Thoughts on the design of a Cooling System

for Solid State Ham Amplifiers

So, having selected our RF amplifier transistors and the boards that will carry them, we now turn our attention to the question of keeping those transistors cool. Unlike tubes, transistors XXX. In the case of our MR-150’s, they will run at between roughly 45% and 75% efficiency. If we want to generate 1500 watts of RF output, we will also generate something like 1800 watts of heat in the worst case. If we are running SSB or CW, we get a tail wind from the fact that we aren’t always transmitting (probably around a 50% benefit), but if we are running RTTY or a digital mode we have to design for a 100% duty cycle and thus for the full 1800 watts of heat being generated.

There are basically two strategies when dealing with a problem like this. One is to just do what everyone else does, and assume it will work. The other is to engineer the solution so that you know it will work. In the case of the EB-104 amplifier board, there is a “copy cat” solution that we could use. Communications Concepts sells a heat sink that they say will serve to cool an EB-104 board, and they also sell a copper heat spreader that they recommend using between the board and the heat sink. But when I began thinking about this problem and started trying to convince myself that the CCI heat sink was the right solution, I was a bit concerned by the lack of supporting information that was available from CCI.

Given that we are trying to design an amplifier that can be built by lots of hams around the world, and given that we want this design to actually work in a high duty cycle contest environment, it seemed prudent to at least try to do the engineering required to design a proper and reliable cooling system. In order to do that, we have to start with some basic concepts of cooling system design – which I knew absolutely nothing about when I started this project. Luckily, both the concepts and the math are easily understood with a small bit of work…

Designing a cooling system for a heat generating transistor can be broken down in to two steps. First, we have to get the heat out of the transistor’s package, and into something else. Second, we have to get the heat out of the “something else” (some kind of heat sink, for example) and into the air of our shack. Some cooling systems just dump the heat into a continuous flow of water (coming out of a faucet, and going down a drain, for example) instead of air, but that didn’t seem like a very useful model for this project.

XXX drawing here

Getting the heat out of the package is a relatively simple problem. The data sheet for the MRF-150 specifies the two things that we need to know. First – what is the maximum allowable operating temperature of the amplifying junction of the transistor? In our case, this is 200° centigrade. Second, how much resistance is there to heat flowing from that junction to the outside of the transistor’s case? The heat resistance tells us how much difference there will be between the temperature at the junction and the temperature at the outside of the case – in this instance that difference (or delta) will be .6° centigrade per watt of heat energy generated. Since we are asking each transistor to generate about 100 watts of RF (1500 / 16 = 94, but we will round up to be safe), and we are designing for a worst case efficiency of 45%, we need to design for each part to generate about 120 watts of heat. Multiplying 120 by .6° per watt we find that the temperature difference will be 72° C. Since we want to keep the junction at or below 200°, and the difference between the junction and the case is 72°, then we need to keep the case temperature at or below 128° C.

Ok, now just one more subtlety. We have to mount the transistor on something – and that joint isn’t going to conduct heat very well. If we use a good thermal grease for the joint we might get the heat resistance for the joint down to .03° C per square inch. The base of an MRF-150 is about .18 square inches, so at 120 watts we are going to see another 20° C temperature difference across the thermal grease joint ([120 x .03] / .18). So now we have to keep the other side of that joint down at 108° C or less.

Ok, so how do we keep the surface of our mounting material (our heat sink, for example) at or below 108° C? Again, we have two realistic choices: forced air cooling, or water cooling.

In theory, it might be possible to use natural convection – that is to say the natural process by which heat moves from a material into the air around it without a fan to push that air around - to cool our amplifier. However, a huge heat sink would be required – and it just isn’t practical.

There are also lots of other, more esoteric, ways to cool things – Peltier coolers, compressor based refrigeration systems, etc. – but they are all too complicated or expensive or small scale for our purposes.

CCI proposes that we should use forced air cooling (they don’t specify that you are supposed to use a fan with their heat sink, but a bit of math will tell you that you need one for any kind of high duty cycle application) – so let’s start there.

CCI wants us to mount our transistors onto a piece of copper 3/8” thick, and then mount that piece of copper onto an aluminum heat sink. The first question that a beginner might ask (I did!) is: what is the copper heat spreader for? The answer is simple, but you have to know a bit about how heat transfer works to figure it out.

Suppose, for example, that we have a transistor that is quite small (the surface area of an MRF-150 is less than a quarter of an inch), and is generating a bunch of heat. We want to dump that heat into some air. The easy way to do that is to flow air over the hot surface - but the amount of heat that we can dump into the air is related to the surface area of the hot thing that the air is flowing over.

The formula, for the mathematically minded amongst us, is

Heat in watts = H * area in square meters * Dtemperature in degrees C

Where H is a constant equal to the heat resistance of our material.

So if H is constant and we want Dtemperature to be small, we need area to be big – or to put it another way: to get a bunch of heat into the air, we need quite a bit of surface area exposed to the air – a quarter of a square inch isn’t going to do it. This is the whole point of heat sinks – the reason that they have fins and bumps and ridges and grooves is to increase the surface area exposed to the air. The problem is that for the heat sink to work as well as possible, we need all of that surface area to be equally hot – or nearly so. If we take a small heat source and bolt it onto huge heat sink – most of the heat sink will be very cool and only the area around the source will be hot. That means that most of our heatsink isn’t doing anything useful. If we want our heatsink to do its job well, we need to transfer the heat from our small transistor into the heatsink in a way that makes all of the surface of the heatsink exposed to the air more or less equally hot.

But moving heat from our transistor to the surface (or fins) of the heatsink can be thought of as using the same formula above – except instead of surface area now we are thinking about the cross-sectional area through which the heat has to move. I realize that the thermal engineers among you are cringing at this rough treatment – but it is close enough for our purposes and much easier to understand than doing calculus!

The reason that bolting our transistor onto a typical heatsink doesn’t work well is that the base of the heatsink is thin – it has a small cross-sectional area, and thus heat doesn’t travel through it easily. The K (analogous to the H above) of aluminum is around 200, so for a .20” x 6” piece of aluminum heatsink base (like the one that CCI sells) we can calculate:

120 watts = [200 * ((.20” * 6”) / 1600) * delta temperature] / distance traveled (meters)

The 1600 is a rough factor to convert from square inches to square meters – figuring 40 inches to the meter.

Do the math, and you realize that for each inch that the heat has to travel in our piece of .20” x 6” aluminum cross section, there will be a temperature change of 20° C. So much for keeping all of our fins at the same temperature! The fin just over the heat source will be 80° C hotter than the fin over at the outside edge 4” away.

But the K for copper is 400, and our piece of copper heat spreader recommended by CCI is 3/8” thick. Do the math again, and you will find that the temperature change in our copper heat spreader is only about 5° C per inch – with half of the difference coming from the fact that it is copper not aluminum, and the rest coming from the fact that it is almost twice as thick. The CCI heat spreader is 6” x 8”, which means that the maximum distance that the heat has to travel from the transistors to the edge is 4”, which will cause a temperature difference of just 20° C, vs. 80° C for our aluminum heat sink base.

So if you want to use a big heat sink to cool a small heat source, you need a heatsink with a thick base, or you need a thick heat spreader under the base of the heat sink. CCI is proposing a heat spreader – and wisely so too.

Now we move on to the question of the heat sink itself – still on forced air cooling remember. Given that we are using a heat spreader, we can simplify the problem by assuming that we are applying the heat evenly across the bottom of our heat sink. That in turn means that we can treat each transistor as if it has its own heat sink, equal in length to a quarter of the total length of heat sink that we are using, and assume an average temperature differential through the spreader of 10° C (half of the 20° to the edge – to be honest I just picked half because it seemed reasonable). Recall that we need the temperature at the copper side of the “top” of the spreader to be 108° C. Subtract 10 for the resistance of the spreader, now we need the bottom of the spreader to be 98° C.

Back to our thermal compound – we have a layer of it between the copper spreader and our aluminum heat sink. It’s thermal resistance is .03° per square inch per watt. So if we have a 6” x 8” spreader, that gives us 48 square inches. This spreader carries four MRF150’s, generating 480 watts worst case – so our heat delta across the joint from the copper to the aluminum is (480 watts * .03 degrees per watt / 48 square inches), or .3° C – not enough to worry about.

So now the problem is rather simple. Recall from earlier that we need to keep the bottom surface of the heat sink at 98° C or less. Let’s assume that the air in our shack is 25° C or cooler (25° C is 77° F, at the high end of most indoor temps). That means that the temperature difference from the base of our heat sink to the air flowing by must be 73° C or less (98° – 25° = 73°). Since we are designing for each MOSFET to create 120 watts of heat, we need a heat sink with a thermal resistance in degrees per watt of .61 or less (73° / 120 watts). Does our CCI heatsink meet this requirement?

Well, CCI is selling a heat sink made by Accel Thermal in Los Angeles. Aavid, another heatsink manufacturer, provides some very useful tools on their web site. If we find an Aavid product similar to the Accel model 99 (Aavid #62285 for example), we can ask the Aavid web site to tell us the thermal resistance of a given length of that heatsink in a given flow of air. Doing so, we learn that a 3” length (12” total length, 4 transistors, so effectively 3 linear inches per transistor) of #62285 has a thermal resistance under .69 as long as the air flowing over it is more than about 200 linear feet per minute. Since a 50 CFM fan blowing through a 6 inch square channel generates 200 linear feet of air per minute, and since 50 CFM muffin fans are available “from stock in quantity” as they say, we conclude that CCI is quite right – their model 99 heatsink with a copper heat spreader will do the job nicely as long as we blow a decent amount of air over it. In fact, any heatsink with a thermal resistance under .61 will do – let’s keep that in mind as we move on to actual construction.