I know this is a classic one, but I have not been able to find a simple answer in this boards archive, neither in the applications notes at Ti.com. I have stumbled across various methods of determining heatsink rating in applications notes on ti.com but all required knowledge and understanding I do not posess (incorrectly spelled?).
I hope someone experienced with chip amps and class a/b amps in generel can provide me with an answer.
I need to heatsink two OPA549 running at 25v rails. They will be powering drivers rated 8 ohm nominally but with impedance ranging from 5.5 ohms to 25ohm. Phase shift for the bass/midrange will not exceed 35 degrees (approx.).
We are not talking party spl.
How much heat should I expect generated by the two ÌCs? The one powering the bass/midrange will be carrying a much greater load than the tweeter amp, since the crossover frequency will be somewhere between 3500-4000hz.
I hope someone is able to provide a simple answer.
I hope someone experienced with chip amps and class a/b amps in generel can provide me with an answer.
I need to heatsink two OPA549 running at 25v rails. They will be powering drivers rated 8 ohm nominally but with impedance ranging from 5.5 ohms to 25ohm. Phase shift for the bass/midrange will not exceed 35 degrees (approx.).
We are not talking party spl.
How much heat should I expect generated by the two ÌCs? The one powering the bass/midrange will be carrying a much greater load than the tweeter amp, since the crossover frequency will be somewhere between 3500-4000hz.
I hope someone is able to provide a simple answer.
Well, lets see now:
If we make some basic assumptions about the chips performance, say that at best it will start to clip when the output is about 3V from the rail voltage, and as this is the heatsink, we will want the worst case (i.e. maximum) output current in your aplications (will be about 4A peak in your case at a minimum 5.5 ohm load).
So, your rough peak output power will be (25-3)*4 = 88W and the RMS will be roughly half of this at about 44W per chip.
Now, assuming the chips are 60% efficient, this means that they will disipate around 30W (this is an absolute maximum here, as it is far better to over estimate this than to under estimate).
If you want to keep the chip so that it will only rise at most say 40 degrees above the ambient temp, then the thermal resistance of the heatsink, and chip/heatsink contact must be equal to or less than; 40/30 = 1.3deg/W.
Now, this is assuming absolute worst case (the amp driving a constant sine wave at the frequency which corresponds to the 5.5 ohm dip in load impedance), so you should be able to get away with a fair bit less than this in real life. Personaly, I would try and go with something around this or a bit better to run the two chips on at the same time (idealy a bit better than this), but I doub't you would ever need to go better than around 0.8-1.0 deg/W for both chips.
One other thing you should note is that this is only how I would go about looking at my own equipment, many other people would consider this as being too littel or too much heatsinking, but hopefuly they will also post their oppinions here (or a more scientific way of looking at things, which can generaly be aproximated if you keep in a bit extra just in case)
If we make some basic assumptions about the chips performance, say that at best it will start to clip when the output is about 3V from the rail voltage, and as this is the heatsink, we will want the worst case (i.e. maximum) output current in your aplications (will be about 4A peak in your case at a minimum 5.5 ohm load).
So, your rough peak output power will be (25-3)*4 = 88W and the RMS will be roughly half of this at about 44W per chip.
Now, assuming the chips are 60% efficient, this means that they will disipate around 30W (this is an absolute maximum here, as it is far better to over estimate this than to under estimate).
If you want to keep the chip so that it will only rise at most say 40 degrees above the ambient temp, then the thermal resistance of the heatsink, and chip/heatsink contact must be equal to or less than; 40/30 = 1.3deg/W.
Now, this is assuming absolute worst case (the amp driving a constant sine wave at the frequency which corresponds to the 5.5 ohm dip in load impedance), so you should be able to get away with a fair bit less than this in real life. Personaly, I would try and go with something around this or a bit better to run the two chips on at the same time (idealy a bit better than this), but I doub't you would ever need to go better than around 0.8-1.0 deg/W for both chips.
One other thing you should note is that this is only how I would go about looking at my own equipment, many other people would consider this as being too littel or too much heatsinking, but hopefuly they will also post their oppinions here (or a more scientific way of looking at things, which can generaly be aproximated if you keep in a bit extra just in case)
bigparsnip, that is very informative and all good info but looks pretty similar to what one would see from reading app notes from national or TI or something... I think what we are looking for here in terms of a simple answer is "a X by X by X aluminum/steel/whatever?" chunch of metal is big enough... anyone know an easy conversion factor or rule of thumb here?
bigparsnip said:
So if the calculation is correct you would look for a heatsink with this rating.
rayban68 said:
I think the issue is that there are a lot of factors in a heatsink that contribute to the ability for it to get rid of heat. Size, number of fins, fin orientation, and the finish all have effects. So it is hard to say without knowing a few things about the heatsink.
If you see a heatsink and it has no specs use the heatsink calculator on the link below to estimate it's deg/W rating. Click on the heatsink.zip link. You will also need the VB40032.dll on the same page I believe.
http://sound.westhost.com/download.htm
There is also a heatsink article on the site and a temperature activated fan circuit in case you undersize the heatsink
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