Hallo James,

when calculating heat transfer, temperature of junction, heat sink and all that stuff, we use the fact, that heat transfer works like the current flow through an ohmic resistor forced by the potential difference between its terminals. But what flows is heat power not electrical current, what forces heat power flowing is temperature difference not potential difference. And finally, not ohmic resistance is what defines how much current is flowing with given potential difference, but thermal resistance is what defines how much heat power is flowing with given temperature difference.
All these many words can be simplified by the following formula:

Rth = (Ta-Tb) / P

Rth is thermal resistance between points A and B, (Ta -Tb) is temperature difference between points A and B, and P is the heat power flowing between points A and B.

Let's have an example:
Assume having a transistor fabricated in a TO220 package. Collector emitter voltage is 4V and collector current of 80mA is flowing along emitter and collector path. This means, that heat power is fabricated inside the package. Heat power is calculated by P = U x I = 4V x 80mA = 320mW. This heat power will heat-up the crystal.
Now assume, that our transistor is fully isolated, so that no heat power can flow away from crystal. Even along terminals of transistor no heat power shall flow away. This we can achieve by soldering very very thin wires from transistor terminals to outer world.

What will happen?
Because there's absolute no heat transfer away from transistor, whole transistor will heat-up, heat-up, heat-up, etc. After some time transistor will get a temperature of more than 150°C ... 200°C and will be damaged. Crystal becomes too hot and breaks.

How to overcome this problem?
We must introduce some path for the heat power to outer world, so that as equally much heat power can be transfered OUT of transistor as is produced INSIDE!
TO220 package provides a very good path for heat transfer. A measure of this goodness is thermal resistance. The lower thermal resistance the lower temperature difference between junction (where all the heat power is produced!) and some point outside. This point outside has to be defined, of course. It can be the case of package or ambient air.

Typical thermal resistance of TO220 package is 4°C/W between junction and case (Ojc) and 54°C/W between junction and ambient (your Oja). In the latter case, case of package is not connected to any heat sink, but is only exposed to still air.

How about junction temperature of our transistor, when transistor is exposed to still air?
From our above formula we can calculate:

Tj - Ta = Rth x P = Oja x P = 54°C/W x 320mW = 17.28K

Tj is temperature of junction and Ta is temperature of ambient. Unit of difference temperature is Kelvin, without '°'.

So, equally at what temperature junction, case or ambient are, temperature difference between junction and ambient is always 17.28K. So, if ambient temperature is 25°C, then junction temperature is 42.28°C. If ambient temperature is 85°C, then junction temperature is 102.28°C.
Ok, Oja is a bit dependent on temperature also, but to assume a constant value is a very good assumption in most cases.

We can also calculate at what temperature case of TO220 package is:

Tj - Tc = Rth x P = Ojc x P = 4°C/W x 320mW = 1.28K

So, at whatever temperature junction is, case will always be 1.28K 'colder' than junction. This is also valid, when case is mounted to heatsink, because thermal resistance between junction and case is not influenced by heatsink.

The fact, that Ojc is so much lower than Oja makes it possible to drastically decrease thermal resistance from junction to outer world: We put our transistor with its case very close (guaranteed by the force of a screw or clip) to a heatsink. Then we get a series circuit of three thermal resisatnces: Ojc (thermal resistance between junction and case), Och (thermal resistance between case and heatsink) and Oha (thermal resistance between heatsink and ambient):

Oja' = Ojc + Och + Oha

Oja' is thermal resistance between junction and ambient of the new arrangement with heatsink, not to confuse with Oja (thermal resistance between junction and ambient of transistor package only)!!

Och seems to be trivial, but in case of using isolating material (silicon, Al2O3 or similar) Och can be in the range of 1°...2°C/W. Having bare metal joints Och is almost neglectable.

Let's assume a heatsink with 10°C/W. At what temperature will junction now be?
Oja' = Ojc + Och + Oha = 4°C/W + 1°C/W + 10°C/W = 15°C/W. This yields:

Tj - Ta = Rth x P = Oja' x P = 15°C/W x 320mW = 4.8K

This thermal resistance is much smaller than that of transistor package only, without any heatsink, namely 17.28K!! So, with the help of rather tiny heatsink, we can drastically improve heat transfer capability. This allows us to have much higher heat dissipating in transistor.


Three points finally:

1. Calculated temperatures above represent behaviour of steady state condition, when heat transfer and temperatures have stabilized after some time.

What does this mean?
Assume our transistor has just started to produce heat power. Then its temperature is still low, like the ambient. When now heat power is produced, temperature difference is zero and no heat power can flow away from junction. But where the heat power then is going? It goes into material of junction and heats it up. This effect is called 'heat capacity': When heat power P is flowing into material for a certain time t, then temperature of this material rises, while this temperature rise is proportional to P and t. Of course, each part of transistor does have this heat capacity, not only junction. The effect of heat capacity is, that in case of not yet having reached thermal stability, MORE heat power may be produced inside transistor, while having same junction temperature.
This explains the introduce of dynamic thermal resistance in certain datasheets of power transistor, where dynamic thermal resistance can be much lower than static thermal resistance. All the calculations above with our example transistor used static thermal resistances!

2. Many SMD packages, which do not have heatsink mounting capability, transfer the most of produced heat power through soldered pins!! So, only wide copper traces of PCB or even large area copper planes can help to decrease thermal resistance between junction and ambient. To fix a heat sink on plastic package will often not satisfy, because thermal resistance between junction and plastic case is just too high.

3. It's not wise to have transistors running for long time at very high junction temperature. This will drastically decrease life time of transistor!! So, normally, junction temperature should be some dozens of degrees lower than specified maximum crystal temperature, which is about +150°C for common purpose transistors and up to +200°C for some high power transistors.
That for your regulator only 125°C is allowed, comes from the introduce of temperature 'fuse': Whenever junction temperature rises above about 125°, additional circuitry of LP2985 reduces output current and limits by this produced heat power inside chip. This threshold temperature shows some tolerance, of course, so that we need a certain headroom for reliable operation. No more than about 100°C should be reached at extreme situations (maximum output current)! And for longer periods junction temperature should be well below 70°-80°C.


Kai