# Difference Between Safe Operating Area(SOA) of two MOSFETs

#### alison_941

Joined Dec 21, 2022
18
Good Day,

I want to compare Safe operating area of two MOSFETs based on graph given by manufacturer in datasheet. Since our application is 24V Inverter so what i concluded based on below information is that device A can handle higher current for DC operation, 10ms operation and 1ms operation and values of current at 24V for these three operating points are 10A, 20A and 100A as compared to 0.22A, 1A and 17A of device B, However for 100us operating point device B has higher current carrying capability as compare to device A and value is 287A as compare to 260A for device A for 24V. Now my question is that is my analysis is correct and what else you can identify from this information when comparing two MOSFETs. Secondly, since I am switching at 20kHZ for my application so I want to see the behaviour at operating point of 50us which is not given in this graph. Can anyone please help me in identifying the difference at 50us operating point. Thank you

#### UweX

Joined Sep 2, 2020
32
I think, you are on a wrong track. The graph shows the power dissipated in the device. So if you are switching, you build your circuit and check with scope and current probe the dynamic power dissipation inside your output stage. That gives you the value you can locate in this graph. And it should be well within the limits. Maybe as a first check you do it by simulation. But using a frequency of 20 kHz gives a period of 50 µs. depending on the switching principle (buck, Boost, totem pole...) the device is off (assume for 50 %) of the time, so high voltage, no current => no power. Some time the device is on, so high current, low voltage => low power in the RDS_on limited part of the curves above. Then we have the switching phase off-> on and on-> off. Here you have high voltage and the high current at the same time. These edges are important for the curve above, but should be short. 1µs or much less. So the 20kHz or 50µs are not related to this.
I am not sure, what you are planing exactly, but another way is looking from the top desing point of view: You want to convert something to or from 24 V. It has some power rating. You need an estimate for the efficiency, from that you have a figure for power dissipation. You want to minimize the power dissipation, because it simplifies cooling. Still you have some power loss in external components like capacitors (ESR) and inductors (series resistance), so also from that point of view you want to have the high power dissipation during turn-on/off limited to a short time.

#### ronsimpson

Joined Oct 7, 2019
2,904
Using numbers from post #1. You have 24V and 100A. You should never have 24V and 100A at the same time. (maybe for the 50nS switching time of the MOSFET) The voltage happens at a different time than the current.

Pick a MOSFET that can handle the voltage, 24V + ring.
Pick a MOSFET that can handle the current for 50uS of on-time.
Look at Safe area for 50nS.
Mostly watch for RDSon & current. At 100A what is the voltage drop across the part? What is the I^2R=power loss?

#### UweX

Joined Sep 2, 2020
32
From the curves above the right device has lower RDSon, at 1 V VDS you have 130A I_Drain, on the left one only 80 A @ 1 V VDS. so the right one looks like a good choice. But take care: if both the devices are made in the same process, the right MOSFET has the bigger area and therefore lower RDSon but also higher input capacitance. That requires a bigger driver circuit to get to the same switching speed. So the MOSFET should be low RDSon, but don't exaggerate. In that case you need a bigger gate control driver (= more effort and more current in the driver stage). So that is the main trade-off.
Even the right Mosfet has conduction losses of 0.6V * 100A = 60 W during on phase. Add to that the short but high losses during turn-on and turn-off and operating the circuit at 100A requires you to use a big heat sink. The left Mosfet has 2V * 100A = 200 W during on-phase. that is a lot of heat, so the trade-off you have to face is maybe using 2 MOSFET from the left side type or one MOSFET from the right side. Now usually the cost has to be taken into account.....