Help Me Read Contradicting MOSFET specs, Wattage and SOA

[Mechatrommer]

digging deeper into transistor's working condition just get me nuts almost everytime.. here for example... IRF540 mosfet only speced to 150W working condition, but looking at SOA, whether you are drawing 5W or 400W, they lie in the same region. given enough heatsinking... 5W should be able to run continuosly, but...

what about 400W?
why it lies in the same region as the 5W? (blue region)
and whats with the 10ms? is it for the blue region or the yellow region?

and green region, what is the meaning of "operation is limited by Rdson"?
how do we decide which area can work continuously, which area need to be pulsed? (in green or/and blue region)
why it (green area) got separated from the blue region?

duh...

[BravoV]

Not an expert, but let me have a few words.

What are you confusing about ? Just use basic power formula for the mosfet's power dissipation.


Power Dissipated by mosfet = Current^2 x Resistance

Resistance of the IRF540 is 0.077 Ohm (RDSOn).

For simplicity sake, lets assume the RDS ON is constant throughout the temperature, although this is NOT the case, check the other chart at "Normalized On-Resistance vs. Temperature" .


So at your 5 Watt which is at 1 Volt at 5 Amps :

  Dissipated power by mosfet = Current ^ 2 x Resistance = 5^2 x 0.077 = 1.925 Watt

  or merely just 2 Watt of heat.


Now at your 400 Watt which is at 80 Volt at 5 Amps :

  Dissipated power by mosfet = Current ^ 2 x Resistance = 5^2 x 0.077 = 1.925 Watt


Same heat generated. 


So for power dissipation at the mosfet, voltage does NOT matter, only the current passed through the mosfet and it's resistance. Not always the case though, read thru the end.

Another important thing to watch, see the note at your chart there, its stated the TC must be at 25C while the TJ is at 125C.


Another better example why you should watch for temperature, as it does matter, here the SOA of the mosfet at different temperatures, put few note in there too.

You will see at higher temperature, operating at higher voltage, the 900 Watt operating state (the tiny red dot)which is at 90 Volt, when heated, it is moved from DC to 100 ms region.


To other experienced fellows, CMIIW.

[Zero999]

So for power dissipation at the mosfet, voltage does NOT matter, only the current passed through the mosfet and it's resistance.

That is true when the MOSFET is operating in the Ohmic region and is connected to, what is effectively, a constant current source i.e. R_L >> R_ON. However, I think the original poster is referring to the saturation region, where the drain-source voltage is much higher and the MOSFET itself is acting as a constant current sink.

[rx8pilot]

Quick note from my cell:

SOA generally has nothing to do with heatsink being used. Its an area where the heat is building to a destructive level much faster than it is being conducted through the package and into the heat sink.

Sent from my horrible mobile....

[Kleinstein]

In the SOA diagram the power dissipation is voltage times current - so the voltage really matters.
The R_limited range is where the R_on is limiting the current than can low at that low voltage. So it is an area that is impossible to reach (at least with the maximum R_on). So for the 77 mOhms R_on, you just don't get 10 A at less than 770 mV. If you get a good sample one might get a little in that range, but not much.

For short times higher power is possible, as the heat capacity can absorb the heat and this way limit the temperature. The P_tot values is for DC and thus limited by thermal conduction. So for a short time higher power is permissible. This is important for switching.

The permissible power gets lower at a realistic higher case temperature (e.g. if you don't have water cooling with chilled water).

At high voltages (e.g. > 50 V), there can be a thermal instability, similar to the second brake-down in BJTs. The current concentrates to the hotter parts of the chip. This leads to the SOA curve to give a lower permissible power at high voltages. Especially the DC curve if present may not have that simple line of constant power, but above a certain voltage the maximum permissible current will drop faster. How strong this effect is depends on the type of FET - it is more common in modern lower voltage FETs. First generation MOSFETs are less effected.
Many FETs are not made for DC operation in the saturation region (e.g. significant current at more than 10 V) and thus don't how a DC curve in the SOA diagram - so they don't tell how well they behave. So without the DC curve chances are that FET type is not suited. This is especially true for second sources from a different manufacturer. Things usually get worse at higher temperature. For the IRF540 it might be just old enough to be not that much influenced, but I would not be too sure about it.

Now comes the really nasty part about the SOA curve of MOSFETs:
There are datasheets around that do show a wrong SOA, ignoring this effect. So they show a straight line for the DC SOA curve, but chances are they will blow if used at there maximum voltage and power for an extended time. So even if you find a DC curve, you don't know for sure it is reliable.

So finding suitable MOSFETs for high power dissipation is a difficult topic.

[Mechatrommer]

So for power dissipation at the mosfet, voltage does NOT matter, only the current passed through the mosfet and it's resistance.

However, I think the original poster is referring to the saturation region, where the drain-source voltage is much higher and the MOSFET itself is acting as a constant current sink.

ditto... here's the circuit i'm working on, still figuring out how the mosfets will exactly work, and some other things that i'm going to run through again after this sleep time... i'll peruse this thread after that "sleep time" my brain is saturated... https://www.eevblog.com/forum/projects/check-my-constant-current-load-(with-mcu-control)-2-days-worth-of-initial-design/msg1191821/#msg1191821

[Zero999]

So for power dissipation at the mosfet, voltage does NOT matter, only the current passed through the mosfet and it's resistance.

However, I think the original poster is referring to the saturation linear SATURATION region, where the drain-source voltage is much higher and the MOSFET itself is acting as a constant current sink.

I was right the first time. I think you're confusing BJTs with FETs.

When a BJT is in its saturation region, increases in the base current have no effect on the collector current.

When a MOSFET is in its saturation region, it behaves as a constant current sink. The drain current hardly changes when the drain-source voltage is varied and can be controlled by varying the gate-source voltage.

https://en.wikipedia.org/wiki/MOSFET#Modes_of_operation

[digsys]

Now comes the really nasty part about the SOA curve of MOSFETs:
There are datasheets around that do show a wrong SOA, ignoring this effect. So they show a straight line for the DC SOA curve, but chances are they will blow if used at there maximum voltage and power for an extended time. So even if you find a DC curve, you don't know for sure it is reliable.
So finding suitable MOSFETs for high power dissipation is a difficult topic.

Agreed
To OP. There's a few past threads on operating power FETs at SOA / linear. I've posted about my findings after destructive testing many 100s devices to create my own charts.
And I agree, pretty much ALL of those SOA curves supplied LIE :-)  Even INXS's specific linear devices aren't perfectly accurate, but MUCH better than most.
Apart from all the main V/I/P.. considerations, hotspotting is rarely mentioned, or even easy to define / predict.
If you're still stuck, I'll dig out the links to the best white papers on the subject. They're in my past history somewhere :-)

[T3sl4co1l]

400W lies near the 10ms curve.

What the graph is telling you:

You can sink this much power, for up to 10ms, single shot (i.e., plenty of cooldown time so TJ returns to 25C before the next event).

You can infer that, hey, 400W isn't quite on the 10ms line, so it can run for longer, right?  Maybe 12ms or something.  Technically, no, you should not make that judgement.  However, refer to the transient thermal impedance curve to get a more accurate measure with respect to time.

Note also, this is a square power pulse, minimal ramp-up, minimal ramp-down.  For other pulse shapes, the peak power will be higher than the average power during the pulse, and a shorter time limit will apply.  It's not obvious how much shorter, for a given shape, would be required.  (Likely, you'd want to integrate the thermal impedance curve with respect to the pulse shape.)

Tim

[sleemanj]

Not an expert, but let me have a few words.

...

400 Watt which is at 80 Volt at 5 Amps :

Dissipated power by mosfet = Current ^ 2 x Resistance = 5^2 x 0.077 = 1.925 Watt

Unfortunately what you said, makes no sense.

The chart explicitly states the Vds is 80v, and 5A is passing through it, that is to say, that if you measure across the drain and source you are seeing 80v difference.

That means that the apparent Rds at that instance is, and must be, 80/5 = 16 Ohms. Not 0.077 Ohms. 

If the Rds at that point was 0.077 Ohm you would see only 0.077 * 5 = 0.385 Vds.

[Mechatrommer]

I was right the first time. I think you're confusing BJTs with FETs.

exactly! the earlier post will be corrected... you can see how much confusion i have here... this is another why? the contradicting term between bjt and mosfet..

btw, here i replotted the 150W line limit, just in case Vishay (datasheet manufaturer) care to have a look.