Understanding the safe operating area for low values of Vds

Question

I am designing a circuit where a logic level signal switches an NPN BJT (Q1A) which turns a P-channel MOSFET (Q2) on as a high-side configuration to power a load (LOAD_100W) of up to 100W at 12V.

My problem is finding a correct SMT MOSFET which will handle the situation without overheating (and preferably without a sink, just by a larger 2oz copper layer where it's sitting on).

The load is e.g. a light bulb with 100W, operating at 12V.

With a current of I = 8A and a MOSFET SQD50P04-13L with an Rds_on = 13mOhm, we calculate a Vds = 104mV and the power dissipation of Pdis = 104mV * 8A = 832mW.

What I am confused about is the safe operating area plot in the data sheet of the SQ50P04-13L. I marked a red dot where I think the MOSFET is operating (I = 8A and Vds = 104mV:

The red dot is not in the safe operating which is in that are "limited by Rds(on)".

My question now is: does this mean that the MOSFET will not operate at all, or will it overheat?

I have difficulties to find out if the package can dissipate almost 1W. According to the table on page 1, the Junction-to Ambient temperature is 50 °C/W, so if I understood correctly, this is actual temperature increase.

What's even more confusing is that if I search for MOSFETs with lower Rds to limit the temperature increase, I always hit the limit of the left edge of the "safe operating are" plot above the Rds limit.

Another question: the same circuit should also be able to switch low-power loads as well (1W or less). Is that even possible? Those are not even fitting in the "safe operating area" plot.

Say e.g. I = 0.1A, can I assume that Rds = 13mOhm? In the datasheet the lowest current I see in the plots is about 5A. which gives a Vds = 1.3mV. It's far left outside of the safe operating area but it makes intuitively little sense that such a "power MOSFET" cannot handle (whatever that means) 0.1A at 12V, isn't it?

Either I don't understand something or I need to search harder ;)

Answer 1

That sloping line that your operating point is just above represents where ID is limited by RDS not by power dissipation. The slope of that line is 22 milliohms which is the RDS at a Vgs of 4.5V. You have a Vgs of around 10V which will move this line higher.

Your operating point on this graph will move to sit on the line think of it as having inserted a 13mOhm resistor in series with your load with the appropriate drop in voltage. You are in a safe operating region as long as your operating point does not go up this line past the dashed lines that represents your duty cycle.

The calculations that you presented earlier in your question mean that you should be safe to operate your transistor there.

Answer 2

I think you're calling it too close on the MOSFET rating. For an actual current of 8A continuous I would be starting with a datasheet "big print" continuous current rating in excess of 60A (often they are misleading anyway, because they exclude package limitations).

Then sort by P-channel, voltage (keep it as low as safe), stock, price and package. That should yield a flood of possible parts. For example the ~5mm x 5mm GSFP03101 from "Good Ark" has a maximum Rds(on) of 3.3mΩ at 25°C so perhaps 5mΩ maximum even when hot, which would drop 40mV, well within the SOA of the transistor. It's also not too expensive. And dissipating about 0.33W worst-case you won't need much copper area to keep Tj reasonable under sensible Ta conditions.

Note that if your load is an incandescent bulb you might have to worry about the safe operating area during turn-on. Fortunately your driver circuit drives the MOSFET 'on' hard but turn-off is much softer (when the resistance of the load would be higher).

Answer 3

does this mean that the MOSFET will not operate at all, or will it overheat?

Well, it's a safe operating area and the graph will use worst case values for the on-resistance of 20 mΩ (and it may even use 22 mΩ): -

I see your point though.