Transistor power dissipation

[tony359]

Hello all,

Can I pick your minds on a simple (I think) issue I am encountering please.

I have repaired a server SMPS and upon testing I have noticed that some components were running VERY hot, over 110 degrees Celsius. I tracked the issue down to two small PNP transistors (2SA1020, https://www.onsemi.com/pub/Collateral/2SA1020-D.PDF) which are wired in parallel - I mean BCE are all linked together on the PCB - and with the emitter connected to the cooling fans which are 12V, 0.4A.

I discovered that only one of the transistor was getting crazy hot - I did a search and I found that you need balancing resistors to connect two transistors in parallel so that behaviour was, unfortunately, expected. The transistors are located on a small air tunnel inside the SMPS so I guess whoever designed that thing thought it would be ok.

However since I ran the SMPS with no cover for a little bit, those transistors overheated and eventually failed. Looking into the circuit I found that those two transistors are only rated 1.5W each (and again, basically only one is working as they are in parallel) but my fans are 12Vx0.4Ax2 = 9.6W (in fact they normally run at 8.1Vx0.3A=4.86W but I guess you want to account for max power as they may speed up when temperatures rise).

I thought that the design was flawed so I decided to replace the small SA1020 with a single BD682G (https://docs.rs-online.com/d230/0900766b816decb9.pdf)  which is rated 40W and since the legs are too big for the PCB, I moved the transistor to the main heatsink with a short extension wire.

Everything works BUT the transistor "in air" still goes over 100 degrees Celsius after a few seconds. After I attached it to the heatsink I measure 50 degrees - with the lid open so no airflow but the heatsink was chunky and cold by then.

I just wanted to know if a transistor rated for 40W is expected to run at 50 degrees on a heatsink (using a heat-rubber pad, which is not the best I guess - and the setup in the picture is temporary as Farnell sent me the wrong item so I improvised with something I had handy) or whether I am missing something here.

To be sure I have measured the current flowing out of the emitter and it's indeed 0.6A as expected.

Thanks all for your help!

[bob91343]

Your question pertains to thermal design.  The thermal resistance, ambient temperature, and power dissipation combine to result in a hot spot temperature.  So you decide the maximum temperature you are willing to have it endure, then the formula shakes out the required thermal resistance.  You accomplish that with good heat sink design.

It seems you are asking for qualitative answers to a problem that is better solved mathematically and then verified experimentally.

[tony359]

Clearly I am not familiar with thermal design but all I am asking is that given the transistor I linked, it is expectable to read 100+ degrees C with no heatsink and 50 degrees C with a heatsink. I'm am expecting that to require some math indeed but unfortunately I am not familiar with that math so I am asking here!

Now, I am always happy to learn so I have found that the Tj is equal to ambient temp plus (Thermal Resistance x power). The actual power I read is 4.86W. The Thermal Resistance "junction-to-case" of my transistor is 3.13C/W. So my Tj should measure 36 degrees. I am reading 100+ instead.

[Kleinstein]

With the transistor in free air, the relevant thermal resistance is not junction to case, junction to ambient. The junction to case part would add to the temperature of the case. So this would come in addition to the 100+ C you measured for the case / though some of the heat may follow the leads directly.

[nigelwright7557]

As the transistors you have used are volumetrically similar then they will always get very hot.
The answer is a heatsink as you have found.
The bigger the heatsink the cooler the transistor will run.

[tony359]

As the transistors you have used are volumetrically similar then they will always get very hot.

Ok but my concern came from the fact that the factory transistor was a 1.5W unit, the replacement is a 40W unit and I was just wondering whether the 40W one was supposed to operate at a reasonable temperature without a heatsink - don't get me wrong, it's going on a heatsink anyways, I am just asking.

With the transistor in free air, the relevant thermal resistance is not junction to case, junction to ambient. The junction to case part would add to the temperature of the case. So this would come in addition to the 100+ C you measured for the case / though some of the heat may follow the leads directly.

Right, so you're saying that the 36 value I found means that the Junction temperature was 136 degrees, given 100 degrees case temperature.

Is there a way to predict the case - or junction - temperature based on the power involved?

[TimFox]

Some heat sinks have a manufacturer’s specification for thermal resistance (K/W) or graphed temperature rise vs. power.  Remember to include extra resistance for any insulation and thermal compound.  This temperature increase is added to the transistor’s specified junction-case resistance, or used with the derating value for the power as a function of case temperature (equivalent calculations).   There are textbooks that cover this in general.  The result depends on airflow and orientation (for convection).

[Audiorepair]

Tony,  the original design appears to be woefully inadequate, as you deduced.

Your idea of a single high powered transistor remotely mounted to the heatsink seems a very elegant solution.

I would try TIP31/32,  these have been a common choice for fans for decades.
I have seen these in free air, but they do get dry joints eventually.


I'm not sure why the other posters here are giving you such a hard time over this.
It is not rocket science to fit such a transistor to a heatsink, and it doesn't require poring over textbooks for hours to accomplish this either.

You just over engineer the solution for less than a quid.
Sorted.

[tony359]

Thanks all

Audiorepair
Cheers for the advice. I’ve already purchased the transistors I mentioned, do you think they can work too? They’re similar specs and the case is slightly smaller which helps in my case as the available room is very limited. I’ll bear the purchase advice in mind for next time!

Indeed I think that once I place the transistor on the heat sink with some mica and heat transfer paste all will be good. However my initial question was trying to understand better rather than just asking for a turnkey solution 🙂 So I still would like to understand how the case temperature of a transistor (or a component for what matters) can be calculated as I thought that with just a few watts the transistor I chose may not need a heatsink at all - but even with a heatsink (and a probably poor heat transfer mat) the temps still stay around 50 degrees, which means 86 degrees C junction temperature as I was explained a few posts above.

[magic]

See RthJC in the datasheet - the thermal resistance from junction to case. Multiply by worst case power dissipation and you know how hotter the inside is than the outside. Don't exceed some 100~150°C junction temperature (see specs) and it should be good.

Power dissipation is current times voltage across the transistor. Worst case will probably occur when the fan is at low RPM (low voltage across the fan, high voltage across the tranny) but still drawing close to normal operating current.

[David Hess]

Ok but my concern came from the fact that the factory transistor was a 1.5W unit, the replacement is a 40W unit and I was just wondering whether the 40W one was supposed to operate at a reasonable temperature without a heatsink - don't get me wrong, it's going on a heatsink anyways, I am just asking.

The larger the transistor is for a given package, the greater the difference is between the heat sinked and open air performance.  With no heat sink for instance, any TO-220 packaged transistor, or IC like a regulator, is limited to about 2 watts.

To put it another way, the thermal resistance is the total of the junction-to-case and case-to-ambient thermal resistance.  Without a heat sink, the case-to-ambient thermal resistance is very high making the junction-to-case thermal resistance from a transistor die with lot of area irrelevant.

[shakalnokturn]

I'm not sure I read your power calculations right.
The power the transistor has to dissipate is the product of Vce voltage and current flowing.
Theoretically at full speed even if current is higher the Vce should be lowest (if the fan power supply is 12V) thus power dissipation in the transistor not the highest.
If I understood right your fans are 12V models, what supply voltage is the transistor's collector linked to?

[tony359]

I'm not sure I read your power calculations right.
The power the transistor has to dissipate is the product of Vce voltage and current flowing.
Theoretically at full speed even if current is higher the Vce should be lowest (if the fan power supply is 12V) thus power dissipation in the transistor not the highest.
If I understood right your fans are 12V models, what supply voltage is the transistor's collector linked to?

Right, I had "forgotten" that the voltage used for power dissipation is the voltage across C and E! So my Vce is actually 4V (supply is 12V) and the fan runs normally at 8V.

Hence the power dissipated by this transistor - when the fan is in low speed I suppose - is 4V x 0.6A = 2.4W. That kind of explains the choice of 2x1x5W transistors from the factory - even though it won't work with no balancing resistors and does not account for faster speeds.

@David Hess - thanks for that number. It's probably what I was looking from the start. This is a TO-225 so a bit smaller than a TO-220, hence I guess the dissipation in free air is even lower than that.

So do you feel that reading 100+ C is ok for a TO-225 in free air with 2.4W? I won't comment on the 50 degrees when attached to a heatsink as chances are that with a proper MICA film things will get MUCH better!

To be honest I recalled that I looked for the Pd meaning for a transistor and I believe I read somewhere that it was the power that a transistor can dissipate WITHOUT A HEATSINK - I clearly misread that or I found a bad source!

[shakalnokturn]

Often transistors that don't have a heatsink mount are rated for maximum power dissipation at a given ambient temperature.

An improvement on the original design would have been to not completely parallel both transistors but at least to run the Emitters separate, one to each fan.

Temperature sould be fine once you've got the transistor correctly heatsinked.

[graybeard]

Discrete bipolar transistors should never be wired in parallel unless each one has it's emitter ballasting resistor.  If the power dissipation gets high enough the one with the lower V_BE will current hog.  Even if the transistor are identical this can happen if there is any difference in thermal resistance.

You can add some emitter ballasting  by putting identical resistors in series with each emitter.  Start by trying a value that drops 50mV at the operating current.

[magic]

To be honest I recalled that I looked for the Pd meaning for a transistor and I believe I read somewhere that it was the power that a transistor can dissipate WITHOUT A HEATSINK - I clearly misread that or I found a bad source!

Usually allowable power dissipation with and without heatsink is listed. Read the absolute maximum ratings table with all the fine prints.

If there is a big number printed at the top of the front page, it's most likely power dissipation with 25°C case temperature and maximum allowed junction temperature (150°C or so). Pretty unrealistic unless you run the transistor in the Arctic.

[Audiorepair]

TIP31 and TIP32 datasheet has a power/temperature derating graph.
So at 25C it is a 40 Watt device, at 100C only 15 Watts can be utilised.


https://www.farnell.com/datasheets/296719.pdf

[tony359]

Thanks all,

I've installed a thin pad under the transistor and now temperatures are more around 38 degrees. Surprisingly, installing a mica film with thermal pad yields a worse result.

@magic Yes, I did notice the 25 degrees case temperature. It's just that at the beginning I recall looking for the meaning of Power Dissipation for a transistor and I am quite confident I found a web page claiming it was the power that could be dissipated in free air. What you - and others - say makes sense of course.

@graybeard
I'm going to write down my trail of thought so you can tell me if I'm making mistakes.
For 50mV at 0.6A I would require a 0.08 ohm resistor - I am not sure it is available? For a 1 ohm resistor, the voltage drop would become 0.6V and the power to be dissipated by the resistor is going to be 360mW, correct?

I didn't investigate on this course of action anyways as the space available was very limited - transistors needed to be upgraded anyways so I thought that going with one, large transistor was the best solution.

[David Hess]

If there is a big number printed at the top of the front page, it's most likely power dissipation with 25°C case temperature and maximum allowed junction temperature (150°C or so). Pretty unrealistic unless you run the transistor in the Arctic.

And do not take the maximum specified junction temperature of 150C or 175C as a requirement to design to.  Continuous operation at that level will compromise reliability.  It is the "absolute" maximum and not the design maximum.

[nigelwright7557]

While you might think a 40 watt transistor would be better than 1.5 watt transistor it all depends on size to expel heat.
A transistor on its own would probably be ok for about 1 or 2 watts.
To soak up any more power would mean a heatsink to reach speced wattage.
It can all be worked out from datasheet.
The transistor will have a die to outside air temperature rise per watt.
After that you need a heatsink with a good degrees C/watt spec.