A MOSFET has the following datasheet: http://www.vishay.com/docs/91015/sihf510.pdf

which states for the max. ratings:

Maximum Power Dissipation = 43W

Continuous Drain Current = 4A (at 100°C)


RDS(on) = 0.54 ohm

What I interpret from this info is: In continuous operation this MOSFET should not exceed 4A. And since RDS(on) = 0.54 ohm, the power at max allowable current:

P = Imax^2*R so P = 8.64W (I think this is the max power of MOSFET can handle without any heatsink?)

So if I obey the datasheet I find 8.64W but the datasheet says the Maximum Power Dissipation is 43W.


Does that mean the maximum power dissipated by a big heatsink cannot be more than 43-8.64 = 34.36W ?

And secondly datasheets says:

"The TO-220AB package is universally preferred for all commercial-industrial applications at power dissipation levels to approximately 50 W" Is 43W approximately 50W? What is this 50W?

  • 2
    \$\begingroup\$ The answers below are fine and explain why power consumption can be greater than 8.64W. There is another misunderstanding in your post, however: 8.64W is not the max power dissipation without heatsink. It is the power dissipated when Vgs=10V and I=4A. But you need a heatsink even with 8W. The max power you can dissipate without heatsink depends on RthJA (given in datasheet). It is 62°C/W, so it means for each dissipated W, the junction temp will raise by 62°C. If you reach Tjmax = 175°C (also given in datasheet), it will fail. So max power without heatsink is (175-ambient)/62 = 2.4W at 25°C. \$\endgroup\$
    – dim
    Nov 4, 2016 at 9:01
  • \$\begingroup\$ Maximum Power Dissipation = 43W means that if we have the proper heatsink whe transistor can dissipate max 43W even in the linear region right? \$\endgroup\$
    – floppy380
    Nov 4, 2016 at 20:41
  • 1
    \$\begingroup\$ Yes. But that is with a hell of a heatsink. Actually, you even need an infinite heatsink, given the values in the datasheet. Max realistic power is Pmax = (Tjmax-Tambient)/(RthJC+RthCS+Rheatsink). All values are given in the mosfet datasheet, except Rheatsink normally given in the heatsink datasheet, and, of course, Tambient. Do the math. \$\endgroup\$
    – dim
    Nov 4, 2016 at 20:54
  • 1
    \$\begingroup\$ why do think they wrote there 43W for max dissipation eventhough they know it requires infinite heatsink? whats the point of confusing people? and they also hide the details of what they mean and make me to beg people to seek an answer.. \$\endgroup\$
    – floppy380
    Nov 4, 2016 at 21:27
  • 1
    \$\begingroup\$ This is the convention. Since they don't know what heatsink you'll use, they assume the biggest possible and let RthCS and Rheatsink be zero (and assume 25°C ambient). You can check: (175-25)/3.5=43. This isn't really confusing to people used to read datasheets. What other value could they assume anyway? \$\endgroup\$
    – dim
    Nov 4, 2016 at 21:39

2 Answers 2


The maximum power dissipation in the case of any transistor corresponds to the multiplication of voltage across drain to source and current flowing through it.

If your application happens to be for the switching purpose, then when you turn off the device, it cannot go off instantaneously. It takes some time to remove the carriers and reach the cut-off state.

According to the datasheet, maximum Vds=100V, Let's say you are operating at 50v at off condition and in the on condition, current flowing through it be 1 A.

When it switches off, Vds has to go from <1v to 50V, in that particular time period, current has to go from 1A to 0A. During this time, the losses in the transistor occurs causing the temperature increase. If your switching frequency increases , switching losses increases. If the working current increases as per the application, switching losses increases proportionally. As the temp increases, degradation of FET occurs causing more dissipation.

Also in the case of amplification(usually BJT are used), Voltage across drain to source is not according to the saturation region but linear region in FET, causing V*I to very high.

Because of this, you would see the dissipation higher in datasheet.

  • \$\begingroup\$ you wrote: " If your switching frequency increases , switching losses increases." is it similar logic also with the switching on off speeds of the MOSFET? i mean the longer the switching on off (like rising falling edges duration) the more is the power loss and heat ?? \$\endgroup\$
    – floppy380
    Nov 4, 2016 at 3:20
  • \$\begingroup\$ Yes, the longer the transition period which happens with max continous ampere condition or high vgs, switching losses increases. \$\endgroup\$
    – Div-lcr
    Nov 4, 2016 at 3:26
  • \$\begingroup\$ Also if you are working with lower Vgs voltage, the FET might go into "active" region causing Vds to be high \$\endgroup\$
    – Div-lcr
    Nov 4, 2016 at 3:42

If you operate the FET with a low Vgs, or if you switch it between on and off with high frequency, it will generate more heat than what's predicted by just \$I_{max}^2 R_{ds}(\rm{on})\$.

So the power consumption limit may be relevant, even though you won't exceed it by keeping the FET fully turned on and respecting the maximum current spec.

  • \$\begingroup\$ 1-) I didn't know that heating would be inversely proportional to Vgs. I thought higher Vgs causes more heating. Do you mean in this case 15V Vgs causes less heating comparing to 10V Vgs? 2-) You say switching pulse will generate more heat but datasheet says 20A for pulsed current which is much more than continuous current limit. I would interpret pulsed current causes much less heat from this info but you say the opposite. I'm confused. \$\endgroup\$
    – floppy380
    Nov 4, 2016 at 2:55
  • \$\begingroup\$ 1. threshold voltage is 2-4 V, but Rds(on) is spec'ed with Vgs=10 V. If you use only 4 or 5 V Vgs, you'll get a higher channel resistance than if you use 10 V. I never said it would be inversely proportional, only that it will be higher with a lower Vgs. \$\endgroup\$
    – The Photon
    Nov 4, 2016 at 2:59
  • \$\begingroup\$ 2. Switching causes more heating because Vgs has to pass through 4 V on its way to 10 V. And you can't switch instantaneously because the gate capacitance isn't 0. If you use a weaker gate driver, you'll ramp up the gate voltage slower and produce more heat as you transition from fully off (0 current) to fully on (fairly low voltage across the channel). \$\endgroup\$
    – The Photon
    Nov 4, 2016 at 3:00
  • \$\begingroup\$ regarding 2-) what i understand from your words is that: Rds(on) is meaningless in switching mode and P = V*I and that can be huge during transition period. is that right what i understand?.. so when we visualize the 20A max rating with pulsed current should we think that the MOSFET can handle for a very short moment 20A? How short I dont know though.. \$\endgroup\$
    – floppy380
    Nov 4, 2016 at 3:16
  • \$\begingroup\$ The 20 A pulse spec is likely based on full turn on. Have a look at Figure 8 (safe operating area) and Figure 10 (thermal response) in the datasheet for more details. The transient during switching is a different phenomenom that's more difficult to characterize. Just realize that transitioning as quickly as possible between off and on will reduce heating. \$\endgroup\$
    – The Photon
    Nov 4, 2016 at 5:09

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