I have an N-Channel MOSFET IRF530 with its datasheet used in the following circuit:

enter image description here

When turned on it passes 2.5A current. So I first wanted to make some calculations to check whether I would need a heat-sink if it was continuously at saturated/on state.

Following the knowledge from this document: For any power dissipation P (in watts), one can calculate the effective temperature differential (ΔT) in °C as:

ΔT = P × θ

where θ is the total applicable thermal resistance

enter image description here

The series thermal resistances at above model shows the total thermal resistance path a device may see. Therefore the total θ for calculation purposes is the sum, i.e.,

\$\theta_{JA} = \theta_{JC} + \theta_{CA}\$. Given the ambient temperature \$T_A\$, P, and θ, then \$T_J\$ can be calculated.

So in my case TJmax = 175°C, Junction-to-Ambient θ given as θJA = 62°C/W

\$R_{DS}{(on)}\$ = 0.16 ohm

Since current is 2.5A

\$P = I^2 R = (2.5)^2(0.16) = 1W \$, so ΔT the temperature rise becomes:

ΔT = P × θ = 1 × 62 = 62°C

Let's say the ambient temperature is 35°C, then the total temperature becomes

35 + 62 = 97°C which is smaller than 175°C

First conclusion was the MOSFET does not need a heatsink.

Until I simulated the circuit...

What I noticed is that, in my application the MOSFET turns on very quickly, passes almost a constant 2.5A through itself for few seconds, and turns-off slowly. And during turn-off there is a time interval where I × V product becomes quite high.

Here is the voltage and current plot in LTspice where it shows how the MOSFET turns-on at time zero, stays on and turns-off slowly:

enter image description here

And here is what LTspice shows for the power during this time interval:

enter image description here

My questions are

1-) What kind of logical reasoning should I follow at this point? Pmax = 30W here. If I use the procedure I wrote at the beginning the temperature rise becomes

ΔT = P × θ = 30 × 62 = 1861°C

But this is insane. If I switch the MOSFET on and off many times I can feel by finger it really gets hot. In my application a push-button switch turns on the MOSFET and an RC delay turns it off. It doesn't repeat continuously I mean. Do I need a heat-sink here?

2-) This is about the power during on time. In LTpice during the on time as you see the MOSFET's power is around 500mW but I calculate the power dissipation as:

\$P = I^2 R = (2.5)^2 ( 0.16) = 1W\$ using data sheet's \$R_{ds(on)}\$. In my case \$V_{gs}\$ is 15V not 10V. Can that be the reason for this difference?


For @jbord39, power at the load R1 shown together with the MOSFET's power below:

enter image description here

  • \$\begingroup\$ Could you show the circuit you used to generate this simulation? \$\endgroup\$
    – jbord39
    Commented Oct 26, 2016 at 0:20
  • \$\begingroup\$ sure i edit right now \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 0:21
  • \$\begingroup\$ It may also be very helpful to show the power in the loads as well, in the same chart as the power in the FET; i.e.: the current through the mosfet times the voltage drop across the two parallel loads (which will be the same). I think this power spike is just due to the higher resistance encountered when the MOSFET has not yet fully turned on, so as current flows through it burns more power; P = I^2*R. Switching it faster should reduce the power, if this is the case (a ~2s turn on time is very slow) \$\endgroup\$
    – jbord39
    Commented Oct 26, 2016 at 0:25
  • \$\begingroup\$ @jbord39 made an edit for you \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 0:30
  • \$\begingroup\$ "turn-off time" is slow due to RC delay, turn-on time is fast. \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 0:33

2 Answers 2


If I use the procedure I wrote at the beginning the temperature rise becomes

ΔT = P × θ = 30 × 62 = 1861°C

But this is insane. ... Do I need a heat-sink here?

You are neglecting to consider the heat capacity of the chip. This acts like a capacitor in the thermal equivalent circuit, connected between Tj and (some arbitrarily defined) ground, which prevents the chip from heating up instantaneously. Unfortunately you are unlikely to find good data on exactly what the value of this capacitor should be.

You can refer to the safe operating area curve for your MOSFET:

enter image description here

Based on this it's probably safest to keep your switching time below 10 ms, instead of close to 1 s as you have currently.

  • \$\begingroup\$ if i push button couple of times, which means if the slow turn-off occurs very rare, then the transistor doesn't get hot i check with my fingers. but i repeat it continuously i can feel it gets really hot. in my application it will be pushed once or two times not more than that. but i still want to be sure if i need a heat sink here. i will wire 20 of this circuit parallel thats why i wanted to be sure if i need heatsink here or i can ignore it. wanted to quantify besides checking with fingers. \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 1:08
  • \$\begingroup\$ how to change switching time? the RC decay causes it \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 1:10
  • \$\begingroup\$ It could still be heating up enough in a small volume internally, so that the chip is being damaged, even if it is not hot enough to burn your finger. \$\endgroup\$
    – The Photon
    Commented Oct 26, 2016 at 1:10
  • \$\begingroup\$ How to change switching time? Reduce R or reduce C. If you're using RC also to delay the turn-off after the switch is opened, try using a one-shot instead. \$\endgroup\$
    – The Photon
    Commented Oct 26, 2016 at 1:11
  • \$\begingroup\$ the thing is i could do it with an opamp or one shot but it requires extra power supply and more components. when i reduce RC the peak power remains the same but i will try to narrow it to 10ms as you suggested \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 1:18

Your simulations look very reasonable. You are switching 120W and are burning about 30W in the MOSFET when turning off. If you want to prevent the FET from heating up so much, you need to switch it off faster.

Right now, when the switch opens, the charge on the gate has to slowly leak through the 1Meg + 470kOhm + 5Meg pot before reaching ground.

So there is more time (compared to if the gate was discharged more quickly) which the MOSFET is weakly conducting/higher-R (compared to on-state). This increases the power. This is exacerbated a bit since the inductor current cannot change instantaneously (and so will force current through the higher resistance FET) as the MOSFET turns off.

Alternatively, the pullup path is just through the 1Meg resistor.

You can replace your SPST switch with a SPDT switch and connect the "off" setting to ground (rather than floating). This should speed up the gate turn-off and prevent you from burning as much power per switching event.

  • \$\begingroup\$ İt is not Spst it is a push button. Could you draw what you mean i still didnt get it \$\endgroup\$
    – user16307
    Commented Oct 26, 2016 at 7:03
  • \$\begingroup\$ @dicksonchargepump, pushbutton switches are available with either SPST or SPDT electrical configuration. \$\endgroup\$
    – The Photon
    Commented Jun 14, 2017 at 17:09

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