_{load}\$ \$\endgroup\$
    – Neil_UK
    Aug 31 '17 at 16:47

You have measured the voltage and current under different circumstances.

  • Your voltmeter acts as a very high impedance and barely loads the generator at all so effectively it is the open-circuit voltage. I = 0. Therefore P = 0.
  • Your ammeter has a very low resistance and so measures the short-circuit current when connected as you have shown. V = 0. Therefore P = 0 again.

You are correct that you need to measure current with a load connected. Try various values, measure the voltage across the resistor and then the current in series with the resistor (a second meter would be handy) and plot them on a graph. You should find that available power peaks at a certain load resistor value.

  • \$\begingroup\$ If you know the value of the resistor you only need measure voltage across it P = V^2/R \$\endgroup\$ Aug 31 '17 at 8:09
  • \$\begingroup\$ but the peltier already has a resistance of it's own. Still I have to connect load?That's why the confusion is being created. \$\endgroup\$ Aug 31 '17 at 8:12
  • \$\begingroup\$ See Maximum power transfer theorem. \$\endgroup\$
    – Transistor
    Aug 31 '17 at 8:18
  • \$\begingroup\$ Will I have to add internal resistance with load resistance while using the formula P=V^2/R ? \$\endgroup\$ Aug 31 '17 at 15:02

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