Mosfet P, for given Vds and Ids

Thread Starter

gdallas

Joined Apr 25, 2012
74
Hi,

I’m trying to understand the value stated for power for a particular mosfet attached given its Vds and Id of 600v and 100mA. It given P tot is 1.7W, and I had incorrectly assumed that any combination of vds up to 600V with any Id up to 100mA would be acceptable, but obviously when you do the calc at the extremes the power would be 60w.



So to ask a daft question!! Which I think I already know the answer to, I presume what it actually means is you are allowed any combination of Id and Vds so long as the product of that does not exceed 1.7W. i.e. the max values of voltage permissible at max Id of 100mA would be 17V. and the max current permissible at the max Vds of 600V would be 2.8mA.
 

alfacliff

Joined Dec 13, 2013
2,458
the power disapated by a mosfet is determined by the current flow and the drop across the mosfet. you should keep that within ratings. just because it is rated for 600 volts dosnt mean much when determining power disapated ina circuit.
used to switch, it has very low actual disapation, it heats more in the linear mode.
 

Jony130

Joined Feb 17, 2009
5,598
so pretty much what i said then?
Yes,
VDS - the maximum voltage between drain and source that the device is guaranteed to block in the off state before breakdown.
ID - the maximum continuous current the device can carry for Ta < 44°C
And don't forget that this Ptot = 1.7W is for Ta = Tjunction = 25°C.
So you can dissipate 1.7W but only if you use ideal heatsink.
Because for every 1W dissipated in the mosfet the junction temperature will rise 72°C above the ambient see (Rthj-a) = 72°C/W in datasheet.
 

MikeML

Joined Oct 2, 2009
5,444
Yes, the FET can only dissipate 1.7W if it is soldered to an infinite heat sink. It must be derated if the tab is soldered to a copper-trace that has already been heated to a temperature above 25degC.
 
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