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Here's a trick that is useful the next time you do thermal testing with your MOSFETs or IGBTs.
Yes, that's right. It's important to make sure your power transistors don't overheat. In the datasheet, you will find some information that you can use to estimate how hot the junction inside the IC will get.
Let's look at an example. Here's a page from the IRF7739 DirectFET datasheet. I like this datasheet because it has almost all the thermal stuff on one page, which we'll get to in a moment.
The IRF7739 is one of IR's "DirectFETs" which are basically metal cans around a power transistor die. This one is one of their lowest on-resistance MOSFETs: it's a 40V N-channel MOSFET rated at 1 milliohm maximum at 10V Vgs with 160A current. On page 1 you will see some absolute maximum numbers that are somewhat scary:
270A at 25°C case temperature! 190A at 100°C case temperature! That is a lot of current. You probably won't be able to get that much out of the part, but here they're telling you that you have no hope of exceeding these ratings without causing damage to it.
Thermal modeling and design involves answering these questions:
Let's take a relatively easy example, and pretend that you need to use this part as a static DC switch, and you have a way of guaranteeing that at most 40A flows through it. The MOSFET will be on a circuit board inside a box, and the box needs to work in environmental conditions from -25°C to +50°C ambient air temperature.
The IRF7739 is specified, as I mentioned earlier, with an absolute maximum on-resistance of 1 milliohm at 10Vgs conducting 160A at 25°C. On-resistance is generally fairly flat, so we should expect it to be about the same at 40A. (If you look at Fig. 2. on page 1 of the datasheet, you'll see that on-resistance decreases somewhat at "only" 40A as compared to its maximum current.) But that's at 25°C. The other important piece of information is later in the datasheet:
MOSFETs are positive temperature coefficient devices: as they heat up, the voltage drop across them increases. Here we can see that at about 135°C, the on-resistance increases by about 50% from its nominal value at 25°C, and at 175°C (its maximum rating) it increases 75%. So to be conservative, we should plan on Rdson being as high as 1.75 milliohms, and then later if we can make sure it doesn't get that high, we might be able to show it's never going to exceed 135°C and plan on Rdson being at most 1.5 milliohms. MOSFETs used in constant-current applications can undergo a mild form of thermal runaway: they get hotter, which causes them to dissipate more heat, which gets them hotter, which causes them to dissipate even more heat, and so on, until POP! it doesn't work anymore.
Anyhow, onto page 3, which is where all the fun thermal stuff is shown:
Okay, here's where we start figuring things out. At 40A, power dissipation, which is solely I2R losses, is at most 40A * 40A * 1.75 milliohms = 2.8 W. Note 3 says that if you get lazy and just solder this thing onto a 1 square inch copper pour, you can dissipate at most 3.8W at 25°C ambient. That's a little too close for comfort, and we're going to be running inside a box at 50°C ambient outside the box. So we'll probably have to use a heatsink and perhaps a fan as well.
Since we have our power dissipation at 2.8W, what we need to do now is estimate the thermal resistance from junction (where the heat is generated) to ambient. You do this by adding up the series thermal elements: junction to case, case to heatsink, heatsink to ambient. If there is more than one thermal path, it's essentially the same idea as resistors in parallel, only with a thermal circuit the thermal resistance causes temperature rise from power flow (Rθ measured in °C/W), whereas in an electrical circuit the electrical resistance causes voltage rise from current flow (R measured in V/A).
The datasheet tells you RθJC = junction-to-case thermal resistance. Actually in this case there are two paths: the thermal resistance to the top case is 1.2°C/W, whereas the thermal resistance to the PCB on the bottom of the part is 0.5°C/W.
So you can go through the exercise of figuring out the thermal resistance from top case to heat sink, and heat sink to ambient, and PCB to ambient, and crunch the numbers to figure out the net junction-to-ambient thermal resistance, multiply it by the 2.8 W max power dissipation, and come up with a temperature rise above the 50°C ambient to get the junction temperature, and make sure you have enough margin below the 175°C maximum junction temperature.
That's a whole exercise in itself, and unfortunately we won't go into it here.
But remember that it's only an estimate. If you really want to make sure your transistors don't go POP!, you should be doing empirical testing. This involves taking the actual part on an actual circuit board inside an actual box, and putting it in a 50°C ambient environment to see how hot it gets.
Now, 40A is a lot of current. (You might be asking, wait a minute, where's all that current going to come from? Perhaps it's a power supply for a new CPU -- the computer I'm writing this on is 2 years old and gulps down up to 50A at 1.3V, from a DC/DC converter.) We can certainly try to find a power supply capable of outputting 40A and feed it into the part, using some thick wires -- 10 gauge wire ought to do it.
But there's an easier way!
OK, here's the trick, because otherwise ultra-low on-resistance MOSFETs are a pain to test if you are going to try to bring in that much current from the outside world.
We're not going to put 40A through the IRF7739. Instead, we'll dissipate 2.8W in the IRF7739. And we'll do it at a higher power supply voltage with less current. In fact, if you have a 12V 25W lab supply capable of running in constant current mode, you can easily do this test.
Here's the circuit we'll use:
One zener diode and a resistor! What's going on here?
The zener diode between drain and gate will start to turn the MOSFET on when the drain-to-source voltage exceeds the zener voltage enough to start conducting current. If you look at the IRF7739 datasheet, you'll see these graphs:
At lower Vgs voltages, a MOSFET acts like a constant current source. Here we can see that at 4.5Vgs, the drain-to-source current is typically 0.2A at 25°C, rising to over 10A at 175°C.
When the MOSFET starts conducting, that will pull current away from the zener, keeping the MOSFET from turning on fully. In effect, we have built a constant voltage sink, i.e. a shunt regulator. The voltage in question is going to be about 4.5Vgs plus the zener voltage, so for a 5.6V zener we might see 10V total across drain-to-source. A very small amount of power is being dissipated in the zener and the resistor (speaking of which -- you'll want a few milliamps of current to flow through the resistor and zener, in order for the zener to work properly, so I'd probably choose a 1K resistor), and the rest is going into the MOSFET. Set your power supply to about 0.28A, and you'll get 2.8W dissipated in the MOSFET.
Now, in reality the shunt voltage isn't that well regulated, and it will change slightly as the MOSFET heats up (the shunt voltage should go down by maybe 0.5-1V, because at constant current the Vgs threshold for conducting that current decreases, as you can tell by the graphs of Fig. 4 and Fig. 5), so you may have to adjust the power supply current to compensate.
But you can run your test, see how hot the case temperature and PCB temperature gets (don't forget to measure that -- I recommend using type T thermocouples, because you can solder one right to the case and another to the PCB next to the MOSFET), and calculate maximum junction temperature using the thermal resistances listed in the datasheet, to make sure you've got enough margin and the MOSFET stays within acceptable limits.
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