[Solved for now]Power dissipation and junction temperature; how to?

Other Hardware General Electronics
1

Simple question :smiley: But at this moment not so simple for me :frowning:

How to calculate the junction temperature of a FET / transistor so I can predict if I will blow it up :wink:

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Using an IRL520 to drive a 12V / 2A LED strip

The relevant specs (as far as I know):
Rds(on): 0.27 Ohm

Thermal
Rtheta(junction to case): 2.5 C/W
Rtheta(junction to ambient): 62 C/W

Tj(max): 175 C

Environmental temperature, worst case 50 C

I got this far:
P = IxIXR = 2 x 2 x 0.27 = ~ 1W
Tj = 50 + (62 x 1) = 112 C

Nice, possibly acceptable. But according to the datasheet, Rds(on) is a function of the junction temperature and at 50 C ambient is around 1.2 C/W Ohm.

So the power dissipation is now 5W and Tj will be 50 + (62 x 5) = 360 C. Far outside the spec and I need serious cooling.

Even if I manage to cool it down to 100 C, Rds(on) will be close to 1.75 Ohm increasing power dissipation and the temperature again.

I feel that I'm in an endless loop and like to know how to calculate correctly. And how will a closed enclosure affect all this?

IRL520 datasheet

Additional question: looking at fig. 4 in the datasheet, where do I find the specified Rds(on) of 0.27 Ohm; looks like that must be at a temperature of 0 Kelvin. I guess I'm totally missing something :frowning:

Note:
I'm aware that there are better FETs, but the question is general.

Thanks in advance for any clarifications, explanations, calculations and so on.

2

See this
http://www.thebox.myzen.co.uk/Tutorial/Power.html
And also
http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html

3

Figure 4 says : "Normalized". That means a value of 1.0 at 20 degrees is normal (0.28 ohm), at 130 degrees it is twice 0.28 ohm.

It is hard to calculate, because the way a heatsink is attached matters a lot.
Many on this forum would say that 2A in a 50 °C environment and 0.27 ohm is not okay.
You gave the answer yourself, you need a lower Rds(on).

This one is 13mohm, that's a lot better. You don't even have to calculate something, because 2A and 13mohm causes very little heat.

Using a 60A mosfet to turn on a 2A led strip is fine, for a low Rds(on) often a mosfet is used that is able to switch a lot more current than it is used for.

4

But according to the datasheet, Rds(on) is a function of the junction temperature and at 50 C ambient is around 1.2 C/W.

That doesn't make sense. You are saying here that the thermal resistance is dependent on the temperature, that is clearly a nonsense.

I feel that I'm in an endless loop and like to know how to calculate correctly. And how will a closed enclosure affect all this?

You have to fix some design parameters, like what is the maximum temperature you want to run the junction at. The box will add another thermal resistance into the mix, with a case to box and box to ambient to consider. These are hard to determin.

5

@Grumpy_Mike, sterretje was confused by the number of 1.2 in Fig. 4 in the datasheet. I have explained that number. The Rds(on) is a function of the junction temperature... period.

6

Thanks for the replies; something to chew on.

Grumpy_Mike:
That doesn't make sense. You are saying here that the thermal resistance is dependent on the temperature, that is clearly a nonsense.

Thanks, after the first two replies I re-read my post, thought that the specific part did not make sense, checked it and changed it; should have been Ohm.

Koepel:
The Rds(on) is a function of the junction temperature... period.

Thanks, at least get some more decent numbers now :wink:

7

Hi,

Here's something else to munch a bunch on :slight_smile:

If we calculate the power then solve for the resistance R, then fit a second degree curve to the resistance change with temperature T, then equate the two, we come out with the following equation:

(T-Tamb)/(CWI^2)=R1((43T^2)/975000+(679T)/195000+1151/1300)

where
Tamb is the ambient temperature in degrees C (such as your 50 deg C),
CW is the temperature rise with power coefficient (such as your 62 deg C/W),
I is the current (such as your 2 amps),
R1 is the initial resistance (such your 0.27 ohms),
and T is the final temperature.

To solve this, simply enter in the values for the above variables, except for T. Then, solve the equation for T. You will get two values, only one is right, the lower one. If you get two values that are complex (contain an imaginary part) then the parameters are invalid, which means the temperature rise would be too high.
Note i would show the explicit solution but it's too long and cumbersome.
Also note this is only for this particular device

Using the following data:
Tamb=50, CW=10, I=2, R1=0.27, i get a temperature of 63.9 deg C.

Using this data:
Tamb=50, CW=20, I=2, R1=0.27, i get a final temperature of 81.6 deg C.

Using this:
Tamb=50, CW=30, I=2, R1=0.27, i get 107.2 deg C

Using this:
Tamb=50, CW=40, I=2, R1=0.27, i get 164.7 deg C

Using this:
Tamb=50, CW=50, I=2, R1=0.27, i get a complex number which means it will not work.

Finally, using this data:
Tamb=50,CW=41, I=2, R1=0.27, i get a final temperature of 181.7 deg C which is beyond the limit of the temperature for this device.

Therefore something around CW=40 or less would be needed, which of course means a heat sink.
Also, might want to leave room for device variation, such as if the resistance for your device is really 0.29 instead of 0.27 for example.

I was going to post a complete graph showing all solutions but i dont see a way to upload a picture file like .gif to this post. Oh i think i found it...
Yes, see the graph for the entire picture in a nutshell.

The limit for CW appears to be close to 41.6, after which the temperature rises indefinitely. This means a value of 62 will never work with this transistor unless the current is decreased.

CW_vs_T-1.gif
CW_vs_T-1.gif886×511 7.95 KB

8

Yeah, well I heard Arduino is just for school kids to fool around with and nobody who knows anything does anything with it. Thanks for the education, MrAl!

9

Grumpy_Mike:
That doesn't make sense. You are saying here that the thermal resistance is dependent on the temperature, that is clearly a nonsense.

Its not clearly a nonsense. It happens that copper has a pretty constant thermal conductivity across the
normal range used in electronics, but the silicon wafer itself does not, conductivity varies significantly.

Thermal data in datasheets usually makes the simplfying assumption that conductivity is constant, but
its just an approximation. For devices with non-copper heatsinking, for instance BeO heatsinks in RF transistors
the constant conductivity assumption is a poor fit to the data.

Anyway back to the point, the way to proceed in these calculations that is simple and reliable is
to start with a maximum temperature you are prepared to tolerate and calculate the dissipation with
the worst case Rds(on), then see what ambient temperature would need to be to prevent the
junction exceeding your self-imposed maximum. If that ambient temperature is unrealistic, reconsider
your device and or heatsink.

The reason to choose a temperature is that hot components have reduced reliability and reduce the
reliability of neighbouring components due to thermal stresses, and to prevent any plastic nearby
parts softening. I would recommend keeping below 75C if at all possible - these days there are many
very high performing MOSFETs available and cooling fans of all sizes are readily available too.

10

A bit of googling turned up this clever technique of determining the junction temperature of a power mosfet. The source to drain parasitic diode can be characterized to determine Vforward vs T at a fixed current (10 mA). Knowing that, a known power can be applied to the mosfet and a switch can then be thrown to measure Vforward and then translate to T. It is a pretty clever bit of engineering (though I think his fig. 2 is wrong, the DUT is an NFET so current IF should be reversed and the gate tied to the drain).

11

terryking228:
Yeah, well I heard Arduino is just for school kids to fool around with and nobody who knows anything does anything with it. Thanks for the education, MrAl!

Hi there Terry,

Well you're welcome :slight_smile:

I hope others use the Arduino too as it is a pretty good platform for getting things going quickly.

I found an even better fit:
(T-Tamb)/(CWI^2R1)=(11T^2+2165T+344000)/405000

and really all we have to do is plug in all the values except for CW, where T then takes the role of the target temperature (in deg C). So if we wanted to look for what CW would be for the max temperature we would just plug in 175 for T and plug in the rest and solve for CW.
When i do this with the current problem, i get a value of CW=44.2 which means CW cant go over 44.2 or else the device burns out. If i was building this though i would stay WELL under 44.2 to make sure it doesnt get too hot, so maybe set the target temperature at 100 C, which brings us to CW=28, so a heatsink that can make that happen would be necessary.

Here's an explicit equation for CW (T from 25 to 175):
CW=(405000*(T-Tamb))/(I^2R1(11T^2+2165T+344000))

Good luck with your projects :slight_smile:

12

Thanks everybody

I will go for better FETs so I don't have to worry (too much). Still chewing :smiley: And might revive this thread later.