Hi I am having difficulty getting my head around the SOA diagrams in MOSFET datasheets. Heres an example:



The part is described as follows:
BUK952R8-60E
N-channel TrenchMOS logic level FET
60V, 120A, 2.8 mOhm, TO-220AB

I dont understand what VDS stands for in this diagram. Initially I thought it meant the voltage across the MOSFET when it is not conducting. But that implies that if there is 60V across the terminals and you turn on the MOSFET to pull current through your load, the DC current can only be 0.6A according to the SOA. 0.6A with a 120A device? That can't be right.

So instead I am thinking that VDS must be the voltage between D and S while conducting, but I imagine that we want to keep the MOSFET out of linear, so we turn it on hard, which means with 100A and 0.0028 Ohms we have a voltage drop of 0.28V across DS, but on the SOA the graph doesnt event extend that far. So that cant be it either ...?

Can somebody please set me straight?
Thanks for any advice.

 

Hello there,

Yes this shows the device in conduction, either for DC or for various pulse times.

It appears that they are showing a relationship between what Vds is and what you can allow as current through the device with that given Vds and with that given pulse width.
There is a dependence on Vds and time, and this graph shows that the level of current must be limited more as the voltage goes up, and limited less as the pulse width goes down

I would think the current extrapolates back to 0.3v Vds, but with DC current we would probably see heating and so the Rds will rise and this rise would be significant so we might only see 0.3v when the device is first turned on (with DC current). You could check that on the data sheet.

So unless they made a mistake, these are the limits. You could compare with other data sheets too.

 

Try read this
http://www.mcmanis.com/chuck/robotics/projects/esc2/FET-power.html

100A and 0.0028 Ohms we have a voltage drop of 0.28V across DS, but on the SOA the graph doesnt event extend that far. So that cant be it either ...?

Pot = 100A^2 *2.8mΩ = 28W
This Power will rise the junction temperature up to ΔT = P*Rthja = 28W * 60°C/W = 1680°C above the ambient temperature. So you need a to use a big heat sink and Rds(on) will also rise with the temperature fig 12.
http://www.nxp.com/documents/data_sheet/BUK952R8-60E.pdf