TO-220 heatsink question about aluminum - potentially dumb idea

[ee851]

Planning to prototype some high-power LED current regulators using an N-Channel MOSFET power transistor in the TO-220 package.    I have calculated for the worst case that the heatsink needs to have a thermal resistance of 7 degC/W or less  (corresponding to 10.4W heat being dissipated by this transistor).

A commercial heatsink that meets this requirement will make my board larger than desireable and approximately as much as all the other components combined.      So am thinking about using a solid piece of aluminum instead.       Does anybody know the approximate size of a piece of solid aluminum that would provide a thermal resistance less than 7 degC/W  for this device?  or how to approximate it ?

This heatsink is for the power transistor only--not the LEDs, which will be mounted separately.

If its size turns out to be larger than the commercial heatsink, then I will know this is a dumb idea.      (my idea's stupidity being in direct proportion to the percent size increase of the solid block compared to the commercial heatsink's size.)

[IanB]

A solid block isn't going to make a good heat sink. For systems that need to run continuously the mass of the heat sink becomes irrelevant; what matters is the rate of heat transfer to the surroundings. This is known as a steady state calculation and depends on the geometry of the heat sink, thermal conductivity, fin surface area, air currents and related factors. One possible choice when a free convection heat sink is inadequate is a fan cooled heat sink. This can make things much smaller (at the cost of ambient noise).

(But for a system that only needs to run for short periods with cool down periods in between, a large massive block can work. It absorbs the heat and then releases it to the surroundings during the cool down periods.)

[robrenz]

Also a huge difference in effectiveness of a heat sink with the orientation of the fins. if oriented to take advantage of convection currents it will be much more effective.

[krish2487]

However, if you are still seriously considering a solid block of aluminium for dissipating heat effectively (due to various constraints of your design - size, shape, profile etc) drilling a number of thermal relief holes would work better than a solid block.

It is the same concept of why most heatsinks have fins ( fins dissipate heat into the surrounding environment)

As ian stated, a solid block has a larger thermal mass, hence it takes longer for it to absorb the heatsink from the device and longer to dissipate it. A finned heatsink has lesser thermal mass owing to a larger surface area to dissipate the same amount of heat.

[SeanB]

Plain heatsink wil  be larger than the finned one made for the task. If you are using a diecast box you often can use it as a heatsink, or design the board to fit in a slotted heatsink.

[grumpygeek]

If your LED's are a series string, adjust the number of LEDs so that the  voltage drop at the current you want will be  close to the DC voltage of your power source,so that you dissipate  the least amount of power. If you use a die cast box, then an aluminum block will transmit the heat to the case nicely. 

[T4P]

I have calculated for the worst case that the heatsink needs to have a thermal resistance of 7 degC/W or less  (corresponding to 10.4W heat being dissipated by this transistor).

Why do you want to run the MOSFET at the max? It's not going to survive much at 102.8C! You need at least 4C/W to get 71.6C (30C ambient) And you also need to take into account convection currents (Standing vertical heatsinks are best for a natural convection heatsink) plus junction-case temperature rise

[ee851]

Yes, thank you for pointing out that I must consider the efflux of heat from the heatsink to the air as well as the influx of heat from the transistor to the heatsink.

I estimated MOSFET power dissipation as
Ic*Vds = (1.5A)*(13.8V/2) where I assumed the MOSFET would drop approximately half the supply voltage at a maximum load current of 1.5A direct current.

Is this equation okay to estimate the power dissipation, AKA heat,  of a MOSFET being driven continuously with DC?

The high-power Cree 7090XR-E LEDs have a max. jcn. temp. of 150C, so am more likely to drive them with 700mA than with one ampere.     Like I said, my calcn. was for worst-case-scenario--no forced air circulation, just a vertically-oriented heat sink in still air at 27C.

Since Dave showed us how to make a Li-ion battery charger, I might switch from a 12V battery to a couple of 7.4V Li-ion cells in series, so that would increase power slightly too.   So the 13.8VDC I used in my calculation is probably intermediate between a 12VDC battery pack and the Li-ion battery output voltage.

[G7PSK]

The greatest possible surface area is what is required, to this end you could use folded sheet aluminium, I have done this in the past, I stacked six or more pieces of 1mm aluminium sheet which were bent with increasingly tighter angles  so that each piece fitted inside the other with power diodes bolted in the middle of each assembly. That way I was able to have 200 amp diodes on a small foot print heat sink.

[nitro2k01]

Say what? How did you arrive at the conclusion that the MOSFET would stand for a voltage drop of half the supply voltage? That's just pure craziness unless you're looking to build a power supply with a built-in heat radiator to warm your room.

If the MOSFET is anything like most modern MOSFETs, it will have an on resistance well below 1 ohm. I have built my own LED driver using MOSFETs and it can switch 1 A without getting noticeably warm. If you do things right, you don't need a heatsink at all for the MOSFET for that kind of current.
What you need to do is connect as many LEDs as you can in series up to the battery/supply voltage and then a resistor to fill the gap. For example, 12 V battery:

You can at least use 3 LEDs of that type in series. 3*3.3 V = 9.9 V. That leaves a 12 - 9.9 = 2.1 V gap to be covered. Say you want to drive the LED with 700 mA. Ohm's law gives R = 2.1 V / 0.7 A = 3 ohms. That's where most of the excess power will go. How much? P = 2.1 V * 0.7 A = 1.47 W. Nothing to write home about. Add many of these branches of 3 LEDs you need in parallel. Once you get up to maybe 5-10 A total in the circuit you want to think about a small heatsink.

[ee851]

Oh, yeah, you are right.    Dropping half the supply voltage across the MOSFET is a very bad idea and completely unnecessary.     

Thank you for pointing out my error.

If I use two Li-ion cells in series for power and four white LEDs, dropping 3.5V each, that means the MOSFET only need dissipate

P=(V_DD-V_LED)*I = (14.8-14)(1.5A) = 1.2W at most

That will drop the thermal resistance down to
(100 degC - 30 degC)/1.2W =  58 degC/W at 30C ambient
temperature.    That means almost any heatsink will do the job.

1.5A is a worst-case scenario.    More likely to run the LED at 700mA.

Much appreciate your comment!

[nitro2k01]

Keep in mind that a fully charged nominal 3.6 Li-ion battery cell (your batteries will contain two of those cells each) will have a voltage of something 4.2 before they reach the flat discharge region. So the combined series voltage may actually be in the 15-17 V region. For this reason, unless you're using some other current control mechanism, it might be a good idea to use additional resistors on each parallel section to avoid blowing the LEDs because of overcurrent. The MOSFET would in this case be in the saturation region and emit practically no heat.

If you cared enough, you could design an simple circuit which uses the MOSFET as a current limiter so that the LED intensity is the same over (almost) the full range. This would obviously take the MOSFET out of the saturation region so it would emit more heat, especially when the battery is freshly charged.

[ee851]

Do you mean 3.6V per cell is the end-of-life (fully discharged condition) for Li-ion batteries  and that I ought to design my series LED string to this condition?

Here's the battery I plan to use:
http://www.hobbyking.com/hobbycity/store/__14843__Turnigy_5000mAh_4S_25C_Lipo_Pack_USA_Warehouse_.html

This battery contains four cells.   So (3.6V)*(4) equals 14.4V.   So I can plan a series LED string to be up to 14.4V minus the voltage drop through the current regulator.   Am I correct ?

[Jimmy the Squid]

The cells will drop fairly quickly from their fully charge voltage of 4+ volts down to the operating "plateau" of 3.6 volts or so for quite some time during discharge. Towards the end of the available capacity, voltage will begin to drop rapidly until the battery protection circuit kicks in and shuts down the output.

Dave did a tutorial on battery capacity at one point, but I couldn't find the right one.

[Jimmy the Squid]

Found it! Here is the link for the battery capacity tutorial I mentioned.

http://www.eevblog.com/2011/01/23/eevblog-140-battery-capacity-tutorial/

[nitro2k01]

The cells will drop fairly quickly from their fully charge voltage of 4+ volts down to the operating "plateau" of 3.6 volts or so for quite some time during discharge.

Of course. My concern though is that he might fry the LED at the peak charge unless he's using some form of current limiting. (If your comment was a reply to me.)

[Jimmy the Squid]

The cells will drop fairly quickly from their fully charge voltage of 4+ volts down to the operating "plateau" of 3.6 volts or so for quite some time during discharge.

Of course. My concern though is that he might fry the LED at the peak charge unless he's using some form of current limiting. (If your comment was a reply to me.)

Sorry, I should have been specific. My response was the the OP's question about whether the 3.6V was at full discharge.

[ee851]

Okay,  after reviewing my calculations several times, I realized that I need a heatsink with thermal resistance of 7.8 degC/W or lower for my N-Channel power MOSFET in a TO-220.

Now I have another question about board layout.  I already laid out the board in EAGLE.     This will be my first homemade PCB.   It is only  20mm x 20mm without the heatsink.     I looked up datasheets for two heatsinks that meet the 7.8C/W spec.   

One measures 35mm wide x 50mm tall.   The other one
measures 42mm wide x 38 mm tall.    I did not make the PCB yet.    Obviously neither of these heatsinks will fit on the PCB, even if mounted along the diagonal.

Does this matter ?    Or can the heatsink be larger than the board ?    Does it make any difference ?   I plan to put the PCB in a small plastic box with one end open to let the heat escape.
I plan to mount the heatsink perpendicular to the PCB.

[nitro2k01]

There isn't necessarily a problem with a bigger heatsink than the board, but there may be a problem if the weight from the heatsink puts mechanical stress on the transistor legs if things are not fixed. However, since your PCB is so small, I don't really see how you could arrange the transistors efficiently. You would have to make them stick out from three sides of the board in a T shape and you would have to have a lot of unused space.

...oh wait, perpendicular to the PCB, so the transistors would be standing. Then you would have to take care that the heatsinks don't get misaligned and touch each other, for reasons mentioned below.

I would personally have used a certain type of prototype PCB, with tracks similar to that of a breadboard (strips for components, as well as long perpendicular strips for the power rails) and cut out a suitable piece. The transistors would face inward, and the power rail would be connected to the sources of each transistor. However, I'm, biased because I both have access to quantities of such boards, as well as a sheet metal cutter which also works excellently for cutting PCB.

One more precaution: If you're considering connecting all three transistor to a single heatsink, be aware that unless the transistors are in a TO-220FP package, (non-metallic tab) or take steps to insulate the tabs, the tabs of all three transistors will be electrically shorted. The tab is connected to pin 2 (the middle pin) which for an N channel MOSFET would probably be drain, so bad idea to short them all together.

[ee851]

Thank you for warning about mounting more than one transistor on a heatsink.   I had not considered that.

For this board, there is only one heatsink.    It will be mounted to the MOSFET in a TO-220 package.   There is only one MOSFET.    The other transistors are BJTs for biasing and don't dissipate power so they don't need heatsinks.   I have to design a way to provide strain relief on the leads of the TO-220, without adversely affecting the free flow of air over the heatsink.