# Negative Thermal Resistance?!?!?

Discussion in 'General Electronics Discussion' started by BlackMelon, Nov 19, 2014.

1. ### BlackMelon

188
5
Aug 7, 2012
Hello,

I'm working with this MOSFET:

http://www.es.co.th/Schemetic/PDF/IRF640N.PDF

This datasheet shows us about thermal resistance in the bottom of page 2. I need this MOSFET to be operated at (Pd)150W dissipation power and its junction temperature(Tj) to be 100 celsius. Ambient temperature(Ta) is 50 celsius so I have to calculate the thermal resistance from junction to ambient (θja) first and then the case to ambient thermal resistance (θca)

so the equation is Tj=Ta+ θja*Pd
substitutes all those stuff, and you'll get θja= 0.333 celsius/W....... seems to be fine

From the datasheet we have θjc=1.0
θca= θja-θjc = 0.333-1.0=-0.667 celsius/W OMG!!!

1) Did I do anything wrong about calculation?
2) Did I misunderstand something in the datasheet?
3) Is this an impossible case?

2. ### Harald KappModeratorModerator

11,175
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Nov 17, 2011
1) Nothing obvious.
2) Tj can be up to 175 °C, no need to limit to 100 °C, but this results in θja= 0.833 celsius/W, still too small a value.
3) As described: yes. But: How do you arrive at 150 W power dissipation? This transistor is meant to be used in switching applications where you have either full voltage or full current, but not both (except during switching). In case you operate this transistor as a linear element, you can try to parallel a few of them to distribute the load.

3. ### BobK

7,682
1,688
Jan 5, 2010
Case to ambient is used when there is no heatsink. All you have discovered when your calculation goes negative is that you cannot dissipate 150W without a heat sink and keep the junction temperature where you want it. Which is no surprise, since the limit for a TO220 for example, is about 1W without a heatsink.

But, as Harald said, you are probably calculating the power wrong. If you really need to dissipate 150W in the MOSFET you are going to need a heatsink comparable to the large CPU ones before Intel started lowering the power and lots of forced air.

Bob

4. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I agree with Harald and Bob. If you really truly need to dissipate 150W you'll need several TO-220-package devices in parallel, or a device in a package with a much lower θjc. And a big heatsink, probably with forced air cooling.

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Another option is to use a peltier device or water cooling. A peltier allows the effective ambient temperature to be lowered, and water cooling provides a very efficient means of carrying away heat. (150W for a peltier would be enormous though)

6. ### BobK

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Jan 5, 2010
Neither of those really helps, since you still have to get rid of the excess heat somewhere. All the can do is move it to another place.

Bob

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Water cooling provides a highly efficient transfer of heat from a small heat source to an arbitrarily large radiator. It helps enormously. Another option is to immerse the entire circuit in a coolant. Both of these techniques have been used in high performance commercial computers (mainframes and supercomputers).

Vapour phase (refrigeration) and heat tubes can also be used. The former are more energy efficient than peltier devices, the latter are really a means of moving heat from one place to another (they are used extensively in laptop computers) and have the advantage of being passive devices.

Efficiently moving heat from one place to another can be a very useful way of increasing dissipation.

Even better though is to have a circuit design that does not require the dissipation of large amounts of heat.

8. ### BlackMelon

188
5
Aug 7, 2012
The MOSFET is used in a boost converter as a switching device. The reason why I set output power so high is that I want the circuit to be able to power a scooter. (It can be used in the other application but scooter is just an idea to imagine the amount of power)

I've ever build a robot which can carry 1 person with two 12Vdc motor. I measured output current and that was 6A. So the power required for carrying 1 person is (12*6)*2=144W (I'm 70kg at that time). Someone might need my converter for boosting 12Vdc to 48Vdc motor scooter at this power rating.

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
As suspected the power dissipated in the mosfet will (or should be ) much lower than the power delivered to the motor.

If we assume your circuit can be 90% efficient, then a power of 144W (which is actually pretty tiny) delivered to the motor will result in dissipation in the mosfet of 15W. Actually the dissipation in the mosfet will be much lower because there are other components contributing to the losses.

10. ### BlackMelon

188
5
Aug 7, 2012
Thank you everyone, I'll recalculate the power again.

PS: If I sum up all dissipation of every component, It'll be equal to the power of the voltage source, won't it?

11. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Only if you're not taking any power out of the power supply!

The total power dissipation in all the components will be equal to the input power minus the output power.

12. ### BobK

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Jan 5, 2010
All true. What I meant is that it does not eliminate the need for a large heat sink of some kind, it just moves it somewhere more convenient.

bob

13. ### BobK

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Jan 5, 2010
Most of the power eventually shows up as heat in the motors and mechanical linkages that drive the scooter. Unless it is going uphill or accelerating, all of the power is eventually converted to heat.

The power dissipated by the MOSFET consists of two components.

When the MOSFET is on, it is I*I*R where I is the current and R is the on resistance of the MOSFET.

The rest of the power is during switching, when the MOSFET is briefly in linear mode, being partially on with a higher resistance.

Bob

14. ### BlackMelon

188
5
Aug 7, 2012
I've recalculated the amount of dissipation power (It takes me the whole day to do :v :v). What I've done is based on the equation
Pd=Psource-Pinductor-Pdiode-Pcap-Presistor (based on Kris' idea)

What I've done is to take a waveform of voltage of each component in the circuit to find an average voltage, current waveform to find an average current. (Need to do some integral stuff) Then Paverage=Iave*Vave..... one by one component. I found that Inductor and capacitor have a minus power which means it supplies small amount of power to the circuit... And finally I've got a dissipation power (Pd) around 39.019W

Do a dissipation calculation again and finally got θsink-to-ambient=1.70357 Celsius/Watt but I don't actually know its dimension. Does anyone has a table presented a relation between dimension and the sink-to-ambient thermal resistance?

15. ### BobK

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Jan 5, 2010
Still wrong. The average power is not the average voltage * the average current. It is the integral of the voltage * current over time.

Say you have a MOSFET that has on resistance of 1Ω (which would be very high for a power MOSFET) switching a 100V supply to a load.

It is on 50% of the time and 10A is flowing so the voltage across it is 10V.
If is off 50% of the time and 0A is flowing and the voltage across it is 100V

If you simply average the voltage and current you get 55V and 5A for 275W.

But that is incorrect. You have to take each segment separately, then average the power.

By that method you get 50% times 100W + 50% times 0W = 50W which is the correct answer.

Bob

KrisBlueNZ likes this.
16. ### BlackMelon

188
5
Aug 7, 2012
Thanks BobK, I will look at the waveform of current and voltage segment by segment carefully to do the integral.

I've found new problem. According to http://www.ti.com/lit/an/slva462/slva462.pdf, there is the equation in page 3:

Volume(heatsink) = volumetric resistance (Cm3°C/W)/thermal resistance θSA (°C/W).

It says that the more thermal resistance heat sink is, the smaller it will be... (sounds unusual)

I really want to test my understanding now.