# what is the tempreture at which the thermal limit of an amplifier is reached

Introduction:

I am upgrading a "power amplifier" that a fellow engineer has designed , my job is to add thermal protection against failure in prolonged use (or operator miss use).

The package of the transistors used is To-247-3, the metal area is 140mm2 according to :infineon the interface material (silicon pad , insulator and grease free) used has thermal impedance of 0.23°C-in2/W (@50 psi) , the thickness, conductivity and other parameters are provided in this table :

The tempreture sensor has a lug mount and will be screwed into the heatsink directly close to the transistor similar to this amplifier.

The power dissipated by each transistor is 5W , the maximum junction temperature is 175 degree Celsius and Rjunc/case is 0.29 degree / W.

I think we should keep the junction temperature below 150 to be a bit away from the absolute maximum.

Now for calculations :

1. 140mm2 = 0.217 inch2 , so sil-pad resistance is 0.23/0.217= 1.05 degree C/ watt.
2. total resistance from junction to sink is 1.05+0.29= 1.34 .
3. delta T = 1.34*5w= 6.74 degree C . So the heat sink is allowed to reach 150-6.74= 143.25 degree Celsius ?!!

EDIT: due to comments and answers i would like to add the following , the transistos used are mosfets used switched 45 A in half bridge topology at 20khz from a decent gate driver to minimize switching losses , the conduction losses are : RxIxI = 0.0022 x 45 x45 = approximatly 5W .

Rds(on) MAX is found from this curve at 125 degree .

QUESTION: I feel the numbers are not adding up (because other parts on the pcb would be toasted if the case is at such a high tempreture) please correct my calculations and tell me what i am missing ? and state what is usually the case temperature at which the thermal limit is reached ??

• How do you know that is only 5 watts ? What is the worst possible ambient temperature around the Heatsink ? What is the thermal resistance of the heatsink ? Jun 30, 2016 at 16:55
• i have added how i got the 5w figure , no regarding the heatsink thermal resistance , i donot have this number unfortuanatly ( but it is a whole enclosure aluminum which weighs 1kg and has some fins , the maximum ambient is 40 to 45 degree celeius . Jul 1, 2016 at 12:08
• but why is the heatsink resistance relevent and the ambient , the tempreture sensor is placed nexted to the power transistor , so we can know exactly the heatink tempreture and go backward to the junction. the whole idea of using the temp sensor is to bypass the heatink thermal caluculations and different ambient conditions . Jul 1, 2016 at 12:11
• The temperature sensor is for safety, but the heatsink must be properly dimensioned for the worst ambient and dissipation conditions. How do you plan to do it ? Jul 1, 2016 at 12:21
• please read the introduction : "I am upgrading a "power amplifier" that a fellow engineer has designed , my job is to add thermal protection against failure in prolonged use ( or if operator miss use)." , another engineer have designed the product and have done his testing and it is supposed to be working correctly. Jul 1, 2016 at 13:19

Going by the picture (if that's the amplifier under question), that's a pretty large heatsink for 4 MOSFETs dissipating 5W each. It looks like there's other heat sources attached to the heatsink, however, and although you don't say what these are dissipating, or what the thermal impedences from other heat sources to MOSFETs-in-question are, I'm tempted to say the they (in combination with the heatsink) are as likely to heat up your MOSFETs as cool them!

Another thing missing in the calculation of temp at the temp sensor is the thermal resistance from furthest transistor to temp sensor (and assuming no other thermal sources are at work here). Heat sinks are good, but not perfect, heat conductors, and at least some of the transistor heat will be flowing past the temp sensor to other parts of the heat sink. However, this in unlikely to alter your conclusions much.

As was pointed out by others, 5W is not a great deal of power to be dissipating in a transistor, and in this case the transistor may well not be the operative thermal limit. You might consider the poor user, who certainly would not want to handle an amplifier with a 150˚C heatsink! Health and safety concerns may well dictate a much lower thermal cutoff than 150˚C, and I would suggest that at any thermal cutoff higher than 100˚C, you are "playing with fire", health and safety-wise.

Other than the above-mentioned 2 unaccounted thermal considerations, I think you're calculations are OK. You are right that you don't want to be cooking the other PCBs with a 150˚C temp, and most of your other components are likely rated at something like a max temp of 75˚C. Given that, and the H&S consideration, you might even consider setting 75˚C as your thermal cutout temp.

• +1 for including ALL heat sources sharing the same heatsink. Ambient temperature also plays a major factor if not controlled. Right side mosfets look like DC-DC converter. Back wall is rectifiers. Left side is audio outputs. That is all accumulated heat.
– user105652
Jul 3, 2016 at 0:56
• Last time I had this problem, 75 degrees was for poor thermal conductors like plastic which where allowed to go higher (same principle as walking on glowing coal) but for good conductors like steel and aluminium, the limit for when a yellow CAUTION - HOT SURFACE sticker was needed was 65 degrees. Jul 3, 2016 at 7:19
• thanks a lot Andy , Sparky and Winny , this exactly the kind of feedback i was looking for . just for clarification , this is not the amplifier i am working with , this is just something similar . so now thanks to you guys i know that the tempreture sensor should cut off at 65-75 degree maximum .That means i can push more power of of the transistor:)) yess :)) the lower limit will be something like 45-55 degree for hysterisis. i will keep the bounty open just incase others want to throw more help and experience . Jul 3, 2016 at 14:09

## Power v. Temperature correlation...

What I think you're missing is that the transistor of that size and power rating is not likely to reach 150deg C at only a 5W output. That is, you will not simultaneously have both of those conditions true.

Therefore, your subtraction is not a true statement. You think:

$$T_{case}(5W) + R_{system}(5W) = 150^oC$$

when in reality, it's probably something like...

$$T_{case}(50W) + R_{system}(50W) = 150^oC$$

Essentially, there is another function in the system of equations (the relationship between power and temperature) and you are asserting that two things with a relationship are independent when they are not.

## Why bother with this?

It seems like you're approaching a complex thermal situation, where you neither have all of the relevant thermal models, nor all of the relevant details about construction and operation, by attempting to apply math over any datasheet parameter that seems likely relevant. This isn't likely to work.

## I propose an alternative: measure it.

1. Create a worst-case ambient environment.

2. Load the amplifier to its maximum designed output power

3. Measure the temperature of the case of the transistors

Repeat this heat-cool process 10 times on 10 units.

## The 3 Regimes

The results of your 100 trials will tell which of these are true:

• No problem -- The maximum temperature observed was well below your design limit of 125oC. In this case set the limit in production to the upper tolerance of your temperature sensor above the maximum observed in the 100 trials. This is probably what you'll see.

• Oh no! -- The maximum temperature observed was well above your design limit of 125oC. You will need to redesign the product. The design doesn't meet spec. This is very likely not the case for you, since you know the product has performed in the field successfully.

• Maybe a problem -- The maximum temperature observed was within range of your upper limit of 125oC. More detailed and precise analysis will need to be done. If this is what you see from your testing, it's now worth the effort of a deeper investigation or a redesign that improves your thermal operating margins.

• Excessive exponentiation? Jun 30, 2016 at 1:25
• To the OP: in other words your calculations are correct for the case where you are only dissipating 5W. But I don't know where you came up with that figure. Unless it is a class D amplifier, or a very quiet one, the transistor will probably dissipate more than 5W. Jun 30, 2016 at 6:31
• It is a Class D amplifier of sorts, as can be infered from ElectronS' question: "used switched 45 A in half bridge topology at 20kHz from a decent gate driver to minimize switching losses " Jul 2, 2016 at 22:16
• @AndyW, two days ago, when I wrote my comment, the bit about the half bridge was not yet included in the question. Jul 3, 2016 at 6:01
• Ah. I see. You have my sympathies. That's the problem with editing questions… the answers and the question begin to diverge. A perfectly good answer becomes wrong after a question is edited and changed… I've had answers downvoted because of this. Jul 3, 2016 at 16:54