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I've used an N-Channel MOSFET (IRF540) rated at 100V 28A as a switch with a 54V load and current source that maxes out at 20A. The switch worked however, using the load for a while resulted in MOSFET burned out. I have heard that one should not use rating too close to the load but if these ratings mean nothing then what is the point of listing such ratings? what is the general guideline and in what theoretical principles is it grounded to choose a MOSFET rating for a load?

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    \$\begingroup\$ Those ratings are given for specified temperatures. Did you have sufficient heat-sink/cooling for the component? \$\endgroup\$
    – Mat
    Nov 27, 2020 at 11:15
  • \$\begingroup\$ (1) Higher Vgs lowers Rds(on) and therefore less power loss. (2) Perhaps try IRL540, which is logic level trigger, so for same Vgs, power dissipation is smaller. You can compare and contrast IRF540 with IRL540. (3) Of course you can use a heat sink for marginal cases, (3) BTW, what is your Vgs? \$\endgroup\$
    – tlfong01
    Nov 27, 2020 at 11:16
  • \$\begingroup\$ What Vgs did you activate the MOSFET(s) with and, why did you close down your earlier question when I had left you the correct answer thus depriving me of much needed (ahem) reputation points? Please create a link to the data sheet. \$\endgroup\$
    – Andy aka
    Nov 27, 2020 at 11:41
  • \$\begingroup\$ @Andyaka Vgs = 5.0V with a 100ohm resister in series. (I found an issue with the circuit in that question, so the question became wrong) \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 11:45

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The 100V rating is indicating the maximum drain-source blocking voltage that the MOSFET can block when turned off. The maximum current you can pass through the MOSFET is mainly a thermal design issue. You need to estimate the power dissipation of the MOSFET and make sure you can provide sufficient cooling using a heatsink.

When using the MOSFET with a continuous current, the power dissipation is mostly related to the on-resistance \$R_{ds,on}\$ and the current: $$P_{loss} = R_{ds,on}I_d^2$$ The on-resistance can be taken from the datasheet. But make sure to use the required gate-source voltage. The on-resistance is also very temperature dependent:

Temperature dependence of Rdson

Source: IRF540 Datasheet

So you should not only take the raw 0.077 Ohm on-resistance from the datasheet for your calculations but be more conservative. Try to find a MOSFET with a considerably lower Rdson or a relay to make your life easier.

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  • \$\begingroup\$ Is there a relation between current dissipation and potential temperature rise of FET? Would be nice to know at what point one needs a heatsink. \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 11:52
  • \$\begingroup\$ To determine whether you need a heatsink or not, you have to take a look at the section "Thermal resistance ratings". The maximum junction to ambient thermal resistance is 62°C/W and the maximum allowed junction temperature is 175°C. Depending on the ambient temperature you can calculate the possible power dissipation e.g. (175°C-50°C)/(62°C/W)= 2W. So you really need a heatsink. \$\endgroup\$ Nov 27, 2020 at 12:02
  • \$\begingroup\$ @azad.parinda. I noted the critical parameters and how I came up with my numbers. Its simple - lower Rds on and heatsinking. Rth ja is a function of the ic package and physics - so there's not much choice there. \$\endgroup\$
    – Kartman
    Nov 27, 2020 at 12:07
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There may be a number of reasons the mosfet failed.

  1. it got too hot. Assuming you drive the mosfet correctly, it has an on resistance (Rds on) of 0.077 Ohms. At 20A, the mosfet is wasting around 30W (P = I * I * R = 20 * 20 * 0.077). Without a heatsink the mosfet will get to around 62 (Rth ja) times 30W = 1860 degrees C. The mosfet will have expired well before that temperature was reached. You need to ensure the mosfet has enough heatsinking at all expected situations such that the temperature of the device doesn't exceed, say 100C. Less is better.

  2. There was a short circuit and it got overloaded. The average fuse is way too slow to protect the mosfet.

  3. the power was connected in reverse.

  4. there was not enough voltage on the mosfet gate to enable it to turn on fully.

  5. There was inductive kickback from the load that exceeded the voltage rating of the mosfet

  6. Lightning

The choice of the IRF540 was not unreasonable if it was adequately protected and heatsinked. To give a more concise answer would involve understanding the application, environment etc more.

Selecting a suitable mosfet without a basic understanding of electronic and semiconductor basics is a bit tricky. Sometimes the humble relay is a simpler and more robust choice.

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  • \$\begingroup\$ No 1. could be the most probable reason. If I use multiple (3) in parallel, do I still need to use the heatsinks? \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 11:48
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    \$\begingroup\$ That would mean at 20A each mosfet would reach 600+C degrees. You need a heatsink! The IRF540 is an older device, there are newer devices with better specs, so a better device will mean a smaller heatsink. \$\endgroup\$
    – Kartman
    Nov 27, 2020 at 12:00
  • \$\begingroup\$ but when used three of them, wouldn't they share the load? Should it cool down all the FETS? \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 12:32
  • \$\begingroup\$ how did you determine 600degrees? Is there a relation between power dissipation and temperature? \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 12:33
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    \$\begingroup\$ @azad.parinda Please see item 1 in the answer. \$\endgroup\$ Nov 27, 2020 at 14:31
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@Andyaka Vgs = 5.0V

Look at the data sheet and explain to me how you would expect this MOSFET to be suitable for 20 amps with a Vgs of only 5 volts: -

enter image description here

The 5 volt Vgs trace is 2nd from the bottom and a typical device cannot even reach 20 amps with Vgs = 5 volts.

what is the general guideline and in what theoretical principles is it grounded to choose a MOSFET rating for a load?

There is no general guideline; just the data sheet and this graph in particular: -

enter image description here

Then, if you extend the graph to show durations of 100 ms, 1 second and 10 second you can "get a feel" that operating at more than 10 amps indefinitely could be problematic without significant heatsinking: -

enter image description here

Without a heatsink, the MOSFET's internal junction is subject to a rise in temperature above ambient of 62°C per watt. It will not survive for very long at all.

With a modest heatsink of 10°C per watt of thermal resistance and, a power dissipation of 10 watts (10 amps with 1 volt across the device), the net thermal resistance is that of the heatsink plus the junction to case thermal resistance of 1°C per watt plus 0.5°C per watt for the heatsink contact resistance.

That's a total of 11.5°C per watt and, with 10 watts being dissipated, the junction temperature will be 115°C above ambient. If ambient is 25°C then the junction will rise to 140°C and the local heat generated could easily increase the localized ambient another 40°C and that takes you over the 175°C limit for the device. With a current of 20 amps, operating this device indefinitely is going to destroy it without heavy and significant heat-sinking.

Be very aware that paralleling devices isn't just a case of simply dividing current by 3 for each MOSFET - they won't share equally and, the local ambient will still rise by the same amount. What you have to do is reduce the power dissipation in your application and this means choosing a better MOSFET.

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  • \$\begingroup\$ That is a good point! I think I should go for 12V then. \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 11:59
  • \$\begingroup\$ @azad.parinda With Vgs at 12 volts the device will still dissipate about 20 watts - look at the graph above - it's all there, the volts drop at 20 amps is about 1 volt hence the power is 20 watts. Do you know what you should consider doing next \$\endgroup\$
    – Andy aka
    Nov 27, 2020 at 12:06
  • \$\begingroup\$ Shouldn't according to this graph at 10V it should theoretically be outputting the max current capacity (i.e.28A)? I am using 100ohm resister which must be causing 0.5 voltage to drop from 5.0V. So now at 12V input, it is about 10.4V Vgs. This should set the FET at its max? \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 12:30
  • \$\begingroup\$ A MOSFET doesn't output current i.e. it doesn't force 28 amps; it can only facilitate that current to flow if the supply voltage and load permit it to flow. If the Vgs is only 10.4 volts then either the 12 volts is incorrect, you are measuring it incorrectly or the MOSFET is damaged. \$\endgroup\$
    – Andy aka
    Nov 27, 2020 at 12:34
  • \$\begingroup\$ This reason doesn't explain why FET got blown though! Limiting the Vgs can only cause less source-drain current to flow but not cause it to blow up and become dysfunctional. Further, you said yourself that at 5.0V Vgs, 20A should have been possible, so, Vgs is not the cause of the problem. \$\endgroup\$
    – azad.wolf
    Nov 27, 2020 at 12:36
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what is the general guideline and in what theoretical principles is it grounded to choose a MOSFET rating for a load?

First check voltage rating, with some margin. For 54V supply voltage, a 60V FET is too tight, a 80V FET would work if you are sure there won't be voltage spikes due to inductive load switching or layout inductance, and a 100V FET would be foolproof. Note RdsON goes up in higher voltage FETs, so using a part with too high voltage spec is a compromise.

Second, the RdsON*Qg compromise. FETs with larger silicon chips have lower RdsON but they are slower, so there is a tradeoff between conduction losses (RI^2) and switching losses. So have a look at the switching frequency and decide if you want a fast FET with higher RdsON, or a slow FET with low RdsON, or a compromise. Here the application is load switching, so frequency is very low, besides if you don't want EMI you'll switch it slowly anyway, which means the optimum compromise is a slow FET with low RdsON.

Next, gate drive voltage. If you find a FET you like with 5V gate drive then it can make the circuit simpler. If you have a 12V rail available, then it is less of a problem. Note what I mean with "5V gate drive FET" is that the datasheet specifies a guaranteed max RdsON for that Vgs at the current you'll use or higher. For example, IRF540 does not support 5V gate drive, as Andy explained. You'll have more options with 10V gate drive and above, especially with higher voltage FETs. There are many excellent FETs with 5V gate drive due to the ubiquity of buck converters on PC and laptop motherboards, but they're low voltage, so unsuitable.

So, I do a digikey search with voltage 80-100V and current>30A, sort by price, to get a feel of what kind of RdsON is available at low cost. Okay, about 5-50 mOhms.

Resistive losses at 20A for 5 mOhms=2W, for 50mOhms 20W.

Since the best heat sink is no heat sink, and at 2W we have an opportunity to get rid of the heat sink, I'm going to go for "big slow FET with low RdsON, possibly paralleled if it's cheaper" and either SMD to use the board as a heat sink, or TO220.

So I filter on RdsON below 3 mOhms and sort by price.

STH240N10F7

SUP60020E-GE3

At around 2.5 mOhms these will dissipate 1 watt, for the SMD one this is nice (no heat sink) and for the TO220 it's also acceptable.

If you want to use a heat sink, you can filter on a higher RdsON value and you'll probably get a cheaper FET.

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