How to address thermal concerns with a mosfet and heatsink?

[FusionITR]

How do you address thermal issues solely from a datasheet? Lets say for example you are using this mosfet:

http://search.digikey.com/scripts/D...g=en&site=US&KeyWords=785-1189-5-ND&x=14&y=22

And the current will be 3A with a 20khz 50% duty cycle switching frequency. If it were DC, the power would be 9W (I^2*Ron), but lets say because of the switching it's 4.5W.

Based off the datasheet, how do I determine if I need a heatsink to dissipate the 4.5W? If so, how do I choose an appropriate heatsink?

 

[shortbus=]

This is a good link that is written to address this question; http://mcmanis.com/chuck/robotics/projects/esc2/FET-power.html

 

[crutschow]

If the MOSFET is being switched on with a Vgs = 10V, then the transistor on resistance is 1.2 ohms. Thus the dissipation will be approximately 3.0² x 1.2 x 50% = 5.4W. You definitely need a heat sink. Such a transistor can only dissipate a watt or so without one.

If you could use a lower voltage rated MOSFET with a lower "ON" resistance, that would reduce the power dissipated and your heat sink requirements.

 

[FusionITR]

Such a transistor can only dissipate a watt or so without one.

How do you determine this?

 

[cr0sh]

How do you determine this?

Get the datasheet from Digikey, and read the page that shortbus= posted; I found that article very useful, so I tried doing the calcs for it (from the section "Real World Applications" - some of the wording is also cut/pasted from McManis' site - thank you, Chuck!):

1. Find Rtheta(ja), which is junction to ambient air. For your device from the datasheet, this is 65 °C/Watt. This tells us that in still air, the junction is 65 degrees warmer than the temperature of the ambient air for each watt of power we're dissipating.

2. Find the Max junction temperature - for your device, it is 150 °C (maximum value from "Junction and Storage Temperature Range" on the datasheet).

3. If we assume the ambient room temperature is 25 °C, then we know that there is a 125 °C difference between ambient temperature and the junction temperature.

4. Using that number (125 °C), we simply divide it by our Rtheta(ja) of 65 °C/Watt to get the "free air wattage dissipation" value:

P = 125 °C / Rtheta(ja) (°C/W)
P = 125 °C/ 65 (°C/W)
P = 1.92 Watts

So this device - sitting in open air (@ 25 °C ambient temperature) without a heatsink - can dissipate 1.92 Watts without failing. If you need to dissipate more watts, you would need a heatsink. The following page gives some help here:

http://www.kpsec.freeuk.com/components/heatsink.htm

"Heat sinks are rated by their thermal resistance (Rth) in °C/W. For example 2°C/W means the heat sink (and therefore the component attached to it) will be 2°C hotter than the surrounding air for every 1W of heat it is dissipating. Note that a lower thermal resistance means a better heat sink."

Based on the earlier calculation - I believe this means you would want a heatsink with an Rth of 1.92 (our max ambient wattage) divided by the number of watts over this value you need to disspate; so if the above calcs showed that the device needed to dissipate 4 watts into 25 °C ambient air, you would want a heatsink with an Rth rating of 0.48 or less - I think (???).

McManis' page goes on further to describe how to take that number and calculate the maximum number of amps the device can handle at that ambient temperature (without a heatsink). Note that for the calculation of the amps, I had to make a guess at the Rds(On) value - the datasheet says it's "1.2|" - I assume that "|" meant mOhms - or .012 ohms.

I've never done those calcs, so the tutorial and this thread was very enlightening (btw for anybody else here who cares - was my math right?).

 

[shortbus=]

@cr0sh- That link really finally made the mosfet data sheets make sense to me. He puts it in plain language!

 

[FusionITR]

Get the datasheet from Digikey, and read the page that shortbus= posted; I found that article very useful, so I tried doing the calcs for it (from the section "Real World Applications" - some of the wording is also cut/pasted from McManis' site - thank you, Chuck!):

1. Find Rtheta(ja), which is junction to ambient air. For your device from the datasheet, this is 65 °C/Watt. This tells us that in still air, the junction is 65 degrees warmer than the temperature of the ambient air for each watt of power we're dissipating.

2. Find the Max junction temperature - for your device, it is 150 °C (maximum value from "Junction and Storage Temperature Range" on the datasheet).

3. If we assume the ambient room temperature is 25 °C, then we know that there is a 125 °C difference between ambient temperature and the junction temperature.

4. Using that number (125 °C), we simply divide it by our Rtheta(ja) of 65 °C/Watt to get the "free air wattage dissipation" value:

P = 125 °C / Rtheta(ja) (°C/W)
P = 125 °C/ 65 (°C/W)
P = 1.92 Watts

So this device - sitting in open air (@ 25 °C ambient temperature) without a heatsink - can dissipate 1.92 Watts without failing. If you need to dissipate more watts, you would need a heatsink. The following page gives some help here:

http://www.kpsec.freeuk.com/components/heatsink.htm

"Heat sinks are rated by their thermal resistance (Rth) in °C/W. For example 2°C/W means the heat sink (and therefore the component attached to it) will be 2°C hotter than the surrounding air for every 1W of heat it is dissipating. Note that a lower thermal resistance means a better heat sink."

Based on the earlier calculation - I believe this means you would want a heatsink with an Rth of 1.92 (our max ambient wattage) divided by the number of watts over this value you need to disspate; so if the above calcs showed that the device needed to dissipate 4 watts into 25 °C ambient air, you would want a heatsink with an Rth rating of 0.48 or less - I think (???).

McManis' page goes on further to describe how to take that number and calculate the maximum number of amps the device can handle at that ambient temperature (without a heatsink). Note that for the calculation of the amps, I had to make a guess at the Rds(On) value - the datasheet says it's "1.2|" - I assume that "|" meant mOhms - or .012 ohms.

I've never done those calcs, so the tutorial and this thread was very enlightening (btw for anybody else here who cares - was my math right?).

Wow, awesome post man. You have really made my day. Its very encouraging someone took the time to explain the details in such a easy to understand way. Thanks! I would rep you if this forum had a rep system.