Hello all,

I have a question regarding thermal dissipation of PCB using copper surface. I have seen couple LED PCBs used FR4 board and top and bottom surface were field with 2oz copper. It was copper trace, but it covered entire circuit board. I thought I understand heat sink application, but I don't get how the thin 2oz copper trace on FR4 board works as a heat sink. knowing 2oz copper is about 70um which is really thin as compared to FR4 1.6mm board.

Please advise me!

Thanks in advance.

 

Think of the total cross-sectional area of copper that is available to carry heat away. It may not be huge but it's significant.

 

Any heatsink helps to remove heat via conduction, convection, and radiation.
A copper surface on FR4 board can help to remove heat by all of those three methods.

 

Thank you for the answer!

And I have another question.. In my mind, it might be very common questions, but bothers me a lot..

We knows that heatsink help to dissipate the heat. But theoretically I'm quite confused. Most of text saying that in thermal calculation should treat like this:

Power = current
Thermal resistance = Resistor
Temperature = Voltage

Now we do the thermal analysis, then we usually start with:

Rth_js
Rth_FR4
Rth_copper

Then now we decide to add a heatsink.

Rth_heatsink

After adding a heatink the overall thermal Resistance is increased. Which means junction Temp must be increased too. According to the following formula.

(Junction Temp - Ambient Temp) = Power x Thermal resistance.

However we all know this is not true... What am I missing here?

Please help me!

 

After adding a heatink the overall thermal Resistance is increased.

That is false.

 

In your analogy, adding a heat sink is like placing conductors (which all have some resistance) in parallel, not serial.

 

You have a small area, on the chip, which is your heat source it will disipate heart faster as it heats up but the amount of heat it can disipate is limmited by its area.

Now add a heat sink ... it has a much bigger area so will disipate more heat at a lower temperature, its a heat sink!
However if you want toi work out how much heat it can disipate then you need some figures.

Rather than thinking about voltage and current as a model think about quantity and flow.

if heat were fluid and you let it flow into a tank with no outlet the tank would simply fill untill it failed or was full and had no further effect.
now considder a tank with a v shaped vertivcal slot in its side.
As you fill the tank fluid leaks out, the higher the level the faster the leak, your tank now disipates fluid at a rate proportional to the amount of fluid in it.
The level in the tank, how much fluid (heat energy) it contains is now a function of how fast it leaks, at a given level, and how fast you are filling it
More flow = more level but the system will ballance provided you dont fill it faster than the maximum raye at which it can empty.

To address your question RE thermal resistance all you need to do is think about the tank as a sponge rather than an empty space.
the slot represents the potential to empty, disipate heat, but it isnt going to hep you if the flow to the slot is impeded by the material that the tank filled with. (made of it is a heat sink containing heat)

A heatsink is described by its ability to move heat from one place to another and that is based on its area, emisivity and thermal resistance.
there will be a charictoristic flow W/S at a given temperature diferencial but they are usually specified by C/W which describes the temperature increase as a function of the energy being disipated.

Its all actually quite complex when you start looking at forced air flow and surface temperatures.

So if you add a heat sink you are adding thermal resistance and if the heat sink was exactly the same size as the part you are truing to cool the part would in fact ger hotter. You were correct.

Hoiwever you wouldnt do that you add a material with some resistance to transfering heat but you add loads of it with a huge area, in conparison to the part you are cooling, and the net result is a cooler part.

Google it more detail and the technical stuff but keep that leaky sponge in mind as a raesonable model.

Hope that helps,
Al

 

Thank you for the answer!

And I have another question.. In my mind, it might be very common questions, but bothers me a lot..

We knows that heatsink help to dissipate the heat. But theoretically I'm quite confused. Most of text saying that in thermal calculation should treat like this:

Power = current
Thermal resistance = Resistor
Temperature = Voltage

Now we do the thermal analysis, then we usually start with:

Rth_js
Rth_FR4
Rth_copper

Then now we decide to add a heatsink.

Rth_heatsink

After adding a heatink the overall thermal Resistance is increased. Which means junction Temp must be increased too. According to the following formula.

(Junction Temp - Ambient Temp) = Power x Thermal resistance.

However we all know this is not true... What am I missing here?

Please help me!

Read up on heatsinks from Texas Instruments....

https://www.google.co.uk/url?sa=t&s...gghMAA&usg=AFQjCNEDYngi1Sb5zWjcPl4BEL2JHIySpQ

 

AI,

Thanks! It was great help. Here is a summary of your comment. Please correct me if I miss understand anything. If a small area of a sponge get wet, and water will spread from the point of the sponge to dried part of sponge. As the water travels further from the origin, the density of the water will be diminished. Also more water would travel faster within the sponge. Eventually the sponge expose the water to the air, so it can evaporate fast, as compare to the water from the small area. The ability of water evaporation is depends on emissivity, thermal resistance, and area of the sponge. If the area of sponge is the same as the water origin, then the density of water will be staying longer than without the sponge.

So I understand larger the area of the heatsink is the small K/W. Then when I calculate the total thermal resistance, then the heatsink being added as parallel to the thermal resistance circuit?

You have a small area, on the chip, which is your heat source it will disipate heart faster as it heats up but the amount of heat it can disipate is limmited by its area.

...

Hope that helps,
Al

 

more or less.... Its only an analogy so dont get tied up thinking about water evaporating, heat cant evaporste so that will not help you.
You would be better served reading the link that DD offered.
I was just trying to give you a mental model with which to visualise the mechenismes.

Heat is energy so it can move or be transformed and has the potential to do work, like any energy, but it cant simply disappear.
Some heat squeezed into a small volume will have a higher temperature than the same amount of heat in a larger volume.
A heat sink isnt working because of its integral volume but because it offers a relitivly low resistance path to an almost limitless volume, IE the surrounding environment.

Al

 

Elaborating on what I said in post #3, heat is dissipated via:

1. conduction
2. convection
3. radiation

Imagine a through-hole resistor mounted flat on a plain FR4 pcb soldered to standard sized resistor pads.

Now think of how the three modes of heat dissipation applies in this example.

1. conduction - heat is conducted away via the leads of the resistor but FR4 is a poor conductor of heat
2. convection - heat is removed if there is air flow around the resistor
3. radiation - the body of the resistor can radiate a small amount of thermal energy

Now imagine that we were to mount the resistor further away from the surface of the board.

1. conduction - there is more thermal mass in the leads of the resistor, hence more heat dissipated
2. convection - much improved air circulation
3. radiation - not much difference from before

Now imagine that the leads of the resistor are soldered to larger areas of copper on the PCB.

1. conduction - there is more thermal mass, more heat conducted away from the leads of the resistor
2. convection - there is increase convection current especially if the pcb is mounted vertically on its edge
3. radiation - more heat is radiated because of the larger surface area

Having large power and ground planes around all components will help in improving heat dissipation.