Why is pulsed drain current higher than continuous drain current in MOSFETs? In MOSFET data-sheets, pulsed drain current is much higher than (by at least 2x) continuous drain current. What is the reason behind this?

  • 3
    \$\begingroup\$ I think this is because of heat transfer. It takes time to heat something up, and for the heat to propogate. A pulse is too fast. It's like waving your finger over a fire. Nothing will really happen to you. Leave your finger in a fire...well... \$\endgroup\$
    – efox29
    Nov 7, 2014 at 7:24

2 Answers 2


The main problem of current is that when it runs through a resistance it drops a voltage and hence generates heat, which causes a rise in temperature. A lot of things will break down at a certain temperature (think of light bulbs, fuses).

When the spot where the heat is generated is thermally-connected to something that can quickly absorb a lot of heat and pass it off to the surroundings at a slower rate, a small 'heat-pulse' will not generate much rise in temperature, hence it will not be a problem, provided that it is not repeated too often. In such a case a high current pulse can be tolerated, but is subject to certain limitations (pulse duration, repetition frequency). This type of limitation is typical for a semiconductor that is intimately coupled to a metal tab.

The bonding wire(s) of a chip or MOSFET have a very different characteristic: they are suspended in air (or some other stuff that does not conduct heat very well), hence they have a hard limit on the current, that is almost independent of the pulse duration.

In a datasheet you will often find a graph that expressed the maximum current under various circumstances. In the graph below the DC and 5ms ... 100uS lines show the average-heat limits of the Safe Operating Area. They depend on the VCE, because the heat is generated in the semiconductor area where this volateg drop occurs. The horizontal line at 5A is the DC limit. It is (for a large part) independent of the VCE, because it is about a bonding wire, which is Ohmic (voltage drop and hence heat is determined only by I * R).

There are other limits, like the maximum emitter-collector voltage, that are also expressed in this diagram.

enter image description here


The pulsed drain current will have some duty cycle or ON time and OFF time specified in the datasheet. The reason that the Pulsed current is much more than the continuous current is that, the average value of the pulsed current becomes equal the continuous current.

For the average value to be equal to the continuous current, the ON time current of the pulsed current is made high, or in other words, ON time of the pulsed current is allowable to a more value.

  • \$\begingroup\$ I think the question is asking "Why is the so", Welcome the EE stackexchange. So I think the answer needs to refer to some of the underlying physics. \$\endgroup\$ Nov 7, 2014 at 8:00

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.