First, I'm still quite a novice in the electronics, so may miss some basic knowledge, but trying to learn anyway. That being said, here's my situation:

Goal: Control of the direction of the heat transport in a chain of Peltier devices. Basically, a possibility to swap hot and cold sides on demand (or turn them off althogether), by electronic means only.

Current situation: To achieve my goal, I created a circuit controlling the direction of the current in the Peltier devices using MOSFETs (the gates are connected to switches for now, to be replaced by a microcontroller later). I drive everything with a 12V, 1.5A, DC power supply:

Peltier devices controlled by MOSFETs

And principially it works as expected - switch SW1A makes the current to flow through Peltier devices PE1-4 "from left to right" (on the picture), switch SW2A "from right to left".

Questions: I've got actually two problems/questions related to my design:

  1. Heat of the MOSFETs (Q1-4). They get very hot very quickly; can't give a number, but basically, after 4-5 seconds from closing the switch, the MOSFETs attached to it get so hot, so you can't really touch them anymore. My understanding is that their source-drain resistance should get very small when on, and while some heat is expected, should they really get that hot? Or do I do something suboptimal there?

  2. Is it actually a good design for this kind of application? I'd be especially interested in reducing the number of transistors needed, but is it possible to build such a circuit without having at least two transistors for each direction of the current? (I know I'd need only one transistor if I was interested in only one direction, but can't find a way to use less than 4 to support both directions).

Note: I've read about PWM modulation, but assuming I understand it correctly, it'd reduce the cooling/heating power of the Peltier devices. What I mean is that I'm happy with how strong they are at the moment and how fast they get hot/cold when switched. My understanding is that PWMing the MOSFETs would inevitably lead to PEs getting less (average) current than they get now, which would reduce their heating/cooling power. Am I right here?


PE1-PE4 Peltier Modules (TEC1-12706)

Q1-Q4 Power MOSFETs (IRF510)

Bottom note: this is actually the first time in my life building anything with FETs, I only used BJTs before. So generally I'm still a bit confused about them.

  • 1
    \$\begingroup\$ Note: This arrangement of transistors is called an "H-bridge". \$\endgroup\$ Feb 3, 2021 at 20:49
  • \$\begingroup\$ When you turn on SW1A, do both Q1 and Q2 get hot, or just Q2? And, you know that you must never turn on SW1A and SW1B at the same time, right? Because that will probably destroy all 4 MOSFETs \$\endgroup\$ Feb 3, 2021 at 20:49
  • \$\begingroup\$ @user253751 : 1) Interesting, I've never heard about an "H-bridge" I must check it out. 2) both get warmer, but Q2 definitely gets hotter 3) Yes, I'm aware of this and consider it a design flaw. My goal is ultimately to replace the switches by a microcontroller, so assuming I won't find a better solution, I'll probably have to ensure that software never activates both paths simultaneously. \$\endgroup\$ Feb 3, 2021 at 21:02
  • \$\begingroup\$ @notsurewhattodo You wont be able to run the peltier devices off of a microcontroller. But get a copy of "the Art of Electronics". It's a fun and "easy" read. They don't talk about Peltier devices, but basically you're trying to quickly switch several amps, and filter out all the harmonics. Also, per my answer, read all of Olin Lathrope's Peltier answers. There's at least six. \$\endgroup\$
    – user11852
    Feb 3, 2021 at 21:09
  • \$\begingroup\$ Thanks for including the datasheet links by the way! \$\endgroup\$ Feb 3, 2021 at 21:11

2 Answers 2


Q2 and Q4 will get particularly hot because you are using them in "source follower" mode, analogous to emitter follower mode of a BJT: the drain is a fixed voltage (12V), and as the MOSFET allows current through, it increases the source voltage, which reduces the gate-to-source voltage - i.e. the MOSFET turns itself off as it turns on! It will find an equilibrium where Vgs is about 5 to 7 volts - since Vg is 12volts, that means Vs is also about 7 to 5 volts. Power = voltage x current. Big power dissipation!

To avoid this problem you want the MOSFET to turn fully on. One thing you could do is to use a higher gate voltage than 12V, so that even after the source comes up to 12V, there's still plenty of Vgs left. However the absolute maximum Vgs for the low-side MOSFET is 20V (and the specifications go up to 15V). At say 19V you would only have Vgs=7V for the high-side MOSFET, which is not enough for a drain current of 6.4A - see the MOSFET datasheet page 3, lower-right graph. So that won't do.

Your other options for the gate voltage problem are to use a higher voltage gate supply for the high-side MOSFETs only (the ones connected to 12V) or to use P-channel MOSFETs for the high-side MOSFETs and give them a 0V signal to turn on. In either case you'll need some extra circuitry to deliver a different gate voltage to each MOSFET.

If you don't have a higher voltage available, it's not terribly difficult to make a circuit called a charge pump to generate one.

Additionally, check the top-left graph on page 3 - "typical output characteristics, TC = 25 degrees". Even in its "on-est" state, at 6A current you are looking at about 3V Vds. That's 18 watts of power dissipation, per MOSFET, and corresponds to an on resistance of about 0.5 ohms.

Mostly based on the latter point, I think these MOSFETs simply will not work for you. I would look for different MOSFETs with a much lower on resistance. It would also be good to have a higher maximum Vgs so that the high-side and low-side MOSFETs can both be supplied with the same high-voltage gate signal.

  • \$\begingroup\$ 4 peltiers in parallel produce a current of 24A, so you might need to adjust or recommend a mosfet per peltier. \$\endgroup\$
    – K H
    Feb 4, 2021 at 8:15
  • \$\begingroup\$ @KH I saw that the asker's schematic has 4 in series, but you are right to point out that they are designed for 14V so putting them in series makes the voltage too low. \$\endgroup\$ Feb 4, 2021 at 21:57

Your problem is that you are using all N-channel MOSFETs. The high side ones are being driven by +12 so they act as source followers and drop several volts.

Q2 and Q4 are the ones in question.

In order to avoid that, either drive them from higher voltage than 12V and add circuitry to make sure the +/-20V Vgs(max) is never exceeded, or use P-channel MOSFETs for the high side and ensure shoot-through is avoided.

They’re not especially low Rds(on) MOSFETs, sub-milliohm ones are available and those are 500x worse.

Edit: I mentioned other MOSFETs (do a parametric search to find them, at any distributor such as Digikey) sub-milliohm Rds(on) above, and your one is ~550m\$\Omega\$ (when cold, much more when hot) so that's >500x worse than 0.001\$\Omega\$.

You won't even notice it until you fix the source follower thing, but 2 MOSFETs running at 1.5A and 0.550 ohms and maybe 40% more because they are hot will dissipate a total of more than 3W. A pair of 0.005 ohm MOSFETs (100+ times better) will dissipate a total of 15mW, which is negligible.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.