# Wiring multiple peltier units

First of all this question relates to powering Peltier units and not the physical relationship. I have two Peltier units and a 320w 12v supply similar to this the units are TEC12715. They are physically arranged in next to one another(parallel) and I have wired them in series; they are sandwiched between two heat-sinks which both have fans(also powered by the same supply). I am trying to cool a sealed box and currently the sealed(cooling) side will only drop a few degrees below ambient while the hot side is hitting 60*C.

Should I wire these units in parallel or series?

Thanks, Mark

• That depends entirely on your supply voltage and the characteristics of the peltiers, which youh haven't told us anything about. Jul 14, 2015 at 9:07
• Please excuse my ignorance. I'm not an electrical engineer my discipline is software engineering. I believe I have supplied the rating of the TEC (15A 12V) and the rating of the power supply(320W 12v 25A)? Can you please clarify on the information you require? Thanks.
– Mark
Jul 14, 2015 at 9:16
• Have you tried more air on the hot side first? Unless you are trying to cool a crate of beers, 170W is a lot for a TEC and just a bunch of air inside a sufficiently isolated box should become cold rather quickly, unless the hot side stays hot, because you're not taking away the heat. (Note: even with a crate of beers, the air should cool down well enough, unless the air gets forced past hot beer a lot) Jul 14, 2015 at 9:39
• I'll try adding a bigger ventilation port its currently 80mm and let you know if that helps after first rewiring the thing.
– Mark
Jul 14, 2015 at 10:57
• How did adding ventilation work out? Jul 12, 2016 at 10:46

Supply and load are both 12V so that means you will need to connect everything in parallel.

BUT since the TEC uses 15A you can only connect ONE to the supply as the supply can only deliver 25 A. When Connecting 2 TECs in parallel you need a supply of: 2 x 15 A = 30 A. Your supply is 25 A which is not enough.

So you can only use ONE TEC with ONE supply. For the second TEC you will need another supply OR ONE more powerfull supply of 12V and at least 30 A and then connect everything in parallel.

• I have measured the current configuration at 5A and 12v when they are run in series. So if the voltage is currently halved am I correct in assuming these devices will consume 20A at 12v when run in parallel. (I don't think the quality of the units is very good).
– Mark
Jul 14, 2015 at 9:38
• 12V/5A = 2.4 ohms for 2 in series = 1.2 ohms for one. 12V/1.2 ohms = 10 A for one x2 = 20 A for two. Indeed, that will work. Jul 14, 2015 at 9:51
• Thanks for the confirmation it appears my a-level in Physics was useful after all haha.
– Mark
Jul 14, 2015 at 10:30

This depend on the actual physical situation as well as the characteristics of the Peltiers. You cannot (successfully) analyze them as simple resistors because the current varies (strongly) with the temperature and temperature differential as well as the type of TEC and the applied voltage. A better way to think of them would be as a (somewhat fixed) resistor in series with a battery that has a temperature differential and temperature dependent voltage. Once the setup is fixed, the current at a given voltage will thus vary with initial conditions and time.

A reputable manufacturer will have a ton of application data - since it appears you did not deal with such a supplier, second best is to piggyback off a major supplier such as Melcor- you can download their analysis software and see what happens in various scenarios. There are also hobbyist-oriented tutorials of various qualities available on the 'net, but it's best to deal with real engineering data if it's not too confusing.

If you don't want to apply engineering to the task, I suggest fiddling with the voltage and current to the extent possible and pay particular attention to cooling the hot side (fans, heat pipes, liquid cooling if possible). If you were to put 300W into the TEC that 300W has to be dissipated along with any other heat it is pumping. At some point the TEC just becomes a heater and the cold side gets warmer than ambient. Even half that is a lot of heat to get rid of - look at the cooler designs for 130W CPUs.

First connect one peltier with power supply and run the sytem and during running collect the information of peak current and voltage and then measure the current and voltage for atleast 10 min and also measure the temp diffrence both the side.If you want to connect to same peltier in paralla then the supply voltage will reamin as it is but the current require is double and if you want to connect them in series then then the voltage will be double and current remain as it is. And make sure your applied voltage is slightly higher than 12 volt(12.3 or 12.5) for one peltier.

I see this topic is quite old, but I wanted to added my few cents worth for posterity.

The effect that Jules is refering is a scenario commonly seen in electronics. In a generalization, experienced electrical engineers will refer to this as "the ballasting issue." The currents in the thermoelectric generators will cause heating. The heating will cause the resistance to decrease. This decrease in resistance causes the current to rise. If the decrease in resistance were second order or higher, we would indicate the device to be voltage driven. Since the heating effect to voltage is 1st order, we will generally speak of the device being current driven. When something is current driven, we "ballast" by operating them from a current source. If something is voltage driven (wherein the current response to thermal energy is 1st order), we operate it from a voltage source. We have to "ballast" to prevent thermal run-away. For the purest, we must perform a multiphysics simulation considering self heating and heat pumping together with the devices own IV characteristics and its own thermal conductivity. The thermal conductivity of a Peltier device changes far less with current than it does with voltage. If you bias with a current source, any changes to the temperature gradient (such as from throttling up of CPU or GPU) will cause very little change in voltage. If, however, we were driving the modules with a fixed voltage source, any abrupt change in temperature would hit the hot side first. The hot side would respond with an increase in current. The increase in current would increase the temperature gradient across the first cell even more, and we would find the process running away with only one unit doing all the heat pumping and risking burning up the other Peltier cells.

For maximum efficiency, as well as thermal stability, it is best to operate the device with a proportional controller of current. Otherwise, we are not transducer matched. In fact, all thermoelectric heat pumps are internally several series connected Peltier cells. (See below for details).

To stop generalizing the problem and speak specifically to thermoelectric modules:

The Seebeck effect is a first-order relationship (close enough for engineers) between the voltage difference across a temperature gradient in a doped semiconductor (or a conductor). Futhermore, it is an inverse relationship:

EMF = -Seebeck Coefficient * delta T = -SdT/drr-hat (direction vector)

If we close the circuit, we allow a current to develop from the resultant EMF. From Ohm's Law, this current generates heat. The current density generated around the junction of dissimilar materials is

The voltage created in the circuit by this current is

dV=-J/sigma

The power density dissipated in the circuit by this current is

pOhmic=-J*dV=J^2/sigma

The difference in the heat flux through the two sides of the junction is

dqPumped=d(kdT)/dL=dk/dLdT+kdT/dL=dk/dJunctiondT+kdT/dL = dkdT+k*dT/dL

The Seebeck effect is different between the p doped and the n doped material

EMFp=-SpdT/dLp=-Sp(T1-T2) EMFn=-SndT/dLn=-Sn(T1-T2)

The different EMF creates a potential across the junction of

dEMF=d(-SdT) = -dSdT

Vjunction=EMF=-dS*T

This voltage sources the current, the power available from this source is

pAvail=VjunctionJ=-JdS/dJunction

The power available from this sourced is distributed as

pAvail=pOhmic+qPumped = (J^2/sigma) + d(k*dT)/dJunction +qExt

If we series connect and drive from a current source then the J is constant between modules.

JdST = (J^2/sigma) + dk*dT + qExt

No change between modules in the stack.

If we drive with a voltage: