Does case temperature of a FET rise linearly with increasing current?

[Jim from Chicago]

I am trying to estimate the maximum case temperature a certain p-FET will experience on a board. I know this FET will have to pass up to 15 Amps in certain applications, but I don't have the test equipment available to load it to 15A. With my equipment I can only load it up to 4.5A. I do have a nice thermal camera though.

So I was thinking to load this FET in say 1A increments and measure its case temperature at each current level, then extrapolate to 15A assuming the increase would be approximately linear. Is this valid for a rough approximation?

Also the FET is Vishay SUD50P08-25L-E3. Datasheet: https://www.mouser.com/datasheet/2/427/sud50p08-1766022.pdf

[Weston]

For a rough approximation, it's better than nothing.

However, the MOSFET on resistance changes with temperature, which can lead to thermal runaway. It's close to double at the upper end of the temp range. This would lead to an under-estimate of the maximum temperature rise.



On the other hand, while thermal conduction to other parts of the PCB is linear, the heat is mostly transferred to the external environment by convection (and possibly some radiative cooling), both of which are non linear, transferring more heat as temperature increases. Its a bit hard to model these effects.


You can pretty accurately calculate the worst case power dissipation given a maximum junction temperature. If the issue is being able to dissipate enough power in the FET with your power supply, you can reverse bias the FET and dissipate energy in the body diode. It has a higher voltage drop, so you can dissipate more power internal to the device with a limited current.

[David Hess]

What you suggest will return good results.  The temperature rise will be proportional to the power, which is roughly proportional to the current.  I have sometimes done this to calculate the junction-to-ambient thermal resistance in a given design.

You could also measure the voltage across the transistor and multiply it by the current to remove the effects of the on resistance changing with temperature.

[TimFox]

Since the ON resistance is relatively low, be sure to connect the voltmeter as close as you can to the FET itself as part of the power measurement.

[Le_Bassiste]

... power, which is roughly proportional to the current. ...

P = RDSON x IDRMS^2 hence, doubling the current will quadruple the power

also, see here:
https://archive.org/details/mospowerapplicat0000unse, Chapter 4, "Thermal Design and SOA"

[Siwastaja]

P = RDSON x IDRMS^2 hence, doubling the current will quadruple the power

Yes, but the question was not about power, but temperature difference. Power lost by thermal conduction is directly proportional to the temperature difference, so power of 2 converts into power of 1.

Power lost by convection is more than linearly proportional because larger temperature difference starts to move the air around. On the other hand, MOSFET Rds_on is not constant but increases with temperature. In some lucky circumstances, these two effects might cancel out each other, resulting in nice linear current ~ temperature difference relation. Better prepare for a bit worse, though, Rds_on at Tj=125degC is often like 1.7x of the 25-degC value (depending on MOSFET, though).

[Jim from Chicago]

In the datasheet in the table it says a max Rds(on) of 29mOhm at Vgs = -4.5V. But one of the graphs shows normalized on resistance vs case temp, and it indicates Rds(on) will be double around 160C case temp. Does this mean that the max Rds(on) is actually around 58mOhm?

[iMo]

The temperature of the fet package is

T = P * RthJA + Tamb

For example with Ids = 15A and Rdson=60mOhm the P=13.5W and with thermal resistivity of the package (see the RthJA in the DS) for example 15C/W the temperature of the fet will be

T= 13.5*15 + 25 = 227.5C

With Rdson = 29mOhm the T = 122.9C.

[ArdWar]

Thread carefully when reading datasheets, often they're set at unrealistic condition.

in the table it says a max Rds(on) of 29mOhm at Vgs = -4.5V.

The table values are specified at T_J=25 °C. You probably aren't going to achieve that unless you're dunking it on ice, or doing very short pulse with low duty cycle, which is exactly what they qualified at with notes "a". (a. Pulse test; pulse width ≤ 300 µs, duty cycle ≤ 2 %)

But one of the graphs shows normalized on resistance vs case temp, and it indicates Rds(on) will be double around 160C case temp. Does this mean that the max Rds(on) is actually around 58mOhm?

Yes. I guess you meant T_J (Junction temperature) instead of case temperature? Also keep in mind that it's also cumulative after you consider the effect of I_DS to R_DS(ON)

[Siwastaja]

In the datasheet in the table it says a max Rds(on) of 29mOhm at Vgs = -4.5V. But one of the graphs shows normalized on resistance vs case temp, and it indicates Rds(on) will be double around 160C case temp. Does this mean that the max Rds(on) is actually around 58mOhm?

Yes, except I doubt the graph is for case temperature, it usually is for Tj. Also, don't go to extreme temperatures even if indicated in the datasheet, leave a bit of leeway. Max design Tj=120degC is a good idea, after that Rds_on rises so much anyway. Unless you are constrained by high-temperature environment (say automotive) and must go a bit higher, but then you need to be extra careful with your Tj calculations and derate.

[Jim from Chicago]

Yes you're right the graph is junction temp vs Rds(on) not case temp

[Siwastaja]

Case temp is irrelevant anyway. You should calculate from Ta to Tj with all intermediate Rths. Datasheet gives you RthJ-C as 1.1 K/W. Add some 0.5 - 1 K/W for thermal inteface material / grease / coupling, then what the heatsink datasheet says, and that will be the total RthJ-A. Then you can calculate die tempearture, which is all the MOSFET cares about. Case temperature for the heatsink is also relevant; case temperature of the MOSFET not that much.