I need to change between heating and cooling with a Peltier, controlling this with PID. I've read a ton of information on Peltiers, looked at their datasheets and despite of this, I still cannot understand some things. First - is the Peltier heat transfer rate linear to the current? I am looking at this chart: enter image description here

This chart is for given Thot, and the curves are for different dT. But since when you change the current dT is changing, in the reality Qc should follow some other curve. But I cannot understand how it should look. Will it be straight? Will it depends on Thot?

I'm not even sure that if I'm interested in Qc linarity, because in an actual system we control the temperature, so probably I should be more interested in the proportionality between the current and dT(hot-cold). But since the heat losses of the cooled volume will be proportional to the difference between the inside and ambient temperature, equal amount of Qc won't change the same amount of dT at different inside temperatures. I feel really confused :)

On the hot side we have enter image description here

So the dissipated heat is the sum of transferred heat and resistance generated heat, which increases with the square of the current. Here too, I cannot understand how will this look in actual operational conditions.

I'm trying to understand all this, for two reasons - first I will have to determine if I have to change PID constants depending on some conditions (as dT, Thot and I). And second - I have to find a way to linearize the dissipated heat at the hot side and to drive the current is a way that gives me equal or at least similar curves on both sides.

  • \$\begingroup\$ A datasheet link would help. And besides, providing the source for these charts is just a good idea anyway. I'm no expert. I've used them once many years ago in a 3-tier arrangement to maintain a cold side at 5 C and a difference of up to 50 C vs ambient. The circuit dissipated only milliwatts on the cold side, in a small package similar to TO-5. We didn't use PWM, but rather a variable voltage-controlled constant current system to drive it. Probably very little of what I remember transfers to your situation. \$\endgroup\$ Commented Aug 1, 2023 at 4:44
  • \$\begingroup\$ I did readily find this general application note on Peltier, which provides some worked examples, too. (There's probably better to be found. But it is a start.) What I found interesting is the performance testing exhibited in this TI app note, where it concludes that PWM performs poorly when compared to constant current drive. This directly opposes the claims made in this white paper. \$\endgroup\$ Commented Aug 1, 2023 at 4:46
  • \$\begingroup\$ So I'll +1 the question in hopes that someone with a comprehensive view on the topic can put things into a simplifying, yet accurate context. \$\endgroup\$ Commented Aug 1, 2023 at 4:48
  • 1
    \$\begingroup\$ @periblepsis - I did some additional research - check my answer below. \$\endgroup\$
    – irilinir
    Commented Aug 1, 2023 at 14:06
  • \$\begingroup\$ @periblepsis consider this simple thought to see that PWM is worse. Consider 50%duty cycle full current, vs. 50% constant current. The average current is the same, so the heat pumped is the same. But the conduction loss I²R is twice higher in the PWM case. \$\endgroup\$
    – tobalt
    Commented Aug 1, 2023 at 19:14

2 Answers 2


I suggest you might feel more comfortable with how Peltier elements behave if you just get a couple of cheap ones (<£5 from the usual online retailers, in the UK at least) and have a bit of a play.

I did some simple experiments by bonding one between two metal water containers and measuring the temperature change in known volumes of water over a set time.

For what it's worth, my experience was that the performance quickly became dominated by the heat 'leaking' backwards across the device from the hot to the cold side. Consequently they worked pretty well for small temperature differences (iirc about 10ºc) but above that the power requirements quickly got out of hand. Whether or not that's an issue will depend on your application of course!

  • 1
    \$\begingroup\$ Actually I have experimented quite a lot and know when they are efficient. I have done experiment to calculate how Q and COP changes when current is fixed and cooled volume temperature gradually goes down. But I haven't designed an experiment to measure the actual relation between the current and the achieved dT. Probably I should do some more. \$\endgroup\$
    – irilinir
    Commented Jul 31, 2023 at 14:21

Here I found two interesting sources that throw some light:Peltier Thermoelectric Modules Modeling and Evaluation

According to this: enter image description here

Constants can be calculated from the Peltier datasheet parameters (in case it is a real one :) ) enter image description here

In the first post I included example charts from this datasheet:PC5-16-F1-4040-TA-W6

For me the equations from (1) look wrong. I can see why the half of the Jaul energy has a minus sign - because it is emitted to the cold side (the other half is emitted to the hot side). But I don't know why the thermal resistance part is added - it should be substracted, because this is a thermal loss for the cold side.

The second source is this: Thermoelectric Technical Reference Here the signs seems correct:

Qc = (Sm × Tc × I) – (0.5 × I^2 × Rm) – (Km × dT)

Qh = (Sm × Th × I) + (0.5 × I^2 × Rm) – (Km × dT)

The first function with some real values looks like: enter image description here we can see that steepness decrease with the increasing of the current.

The second function looks mirrored to the zero and the steepness increases.

enter image description here

But again, this is at given Tc and Th. When changing these values the steepness of the curves changes. So, I think that the complexity of understanding the Peltier is because there are actually 5 independent variables - current, voltage, transferred heat, Th and Tc. In order to solve the system, at least 3 of them should be fixed. But in the real world it doesn't work like this, because Th and Tc (thus delta T) depend not only on the applied current, but also on: hot side air temperature, hot side thermal resistance between the Peltier and the air (how well is cooled), thermal capacity of the cooled volume, thermal resistance from the Peltier to the cooled volume and thermal resistance of the cooled volume to the outside air. So it goes like this: when you apply constant current and delta T is zero, the Peltier is most efficient. It starts to pump energy from the cooled volume to the outside. Since there is no temperature difference between the outside and the inside volume, all the energy goes to cool down the volume. The temperature starts to drop down. This increases the heat losses of the volume, which means that some part of the pumped energy will compensate the losses and the other part will further cool down the volume. Since delta T is increased, the Peltier pumps less energy. In time the energy Pumped out by the Peltier decreases, the heat losses increase and at some point when they become equal, the system will stabilize and this will define the final delta T at the given current. On the heating side will be similar, but the balance will be achieved at different delta T.

So to solve my question probably I have to equalize the heating and cooling heat flow and solve for I1 at the same Th and Tc:

(Sm × Tc × I1)– (0.5 × I1^2 × Rm) – (Km × dT) = (Sm × Th × I2) + (0.5 × I2^2 × Rm) – (Km × dT)

Which if I haven't made a mistake will looks like this: enter image description here

And with the real values (x is I2) enter image description here strong text

So, again, the actual curve will depend on Th and Tc.


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