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I have a 12v/22A DC switching power supply and 9 peltier elements rated for 12v/6A.

I would like to limit current to 2A per peltier element.

I don't know how to calculate the impedance (in ohms) that I need.

Intuitively, by ohm's law, every peltier element's impedance is 12v÷6A=2Ω and the impedance I'm aiming for is 12v÷2A=6Ω. Does this mean I need a 4Ω current limiting resistor in series with the peltier element?

Also, what power (watt) rating does my resistor need?

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If the Peltiers are rated at 12v/6A, then I would try measuring the current taken by 3 in series. If you can tolerate that your resulting current isn't exactly on target, then job done!

This is far more efficient than burning excess power in resistors, especially as you're starting from an SMPS.

While the Peltier current varies a little with its thermal conditions, they are so inefficient that it's not unreasonable to model them as a resistor which changes a bit. If you can turn the voltage down on your SMPS, this will reduce the current. If not, then a DC-DC buck converter is reasonably inexpensive, if you want finer control of the current than just splitting the voltage across several in series.

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The Peltier Device has a nonlinear rise in resistance with lower power applied, which you must derive from datasheet graphs. So it must be regulated with a constant current source or an active current limiter.

The reason for this is the resistance and power absorption changes with temperature difference.

It may mean the device has to dissipate the same power as the Peltier Device, so separate heatsinking is necessary.

Current sensing can be done with a 50mV to 100 mV shunt and scaled down Vref to a comparator driving a BJT or FET switch.

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    \$\begingroup\$ @thwd resistor power = I x I x R , so 2A of current passing through 4 Ohm resistor will require a 16 watt of power resistor, which is not easily available. You can use four 16 Ohm 5 watt resistors in parallel. \$\endgroup\$ – Ahmed M.Zahran Jul 10 '18 at 6:04
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    \$\begingroup\$ You might never achieve 2A with a 4 Ohm resistor as the Peltier device could start at >10 ohms thus an active current power current limiter with low voltage drop is needed. \$\endgroup\$ – Sunnyskyguy EE75 Jul 10 '18 at 6:10
  • \$\begingroup\$ Even if peltiers are connected in parallel with this switching power supply ? \$\endgroup\$ – Ahmed M.Zahran Jul 10 '18 at 6:19
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    \$\begingroup\$ Each device may have a different thermal different and thus never share current equally due to thermal resistance of a common heat sink. So each device must be regulated. But go ahead and find out for yourself. \$\endgroup\$ – Sunnyskyguy EE75 Jul 10 '18 at 6:22
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    \$\begingroup\$ Yes , and define dV/dt , dB compression curve ( many types ) linear, log, non linear Curve \$\endgroup\$ – Sunnyskyguy EE75 Jul 10 '18 at 10:25
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Short answer- the 4Ω is based on assumptions that aren't true, but it probably wouldn't be too far off.

The 6A rating of the Peltier device will simply be a maximum allowable current based on device construction, ie any more and you could damage it.

Add to that the fact that a Peltier device can also generate a voltage which may be positive or negative depending on the temperatures of the two sides (see https://en.wikipedia.org/wiki/Thermoelectric_effect). Plus it will also have a small internal resistance.

A 4Ω resistance will provide at most 3A from 12V to a Peltier device at 'neutral' conditions (ie no temperature difference, but note that will change after power is applied).

If it were me, I would do some tests with an ammeter and an assortment of resistors. Keep in mind that 3A at 12V is 36W which needs to be dissipated somewhere, so the resistors will get hot. You'll probably need a handful of 10W or higher resistors.

EDIT: Because you have 9 peltier devices, another option is to put them in series/parallel. Perhaps start with three strings of three. Then you avoid losses in the dropping resistor.

   +--P--P--P--+
   |           |
---+--P--P--P--+---
   |           |
   +--P--P--P--+
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