As your embedded project grows in scope and complexity, power consumption becomes an ever more apparent issue. As power consumption increases, components like linear voltage regulators can heat up during normal operation. Some heat is okay, however when things get too hot, the performance of the linear regulator suffers.
How much is too much?
A good rule of thumb for voltage regulators is if the outer case becomes uncomfortable to the touch, then the part needs to have an efficient way to transfer the heat to another medium. A good way to do this is to add a heat sink as shown below.
A heat sink attached to a linear voltage regulator on the Breadboard Power Supply.
A heat sink is often times just a big hunk of metal that helps pull heat from the part under load. By increasing the surface area of the heat sink, more of the heat is exposed to cooler air, thus cooling the part more efficiently. This is why you see “fins” on some heat sinks, as shown in the above picture.
If you use a heat sink, it is a good idea to add heat sink compound or thermal tape to the physical contact area between the voltage regulator and heat sink. The heat sink compound or tape allows for proper heat transfer from the voltage regulator to the heat sink. Remember you only need a tiny amount!
Your layout can also use copper planes as heat sinks.
Sometimes the copper pours on PCBs are used as heat sinks. In the above picture, the MCP73831 LiPo battery charging IC needs to dissipate heat on the PCB. The gray areas are copper planes and the black dots are vias (copper-plated holes to the bottom layer). All of this copper amounts to a greater area of radiative thermal mass, which will efficiently dissipate heat to the outside air.
Why is the voltage regulator heating up?
For this brief discussion we will talk about linear regulators (compared to SMPS). The efficiency of a linear regulator depends on the difference between the input and output voltages and how much current is being drawn by your circuit. The greater the difference between input and output voltage or the greater the current, the more heat will be dissipated by the regulator. This means linear power regulators are not very efficient at regulating voltage, since so much energy is wasted as heat! Switched mode power supplies (SMPS) are much more efficient and are becoming more common, however they can be difficult to use since they sometimes are susceptible to generating noise if not used correctly.
We can calculate the average amount of power a regulator will dissipate, which is directly related to the heat generated by the regulator.
.In order to calculate the power used by regulator in the above circuit, we need to know:
Vin, the voltage on the input of the regulator.
Vout, the output of the regulator and the voltage that is used to power the external devices.
I, the maximum amount of current the system can draw. Add up the specified (RTFM) maximum current draw from all of the devices (MCU, GPS, LEDs, etc) for a safe estimate.
We can now use use the power equation and plug in three values to calculate the power used by the regulator.
EXAMPLE 1
How much power is used by the regulator in the above picture? Here are the given values:
Vin. Let’s say we are using a fully charged 9V battery.
Vout. In our example, it is 5V.
I. Let’s assume the maximum current draw from all of the devices is 2.5 Amps.
Use the power equation:
Power = power in Watts
V = voltage in volts
I = current in amps
10 Watts is a lot of energy to be dissipated through a small electronic component! This is why heat sinks may need to be used with linear voltage regulators.
An important point to keep in mind: Our calculation can be considered the peak power dissipation of the regulator, because in reality, the system does not draw 2.5 Amps, continuously. The MCU, GPS, and CELL modules generally pulse with current which averages out to a much lower value. But it is always a good idea to assume worst-case-scenario values!
Heat sinks and thermal dissipation
In above tutorial, we discussed the need for a heat sink in situations where a device is expected to dissipate a large amount of heat. But how much heat sinking is needed? When do you need a modest folded metal heat sink versus a whopping milled-out-of-aluminum heat sink versus a mondo great fins-and-fan heat sink?
If you’re like me, at first glance, this problem seems intractable and you really start wishing you’d taken that thermodynamics course in college. But fear not! It’s not as bad as all that. Most datasheets will tell you what you need to know to come up with a rough estimate (accurate to within perhaps 50%) of how much heat sinking you really need for your application.
The problem defined
As an example, I’m going to use a circuit that I’ve developed for another tutorial I’m working on- a digital constant current load for discharging batteries so I can characterize their power capacity. The core of the load is an RFP30N06LE N-channel MOSFET. I’m not going to get into the circuit design required to use that as a constant current load here- if you’re really interested you can flip over to the digital load tutorial for more details.
At any rate, for this application, I’ve set my goal as being able to discharge a two-cell LiPo battery at up to 2A constantly. Freshly charged, that two-cell battery will be at approximately 8.4V, which means I’m setting myself up to dump 16.8W (P = I*V where I = 2A and V = 16.8V) into my FET. Note that this is worst case- that 8.4V is going to drop fast as the battery discharges, and if I’m NOT using a two-cell battery, the load will be even less.
Let’s do a little derating, too- if our worst case is 16.8W, if we target, say, 30W, then surely we’ll be in the clear- that’s almost a 100% slop factor.
Collecting the necessary information
Remember how I said most datasheets would tell you what you need to know to solve this problem? Let’s paw through the RFP30N06LE datasheet and the milled aluminum heat sink datasheet and see what we can come up with.
Fortunately, device manufacturers know that the datasheet target audience is people who are familiar with Ohm’s law. The information we need is described in terms of ºC/W and is referred to as “thermal resistance” (hereafter “Rt”). In the RFP30N06LE datasheet it’s at the bottom of the second page, in the “Electrical Specifications” section:
Note that there are TWO values here- one for “Junction to Case” and one for “Junction to Ambient”. We can basically ignore the “Junction to Ambient” value: that’s information for the device when there’s no heat sink in place. The “Junction to Case” value is the thermal resistance between the actual semiconductor die inside the plastic and the big fat metal tab hanging off the back. This is the critical value- if TJ gets too high, the device will fail.
In the heat sink datasheet, there’s a value given in a box on the second page:
In other words, for this heat sink, with no fans or forced air or other cooling shenanigans (“natural convection”), we can expect the temperature of the heat sink to rise 2.70°C above ambient for every watt it’s trying to dissipate.
Thermal circuits
How does this relate to Ohm’s law? Well, if we can write Ohm’s law as
R = V/I
then, based on the units of thermal resistance (°C/W), we can write the thermal resistance version of Ohm’s law as
Rt = T/P
What does that mean, in practical terms? It means that you can draw a thermal circuit where thermal resistances are equivalent to resistance, temperature is equivalent to voltage and power dissipation is equivalent to current.
Confused yet? Let’s take it step by step.
Doing the math
Let’s start by drawing out a circuit diagram of our system, with voltages replaced by temperatures and resistances replaced by thermal resistances:
where
TJ is the temperature of the semiconductor junction
RJC is the thermal resistance between the junction and the case (1.55°C/W as given in the datasheet)
TC is the temperature at the surface of the package
RCH is the thermal resistance between the case and the heat sink (.25°C/W; see below)
TH is the temperautre at the surface of the heat sink
RHA is the thermal resistance between the heat sink and the ambient air (2.70°C/W as given in the datasheet)
TA is the ambient temperature (25° is a pretty standard value for room temp, but this can go up if your device is in an enclosure or a more trying environment, such as outside in the sun)
P is the power from which we are attempting to dissipate heat (30W)
RCH is going to vary depending on what method you use to join the case and the heat sink. If you just screw them together, it won’t be very good- perhaps as high as several °C/W. With a little dab of thermal grease, however, you may be able to safely ignore RCH. For our example, I’m using a conservative value based on values I’ve seen in a few white papers on the topic.
So, now what? What’s our goal, here? Well, we need to calculate TJ based on the power dissipation of the circuit (30W) and decide whether TJ is a safe value. If this were an electrical circuit, we could simply add up the resistances and multiply those by the current to find out what the total voltage across the three resistors is:
V = I(R1+R2+R3)
In a thermal circuit, Rt is analogous to R, T is analogous to V, and P is analogous to I. Therefore, we can write:
TJ = P(RJC+RCH+RHA) +TA =30(1.55+.25+2.7)+25 =160°C
Ouch! That’s hot! But, is it safe for the device? Let’s refer to the datasheet. On page two there’s a section labeled “Absolute Maximum Ratings”, which provides a maximum operating temperature of 175°C. Well, we aren’t violating the datasheet maximum value, but there’s not a lot of overhead to spare. Remember, though, that we built in almost 100% slop factor on our power dissipation. If we re-run the calculation with our ACTUAL intended power dissipation of 16.8W, we’ll get a temperature just above 100°C- hot enough to boil water (or burn your finger badly) but quite a ways below our datasheet maximum.
What about the heat sink? How hot will it be, here? After all, it’s this big honking thing sticking out into space right where you’re very possibly going to touch it- how bad a thing is that going to be? Well, that’s simply a matter of taking the power (16.8W) times the thermal resistance of the heat sink (2.7°C/W) and adding that to the ambient temperature (25°C). That tells us that the heat sink, at 16.8W, will rise to about 70°C- about 160°F. Too hot to hold, but probably not instantly skin blistering.
The next steps
What if I hadn’t been happy with the answer? How could I improve matters?
Well, looking back at our system equation, the parameters that we can control to some degree are TA (running it in a cooler room or using a Peltier cooler perhaps), RHA (use a better heat sink, or forced air cooling) and RCH (use a better grade of thermal paste).
We could also try to reduce the power dissipation by, say, using multiple loads in parallel to share the power dissipation and decrease the temperature rise of each one. In fact, for the actual circuit, I elected to do just that- placing two MOSFETs in parallel to reduce the temperature dissipation of each one.