I have a TEC1-12712 Peltier element, which I am driving using Arduino microcontroller through IRFZ44N and buck step down converter (to smooth PWM).

Overall, circuit consists of:

  • 12V, 8.5A max DC power supply
  • 5V, 40mA, 1kHz PWM square wave signal from Arduino microcontroller
  • Totem pole gate driver to drive IRFZ44N gate (stolen from here)
  • IRFZ44N power mosfet with heatsink and ventilator (49A Id)
  • Buck mode step down converter (stolen from here)
  • TEC1-12712 Peltier element (1 Ohm, non inductive load)

After powering on and giving 5V to the gate driver input, the MOSFET becomes immediately hot (to touch), even when PWM is in full cycle (no pulse). Is that normal? It is rated to handle high currents (Id = 49A) and gate should be fully open (totem pole gate driver).

I was able to keep it running a few times already for 30 minutes without burning it but I am worried if there is something wrong and if it survives for a long time.

BTW: When I apply PWM signal (50% duty cycle, 1kHz) the buck converter gives interesting, quiet buzz. I was unable to record it since fans cover it up. Is that normal or a sing of something being wrong?

Could you please suggest an improvements or point to a mistake I could have done?

Disclaimer: I am an absolute & total noob and this is the first circuit I ever build. So please forgive me if I even phrased something inaccurately.

  • \$\begingroup\$ The peltier junction you are using here is an ohmic load. It's basically just a big resistor. My answer below explains your problem I believe. If you want to vary the amount of power your peltier junction consumes, there is no reason to use this circuit. Just use the MOSFET on the low side of the peltier junction and use it to vary the duty cycle of the junction. This will require less components and achieve the same thing. \$\endgroup\$
    – Eric Urban
    May 14, 2016 at 13:00
  • \$\begingroup\$ I'd love to use less components, but it is not recommended to drive Peltier using PWM directly due to mechanical wear caused by temperature cycles. Therefore I must smooth it out. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 13:02
  • \$\begingroup\$ You are using a N Channel FET. A P Channel would be better here with drain and source swapped. With a FET the gate must be greater than source by Vgsth + a bit more. This is +ve for N Channel and -ve for P channel. In your case the gate cannot go above 12V so the source must be lower by at least Vgsth so you have a few volts across the FET when on. Fex x Load_amps = hot. I don't think I like the driver cct with high side cap. Need to think. | IF FET is NOT fully on when on it WILL get hot. \$\endgroup\$
    – Russell McMahon
    May 14, 2016 at 13:44

4 Answers 4


Olin Lathrop has identified your problem, but the solution is pretty simple


simulate this circuit – Schematic created using CircuitLab

A simple level shifter (Q1) will provide a good 12 volt swing on MOSFET gate. With a gate capacitance in the range of 1500 pF, transition time will be less than a microsecond, so at 1 kHz the power wasted will be very small. The transition time will be great enough, though, that there is no need for a separate gate resistor.

With about 8 mA (4V/500 ohms) base drive and 60 mA collector current (12V/200 ohms) Q1 will be adequately saturated.

With a greater PWM frequency you'd need to start looking at more sophisticated drivers, but I don't see the need here.

ETA - Also, you asked about switcher noise. On the one hand, your inductor is doing a pretty good job of keeping peak current within the power supply limits. On the other hand, the 470 uF capacitor is producing 100 amp plus current spikes (for an ideal cap), so that probably has something to do with it.

FURTHER EDIT - The OP asked how to size the components for a level shifter.

First, the BC639 data sheet gives a maximum collector current of 1 amp. Start with an assumption of 1/10 to 1/20 of that, or 50 to 100 mA. The transistor will be used as a switch, not an amplifier, either full on (zero voltage across it, more or less) or full off (no current). This mode is called saturation, and you should assume a current gain of about 10. For amplifiers, a BC639 can run with a gain of about 40 to 250, but this does not apply in this case. A gain of 10 which produces a collector current of 50 to 100 ma implies a base current of 5 to 10 mA. Assuming the Arduino can supply its rated current at nearly a full 5 volts, this says that the base resistor will drop about 4 volts, since the base-emitter junction will drop about 0.7 volts. 500 ohms is a convenient value, and this will provide 8 mA of base drive. 8 mA of base drive times 10 gives a collector current of 80 mA when the transistor is on. Since we're driving the transistor hard on, this says that the collector resistor (the load resistor) will drop about 12 volts. If R2 were 1k, you'd get 12 mA, so 200 ohms (another convenient number) will give about 60 mA. This says that the transistor will operate with a gain of 7.5. Close enough. Note that none of these values is real picky. You can easily ignore 20% slop.

If a 2-stage level shifter is needed in order to maintain the phase relationship between the input voltage and the load current, this is easily done. However, the resistor values can be a bit higher on the first stage in order to cut down of power dissipation. This is because the load resistor of the first stage will provide the base drive for the second, and this is fed by 12 volts rather than 5. If you do this, it's a good idea to place a resistor from each base to ground, with a value of about 10 times the base resistor. This is just to make sure that the transistors stay turned off with no input.

However, it's not clear at all why you'd need a second stage. All you need to do is adjust the PWM timing in software to give the right result, and this should be very straightforward.

  • \$\begingroup\$ Can I try it with BC639 I already have (instead of BC539)? And isn't that inverting the signal (which is something I really don't want and that is why I choose totem pole in the first place). \$\endgroup\$
    – Lukasz
    May 14, 2016 at 16:19
  • \$\begingroup\$ @Lukasz - Oops. Typo. Yes. I've corrected the circuit. Sorry. And yes, it inverts the signal. If that is really a problem (and I frankly don't see why - it's a simple software issue) just use two shifters in series. \$\endgroup\$ May 14, 2016 at 16:34
  • \$\begingroup\$ I will try that. As I am a beginner, can you explain me if the second shifter should look the same and where do those resistor values come from? And could you recommend capacitor value for a buck converter when working with 1khz frequency? \$\endgroup\$
    – Lukasz
    May 14, 2016 at 19:27
  • \$\begingroup\$ @Lukasz - See edit. And no, I can't recommend a capacitor. \$\endgroup\$ May 14, 2016 at 20:38
  • \$\begingroup\$ I tried your 2-stage without success and got desperate to the level I decided to create dedicated question for that: electronics.stackexchange.com/questions/234403/… \$\endgroup\$
    – Lukasz
    May 15, 2016 at 20:28

To protect against possible future edits, here is the circuit we're talking about:

Major problem are:

  1. There are no component designators in the schematic, making the circuit hard to talk about.

  2. The 1 mF cap makes absolutely no sense. You have a double emitter follower driving the FET gate. This provides substantially more current capability than the digital output. OK so far. However, the cap prevents sustained DC into the power input of this current amplifier. I can't even guess what you think it's purpose is, but it keeps the current amplifier from working properly.

    There is also no need for the 3.3 Ω resistor. Just connect the collector of the NPN directly to the 12 V supply.

  3. The FET isn't getting nearly enough gate drive voltage. The double emitter follower current amplifier is a reasonable idea in itself, but it looses two junction drops of voltage. You start with a 0-5 V digital signal, and the FET is driven with about 700 mV to 4.3 V.

    That would be OK if the FET was solidly on with 4.3 V on the gate, but it's not. You have to actually read the datasheet. On page 2 it shows the on resistance as being 17.5 mΩ, but at 10 V gate drive. On the very next line down, it shows the gate threshold can be as high as 4.0 V, at which point you only get 250 µA.

  4. The diodes is labeled "Schottky", but the symbol says ordinarly silicon diode. Which is it? It should be a Schottky.

  • \$\begingroup\$ I did the datasheet and I know this FET requires 10V. I believed that gate driven is doing that actually. And diode is shottky - it is just that the Mac editor I use (iCircuit) does not have shottky diode symbol. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 15:15
  • \$\begingroup\$ And purpose of the capacitor is described in the blog page: "When the PWM signal is high the NPN transistor is on and the PNP transistor is off. This allows a large current to flow from the 5V power supply and the tank capacitor C into the mosfet gate. When the PWM signal is low the PNP transistor is on and the NPN transistor is off allowing for a quick sink of the charge on the mosfet gate into ground." So basically it seems that it helps in charging up / discharging MOSFET gate faster. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 15:15
  • \$\begingroup\$ And you are right, the capacitor is connected wrong way. My editor just automatically made a connection and I did not notice. As I said - I am a complete noob. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 15:21
  • \$\begingroup\$ @Lukasz Emitter voltage of BC639 will be always below base voltage, so no 10V there. Note that the blog uses 5V on collector, because any more is of no use. I did not read the docs on the MOSFETS the blog and you are using to see how much woltage they need to really open. Those are different MOSFETs, so might differ. \$\endgroup\$ May 15, 2016 at 6:06

Looking at your gate driver circuit, the MOSFET is not turned on all the way. I believe this is called "enhancement" in MOSFET terminology.

Measure the voltage drop across your MOSFET's drain-source, and then measure the current flowing through it. You can then calculate the power dissipation in your MOSFET. It is likely to be very high. The MOSFET you have chosen requires the gate voltage to be above the source voltage to be completely on. Almost all power MOSFETs work this way. Based on the datasheet, the gate voltage needs to be about 4 volts higher than the source. You're using some transistors to control the gate however, so your gate control voltage needs to be even higher due to the voltage drop in the transistors.

  • \$\begingroup\$ gate voltage should be much more than 4V above source, since gate driver takes its voltage from 12V power supply. Or is there something I do not get? BTW: you says that it is more likely that it is the gate driver problem, not the buck's inductor reverse current? \$\endgroup\$
    – Lukasz
    May 14, 2016 at 13:00
  • \$\begingroup\$ The source of your MOSFET is not connected to ground based on your circuit. You can't have a Gate-Source voltage of +4 volts with this circuit. \$\endgroup\$
    – Eric Urban
    May 14, 2016 at 13:04
  • \$\begingroup\$ Sorry, you are right - my circuit seems actually wrong. I did it for the purpose of the question. Need to correct it. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 13:07
  • \$\begingroup\$ I corrected the circuit. \$\endgroup\$
    – Lukasz
    May 14, 2016 at 13:30
  • 1
    \$\begingroup\$ Go review the circuit in the blog you took it from. You have the capacitor connected wrong. \$\endgroup\$
    – Eric Urban
    May 14, 2016 at 13:40

I know this is an old thread but I wanted to clarify something every one else missed. The IRFZ44N rating of 49A is a single pulse rating of a few microseconds. You need to look at the Safe Operating Area graphic (SOA) of the FET. In your case you almost go to DC rating, which is not in the IRFZ44N datasheets but is given in other FET manufacturer datasheets, look for example at the IXTP182N055T datasheets SOA, you will see how low the DC curve is from the 10ms pulse curve. You need to respect that or you will eventually burn your device.


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