I'm using low power MOSFETs (N channel-STP4NK60ZFP and P channel- IRF9530). The aim is to build a demo kit for students to study the thermal characteristics of MOSFET at variable frequencies and duty cycles. P channel : threshold voltage = -4V ;Drain current (at 25 deg C) = -12 A N channel : Vt = -4.5 V , Id= 4 A demo kit design for students

With a supply voltage of 9 V(sorry ,not 12v) and gate voltage of 10 v, i'm trying to heat up the transistor as fast as I can to save time. I'm using a higher watt resister. But what should be the value of the resister in order to heat up both the transistors?

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    \$\begingroup\$ I believe you have your source and drain swapped on your P-channel MOSFET in your schematic. \$\endgroup\$ – Tut Jul 15 '15 at 11:59
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    \$\begingroup\$ What will the voltage out of your square-wave generator be when the square-wave is low? Is the voltage on the source of the N-channel MOSFET -12V as shown? \$\endgroup\$ – Tut Jul 15 '15 at 12:27
  • \$\begingroup\$ No, the voltage at the source of the N channel is supposed to be 9V as per my calculation. sorry for the error. Will this be enough to heat up both of my transistors fast enough? \$\endgroup\$ – Gokul Raju Jul 15 '15 at 12:44
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    \$\begingroup\$ Understand what brhans said. | If FETs are on or off then heating is Rdson x I^2. Rload affects I but not Rdson unless you tailor gate voltage to FET curves. I could go and look up data sheets but YOU should provide links to them (where "should" is a complex function of likelihood of good answers) and you need a better description of how and why you think heating occurs and need to show you understand what brhans said as it is central to what you want to do. \$\endgroup\$ – Russell McMahon Jul 15 '15 at 14:21
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    \$\begingroup\$ Btw, if you operate the FETs in the linear region I doubt you will be able to measure any thermal effect related to switching frequency. \$\endgroup\$ – JimmyB Jul 15 '15 at 14:48

At low switching frequency, heating is mostly determined by RDSon (P=I2R). The STP4NK60ZFP is typically 1.7Ω while the IRF9530 is <0.3Ω. Both FETs are in a TO220 package, so the STP4NK60ZFP will heat up much faster than the IRF9530 at the same current.

Thermal resistance of the STP4NK60ZFP is 62.5°c/W junction to air, which at 20°c ambient corresponds to 82.5° die temperature at 1W dissipation. You probably want a considerably lower temperature at low PWM frequency, rising to a maximum of perhaps 80~100°c at higher frequencies (for safety you don't want it to get too hot).

Therefore I suggest making R1 = 18Ω, which sets the current to 0.5A and should raise the STP4NK60ZFP's temperature to about 45°c when turned on continuously. At this current the IRF9530's temperature will only go up about 5°c.


If you're driving those transistors hard with your 10V signal then they'll probably never heat up significantly on their own - you're using them as switches so there'll be no significant power dissipation.

If you mount the resistor on the same heatsink as the transistors, then it will cause them to heat up.

If you really want the transistors to heat up themselves, then adjust your drive signal so that it keeps them in their linear region.
You need both voltage drop across a transistor and current through it at the same time for it to dissipate lots of power & heat up.
Using it as a switch where you have either one or the other won't get you there.

  • \$\begingroup\$ well I'm not using any heatsink for the transistors because I have to demonstrate the heat developed and other parameters across the device.But I can try to connect another resister between the transistors. Eventually this will again reduce the current through the n channel and will reduce the chance of heating up the transistor. :( \$\endgroup\$ – Gokul Raju Jul 15 '15 at 12:16
  • \$\begingroup\$ I'm not sure if you got my main point - if you're operating them as switches (by driving them hard on/off with your input signal) they may never heat up significantly (depending on your load and capabilities of your main power supply). \$\endgroup\$ – brhans Jul 15 '15 at 12:18
  • \$\begingroup\$ Su mean to say if the duty cycle if made high will not heat up the MOSFETs? This depends on the Load resistor. If its a high impedance the whole current will be soourced and sunk via the mosfets. They have only 0.3Ohms Rds. \$\endgroup\$ – Board-Man Jul 15 '15 at 15:17
  • \$\begingroup\$ Nothing to do with duty-cycle. A switch which is on does not dissipate power in itself. Power requires Voltage across the device as well as Current though it. A switch turned on has 0V across itself -> therefore 0W of power dissipation. In order for either of your MOSFETs to dissipate any power you can not operate them as switches. You have to operate them in their linear region between fully on and fully off so that there is simultaneously a voltage across and a current through either one of them. \$\endgroup\$ – brhans Jul 15 '15 at 15:50

It'll heat up real fast if you have no resistor at all.

But of course you don't want to cause a fire ...

I would recommend placing the resistor between the transistors, since you are allowing current to flow through the transistors with PWM. Also, if both transistors are "on", current will flow from +12V to -12V, so pick a resistor value that 24/R=Imax of the transistors. Imax will be the maximum current allowable, it's probably in their datasheets.

I would also be sceptical of doing such a demonstration, large chance of blowing up a really hot transistor in someone's face. I would recommend wearing googles or standing back. I have blown a few in the past and they hurt!

  • \$\begingroup\$ I suggest that you re-think part of your answer. It is not sure from the information given by the OP that both FETs would be on at the same time for any significant amount of time. If the square wave drive goes from rail to rail then only during the transition time would there be any through current going through both FETs at once. There is good reason to keep the load where the OP shows it. \$\endgroup\$ – Michael Karas Jul 15 '15 at 11:52
  • \$\begingroup\$ yes, since the PWM is connected to both the transistors, they will not be turned on simultaneously. But do you think it is better to go for higher ohm resister as this might decrease the current flow in the N channel, which will noyt heat up the transistor. also i'm pretty sure it will not blow up as the max is 175 and 150 deg C respectively @josh jobin \$\endgroup\$ – Gokul Raju Jul 15 '15 at 12:00

This circuit is a disaster. Imagine the square wave is given by a MCU. When the MCU is high (5V), both the PMOS and NMOS will be turned on effectively shorting Vcc and GND.

The best way for you would be 1) Control the 2 MOSFETs via 2 seperate pins of the MCU and implement an inverse operator on the bottom or upper MOSFET.

2) Use an inverter at the gate of the Lower or upper mosfet thereby making one of them complement of the other.

Note - the top PMOS must be flipped. The source and drain should be reversed.

  • \$\begingroup\$ The OP specified "gate voltage of 10v". I read that to mean that his square-wave input signal alternates between 0v and 10v which should be perfectly acceptable - assuming he flips his P-FET the right way around. \$\endgroup\$ – brhans Jul 15 '15 at 15:03
  • \$\begingroup\$ @brhans ... 0-10V square wave is only perfectly acceptable if the source of the N-channel MOSFET is connected to 0V. If it is connected to -9V (or -12V), it would never turn off. \$\endgroup\$ – Tut Jul 15 '15 at 15:35
  • \$\begingroup\$ Just saw the -12V now. But then this means the current would flow through the 2 MOSFETS and has the potential to short Vcc and GND: \$\endgroup\$ – Board-Man Jul 15 '15 at 15:37
  • \$\begingroup\$ I also missed the -12V. That will cause plenty of trouble too ... \$\endgroup\$ – brhans Jul 15 '15 at 15:47

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