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MJ15003 Power Transistor

15 Jan 2012

SG1

Australia

I am using a MJ15003 power transistor as a voltage regulator, it is drawing 12 volts at 2.5 amps with a heat sink.

My question is the transistor is running hot to touch at 60 degrees centergrade is this acceptable within the manufactures specifications.

I have included the data sheet, on page 1 it metions that "Operating and Storage Junction Temperature Range is from -65 to +200 degrees centigrade, so I believe I am well within specifications.

 

Thanks Craig

 

 

 

 

 

Supporting materials

Replies

Bugs Bunny

Belgium
15 weeks ago

Hi Graig 

The oxide layer and air gap (no paste) is of little importance in your design, because the transistor operates far inside its SOAR: the dissipated power, (as well as supplied current and supported voltage) are quite tiny with respect to the ratings of your transistor.

Your thermal resistance is about 5.1 (Rth a-h) + 0.4...0.8 (Rth h-c) + 0.7 (Rth c-j) or 6.2...6.6°C/W, with a (very) well fastened heatsink. The biggest part of the total Rth comes from the heatsink, the other two values being quite smaller. The matter is quite different, however, when a large amount of heat has to be dissipated, with a big heatsink like this one: have a look at http://be01.rs-online.com/web/p/heat-sinks/0158556/. Here, it's mandatory to get an as low as possible Rth (h-c), as this parameter may become the most important in the thermal chain! Removal of oxide layer and addition of thermoconductive paste are crucial., mica sleeves or similar stuff are forbidden.

But. But.

After estimation of the dissipated power, something has begun worrying me: the filtering capacitor(s). Here is how I succeeded to calculate their values. I first assumed that, following a good rule of thumb, you have chosen a transformer delivering 12Veff at 2.5Amp. Peak value : 12 * sqr2 = 17V. Peak value at capacitor : 17 – 1.4 (2 diode voltage drop, bridge rectification) = 15.6V. The voltage between C and E of the transistor can be seen as a trapezoidal waveform, the highest value being Vi (15.6 – 12) or 3.6V, decreasing with a slope (dV/dt) α during a time t until the next recharge, 10 ms later. The formula of the voltage waveform is then V(t) = Vi + α t. The dissipated power is equal to 1/t * I * ƒ (0→t) (Vi + α t) dt. This becomes 1/t * I * [Vi * t + 0.5 * α t↑2] (0→t).

The dissipated power, with a very well fastened heatsink, is estimated as in my previous post: 35°C/ (5.1 +0.4 + 0.7)°C/W or 5.6W. Let's now insert all the numbers in the last formula. 5.6 = 1/10↑-2 * 2.5 * (3.6 * 10↑-2 + 0.5 * α  * 10↑-4). Next step : 5.6 = 9 + α * 1.25 * 10↑-2. The solution is α = - 272 V/s. This means that after 10 ms, a voltage drop of 2.72V will occur, leaving a collector to emitter voltage of only 15.6 – 2.72 – 12 = 0.88V, too low for the transistor to operate properly. From the relationship describing the behaviour of a capacitor C * dV = I * dt, we write C = I / (dV/dt) or 2.5 / 272 = ≈ 9100 mF. Did you use an approaching value, such as 2 x 4700 or 1 x 10000 mF? If my reasoning is correct, you should add a third capacitor, whose value would be 4700 mF. The slope would then decrease to |α (dV/dt)| = I/C or 2.5 / (3 * 4700 * 10↑-6) = - 177V/s, and the collector voltage will consequently drop at 15.6 - 1.77 = 14.4 V, 2.4 V above the emitter voltage, which is now acceptable.

Tell me if I'm near the reality?

Kind regards,

B.B.

 

SG1

Australia
15 weeks ago

Yes you are spot on, that explains it all.

Thanks for the indepth description.

 

Craig

SG1

Australia
15 weeks ago

Thanks for the reply.

 

You are correct, I am using a linear regulator  12V at 2.5A  output.

 

The heat sink I am using is

http://australia.rs-online.com/web/p/heat-sinks/0402973/

I have not removed the paint coating on the heat sink or applied a paste, dose this matter.

 

Craig

Bugs Bunny

Belgium
17 weeks ago

 Hi Graig!

Boss is right : you are well within the Safe Operating Area of your transistor!

If you are worrying about the junction's temperature, here is a little trick to estimate it.

Remember that the total thermal resistance of a “heat circuit” is just the same as the total electrical resistance of resistors connected in series, where the current flow is replaced by the thermal flow (the power to be dissipated), and voltage drops by successive increases in degrees.

Now I assume your small heatsink has a thermal resistance to ambient Rth(h-a) of 2.5°C/W, the thermal resistance between case and heatsink is in the order of 0.2°C/W -Rth(c-h), provided the thin aluminium oxide layer on the heatsink was gently removed and a small drop of thermoconductive paste added-, and thermal resistance between junction and case Rth(j-c) is known : 0.7°C/W. The total thermal resistance Rth(j-a) is equal to the sum of the three values, or 3.4°C/W.

Your heatsink is roughly at 60°C, which is 35°C above common 25°C ambient temperature. According to its thermal resistance to ambient, the power flow is in the range of 35/2.5 or 14W. Translating Ohm's law to thermal flow, we are now able to “climb up” to the junction, and it will give us an increase of (0.2 + 0.7)°C/W x 14 W =~ 13°C, and a junction temperature of about 75°C, which is quite acceptable.

Note that my estimation of power dissipation (14W), times the estimated total thermal resistance (3.4°C/W), added to an ambient temperature of 25°C, approaches rather well 75°C. It's coherent.

This reasoning the wrong way round is valid, but it's of course evident that a maximum thermal resistance has to be estimated starting from junction towards ambient, to avoid overheating the junction in the worst case of maximum power dissipation at the highest allowed ambient temperature.

Just hope these few words will help you.

Regards,

B.B.

SG1

Australia
15 weeks ago

Hi

Thanks  for the well explained report.

 

The heat sink I am using is

http://australia.rs-online.com/web/p/heat-sinks/0402973/

 

I have not removed the aluminium oxide layer on the heatsink and I have not added any thermoconductive paste.

 

Thanks Craig

 

 

 

Boss

United Kingdom
17 weeks ago

Assuming your using it in a linear regulator and your 12V at 2.5A is the output then the power dissipation in your transistor is 2.5Watts for every volt being dropped across it, e.g. 15V input gives 7.5W; 20V input gives 20Watts.

Without knowing your circuit, lets assume that 10W is being dissipated.  The transistor thermal resistance junction to case is 0.7C/W so at 10W only an extra 7C rise is expected so the junction is at lessthan 70C, well within spec and you heatsink is suitable. Check for your actual design!

Make sure you are measuring at the case as there is also the case to heatsink and heatsink to air thermal resistances to be accounted for.

As a rule of thumb I was told many years ago that if you can just about touch the case then the temperature is about 60C and the junction temperature should be OK.