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MOSFET power dissipation definition?

Discussion in 'General Electronics Discussion' started by BlinkingLeds, Jul 3, 2013.

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  1. BlinkingLeds


    Feb 23, 2013
    Hi i was in the middle of calculating a heat sink for a fet but i don't know what max power dissipation is. I know it means losses and not the max load voltage but does this mean it can dissipate 300w without a heat sink or it is the maximum value it can dissipate with a heat sink?
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    I don't know of a transistor case that allows 300W dissipation without heatsink.
    There are several limits:
    - max. voltage
    - max. current
    - max. junction temperature

    In your case 300W most probably is he max. I*V product provided the max. junction temperature is not exceeded. You will need a heatsink. Here's some more info.
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Transistors, MOSFETs etc have a specificaiton for maximum power dissipation. It's normally in the "absolute maximums" section, early on in the data sheet. Exceeding this specification, even briefly, can result in the immediate loss of the Magic MOSFET Smoke (TM).

    This is the reason why you need to hit the gate with a high positive or negative current, to make the MOSFET switch quickly, if you're switching a high-current load.

    If the MOSFET is designed for high power dissipation, you will usually find graphs in the data sheet that will help you understand the nature of the limits that you must not exceed. There are limits on voltage, current, power, and often a Safe Operating Area (SOA) limit as well. Look at some data sheets for high-power transistors and MOSFETs for guidance.

    This is a separate issue from heatsinking. Because of thermal inertia, the heat issue relates to the average power dissipation over a period of time. For example if your MOSFET dissipates 10W when it's ON, but it's only ON for 1 ms every second, the average dissipation is only 10 mW, so that determines the amount of heatsinking you need.

    On the other hand, if it's dissipating 10W for one minute every 1000 minutes, it's still running a 0.1% duty cycle, but you have to heatsink it based on 10W dissipation to prevent it from overheating during that one minute period.

    You need to calculate the heatsink based on the maximum junction temperature, and the thermal resistance at various interfaces. There is thermal resistance between the junction and the package metal, then between the package metal and the heatsink, and finally, between the heatsink and the ambient temperature. These thermal resistances all add together.

    You should also consider that high junction temperatures will reduce the life expectancy of the component significantly, and that high case temperatures can cause discolouration and loss of insulating quality in the PCB.
  4. BlinkingLeds


    Feb 23, 2013
    ok. Is there a way to calculate how much a fet can dissipate without a heatsink
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Yes. First you have to decide how much junction temperature rise is acceptable. Then you look at the data sheet for the case-to-ambient thermal resistance.

    This figure may be specified without any heatsinking; this is common for through-hole packages such as TO-92 where the component body just stands by itself. Or it may be specified with a certain amount of copper on the PCB; this is common for SMT devices that have a metal tab or pad that is meant to be soldered onto an area of copper, such as SMT ICs that have a thermal pad underneath them, and other SMT packages such as DPAK, used for small MOSFETs.

    Let's look at the NTD4906NT4G, a high-current MOSFET in a DPAK package. The data sheet is available at

    At the top of page 2 of the data sheet, four thermal resistance values are given:

    Junction to case: 4.0 deg. C per watt
    Junction to tab: 4.3 deg. C per watt
    Junction to ambient, steady state, mounted on FR4 PCB with 1x1 inch pad, 1 oz copper: 58 deg. C per watt
    Junction to ambient, steady state, mounted on FR4 PCB with minimum recommended pad size: 109 deg. C per watt.

    The third line tells us that if we mount the DPAK package on an FR4 (fibreglass) board that has 1 oz copper thickness with a heatsinking pad that's 1 inch by 1 inch, the junction temperature will rise 58 degrees Celsius above the ambient temperature for every watt the device dissipates.

    If we want to limit the junction temperature to, say, 100 deg. C at a maximum ambient temperature of, say, 35 deg. C, then it can rise by 65 deg. C. Divide 65 by 58 and you get 1.12 watts, the maximum amount of continuous power dissipation you're allowed under those circumstances.

    BTW the "1 oz copper" specification defines the copper thickness; 1 oz copper means that the weight of the unetched copper (on one copper-clad side) is 1 oz per square foot. This is common for higher-power boards; for standard low-power circuitry, 0.5 oz copper is normal.
  6. BlinkingLeds


    Feb 23, 2013
    thanks for that explanation.
    Till now i was guessing how much a fet can dissipate without a heatsink but now i can calculate it precisely :cool: .
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    You're welcome.
  8. gorgon


    Jun 6, 2011
    1oz or 35um copper thickness is the norm for standard PCBs. If you use high density and small geometrics, I suppose 0.5oz or 18um would be used.
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