NPN Thermal Calculation Review

[wuchy143]

Hi All,

I have gone through a couple application notes(this forum as well) and what not on thermal considerations for a transistor(or any electrical device that absorbs energy and dissipates it through heat transfer via heatsink, air, copper, etc). I don't need to get down to 128-bit resolution to see if this device is going to operate within it's derating but would like a ball park idea.

Can anyone review and tell me if I'm right or at least in the ball park. I haven't done many thermal stuff due to never doing much high current so please any tips and what not please say. I"m using 2N6388 as my darlington pair npn configuration.


Here are my calculations/thoughts.

My amplifer(2N6388) will at worse case scenario see 10.5V(DC) across it's collector and emitter while driving 500mA. This is fact. So the power that the transistor is absorbing is:

P = V * I = (10.5V * 500mA) = 5.25 Watts dissipated in transistor as heat.

From ON Semi's 2N6388 datasheet (https://www.onsemi.com/pub_link/Collateral/2N6387-D.PDF) you will see that the thermal resistance to the case is 1.92*C/W

So the 1.92*C/W multiplied by the amount of power(5.25W) you will get a temperature increase of 10*C from ambient. In my case ambient is 20*C(68*F) which will increase my junction temperature to 30*C(86*F)

This means that the junction of my transistor will rise up to 30*C when driving 500mA with a 10.5V potential difference between the collector and emitter.

I know from a seat of the pants feeling(I"m about to prototype my circuit anyway but wanted to do out the calculation as well) that I'm fine. The part might get a little warm but wont even requite a real heatsink.

I have two questions:

1. How close am I to the correct rise in temp?
2. In the datasheet which I supplied above in figure 1 there is a power derating graph. I've stared at it for a while and don't "get it". Can anyone de-mystify this graph to me? You can make fun of me for being a dummy

Thanks again all!!!!

 

[Norator]

You need the case to ambient resistance for a package the size and shape and material and surface finish of the 6388.

 

[kubeek]

Junction to case means that you have infinitely large heatsink attached to the device, i.e. the heatsink stays at constant ambient temperature no matter what.
5W is not a small amount of power, look at the size of a 5W resistor, and this gets quite hot when dissipating full rated power.
For using the device without heatsink you need to look at the junction to ambient figure, which is 62.5K/W, so your transistor would rise to 20+62.5*5.25 = crispy 350°C. This means you definitely need some heatsink.
Maximum junction temperature is 150°C, so you have 150-20-10=120K difference to spare on the heatsink for the transistor to still survive. 120K/5.25W=22.8K/W so you need a heatsink with this or better thermal resistance to ambient.

Some other thoughts: 20°C ambient is quite unrealistic unless you live in polar tundra, 35°C would be better but this mostly depends on the enclosure the part will be in and how well it is ventilated.
150°C is the absolute max temperature, so you´d better aim for no more than 120°C max to have better reliability. (and also account for dust buildup etc. and still be on the safe side)

 

[kubeek]

The power derating is a simpification of the thermal calculations, and it basically tells you that for example if you can hold the case temperature below 80°C you can safely dissipate 25W. This has to do mainly with junction to case resistance for the Tc curve and junction to ambient for the Ta curve. But still it is better to do the proper calculations to be sure.

 

[wuchy143]

Kubeek,

Howdy. I'm reading through your guidance and just want to make sure I'm understanding things.
When you say:

Maximum junction temperature is 150°C, so you have 150-20-10=120K difference to spare on the heatsink for the transistor to still survive. 120K/5.25W=22.8K/W so you need a heatsink with this or better thermal resistance to ambient.

Can you explain why you are subtracting 150-20-10? Actually this whole paragraph I'm having a hard time grasping 100%. Could you maybe explain a little more what's going on. How you are getting your calculations. Also, what is K?

 

[Norator]

It's a DC circuit, of sorts
https://en.wikipedia.org/wiki/Thermal_resistance

 

[kubeek]

Can you explain why you are subtracting 150-20-10?

150-20-10 means: 150°C max junction temp minus 20°C ambient temp minus junction-to-case-thermal_resistance times power_dissipated, and the junction to case difference is from the older posts 10K or 10°C.
Historically absolute tempreatures are expressed in degrees Celsius, but the difference between two temperatures is expressed in Kelvin. Difference of 1K is the same as rise by 1°C.
Well this is the standard scientific notation, so sorry if that I´m being too demanding. But anyway most of heatsink datasheet will state thermal resistance in K/W so this may be my way of pointing out how this all works.