I've been doing some research on submerging circuits in oil, but I can only find info about computers. I have a small (20 watt) amplifier that I am experimenting with and since I upped the input voltage the small heatsink it's attached to is getting uncomfortably hot. I don't want to look for a new heatsink, instead I want to remove the old one and put the whole thing in a small bowl of oil. My question is, will a chip about 2.5x the size of a to-220 transistor case dissipating maybe 10 watts or so of heat or so be able to transfer that heat into the oil without a heatsink? Since I only use it for short runs I won't have to worry about all of the oil overheating, and if I start using it more I can always point a fan at the surface.

In the future I'd like to use this technique to cool my home made 7.1 surround sound system amps that I am currently building.

I don't know the part # of the chip in the amp I'm messing with now but most of my basic surround sound system will be built around tda2030 and tda2050 chips if it makes a difference.

 

Depends on the size of the heatsink. IMO

Oil is ~6 times better in conductivity.

So if the heat sink had 6 times more surface than the chip it would be a wash.

Using the same heatsink-6X better.

I have a feeling the heatsink was much larger than that.

I'm sure there are many more variables.

 

Depends on the size of the heatsink. IMO

Oil is ~6 times better in conductivity.

So if the heat sink had 6 times more surface than the chip it would be a wash.

Using the same heatsink-6X better.

I have a feeling the heatsink was much larger than that.

I'm sure there are many more variables.

The heatsink is only 2-3x the size of the chip. Are you saying the thermal conductivity of the heatsink (the chip to heatsink junction) is 6x better or the heat dissipation per square inch in still air is 6x better? Because the heatsink has no airflow and it wouldn't be easy to fit a fan in, where it would be easier than ever to drop the whole thing in oil. I only want to remove the heatsink because I like the looks of a bare chip better.

 

The only reason why the heatsink won't work now is because during mormal use it is in an inclosed space with no airflow, and when I pull the cover to check the temp I have to remove the speaker ports and volume/tone knobs.

 

No, the conductivity of oil is 6 times better than air.
That ignores air or oil movement.

So if the surface area of the heatsink was only 3X the bare chip then oil would be twice as effective as the heatsink..

There is usually a lot more surface area than that on even a small heatsink.

Chip to heatsink could be safely ignored in this case. Aluminum is 1300X better than oil.

This is just my opinion after looking up the thermal properties of materials.

We'll have to see what the real expert say's.

 

The only reason why the heatsink won't work now is because during mormal use it is in an inclosed space with no airflow, and when I pull the cover to check the temp I have to remove the speaker ports and volume/tone knobs.

Even so, it's surrounded by air.
Just as it will be surrounded by (only 6X better) oil in it's new home.

I would suggest the same heatsink in oil.

Just as air movement will help cool, oil movement could greatly improve the 6/1 advantage.

 

Even so, it's surrounded by air.
Just as it will be surrounded by (only 6X better) oil in it's new home.

I would suggest the same heatsink in oil.

Just as air movement will help cool, oil movement could greatly improve the 6/1 advantage.

OK I'll keep the heatsink and move it to oil. Thanks!

 

You still have to get the heat out of the oil, otherwise the oil will get hotter, and hotter, and hotter...

In the final analysis, the transfer of heat from oil to ambient surroundings must be greater than the transfer of heat from heatsink to ambient surroundings used to be...

 

You still have to get the heat out of the oil, otherwise the oil will get hotter, and hotter, and hotter...

In the final analysis, the transfer of heat from oil to ambient surroundings must be greater than the transfer of heat from heatsink to ambient surroundings used to be...

+1

Yet, it would be a very small (or insulated), oil container to not transfer more heat to air than the small heatsink.

 

+1

Yet, it would be a very small (or insulated), oil container to not transfer more heat to air than the small heatsink.

It'd be about 6inx6inx2in with open air to the top. Like I said earlier I don't plan on running it long enough to heat up that much oil more than a couple degrees.

 

Hi,


It's all about surface area and heat conduction. Know your surface area and figure out if your heat conduction is ok or not. If you have air movement that's a third consideration but that's not always present.

With a regular heat sink designed for transistors, the heat conduction is usually ok, so it then boils down to surface area and air movement.
If you have 8 square inches of surface area, that will dissipate more heat than 4 square inches. If you have air movement (small fan) that helps a lot too but you must ensure that the fan is checked to make sure it does not stop as fans often fail by not turning anymore, then the part overheats.

So we are down to one thing now: surface area. This affects almost all heating problems.

If your chip has 1 square inch of surface area, then the temperature rise will be roughly 60 degrees C per watt of chip dissipation. With modern chips that would mean it should be ok to run it at 1 watt but not any higher.
If you intend to run it at say 2 watts then you should attach it to a heatsink with surface area roughly two square inches. At 10 watts, you'd need 10 square inches or better, but it would have to be a heatsink not just a flat plate.

If the chip is instead immersed into a container of some fluid like oil or water, then the heat dissipation characteristics depend on the size of the vessel and the heat dissipation properties of the sides of the vessel.

Assuming the container has air on all six sides, the surface area now is equal to:
Sa=L*W*2+H*W*2+H*L*2

so for a 6x6x2 inch container the area on top and bottom is 36*2=72 square inches, and the area on both sides is 12*2=24 square inches, and the area on front and back is 12*2=24 square inches, so the total surface area is:
36+24+24=84 square inches.
So we now have 84 square inches which is a lot more than before.

Ideally, the container should be in the shape of a sphere with the chip placed at the center of the sphere, but that's not usually possible so a rectangular container is used. The next choice is a container that is in the shape of a cube, with the chip placed in the center of the cube.
Since the container is neither of these, it is not optimum, but it is still better than before. The heat dissipation to the sides and front and back will not be as good as top and bottom (container held horizontally).
Also, if one side of the container is blocked by an insulator (such as when it is standing up on another surface) then that side dosnt count anymore. Eliminating one side still leaves us with 60 square inches, which is much more than needed for 10 watts. The container shape isnt ideal so it will not be prefect, but it should be pretty good because it's so much bigger than needed theoretically.

Distilled water also works, but other issues come up using that.

If the container is sealed there has to be room left for thermal expansion (or contraction) of the fluid.

Back to the air movement consideration...
If you have a small heatsink 1 square inches surface area then the thing will rise in temperature by about 60 degrees C with 1 watt of power in the chip, but if there is a small fan blowing on that heatsink it can dissipate a lot more heat. To find out how well this works you have to look up the manufacturers data sheet on the heat sink though.

For those interested, the equation for this is a second order partial differential equation that looks something like this:
U_t=U_xx+U_yy+K
where
U_t is the first partial of the function with time, and U_xx and U_yy are the second partials with their respective dimension, and K is the constant heating term.

 

Lol modern technology doesn't really use heatsinks. Switchmode power supplies. Class D amplifiers. Where's the heatsink?

 

Hi,


It's all about surface area and heat conduction. Know your surface area and figure out if your heat conduction is ok or not. If you have air movement that's a third consideration but that's not always present.

With a regular heat sink designed for transistors, the heat conduction is usually ok, so it then boils down to surface area and air movement.
If you have 8 square inches of surface area, that will dissipate more heat than 4 square inches. If you have air movement (small fan) that helps a lot too but you must ensure that the fan is checked to make sure it does not stop as fans often fail by not turning anymore, then the part overheats.

So we are down to one thing now: surface area. This affects almost all heating problems.

If your chip has 1 square inch of surface area, then the temperature rise will be roughly 60 degrees C per watt of chip dissipation. With modern chips that would mean it should be ok to run it at 1 watt but not any higher.
If you intend to run it at say 2 watts then you should attach it to a heatsink with surface area roughly two square inches. At 10 watts, you'd need 10 square inches or better, but it would have to be a heatsink not just a flat plate.

If the chip is instead immersed into a container of some fluid like oil or water, then the heat dissipation characteristics depend on the size of the vessel and the heat dissipation properties of the sides of the vessel.

Assuming the container has air on all six sides, the surface area now is equal to:
Sa=L*W*2+H*W*2+H*L*2

so for a 6x6x2 inch container the area on top and bottom is 36*2=72 square inches, and the area on both sides is 12*2=24 square inches, and the area on front and back is 12*2=24 square inches, so the total surface area is:
36+24+24=84 square inches.
So we now have 84 square inches which is a lot more than before.

Ideally, the container should be in the shape of a sphere with the chip placed at the center of the sphere, but that's not usually possible so a rectangular container is used. The next choice is a container that is in the shape of a cube, with the chip placed in the center of the cube.
Since the container is neither of these, it is not optimum, but it is still better than before. The heat dissipation to the sides and front and back will not be as good as top and bottom (container held horizontally).
Also, if one side of the container is blocked by an insulator (such as when it is standing up on another surface) then that side dosnt count anymore. Eliminating one side still leaves us with 60 square inches, which is much more than needed for 10 watts. The container shape isnt ideal so it will not be prefect, but it should be pretty good because it's so much bigger than needed theoretically.

Distilled water also works, but other issues come up using that.

If the container is sealed there has to be room left for thermal expansion (or contraction) of the fluid.

Back to the air movement consideration...
If you have a small heatsink 1 square inches surface area then the thing will rise in temperature by about 60 degrees C with 1 watt of power in the chip, but if there is a small fan blowing on that heatsink it can dissipate a lot more heat. To find out how well this works you have to look up the manufacturers data sheet on the heat sink though.

Very detailed explanation, thanks. The only thing that I'm not 100% sure about is whether my chip (about 1 sq. in., maybe a bit less) will be able to dissipate 10 watts into the oil without some kind of heatsink installed. You said in open air 1 sq. in. will heat up 60*c/watt, what about that same 1 sq. in. in the oil?

 

Lol modern technology doesn't really use heatsinks. Switchmode power supplies. Class D amplifiers. Where's the heatsink?

I have a 12 volt 50 amp switching power supply, it still has a large heatsink.

 

Hi,

Well, lets look at this as a purely electrical problem for a minute.

If we have a voltage source of 10 volts powering a 1 ohm resistor, we get a current flow of 10 amps. That's like using an aluminum heat sink of the proper area.
If we have a voltage source of 10 volts powering a 1000 ohm resistor, we get a current flow of 0.010 amps which is 1000 times less (oil). But double the thickness of the resistors body and the resistance goes down by 2 times, to 500 ohms. Double it again and it goes down to 250 ohm, double again and it goes down to 125 ohms, then 62, then 31, then 16, then 8, then 4, then 2, then 1. Finally we are down to 1 ohm, but it took a lot more surface area.

The best bet then is to increase the surface area contact between the device and the oil, and that means using the heat sink with the oil will always be better.

The heat from the device sets up a thermal gradient from the device to the surfaces of the container. The larger this gradient is the higher the device temperature is. The conduction affects the gradient so we try to get as much conduction as possible.

It is unfortunate that the oil has such low conductivity, and that is what matters the most and that has to be figured into the equation as well. That means that for continuous operation maybe the oil wont even work at all even with the heat sink, depending on the size of the heat sink. You said that you would not be running it continuously though, so if there is enough cool down time it may have enough time to cool back down between uses.

To be sure, you should try to measure some property of the device that reflects it's internal temperature. For example, if the device was an LED you could measure it's forward voltage drop with a constant current and compare readings to an LED on a real heat sink. Alternately you may be able to epoxy a small thermistor to the device and break out two wires for measurements. Measuring the resistance of the thermistor would tell you how hot the device case was getting and that would tell you right away if it was ok or not. You could even incorporate that into the circuit to shut down if it gets too hot.

BTW we are assuming that the container is made out of thin metal not some insulating material.

 

I have a 12 volt 50 amp switching power supply, it still has a large heatsink.

Really? Gee whiz I tell you that our state-of-the-art Sony class D amplifier is too light weight and compact to have much of an heatsink. Smart phones 1GHz MPU -- no heatsink!

 

Really? Gee whiz I tell you that our state-of-the-art Sony class D amplifier is too light weight and compact to have much of an heatsink. Smart phones 1GHz MPU -- no heatsink!

Hi,

Unfortunately linear amplifiers still need heat sinks because of the shape of the waveform.

Heatsinks have gotten smaller over the years but many applications still need them. A good example is a high powered flashlight where the LED gets very hot. These are modern devices yet they still need heatsinks, and often the body is used as part of the heatsink.

 

Hi,

Well, lets look at this as a purely electrical problem for a minute.

If we have a voltage source of 10 volts powering a 1 ohm resistor, we get a current flow of 10 amps. That's like using an aluminum heat sink of the proper area.
If we have a voltage source of 10 volts powering a 1000 ohm resistor, we get a current flow of 0.010 amps which is 1000 times less (oil). But double the thickness of the resistors body and the resistance goes down by 2 times, to 500 ohms. Double it again and it goes down to 250 ohm, double again and it goes down to 125 ohms, then 62, then 31, then 16, then 8, then 4, then 2, then 1. Finally we are down to 1 ohm, but it took a lot more surface area.

The best bet then is to increase the surface area contact between the device and the oil, and that means using the heat sink with the oil will always be better.

The heat from the device sets up a thermal gradient from the device to the surfaces of the container. The larger this gradient is the higher the device temperature is. The conduction affects the gradient so we try to get as much conduction as possible.

It is unfortunate that the oil has such low conductivity, and that is what matters the most and that has to be figured into the equation as well. That means that for continuous operation maybe the oil wont even work at all even with the heat sink, depending on the size of the heat sink. You said that you would not be running it continuously though, so if there is enough cool down time it may have enough time to cool back down between uses.

To be sure, you should try to measure some property of the device that reflects it's internal temperature. For example, if the device was an LED you could measure it's forward voltage drop with a constant current and compare readings to an LED on a real heat sink. Alternately you may be able to epoxy a small thermistor to the device and break out two wires for measurements. Measuring the resistance of the thermistor would tell you how hot the device case was getting and that would tell you right away if it was ok or not. You could even incorporate that into the circuit to shut down if it gets too hot.

BTW we are assuming that the container is made out of thin metal not some insulating material.

Thanks a bunch, lots to think about. If I decide to use this amp for long periods of time I can either blow cool air over the surface of the oil or put a small computer fan in the oil to circulate it. Either way I'm not seeing this working like I had hoped, so I may have to cut a new heatsink from some scrap aluminum. I'll still try it, after all the amp I'm messing with only cost $10 so if it doesn't work and the chip fries I haven't really lost anything.

Also the container would be a small plastic box normally used to store food, with the lid removed.

 

Hi,

Unfortunately linear amplifiers still need heat sinks because of the shape of the waveform.

Heatsinks have gotten smaller over the years but many applications still need them. A good example is a high powered flashlight where the LED gets very hot. These are modern devices yet they still need heatsinks, and often the body is used as part of the heatsink.

Linear is obsolete though buddy. What you are saying will work though. I was messing around with class AB amplifiers 20-years ago. So loud that the bloke over the road wanted to murder me! I tried some LMxxx linear ICs in water. They stayed cool enough.

 

Linear is obsolete though buddy. What you are saying will work though. I was messing around with class AB amplifiers 20-years ago. So loud that the bloke over the road wanted to murder me! I tried some LMxxx linear ICs in water. They stayed cool enough.

With my budget linear is about all I can afford. Also easier to work with.