Sorry for the lame question, but I am a software developer, and have only a limited understanding of analog circuits.

The ODrive uses the MOSFET NTMFS4935NT1G (since v3.2) in pairs. enter image description here

This test says that you can drive a motor with ODrive continuously at 40A with using only the heatsink, and 75-90A when using fans.

However, when I look at the datasheet, it has the following picture (I have already learned on this site that I cannot just take the V and A from the datasheet and multiply them blindly): enter image description here

(Note that the test is for the 48V variant of ODrive, and this is used in the 24V variant, but I expect at least the same order of magnitude with both ODrive boards.)

So if I look at 20A (because 40A is driven by a pair - red line), then this MOSFET cannot work in DC mode at all, and even a 10ms pulse can have only 2.5V. If I look at 24V, then it can drive only 0.005A, which I find hard to believe.

I would find it very strange if the datasheet would be wrong, as I can see similar values with the MOSFETs used by VESC for example. So what I can think is that 40A is the average current through all the 3 phases (all 12 MOSFETs), and so they receive a lower load. Another is that a heatsink matters so much, and that would mean that I have to do a thermal analysis for selecting MOSFETS.

So what is a quick rule of thumb selecting MOSFETs? Am I reading this datasheet right? I would need just a little more scientific method than ordering and building stuff and then touching the MOSFETs to check if they are too hot...


Thank you for your answers, it seems that I "complected" the breakdown voltage with the drain-to-source voltage. I was thinking that the more voltage I use as VCC to drive the load, the less current the MOSFET can survive. But now, that you have explained that VDS is the drain-to-source voltage drop, suddenly a lot of things make sense...

So, to check if I understand it now, lets assume that I want to drive a single MOSFET NTMFS4935NT1G to source 40A. Because I am a cheap bastard, I will not use a gate driver, but connect the Gate directly (with a current limiting resistor) to an MCU, which can supply 5V/10mA.

enter image description here

It shows that at VGS = 5V, at 40A there is a voltage drop of around 0.15V only. This is clearly in the ohmic region of the MOSFET, so I guess the power dissipated will be Rds(on) * Id^2.

enter image description here

This one shows that the on-resistance with be around 3.7 mOhm, which is very close to the values which are in tables in front of the PDF. 40^2 * 0.0037 = 5.92W, it needs very beefy cooling.

enter image description here

Finally the last graph shows that on 0.15V voltage drop, the maximum current is around 6A DC only. If I want to do 40A DC, then it goes over the Rds(on) limit, which I guess means that there is no way to cool the package so much to reach that. But at 20A it is below the Rds(on) limit, so maybe I can do 20% PWM or similar. (This result seems to be closer to the measured 20A maximum per MOSFET while driving a motor, which is not DC.) <- here I am not sure what does it mean physically to go above the Rds(on) limit...

Is it a correct reading of the datasheet now? (Ignoring that at 10mA, and with Total Gate Charge = 22nC, it will take several microseconds to switch on the transistor, where I think it is in the linear region, heating like crazy.)


  • 2
    \$\begingroup\$ Surely you're not expecting there to be 24 volts across the FET at the same time there's 20 amps through it! \$\endgroup\$
    – Hearth
    Commented Apr 27, 2021 at 5:26
  • 1
    \$\begingroup\$ I think your reading is now correct. The Rds(on) limit in the safe operating area reflects the on resistance at the maximum temperature (150C). The other graphs show the behavior at 25C, where the resistance is lower. This app note can help you interpret the graph: infineon.com/dgdl/… \$\endgroup\$
    – Adam Haun
    Commented May 4, 2021 at 19:12

2 Answers 2


You are misreading the graph. The horizontal axis is the drain to source voltage -- the voltage drop across the MOSFET along the high-current path. It is not the gate to source voltage -- the control voltage. You can think of the graph as showing the safe upper limit on the power lost due to the MOSFET's on resistance. It tells you that you should be dissipating well under 1 watt at DC.

The other graphs in the datasheet will tell you more about what kind of on resistance and VDS you'll get with different gate voltages. For example:

ID vs. VDS and RDSon vs. VGS from the datasheet

These graphs tell you that you need to drive the gate with at least 4-5 volts to keep the on resistance low, and that 7-10 volts would be even better.

EDIT: Looking at the ODrive in more detail, I see that it's driving a three-phase motor with PWM control, so the DC limits on the transistors aren't particularly meaningful by themselves.


Safe Operating Area (SOA) limitations could apply during the switching operation. While it is switching the current from the inductive load continues to flow through the MOSFET as it tries to turn off, meanwhile as Vds increases, it is "fighting" the gate driver trying to suck gate charge out of the part via Miller capacitance. It may take several amperes to drive the current in and out of the gate for each switching operation, and the transistors will heat somewhat as a result (think of a quantum of heat being dissipated each time the transistor switches from off to on or vice versa under load). So that loss increases proportionally to the switching frequency. SOA is not usually limiting when the gate is being driven hard. You want to stay well away from those limits. The schematic you linked has gate drivers with rather low value (2.2 ohm) gate resistors so presumably they're pretty powerful. I don't see a part number on the schematic.

While it is conducting, there is loss due to Rds(on) (which is a function of gate voltage, and die temperature) and varies a bit from one part to the next. You can estimate that rather easily from the datasheet values. The switching losses require a bit more work (or simulation).

It's also possible to have shoot-through losses, which are very bad. That's when the high-side and low side transistors are both "on" for a period of time. A well designed circuit will have dead time inserted that is sufficient to prevent any shoot through.

Then, of course, you have to make sure that the power dissipated is safe under all operating conditions (such as high ambient temperature) and that also requires a bit of work especially for SMT parts.

Here is an application note that describes how to do the calculations. You can find other similar references.

Typically if one was designing a commercial product, one would do the calculations, do simulations, build a test unit and verify the expected performance, probably iterating along the way. This might constitute a significant part of the total design effort.


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