**CURRENTLY UNDER DEVELOPMENT**

**0. Introduction**- Did you burn your finger on a transistor?
- Are you wondering why your voltage regulator drops to a low voltage after a while?
- Did the smoke come out of a 100W transistor when it was only dissipating 30W?
- Did someone say "You need a heatsink?

- You don't know why.

- What is heat and how does it relate to temperature?
- What a heatsink does
- Why we need heatsinks
- How to calculate if you need one
- How to calculate the appropriate "size" heatsink

**1 Is heat is the enemy?**It might be obvious that things get hot, and hot things are bad, but we need an understanding of what heat is, and why temperature not heat is the enemy.

**1.1 Where does the heat come from?**Whenever current flows through a resistance there is a voltage drop. This voltage drop and the current determine the power lost or dissipated at that point.

The simple equations are:

The voltage dropped (in Volts) is equal to the current (in Amps) multiplied by the resistance (in Ohms).

And the power lost is:

The power lost (in Watts) is equal to the voltage dropped (in Volts) multiplied by the current (in Amps).

We can substitute the first equation into the second to get:

You may hear of these losses being called "

This power has to go somewhere, and typically most of it ends up as thermal energy which heats up the path through which the current flows.

The simple equations are:

V = I × R

And the power lost is:

P = V × I

We can substitute the first equation into the second to get:

P = I × R × I

orP = I²R

*I squared R losses*", and the last equation demonstrates why.This power has to go somewhere, and typically most of it ends up as thermal energy which heats up the path through which the current flows.

**1.2 Power and energy**I've been speaking of power (in Watts) and energy, but what is the relationship?

Power measured in watts is a measure of a rate of energy transfer in Joules per second.

So if something is dissipating 10 Watts, it is dissipating 10 Joules of energy every second.

Power measured in watts is a measure of a rate of energy transfer in Joules per second.

So if something is dissipating 10 Watts, it is dissipating 10 Joules of energy every second.

**1.3 Heat and temperature.**In common usage, heat and temperature might seem to be interchangeable. But they are quite difference.

Heat is a measure of the total amount of thermal energy that something has.

Temperature is the average amount of thermal energy each constituent part (think molecules) has.

So a cup of boiling water has a high temperature, but a tepid bath has far more heat.

We can verify this by pouring the boiling cup of water into a tepid bath -- the temperature will hardly change.

I said earlier that temperature was the average energy per molecule. Whilst this is correct, it hides the fact that some substances take a lot more energy to heat them by a degree Celsius. This is called the specific heat of a substance. and is measures in Joules per degree Kelvin (or Joules per degree Celsius -- it's the same thing). As an example 1 gram of water requires 4.2 joules of energy to heat it by 1 K or 1 degree C.

From the above, you can see that if I had a kilogram of water and I was dissipating 100W of energy into it, it would experience a rise in temperature of about 0.42 degrees C per second (assuming all the energy was contained in the water).

Heat is a measure of the total amount of thermal energy that something has.

Temperature is the average amount of thermal energy each constituent part (think molecules) has.

So a cup of boiling water has a high temperature, but a tepid bath has far more heat.

We can verify this by pouring the boiling cup of water into a tepid bath -- the temperature will hardly change.

I said earlier that temperature was the average energy per molecule. Whilst this is correct, it hides the fact that some substances take a lot more energy to heat them by a degree Celsius. This is called the specific heat of a substance. and is measures in Joules per degree Kelvin (or Joules per degree Celsius -- it's the same thing). As an example 1 gram of water requires 4.2 joules of energy to heat it by 1 K or 1 degree C.

From the above, you can see that if I had a kilogram of water and I was dissipating 100W of energy into it, it would experience a rise in temperature of about 0.42 degrees C per second (assuming all the energy was contained in the water).

**1.4 Where does the heat go?**This power measurement in Watts tells us how fast heat is being pumped into the device. If this heat could not escape, the device would get steadily hotter and hotter without end. But obviously heat does escape. The ways it can escape are:

As the component gets hotter and hotter relative to the surroundings, the rate of the first three (conduction, radiation, and convection) increase. This tends to cause the device to reach a thermal equilibrium where the rate at which heat is being pumped into the device is equal to the rate at which the heat is leaving the device into the surrounds.

This equilibrium temperature is the temperature we want to hold to a reasonable value.

If we don't, the device itself may undergo a phase change. A phase change is where the heat energy breaks molecular bonds. A common example is boiling water. In this case the temperature does not actually change, but the change in state from water at 100C to steam at 100C takes a lot of energy.

A similar thing can happen inside a transistor, the obvious signs being the loss of the magic smoke. Whilst this is a catastrophic final stage, other more subtle changes can occur before this, changing the device's characteristics, possibly rendering it non-functional.

Phase changes can result in changes in pressure. This can result in parts of the component's package blowing apart.

- By conduction (warming adjacent objects)

- By radiation (infra-red radiation)
- By convection (heating air)
- Phase change

As the component gets hotter and hotter relative to the surroundings, the rate of the first three (conduction, radiation, and convection) increase. This tends to cause the device to reach a thermal equilibrium where the rate at which heat is being pumped into the device is equal to the rate at which the heat is leaving the device into the surrounds.

This equilibrium temperature is the temperature we want to hold to a reasonable value.

If we don't, the device itself may undergo a phase change. A phase change is where the heat energy breaks molecular bonds. A common example is boiling water. In this case the temperature does not actually change, but the change in state from water at 100C to steam at 100C takes a lot of energy.

A similar thing can happen inside a transistor, the obvious signs being the loss of the magic smoke. Whilst this is a catastrophic final stage, other more subtle changes can occur before this, changing the device's characteristics, possibly rendering it non-functional.

Phase changes can result in changes in pressure. This can result in parts of the component's package blowing apart.

**1.5 So what is the enemy?**What we want to avoid is energy being dissipated by phase change. This is not a really easy thing to measure, and the gross phase changes occur at points well beyond the point where the device has ceased operating.

The manufacturer will give us a maximum temperature. This is something we want to avoid reaching, and NEVER want to exceed. This is a temperature, and that becomes our enemy.

Temperature is actually an easy thing to work with. All sorts of semiconductors ranging from devices with a power rating in milliwatts, to those in kilowatts will all have a similar maximum temperature.

This means that the same calculations can be used for everything.

Our aim is to keep the junction temperature (we'll cover that later) lower than the absolute maximum, and frequently lower than some lower limit that we impose on ourselves.

The manufacturer will give us a maximum temperature. This is something we want to avoid reaching, and NEVER want to exceed. This is a temperature, and that becomes our enemy.

Temperature is actually an easy thing to work with. All sorts of semiconductors ranging from devices with a power rating in milliwatts, to those in kilowatts will all have a similar maximum temperature.

This means that the same calculations can be used for everything.

Our aim is to keep the junction temperature (we'll cover that later) lower than the absolute maximum, and frequently lower than some lower limit that we impose on ourselves.

**1.6 More about this enemy**As a very rough measure, the life of a semiconductor will double for each 10 degrees C you can cool it.

Thus a semiconductor which might have a life of 1000 hours at 125ºC will have a life for 2000 hours at 115ºC, 4000 hours at 105ºC, 8000 hours at 95ºC, and 16,000 hours at 85ºC.

A year is about 8000 hours, so reducing the temperature of a semiconductor from 125ºC to 85ºC will mean that it is likely to last about 2 years rather than about 6 weeks.

This enemy is clearly one to be dealt with very seriously in devices that are powered up continuously. One prime example is in power supplies.

Thus a semiconductor which might have a life of 1000 hours at 125ºC will have a life for 2000 hours at 115ºC, 4000 hours at 105ºC, 8000 hours at 95ºC, and 16,000 hours at 85ºC.

A year is about 8000 hours, so reducing the temperature of a semiconductor from 125ºC to 85ºC will mean that it is likely to last about 2 years rather than about 6 weeks.

This enemy is clearly one to be dealt with very seriously in devices that are powered up continuously. One prime example is in power supplies.

__2 What does this mean for my small transistors?__Have you ever felt a small transistor and almost burnt your finger? Maybe you've let the smoke out of a few.

Let's look at the

For this transistor we are also given the thermal resistance between the junction and the case. It is always lower than the thermal resistance between the junction and ambient.

This is described in the referenced datasheet as R

R

This figure is used when you are heatsinking the device. In this case you'll be providing an easier path for the heat to transfer (typically by conduction) away from the case.

A transistor like this is not really designed to have a heatsink, so we won't discuss heatsinking for these just yet...

Let's look at the

__2N3904__. This is a small signal NPN transistor similar to a BC548, or the more meaty 2N2222.**2.1 Absolute maximum ratings**The first thing to look at is the Absolute Maximum Ratings.

I've highlighted the really important ones.

So, what if I have 4V across my transistor and 175mA passing through it? From the calculations describes above, you can calculate the rate at which we are forcing the transistor to dissipate heat energy.

Now, 0.7W (700mW) exceeds the absolute maximum of 625mW. The transistor would be very unreliable, and would likely be operating above it's maximum temperature of 150ºC.

I've highlighted the really important ones.

- V
_{CEO}is effectively the highest voltage you can switch with this transistor. - I
_{C}is the maximum current you can switch. - P
_{tot}is the maximum total dissipation -- the largest amount of heat it can dissipate. - T
_{j}is the maximum temperature the silicon die that is the actual transistor can be allowed to get. Note that this is more often seen as T_{j(max)}, since T_{j}literally refers to a junction temperature, not the maximum junction temperature.

So, what if I have 4V across my transistor and 175mA passing through it? From the calculations describes above, you can calculate the rate at which we are forcing the transistor to dissipate heat energy.

P = I × V

= 4 × 0.175

= 0.7 W

= 4 × 0.175

= 0.7 W

**2.2 Thermal characteristics (junction to ambient)**Most datasheets will describe the thermal characteristics of the device. Here they are for the 2N3904:

The figure we're interested in is the "thermal resistance junction to ambient". Note that this datasheet abbreviates it as R

This thermal resistance allows us to determine the junction temperature for a given power and ambient temperature. Let's see what we get for the absolute maximum power (P

It should come as no surprise at all that the maximum power dissipation at the standard ambient temperature results in the absolute maximum junction temperature.

This relationship allows you to determine R

The figure we're interested in is the "thermal resistance junction to ambient". Note that this datasheet abbreviates it as R

_{thj-amb}. It is more often abbreviated as R_{θJA}. For ambient, think air. This figure represents how hard it is for heat released at the junction to reach the still air outside a free standing device.This thermal resistance allows us to determine the junction temperature for a given power and ambient temperature. Let's see what we get for the absolute maximum power (P

_{tot}from the datasheet)T = 25 + (P × R

= 25 + (0.625 × 200)

= 25 + 125

= 150

_{θJA})= 25 + (0.625 × 200)

= 25 + 125

= 150

This relationship allows you to determine R

_{θJA}given P_{tot}and t_{j(max)}.

**2.2 Thermal characteristics (junction to case)**For this transistor we are also given the thermal resistance between the junction and the case. It is always lower than the thermal resistance between the junction and ambient.

This is described in the referenced datasheet as R

_{thj-case}. It is more often abbreviated as R

_{θJC}.

R

_{θJC}is always lower than R

_{θJA}because the heat must first get to the outside of the case before it can radiate or convect away into the surrounding air.

This figure is used when you are heatsinking the device. In this case you'll be providing an easier path for the heat to transfer (typically by conduction) away from the case.

A transistor like this is not really designed to have a heatsink, so we won't discuss heatsinking for these just yet...

**2.3 Traps for young players**We've seen that the for a small signal transistor like this (one that's not designed to be connected to a heatsink) that the P

However it is important to point out that you can't multiply the transistors maximum voltage (V

It is also worth noting that P

The major traps are as follows:

The amount you need to de-rate the transistor can be looked up on a graph in the datasheet:

or calculated as:

P

So, if we assume:

The third thing you might see is a derating factor given in mW/ºC above some temperature. Imagine you are told that this is 2mW/ºC above 25ºC.

Again, we can calculate the de-rated P

Note that both the calculations and the graph yield the same results.

This calculation is important if the device will be operated in an environment that might get hot (automotive applications are a good example), or where the device is physically close to something hot (say another transistor), or where there is poor air circulation.

This brings us to the second major trap, that is, even with de-rating the transistor may not have a long service life.

As mentioned earlier, each 10ºC that you can lower the junction temperature will result in an approximate doubling of the life of the transistor.

This means we first need to be able to predict the actual temperature of the junction. Fortunately this is a very simple calculation:

We've seen this before.

_{tot}for the transistor can be used as an upper limit.However it is important to point out that you can't multiply the transistors maximum voltage (V

_{CEO}) and maximum current (I_{C}) to get a maximum dissipation. That is, the transistor can either have the maximum voltage across it, or the maximum current flowing through it, but not both.It is also worth noting that P

_{tot}includes the base current I_{B}multiplied by the base-emitter voltage (V_{BE}). For small signal transistors this is often quite small due to the relatively high gain of the transistor.The major traps are as follows:

- Operating under P
_{tot}does not guarantee operating under T_{j(max)} - Operating under T
_{j(max)}does not guarantee a long operational life.

_{tot}) to keep the junction temperature below T_{j(max)}.The amount you need to de-rate the transistor can be looked up on a graph in the datasheet:

or calculated as:

P

_{tot(tºC)}= P

_{tot}× (T

_{j(max)}- t) ÷ (T

_{j(max)}- 25)

P

T

t

then:_{tot}= 300mWT

_{j(max)}= 175ºCt

_{amb}= 100ºC

P

= 0.3 × (175 - 100) ÷ (175 - 25)

= 0.3 × 75 ÷ 150

= 0.3 × 0.5

= 0.150 W

And so at 100ºC this transistor would have a de-rated P_{tot(100ºC)}= P_{tot}× (T_{j(max)}- 100) ÷ (T_{j(max)}- 25)= 0.3 × (175 - 100) ÷ (175 - 25)

= 0.3 × 75 ÷ 150

= 0.3 × 0.5

= 0.150 W

_{tot}of 150mW.

Again, we can calculate the de-rated P

_{tot}at 75ºC as:P

= 0.3 - (0.002 × (100-25))

= 0.3 - (0.002 × 75)

= 0.3 - 0.15

= 0.150 W

_{tot(100ºC)}= P_{tot}- (factor × (T_{amb}× T_{factor}))= 0.3 - (0.002 × (100-25))

= 0.3 - (0.002 × 75)

= 0.3 - 0.15

= 0.150 W

This calculation is important if the device will be operated in an environment that might get hot (automotive applications are a good example), or where the device is physically close to something hot (say another transistor), or where there is poor air circulation.

This brings us to the second major trap, that is, even with de-rating the transistor may not have a long service life.

As mentioned earlier, each 10ºC that you can lower the junction temperature will result in an approximate doubling of the life of the transistor.

This means we first need to be able to predict the actual temperature of the junction. Fortunately this is a very simple calculation:

T

_{j}= T_{amb}+ (P × R_{θJA})

Let's assume:

145ºC is very hot for a transistor. We would really want to keep it cooler. How can we do that? Read on...T

P = 0.5W (again, the maximum you expect)

R

T

= 45 + (0.5 × 200)

= 45 + 100

= 145ºC

_{amb}= 45ºC (this should be the maximum you expect)P = 0.5W (again, the maximum you expect)

R

_{θJA}= 200ºC/WT

_{j}= T_{amb}+ (P × R_{θJA})= 45 + (0.5 × 200)

= 45 + 100

= 145ºC

**2.4 Keeping your cool**So, you've calculated the T

Alternative packages are worth looking for, but unless you're working with surface mount components, you've likely got the largest package already.

The 2N2222 comes in a variety of packages. Each has a different set of thermal characteristics.

The important thing to note here is that each package has a different P

In this case, if you're using the TO-92 version, switching to the SOT-223 package may yield a significant reduction in junction temperature.

The final option is to use a heatsink. Note that only the TO-92 package has a rating for R

_{j}and it's higher than T_{j(max)}, or just higher than you'd like it to be. What can you do?- Can you reduce the power dissipation?
- Is there an alternative package with a lower R
_{θJA}? - Is there an alternative device with a lower R
_{θJA}? - Can you use a heatsink?

Alternative packages are worth looking for, but unless you're working with surface mount components, you've likely got the largest package already.

The 2N2222 comes in a variety of packages. Each has a different set of thermal characteristics.

The important thing to note here is that each package has a different P

_{D}(which is another way of saying P_{tot}) and R_{θjA}.In this case, if you're using the TO-92 version, switching to the SOT-223 package may yield a significant reduction in junction temperature.

Let's assume:

Just by changing the package from a TO-92 to a SOT-223, we have achieved a reduction in TP = 500mW

T

For the TO-92 package:T

_{A}= 35ºC

T

= 35 + (0.5 × 200)

= 35 + 100

= 135ºC

For the SOT-223 package:_{j}= T_{A}+ (P × R_{θjA})= 35 + (0.5 × 200)

= 35 + 100

= 135ºC

T

= 35 + (0.5 × 125)

= 35 + 62.5

= 97.5ºC

_{j}= T_{A}+ (P × R_{θjA})= 35 + (0.5 × 125)

= 35 + 62.5

= 97.5ºC

_{j}of 37.5ºC. Using the metric of a doubling of life for each 10ºC reduction in T_{j}, the transistor in the SOT-223 package will last about 13 times longer.The final option is to use a heatsink. Note that only the TO-92 package has a rating for R

_{θjC}. I'll get into the maths of how this is used later, but suffice to say that the thermal resistance will be higher than R_{θjA}. With a package like a TO-92 that has no obvious way to attach a heatsink the overall thermal resistance will be hard to calculate.

**3 Why can't my 115W transistor dissipate 30W without failing?**Most of us are used to ½ watt resistors that can safely dissipate ½ a watt, and larger 5 watt resistors that can safely dissipate 5 watts. From the last section we've seen that a 600mW small signal transistor can dissipate 600mW.

Then we look at specifications for power transistors and see things like 60V, 15A, 115W. We subject the transistor to something far less than this (say 5A with 6V across it) and the smoke comes out...

Why? It's not like we got even close to the specified maximums!

Well,

Then we look at specifications for power transistors and see things like 60V, 15A, 115W. We subject the transistor to something far less than this (say 5A with 6V across it) and the smoke comes out...

Why? It's not like we got even close to the specified maximums!

Well,

__here__is that transistor. It's the venerable 2N3055.**3.1 What does the datasheet say?**

Let's look at that datasheet.

Well it sure seems to say all those things, but wait, that doesn't look like the package I have. Mine has three legs and a tab! Will that matter?

Yeah, that's what it looks like!

Notice that this package has a lower maximum current and maximum power.

But we were still under that limit, so it's fine, right?

No, These transistors are designed to be connected to a heatsink. Without one they can only achieve a small fraction of the power they're rated for.

And you've already found out that the same device in a different package will have different limitations. Smaller packages often have a lower power rating, and sometimes a lower current rating.

So what happened when this device had 6V across it and 5A flowing through it?

From our previous section, this means that it was being asked to dissipate 30W of thermal energy (that's 30 Joules every second -- a camera flash is about 10J).

So how fast could it escape? For this we need to look at the datasheet again.

Let's look at

The other figure, R

Note that both are in units of degrees C per Watt (ºC/W).

Well it sure seems to say all those things, but wait, that doesn't look like the package I have. Mine has three legs and a tab! Will that matter?

Yeah, that's what it looks like!

Notice that this package has a lower maximum current and maximum power.

But we were still under that limit, so it's fine, right?

No, These transistors are designed to be connected to a heatsink. Without one they can only achieve a small fraction of the power they're rated for.

And you've already found out that the same device in a different package will have different limitations. Smaller packages often have a lower power rating, and sometimes a lower current rating.

So what happened when this device had 6V across it and 5A flowing through it?

From our previous section, this means that it was being asked to dissipate 30W of thermal energy (that's 30 Joules every second -- a camera flash is about 10J).

So how fast could it escape? For this we need to look at the datasheet again.

Let's look at

__this__datasheet. This is for another package type, but it contains the information we need (which -- surprisingly -- is not in all datasheets!!)

There are two figures here, R_{θJC}gives the thermal resistance between the junction (the silicon wafer inside) and the case. This is used when we attach the device to a heatsink.The other figure, R

_{θJA}gives the thermal resistance between the junction and ambient (i.e. the surrounding air).Note that both are in units of degrees C per Watt (ºC/W).

**3.2 What does the datasheet mean?**

The thermal resistance figures measure how hard it is for heat to escape the device.

You will remember from the previous section that the amount of heat that can escape will increase as the temperature differential rises.

Decrees C per Watt tell you the temperature differential required to dissipate heat at a certain rate (remembering that a Watt is a rate).

In this case we have 30W of energy to dissipate (or more accurately we need to dissipate heat at a rate of 30W). The thermal resistance between the junction and ambient is given as 35.7ºC.

So, if the ambient temperature id 25ºC, the temperature if the junction will rise to:

That's 1000º hotter than the maximum junction temperature. No wonder the smoke got out!

An interesting question is, approximately how much power could we dissipate without the junction exceeding the maximum temperature?

That's easy:

3.5 Watts! That's nothing. And remember that this device is rated for 75W!

You will remember from the previous section that the amount of heat that can escape will increase as the temperature differential rises.

Decrees C per Watt tell you the temperature differential required to dissipate heat at a certain rate (remembering that a Watt is a rate).

In this case we have 30W of energy to dissipate (or more accurately we need to dissipate heat at a rate of 30W). The thermal resistance between the junction and ambient is given as 35.7ºC.

So, if the ambient temperature id 25ºC, the temperature if the junction will rise to:

25 + (30 × 35.7) ºC

= 25 + 1125 ºC

= 1150ºC

= 25 + 1125 ºC

= 1150ºC

An interesting question is, approximately how much power could we dissipate without the junction exceeding the maximum temperature?

That's easy:

P = (150 - 25) ÷ 35.7

= 125 ÷ 35.7

= 3.5W

= 125 ÷ 35.7

= 3.5W

**3.3 What does this mean?**

The big thing is that the package is important. Some packages are better than others at transferring heat to the environment (in general, bigger is better).

The next thing to take home from this is that power transistors cannot dissipate anything like their rated power without a heatsink.

In fact, we'll discover that even with a heatsink it may be very difficult to get near the rated maximum.

The next thing to take home from this is that power transistors cannot dissipate anything like their rated power without a heatsink.

In fact, we'll discover that even with a heatsink it may be very difficult to get near the rated maximum.

**4 How does a heatsink help?**A heatsink helps us in three ways.

One factor that s not always appreciated is the increased thermal mass of a system which includes a heatsink.

This comes back to the heat Vs temperature discussion back in section 1.3. Think of the device junction as being the hot cup of tea and the heatsink as being the tepid bath.

If the transistor junction dissipates a certain amount of heat energy, then stops, the temperature of the device will be lifted by some temperature as that heat energy causes an increase in the average thermal energy of all the parts of the transistor.

If we add a heatsink, that thermal energy is now averaged over the device and the heatsink. The temperature rise will be much smaller.

This applies generally. Even if the dissipation is constant, the heatsinked system will rise in temperature more slowly.

You might think, "aha! but won't it get cooler more slowly too?" This would be true if you were removing thermal energy at a constant rate, but remember from above that a major advantage of a heatsink is that it gets rid of energy faster.

So what's the use of this?

In electrical terms, the thermal resistance (which we can think of as resistors for heat) are accompanied by thermal capacity (which we can think of as capacitors for heat). If you imagine thermal resistances and capacitances like this:

This fact means that we can assume that pulses of heat being injected into the system will behave very similarly to heat being injected at a rate that is the average of these pulses.

In electrical terms, the thermal resistance (which we can think of as resistors for heat) are accompanied by thermal capacity (which we can think of as capacitors for heat).

This is clearly a low pass filter made up of the various thermal resistances, and the thermal capacities of the various components of the system.

Here we see that incoming thermal energy heats up the junction, then passes through the thermal resistance R

Then the heat travels to the heatsink through another thermal resistance (the transfer of heat to the heatsink will never be perfect). Then the heatsink has a thermal capacity (typically larger) here called C

Finally there is the thermal resistance between the hatsink and the air (shown as R

With a constant amount of heat flowing into this circuit, a constant amount must leave it as well. This is the steady state where we can use the thermal resistances to calculate the junction temperature given the power dissipation and thermal resistances.

With dissipation rising and lowering, as long as the frequency is fast enough that the entire system can't react and get near a steady state, we can average the dissipation and do our calculations based on that.

The trick is determining how quickly the system reacts...

- Increases the rate at which heat is transferred away from the device
- Increases the rate at which heat is transferred to the environment
- Increases the amount of heat required to increase the temperature of the device (separate from (2))

**4.1 Transferring heat from the device**We have seen that the three (good) methods of transferring heat from the device are conduction, convection, and radiation.

A heatsink is (generally) a piece of metal brought into intimate contact with the case of the device. This provides a relatively large surface ares where heat can be conducted away from the device. Generally speaking, conduction is faster and more effective than either convection or radiation.

This increase in efficiency is partially indicated by the difference between T

Here is an example:

Let's state what those figures mean in a way that makes the difference very clear.

A T

A T

The increase in efficiency of the process of dissipating heat from the device is approximately 26 times! (35.7 / 1.39 = 25.69)

As you recall from section 3.2 above, without a heatsink, the maximum dissipation for the device at an ambient temperature of 25ºC was about 3.5W.

Now let's repeat that for a case temperature of 25ºC.

The equation used is almost exactly the same as the one used previously. We simply change references to ambient to the same reference to the case (temperature or thermal resistance). Was that last sentence confusing? Let's look at the equation for Junction to ambient:

Now, for a case temperature of 25ºC, a T

There are two interesting things about this.

The second might lead you to think that it would be safe to dissipate almost 90W with this transistor. But you may remember this:

So, we can just attach a heatsink and we'll be fine for 75W dissipation, right?

WRONG!

We have just calculated a theoretical maximum, showing that the device is physically capable of dissipating 75W from the junction to a heatsink while remaining within its absolute limits for junction temperature. Section 5 will discuss the practicalities.

A heatsink is (generally) a piece of metal brought into intimate contact with the case of the device. This provides a relatively large surface ares where heat can be conducted away from the device. Generally speaking, conduction is faster and more effective than either convection or radiation.

This increase in efficiency is partially indicated by the difference between T

_{θjA}(thermal resistance from the junction to the ambient) and T_{θjC}(the thermal resistance from the junction to the case).Here is an example:

Let's state what those figures mean in a way that makes the difference very clear.

A T

_{θjA}of 35.7ºC/W means that if the__ambient__temperature remains constant, the junction temperature will increase by 35.7ºC for every watt of power dissipated in the junction.A T

_{θjC}of 1.39ºC/W means that if the__case__temperature remains constant, the junction temperature will increase by 1.39ºC for every watt of power dissipated in the junction.The increase in efficiency of the process of dissipating heat from the device is approximately 26 times! (35.7 / 1.39 = 25.69)

As you recall from section 3.2 above, without a heatsink, the maximum dissipation for the device at an ambient temperature of 25ºC was about 3.5W.

Now let's repeat that for a case temperature of 25ºC.

The equation used is almost exactly the same as the one used previously. We simply change references to ambient to the same reference to the case (temperature or thermal resistance). Was that last sentence confusing? Let's look at the equation for Junction to ambient:

T

...and compare if to the same equation for junction to case:_{j}= T_{A}+ (P × R_{θj}_{A})

T

That's it! (simple eh?)_{j}= T_{C}+ (P × R_{θj}_{C})

Now, for a case temperature of 25ºC, a T

_{θjC}of 1.39ºC/W, and a T_{j(max)}of 150ºC, what power can be dissipated?T

150 = 25 + (P × 1.39)

150 - 25 = P × 1.39

P = (150 - 25) ÷ 1.39

= 125 ÷ 1.39

= 89.9 W

_{j}= T_{C}+ (P × R_{θjC})150 = 25 + (P × 1.39)

150 - 25 = P × 1.39

P = (150 - 25) ÷ 1.39

= 125 ÷ 1.39

= 89.9 W

- The calculated power is vastly higher than if the device has no heatsink.
- The calculated power exceeds the maximum power for the device.

The second might lead you to think that it would be safe to dissipate almost 90W with this transistor. But you may remember this:

The total power dissipation has a maximum of 75W. As a maximum you should not exceed it, even if other figures may suggest that you can.So, we can just attach a heatsink and we'll be fine for 75W dissipation, right?

WRONG!

We have just calculated a theoretical maximum, showing that the device is physically capable of dissipating 75W from the junction to a heatsink while remaining within its absolute limits for junction temperature. Section 5 will discuss the practicalities.

**4.2 Transferring heat to the ambient environment**The second major advantage of a heatsink, and a necessary condition for it to be effective is that it can transfer the heat which is conducted into it to the ambient environment more rapidly than just the device alone.

So this means that the heatsink needs to conduct, convect, or radiate the heat away faster.

Heatsinks often start out fairly think neat the device and have thinner fins. This allows the heat to be efficiently moved to the fins, and thinner fins allow more surface area for a given weight (and therefore cost) of heatsink.

In the final step of convecting the heat away, the orientation of the heatsink often matters greatly, and forced air (fans) can multiply the efficiency manyfold.

So this means that the heatsink needs to conduct, convect, or radiate the heat away faster.

- Conduction -- yes, it is possible to connect the heatsink to an even bigger heatsink, but eventually you need to use another method.
- Convection --heatsinks usually have fins that allow a large surface that air can pass over to be heated. This larger surface area is the largest contribution to loss of heat.
- Radiation -- some heatsinks are coloured black to enhance radiation, but radiation is a very poor method of dissipation of heat at low temperatures. The surface treatment may impair conduction or convective losses, so be careful!

Heatsinks often start out fairly think neat the device and have thinner fins. This allows the heat to be efficiently moved to the fins, and thinner fins allow more surface area for a given weight (and therefore cost) of heatsink.

In the final step of convecting the heat away, the orientation of the heatsink often matters greatly, and forced air (fans) can multiply the efficiency manyfold.

**4.3 Thermal mass**One factor that s not always appreciated is the increased thermal mass of a system which includes a heatsink.

This comes back to the heat Vs temperature discussion back in section 1.3. Think of the device junction as being the hot cup of tea and the heatsink as being the tepid bath.

If the transistor junction dissipates a certain amount of heat energy, then stops, the temperature of the device will be lifted by some temperature as that heat energy causes an increase in the average thermal energy of all the parts of the transistor.

If we add a heatsink, that thermal energy is now averaged over the device and the heatsink. The temperature rise will be much smaller.

This applies generally. Even if the dissipation is constant, the heatsinked system will rise in temperature more slowly.

You might think, "aha! but won't it get cooler more slowly too?" This would be true if you were removing thermal energy at a constant rate, but remember from above that a major advantage of a heatsink is that it gets rid of energy faster.

So what's the use of this?

In electrical terms, the thermal resistance (which we can think of as resistors for heat) are accompanied by thermal capacity (which we can think of as capacitors for heat). If you imagine thermal resistances and capacitances like this:

This fact means that we can assume that pulses of heat being injected into the system will behave very similarly to heat being injected at a rate that is the average of these pulses.

In electrical terms, the thermal resistance (which we can think of as resistors for heat) are accompanied by thermal capacity (which we can think of as capacitors for heat).

This is clearly a low pass filter made up of the various thermal resistances, and the thermal capacities of the various components of the system.

Here we see that incoming thermal energy heats up the junction, then passes through the thermal resistance R

_{θjC}to the case of the device. But the device has some thermal capacity (shown here as C

_{C}, so it doesn't heat instantly.

Then the heat travels to the heatsink through another thermal resistance (the transfer of heat to the heatsink will never be perfect). Then the heatsink has a thermal capacity (typically larger) here called C

_{S}. This means that the heatsink will take time to heat up.

Finally there is the thermal resistance between the hatsink and the air (shown as R

_{θSA}.

With a constant amount of heat flowing into this circuit, a constant amount must leave it as well. This is the steady state where we can use the thermal resistances to calculate the junction temperature given the power dissipation and thermal resistances.

With dissipation rising and lowering, as long as the frequency is fast enough that the entire system can't react and get near a steady state, we can average the dissipation and do our calculations based on that.

The trick is determining how quickly the system reacts...

__5 Calculating the required heatsink size__Section 4 (especially the diagram in section 4.3) will give you some clues as to how we might calculate the size of heatsink that is required.

**5.1 What is "size" when we refer to a heatsink?**When we refer to a heatsink, we often talk colloquially about a larger or smaller, or a huge, or a tiny heatsink. What does this mean?

Firstly, it does have a strong relationship with physical size, and weight.

But primarily, the "size" of a heatsink refers to its thermal resistance. This is a measure of how fast it can get rid of heat that is transferred to it (primarily by conduction from some device).

It will not surprise you that this figure is given in units of degrees C per watt (ºC/W). This is exactly the same as for the device itself, because it is a measurement of exactly the same thing.

A larger heatsink will have a SMALLER thermal resistance. e.g. a heatsink with a thermal resistance of 10ºC/W is considered "smaller" than one having a thermal resistance of 5ºC/W. So when we say "larger" we really mean "more efficient". Why do we say larger?

If we consider a heatsink having a thermal resistance of 5ºC/W as being twice as effective as one having a thermal resistance of 10ºC/W, what is the relationship in physical size? There are many variables, but typically you would expect about double the surface area, and about double the amount of material. So, roughly speaking, double the efficiency does mean double the size. (Practically, doubling the size -- say the length -- will result in something a little less than a doubling of efficiency, but the relationship is strong enough that the colloquial usage of "size" is not stretched too far by reality).

This thermal resisthace (the "size") of the heatsink is called R

Firstly, it does have a strong relationship with physical size, and weight.

But primarily, the "size" of a heatsink refers to its thermal resistance. This is a measure of how fast it can get rid of heat that is transferred to it (primarily by conduction from some device).

It will not surprise you that this figure is given in units of degrees C per watt (ºC/W). This is exactly the same as for the device itself, because it is a measurement of exactly the same thing.

A larger heatsink will have a SMALLER thermal resistance. e.g. a heatsink with a thermal resistance of 10ºC/W is considered "smaller" than one having a thermal resistance of 5ºC/W. So when we say "larger" we really mean "more efficient". Why do we say larger?

If we consider a heatsink having a thermal resistance of 5ºC/W as being twice as effective as one having a thermal resistance of 10ºC/W, what is the relationship in physical size? There are many variables, but typically you would expect about double the surface area, and about double the amount of material. So, roughly speaking, double the efficiency does mean double the size. (Practically, doubling the size -- say the length -- will result in something a little less than a doubling of efficiency, but the relationship is strong enough that the colloquial usage of "size" is not stretched too far by reality).

This thermal resisthace (the "size") of the heatsink is called R

_{θSA}, the thermal resistance between the heatsink and the ambient environment (or air).

**5.2 Calculating the maximum R**

_{θSA}Calculation of the Maximum allowable R

First we need to calculate the maximum system R

Here are the major components of R

There's one thing there you may have seen before, and may recognise (R

R

We've seen this all before (most notably in section 2.4 above):

So for this set of values we need a system R

Wow! Is that all? So I need a heatsink of around 2.55ºC/W?

No. Remember we now need to calculate R

R

Easy, so we just have to find a 0.66ºC/W heatsink?

Well, kinda. That would be the absolute minimum size heatsink, and if you look for heatsinks rated close to this you'll find they're pretty big.

If you look back at section 4.1, we've gone from thinking that we might be able to achieve almost 90W dissipation from this transistor to realising it may be hard to achieve 45W of dissipation. And we're not finished yet! It only gets worse...

_{θSA}is pretty easy.First we need to calculate the maximum system R

_{θjA}.Here are the major components of R

_{θjA}R

So_{θjA(max)}= R_{θjC}+ R_{θCS}+ R_{θSA}

R

_{θSH}= R_{θjA(max)}- (R_{θjC}+ R_{θCS})

_{θCS}), but won't know how to calculate. Relax, we'll come to that.R

_{θjA(max)}is simply the maximum allowed thermal resistance between the junction and air at some stated power level, ambient temperature, and T_{j(max)}.We've seen this all before (most notably in section 2.4 above):

T

Which we will subtly alter to read as:_{j}= T_{A}+ (P × R_{θjA})

T

Now we just need to rearrange it so we can calculate R_{j(max)}= T_{A}+ (P × R_{θjA(max)})

_{θjA}, which will be R

_{θjA(max)}

R

Let's use the 2N2055 we've been discussing up until now, and assume a P of 45W and an ambient temperature of 35ºC._{θjA(max)}= (T_{j(max)}- T_{A}) ÷ P

R

= (150 - 35) ÷ 45

= 115 ÷ 45

≈ 2.55ºC/W

_{θjA(max)}= (T_{j(max)}- T_{A}) ÷ P= (150 - 35) ÷ 45

= 115 ÷ 45

≈ 2.55ºC/W

_{θjA}of no more than 2.55ºC/WWow! Is that all? So I need a heatsink of around 2.55ºC/W?

No. Remember we now need to calculate R

_{θSH}, the "size" of the heatsink.R

= 2.55 - (1.39 + R

Bugger! there's that R_{θSH}= R_{θjA(max)}- (R_{θjC}+ R_{θCS})= 2.55 - (1.39 + R

_{θCS})

_{θCS}. What's that?

R

_{θCS}) is the thermal resistance between the case of the device and the heatsink itself. We'll talk more about this later, but for now, let's assume it is 0.5ºC/W

R

= 2.55 - (1.39 + R

= 2.55 - (1.39 + 0.5)

= 2.55 - 1.89

= 0.66ºC/W

_{θSH}= R_{θjA(max)}- (R_{θjC}+ R_{θCS})= 2.55 - (1.39 + R

_{θCS})= 2.55 - (1.39 + 0.5)

= 2.55 - 1.89

= 0.66ºC/W

Well, kinda. That would be the absolute minimum size heatsink, and if you look for heatsinks rated close to this you'll find they're pretty big.

__Here__is one that's closer to 0.8ºC/W than 0.6ºC/W (so smaller than you'd need).If you look back at section 4.1, we've gone from thinking that we might be able to achieve almost 90W dissipation from this transistor to realising it may be hard to achieve 45W of dissipation. And we're not finished yet! It only gets worse...

**5.3 Injecting some realism**

We need to move from the theoretical to the practical. We need to consider several additional things:

All of these factors are going to work against us. Some require fudge factors (i.e. they are hard to calculate) but others can be calculated.

(1) and (2) go hand in hand. We might want the normal temperature of the junction to be no more than 125ºC, and we think that maybe the temperature could rise to 55ºC on occasion. We might decide that we will allow T

To make things easier, let's also assume that we were able to reduce the power dissipation of the transistor down to 20W.

We now need to do the thermal calculation twice, once for "normal" conditions, and again for "hot" conditions:

We now have 2 figures for the minimum heatsink size, 4.5ºC/W and 4.75ºC/W. Pick the SMALLEST -- that means the largest heatsink. The larger heatsink will work for either case.

In this case 4.5ºC/W is the smallest value (largest heatsink).

Now we can progress to calculating R

A 2.6ºC/W heatsink is a lot smaller, but what about 3) and (4) above?

The rated efficiency of a heatsink will be a theoretical maximum (unfortunately). It will depend on the heatsink being in the right orientation and in free air. This may even mean (for smaller heatsinks designed to be mounted on a PCB) that there is no PCB under it! In addition to this, the theoretical figure assumes there will be no dust, no rise in ambient temperature, no nearby sources of heat, ... In short, you need to apply some sort of fudge factor to allow for reality.

For small (i.e. PCB mounted, or completely supported by the package) heatsinks, I would get a heatsink at least twice the size calculated. For larger heatsinks I might get one 25% to 50% larger.

In this case the heatsink will be large, so instead of 2.6ºC/W, I'll look for something of about 1.75ºC/W or larger (that's an additional factor of 50% (1.75 × 1.5 = 2.625).

I managed to find a 1.1ºC/W heatsink for a reasonable price

- We would prefer a T
_{j}substantially lower than T_{j(max)} - What if the ambient temperature rises?
- Is the heatsink
__really__that efficient? - What if the heatsink gets dusty?

All of these factors are going to work against us. Some require fudge factors (i.e. they are hard to calculate) but others can be calculated.

(1) and (2) go hand in hand. We might want the normal temperature of the junction to be no more than 125ºC, and we think that maybe the temperature could rise to 55ºC on occasion. We might decide that we will allow T

_{j}to rise above 125ºC with an ambient temperature above 35ºC, but at 55ºC it still has to be less than 145ºC.To make things easier, let's also assume that we were able to reduce the power dissipation of the transistor down to 20W.

We now need to do the thermal calculation twice, once for "normal" conditions, and again for "hot" conditions:

R

Firstly, with_{θjA(max)}= (T_{j(max)}- T_{A}) ÷ P

T

T

P = 20W

R

= (125 - 35) / 20

= 90 / 20

= 4.5ºC/W

Now, with_{A(35)}= 35ºCT

_{j(max@35)}= 125ºCP = 20W

R

_{θjA(max@35)}= (T_{j(max@35)}- T_{A(35)}) ÷ P= (125 - 35) / 20

= 90 / 20

= 4.5ºC/W

T

T

P = 20W

R

= (150 - 55) / 20

= 95 / 20

= 4.75ºC/W

_{A(55)}= 55ºCT

_{j(max@55)}= 150ºCP = 20W

R

_{θjA(max@55)}= (T_{j(max@55)}- T_{A(55)}) ÷ P= (150 - 55) / 20

= 95 / 20

= 4.75ºC/W

In this case 4.5ºC/W is the smallest value (largest heatsink).

Now we can progress to calculating R

_{θSA}for the heatsink.R

= 4.5 - (1.39 + 0.5)

= 4.5 - 1.89

= 2.61ºC/W

_{θSH}= R_{θjA}- (R_{θjC}+ R_{θCS})= 4.5 - (1.39 + 0.5)

= 4.5 - 1.89

= 2.61ºC/W

The rated efficiency of a heatsink will be a theoretical maximum (unfortunately). It will depend on the heatsink being in the right orientation and in free air. This may even mean (for smaller heatsinks designed to be mounted on a PCB) that there is no PCB under it! In addition to this, the theoretical figure assumes there will be no dust, no rise in ambient temperature, no nearby sources of heat, ... In short, you need to apply some sort of fudge factor to allow for reality.

For small (i.e. PCB mounted, or completely supported by the package) heatsinks, I would get a heatsink at least twice the size calculated. For larger heatsinks I might get one 25% to 50% larger.

In this case the heatsink will be large, so instead of 2.6ºC/W, I'll look for something of about 1.75ºC/W or larger (that's an additional factor of 50% (1.75 × 1.5 = 2.625).

I managed to find a 1.1ºC/W heatsink for a reasonable price

__here__.

**5.4 What about the rated maximum power? Can we achieve that?**The normal answer to this is "only downhill, with a following wind".

In practice you may find that it's very hard to achieve continuous power dissipation anywhere near the rated maximum for power transistors using any practical sized heatsink in still air.

So what's the use of this rating? One important answer comes from section 4.3 above. Whilst we may not be able to design a simple heatsink capable of continuous operation at the maximum rated power, if we can average the power over a few seconds, we may have an easier problem.

Let's say that the transistor is called on to dissipate 70W for 1/10 of a second every 3 seconds. In this case, 70W is under the maximum of 75W for this device, but (you should realise) above what we can comfortably heatsink for. However we can average out the dissipation:

This average power is under the maximum dissipation we calculated for this device without a heatsink at all! However, it would probably be wise to attach a small heatsink just to increase the thermal mass to the point where it would take much more than 3 seconds to reach equilibrium.

I'm sure there's a calculation somewhere that allows you to arrive at a good fudge factor, but for me I just multiply the power by the number of seconds over which I averaged it. (This isn't scientific!). For me, I'd assume it was dissipating 7W. If this was greater than the peak dissipation, I'd go with the peak (in this case 7W < 70W, so I go with 7W).

Making only the simple (normal) assumptions from the example above:

R

= 12.9 - (1.39 + 0.5)

= 12.9 - 1.89

≈ 11ºC/W

This heatsink size has been calculated for thermal mass, so we don't have to fudge it for efficiency etc. as the average rate at which it needs to dissipate power is quite a bit lower.

Let's calculate the heatsink size for the actual dissipated power, and see what happens if we add a fudge factor to that.

R

= 39 - (1.39 + 0.5)

= 39 - 1.89

≈ 37ºC/W

A 37ºC/W heatsink is tiny (remember that the TO-220 package could do this without a heatsink at all). Our fudge factor would be to double the size, so 18ºC/W.

For average dissipation we need 20ºC/W, for the thermal mass we need 11ºC/W. We choose the larger (11ºC/W) knowing we don't need to further fudge it.

Here is a 12ºC/W heatsink. It's pretty small and it's pretty cheap. Note this is slightly smaller than we calculated, but since we have quite a number of fudge factors, this approximation is probably fine. If you were worried, a 10ºC/W heatsink would only be slightly larger and of ample size.

In practice you may find that it's very hard to achieve continuous power dissipation anywhere near the rated maximum for power transistors using any practical sized heatsink in still air.

So what's the use of this rating? One important answer comes from section 4.3 above. Whilst we may not be able to design a simple heatsink capable of continuous operation at the maximum rated power, if we can average the power over a few seconds, we may have an easier problem.

Let's say that the transistor is called on to dissipate 70W for 1/10 of a second every 3 seconds. In this case, 70W is under the maximum of 75W for this device, but (you should realise) above what we can comfortably heatsink for. However we can average out the dissipation:

P

= ((70 × 0.1) + (0 × 2.9)) ÷ (0.1 + 2.9)

= 7 ÷ 3

= 2.33 W

_{avg}= Σ(P_{i}×T_{i}) ÷ Σ(T_{i})= ((70 × 0.1) + (0 × 2.9)) ÷ (0.1 + 2.9)

= 7 ÷ 3

= 2.33 W

I'm sure there's a calculation somewhere that allows you to arrive at a good fudge factor, but for me I just multiply the power by the number of seconds over which I averaged it. (This isn't scientific!). For me, I'd assume it was dissipating 7W. If this was greater than the peak dissipation, I'd go with the peak (in this case 7W < 70W, so I go with 7W).

Making only the simple (normal) assumptions from the example above:

R

Firstly, with_{θjA(max)}= (T_{j(max)}- T_{A}) ÷ P_{avg}

T

T

P

R

= (125 - 35) / 7

= 90 / 7

= 12.9ºC/W

Then to calculate the R_{A(35)}= 35ºCT

_{j(max@35)}= 125ºCP

_{corrected}= 7WR

_{θjA(max@35)}= (T_{j(max@35)}- T_{A(35)}) ÷ P= (125 - 35) / 7

= 90 / 7

= 12.9ºC/W

_{θSA}

R

_{θSH}= R

_{θjA}- (R

_{θjC}+ R

_{θCS})

= 12.9 - (1.39 + 0.5)

= 12.9 - 1.89

≈ 11ºC/W

Let's calculate the heatsink size for the actual dissipated power, and see what happens if we add a fudge factor to that.

R

Firstly, with_{θjA(max)}= (T_{j(max)}- T_{A}) ÷ P_{avg}

T

T

P

R

= (125 - 35) / 2.33

= 90 / 2.33

= 39ºC/W

Then to calculate the R_{A(35)}= 35ºCT

_{j(max@35)}= 125ºCP

_{avg}= 2.33WR

_{θjA(max@35)}= (T_{j(max@35)}- T_{A(35)}) ÷ P= (125 - 35) / 2.33

= 90 / 2.33

= 39ºC/W

_{θSA}

R

_{θSH}= R

_{θjA}- (R

_{θjC}+ R

_{θCS})

= 39 - (1.39 + 0.5)

= 39 - 1.89

≈ 37ºC/W

For average dissipation we need 20ºC/W, for the thermal mass we need 11ºC/W. We choose the larger (11ºC/W) knowing we don't need to further fudge it.

Here is a 12ºC/W heatsink. It's pretty small and it's pretty cheap. Note this is slightly smaller than we calculated, but since we have quite a number of fudge factors, this approximation is probably fine. If you were worried, a 10ºC/W heatsink would only be slightly larger and of ample size.

**6 Forced air and other magic**

**7 What could possibly go wrong?**