Determine if a heatsink is required (Darlington TIP120)

Question

I'm planning to switch 1A load at 5V with a TIP120.

Looking at it's power derating graph I see that at 5W case can be around 140C, but how can I tell if it will get hotter than that?

In the datasheet I see that thermal resistance junction to ambient is 62.5C/W so does this mean that it will dissipate 5 times that = 312.5C. That seems like a bit too much for such load.

If power dissipated is VCE * Ic (VCE is 4V for TIP120 at 5V collector voltage, right?), then this means 4 * 62.5 = 250C which is still a lot. I'm getting a feeling that this can't be right.

Also if I dissipate 4W and still provide 5W for my load, then that's almost 50% efficiency, thats horrible. I'm probably off on my calculations, I could really use some help..

Answer 1

You want to refer to Vce(sat) at 1A. Which is around 0.8V typically, so you're looking at 800mW. By my (conservative) standards that's a bit on the high side, but it might be okay for you. A square inch or so of PCB copper would deal reduce it quite a bit.

But really, you want to use a MOSFET and not that darlington with the high voltage drop.

Answer 2

I believe you need to calculate the collector-emitter voltage before multiplying by (current) and (Thermal Resistance, Junction−to−Ambient)

Answer 3

No, 312.5C is the temperature that the TIP120 would reach if the heat pathway was only through the thermal resistance from the case to air. 312C is well above the absolute maximum rating, so at 5W you'll need a heatsink. You also need to consider other exit pathways for heat, like the Junction to case which is 1.92C/W this means 5W*1.92C/W = 9.6C above ambient, which seems more reasonable.

However Junction to Case means that the part temperature will be ~10C above the PCB temp, and that can be difficult to determine because then you have to calculate copper thermal resistances and estimate how much of the heat can go back to air ect.

Do yourself a favor and slap a heatsink on the TIP120. 1-3W you might be able to get away with, 5W will probably require a heatsink. Your application will be happier also as temp swings show up in signals.

In the past I've even used small pieces of aluminum or coins and taped them to parts with a small amount of thermal paste, you'd be surprised at how much a little more surface area will increase the heat dissipation of a part.

Answer 4

In the TO-220 package, you can dissipate around \$2W\$ without the need for a heatsink.

$$P = \frac{150°C - 25°C}{62.5 \:C/W} = 2W$$

But I suspect that the load will be connected between supply rail and TIP120 collector and also the BJT will be working as a switch (in saturation).

So in this case power dissipated will be equal to:

$$P = I_C \times V_{CEsat} \approx 1A \times 0.7V = 0.7W$$

And the TIP120 junction temperature will be \$0.7W \cdot 62.5 \:C/W = 44°C\$ above the ambient temperature.

If this is the case.