After reading posts that talked about chipamps running hot being answered with: "It's probably oscillating" I have wondered many times how could that be? Can an amplifier, which is obviously working and sounding well be oscillating so strongly that it heats up significantly?
After a few threads about heatsinking during the last days, I found there may be a different reason for hot chipamps than oscillation. And while I myself have counseled other members to use the Overture Design Guide to determine their heatsinks, I am herewith warning everybody to simply take the numbers from that tool without further thought. There are two reasons for that.
So let me try to shed some light on heatsinking and the correct heatsink choice.
It is probably not necessary to explain, why a heatsink is needed, so I shall start with how much power a heatsink has to dissipate.
This is the power dissipation formula
Pd = Vrailtot²/(2*PI²*Rl]
followed by an example to make things more clear. If you have 24 V rails and an 8 Ohm speaker your heatsink has to dissipate
Pd = 48²/(2*PI²*8) = 14,6 W
Now there are two temperatures that are important. The IC junction temperature and the heatsink temperature. No, they are not the same.
The LM3886T with 24 V rails, 8 Ohm load and a 2 K/W heatsink may serve as example. Its thermal resistance is 1 K/W. Thermal grease and washers can be assumed as 0,2 K/W.
The heatsink's DeltaT would be
2 K/W * 14,6 W = 29,2 K
The heatsink temperature would be
Tamb + 29,2 K, e. g. 49,2 °C, if the ambient temperature is 20 °C.
The limit should be 60 °C, if you use the heatsink outside of the case. This heatsink would be fine up to an ambient temperature of 30,8 °C. Just okay for a not too hot summer day.
The IC's Delta T would be
(1 K/W + 0,2 K/W + 2 K/W) * 14,6 W = 46,72 K
The IC temperature would be
Tamb + 46,72 K, e. g. 69,72 °C, if the ambient temperature is 20 °C.
The limit is 150 °C before the SPiKe protection system starts to react. It is wise to reduce that temperature by a safety margin of 25 °C. On that conditions this heatsink would still be fine up to 78,28 °C.
Take into account that the temperature inside of a case will be higher than outside of it. So mounting the heatsink inside will allow you to use a higher heatsink temperature at the cost of having to calculate with a higher ambient temperature.
Now compare that to the recommendation from the Overture Design Guide: 6,92 K/W.
Heatsink's DeltaT 6,92 K/W * 14,6 W = 101,032 K -> 121,032 °C for 20 °C ambient temperature. Far too hot to touch.
IC's DeltaT (1 K/W + 0,2 K/W + 6,92 K/W) * 14,6 W = 118,552 K -> 138,552 K for 20 °C ambient temperature. Above 31,448 °C the SPiKe protection system will start to bother you. And there is reason to believe that it will reduce power even before that. Transients may heat the IC up to above average for short moments, tripping SPiKe to reduce power.
Now you may wonder, why I repeatedly used the term "temperature would be" before the above calculations. The reason is that a heatsink with 2 K/W at 14,6 W Pd may not exist. The following diagram from Alutronic's catalog 2008 belongs to a heatsink that is sold by an electronic dealer as 1,8 K/W with a cross section of 100 x 40 mm and 50 mm length.
You will not find the thermal resistance for 14,6 W, because the table only goes from 20 to 60 W. At 40 K it is specified with 1,81 K/W. Let us pretend the characteristic were a straight line. That is not exactly correct, but sufficiently exact for out purpose. Just to get the idea.
Thermal resistance at 20 W is 2,07 K/W.
Thermal resistance at 40 W is 1,64 K/W.
So the change in thermal resistance per W is (2,07-1,64)/(60-20) = 0,01075.
20 W - 14,6 W = 5,4 W.
So the thermal resistance of this heatsink at 14,6 W is approximately 2,07 + 5,4 * 0,01075 = ~2,13 K/W.
While you must have expected a heatsink 10 % better than needed, you actually got one that is 6,5 % worse. So instead of decreasing the calculated temperatures about 3 K as expected, they would actually be increased about 2 K. And the bigger the heatsink, the more it will deviate from the published specs, because bigger heatsinks will be measured and specified with higher Pd.
Conclusion.
After a few threads about heatsinking during the last days, I found there may be a different reason for hot chipamps than oscillation. And while I myself have counseled other members to use the Overture Design Guide to determine their heatsinks, I am herewith warning everybody to simply take the numbers from that tool without further thought. There are two reasons for that.
- - The heatsinks derived from the Overture Design Guide or from the tables in the datasheet are absolute maximum ratings. An adequate heatsink must be significantly bigger, i. e. have a lower thermal resistance than that.
- If you buy a heatsink, a thermal resistance is given. Yet this number may not be significant for the task at hand, because a heatsinks thermal resistance decreases with increasing temperature. The trouble with heatsinks is, there is no norm on how to measure or specify their thermal resistance. So, as long as you do not know at which temperature the thermal resistance was measured, that figure is useless.
So let me try to shed some light on heatsinking and the correct heatsink choice.
It is probably not necessary to explain, why a heatsink is needed, so I shall start with how much power a heatsink has to dissipate.
This is the power dissipation formula
Pd = Vrailtot²/(2*PI²*Rl]
followed by an example to make things more clear. If you have 24 V rails and an 8 Ohm speaker your heatsink has to dissipate
Pd = 48²/(2*PI²*8) = 14,6 W
Now there are two temperatures that are important. The IC junction temperature and the heatsink temperature. No, they are not the same.
The LM3886T with 24 V rails, 8 Ohm load and a 2 K/W heatsink may serve as example. Its thermal resistance is 1 K/W. Thermal grease and washers can be assumed as 0,2 K/W.
The heatsink's DeltaT would be
2 K/W * 14,6 W = 29,2 K
The heatsink temperature would be
Tamb + 29,2 K, e. g. 49,2 °C, if the ambient temperature is 20 °C.
The limit should be 60 °C, if you use the heatsink outside of the case. This heatsink would be fine up to an ambient temperature of 30,8 °C. Just okay for a not too hot summer day.
The IC's Delta T would be
(1 K/W + 0,2 K/W + 2 K/W) * 14,6 W = 46,72 K
The IC temperature would be
Tamb + 46,72 K, e. g. 69,72 °C, if the ambient temperature is 20 °C.
The limit is 150 °C before the SPiKe protection system starts to react. It is wise to reduce that temperature by a safety margin of 25 °C. On that conditions this heatsink would still be fine up to 78,28 °C.
Take into account that the temperature inside of a case will be higher than outside of it. So mounting the heatsink inside will allow you to use a higher heatsink temperature at the cost of having to calculate with a higher ambient temperature.
Now compare that to the recommendation from the Overture Design Guide: 6,92 K/W.
Heatsink's DeltaT 6,92 K/W * 14,6 W = 101,032 K -> 121,032 °C for 20 °C ambient temperature. Far too hot to touch.
IC's DeltaT (1 K/W + 0,2 K/W + 6,92 K/W) * 14,6 W = 118,552 K -> 138,552 K for 20 °C ambient temperature. Above 31,448 °C the SPiKe protection system will start to bother you. And there is reason to believe that it will reduce power even before that. Transients may heat the IC up to above average for short moments, tripping SPiKe to reduce power.
Now you may wonder, why I repeatedly used the term "temperature would be" before the above calculations. The reason is that a heatsink with 2 K/W at 14,6 W Pd may not exist. The following diagram from Alutronic's catalog 2008 belongs to a heatsink that is sold by an electronic dealer as 1,8 K/W with a cross section of 100 x 40 mm and 50 mm length.
You will not find the thermal resistance for 14,6 W, because the table only goes from 20 to 60 W. At 40 K it is specified with 1,81 K/W. Let us pretend the characteristic were a straight line. That is not exactly correct, but sufficiently exact for out purpose. Just to get the idea.
Thermal resistance at 20 W is 2,07 K/W.
Thermal resistance at 40 W is 1,64 K/W.
So the change in thermal resistance per W is (2,07-1,64)/(60-20) = 0,01075.
20 W - 14,6 W = 5,4 W.
So the thermal resistance of this heatsink at 14,6 W is approximately 2,07 + 5,4 * 0,01075 = ~2,13 K/W.
While you must have expected a heatsink 10 % better than needed, you actually got one that is 6,5 % worse. So instead of decreasing the calculated temperatures about 3 K as expected, they would actually be increased about 2 K. And the bigger the heatsink, the more it will deviate from the published specs, because bigger heatsinks will be measured and specified with higher Pd.
Conclusion.
- - Don't rely on the Overture Design Guide or the datasheet diagrams alone.
- Try to find out the heatsink's thermal resistance for the actual Pd you need.
- Don't save on heatsinks, even if you don't always run the amplifier on worst case conditions. Some day you will.
I very humbly submit additional reasons: The nature of the load, and the operating environment.
Confining this to audio amps - They usually are loaded with reactive circuits, not resistive loads. Reactive circuits introduce current phase shift and varying current demands with frequency. It is no secret that some amps do not operate well with some speaker loads. Owners of speakers that present a "capacitive" can have serious amp issues.
* The discussion of work of Matti Otala and the subject of EPDR ( on another forum ) between myself and an amp designer caused a large debate, which I don't want to cause here. But I feel it definitely is related to the subject at hand: Amp stability and heatsink requirements.
See:
http://stereophile.com/reference/707heavy/index.html
http://stereophile.com/reference/707heavy/index1.html
http://stereophile.com/reference/707heavy/index2.html
http://sound.westhost.com/patd.htm
http://sound.westhost.com/soa.htm
Hi,
that Stereophile report is an eye opener.
I thought that using maximum current <=Vpk/RLoad/0.35 was a reasonable maximum that covered severe speaker loads.
There's evidence that the 0.35 could be as bad as 0.2 to 0.25 for severe reactance speakers and much worse for electrostatic speakers.
that Stereophile report is an eye opener.
I thought that using maximum current <=Vpk/RLoad/0.35 was a reasonable maximum that covered severe speaker loads.
There's evidence that the 0.35 could be as bad as 0.2 to 0.25 for severe reactance speakers and much worse for electrostatic speakers.
AndrewT
Over the course of years Jon Atkinson presented his measurements of speaker systems that included impedance/phase graphs. He usually provides a synopsis of it. There is a striking relationship between reactive loads that present extreme - phase values @ low values of impedance and how severe a load it is deemed to present to an amp.
Low impedance and extreme neg shift, you need lots of current. Current equals heat. A 10 degree ( C) rise in component temp results in reduction of life by 50%.
Perhaps the biggest point of contention to some, is not If EPDR is an issue, but the degree to which it should be considered an issue.
Having owned the same amps for years and running many different loads on them you notice that the amp may run hotter on some loads, which I attribute in large part to the widely varying reactive loads, even though they maintain the same nominal impedance.
I contended that if I want to torture an amp I could place a large woofer in a tuned cab and tune it to produce a large neg phase with a low impedance at close to line frequency, and feed it a signal rich in content @ that frequency 50/60Hz. Also some amps should not be used with piezos, or at least very carefully
Some argue that this doesn't represent "reality". But since all complex waves can be assembled by individual sine waves, It then becomes a matter of how much of the wave is at those "danger frequencies".
The same large woofer in a closed box presents a very easy load with it's soft phase curve that has much less shift, and less drift from zero degrees, closer to a stable resistive load.
Ported and tuned properly the neg phase shift doesn't occur at an impedance minima.
The super long straight horns 30' + can have very little phase shift.
Syd
Over the course of years Jon Atkinson presented his measurements of speaker systems that included impedance/phase graphs. He usually provides a synopsis of it. There is a striking relationship between reactive loads that present extreme - phase values @ low values of impedance and how severe a load it is deemed to present to an amp.
Low impedance and extreme neg shift, you need lots of current. Current equals heat. A 10 degree ( C) rise in component temp results in reduction of life by 50%.
Perhaps the biggest point of contention to some, is not If EPDR is an issue, but the degree to which it should be considered an issue.
Having owned the same amps for years and running many different loads on them you notice that the amp may run hotter on some loads, which I attribute in large part to the widely varying reactive loads, even though they maintain the same nominal impedance.
I contended that if I want to torture an amp I could place a large woofer in a tuned cab and tune it to produce a large neg phase with a low impedance at close to line frequency, and feed it a signal rich in content @ that frequency 50/60Hz. Also some amps should not be used with piezos, or at least very carefully
Some argue that this doesn't represent "reality". But since all complex waves can be assembled by individual sine waves, It then becomes a matter of how much of the wave is at those "danger frequencies".
The same large woofer in a closed box presents a very easy load with it's soft phase curve that has much less shift, and less drift from zero degrees, closer to a stable resistive load.
Ported and tuned properly the neg phase shift doesn't occur at an impedance minima.
The super long straight horns 30' + can have very little phase shift.
Syd
Great Post
Above you wrote:
I seem to remember that from the National 3886 pdf. But what I found odd is that I couldn't find a distinction made in the thermal sections of that document concerning the different packages. In other words, do 1 and .2 go equally for the isolated and un-isolated packages? I think not.
I would be willing to bet that Theta jc is higher than 1 for the isolated package.
However, the 4780 which only comes in a single package has a Theta jc of .8, so ...?
Any word on this; please correct me if I'm wrong.
Again, thanks for the great post.
Above you wrote:
I seem to remember that from the National 3886 pdf. But what I found odd is that I couldn't find a distinction made in the thermal sections of that document concerning the different packages. In other words, do 1 and .2 go equally for the isolated and un-isolated packages? I think not.
I would be willing to bet that Theta jc is higher than 1 for the isolated package.
However, the 4780 which only comes in a single package has a Theta jc of .8, so ...?
Any word on this; please correct me if I'm wrong.
Again, thanks for the great post.
About inside mounted heatsink
I have always mounted heatsinks outside the box, wouldnt have it any other way
But with my last build I had to consider inside heatsinks due to the layout I wanted to do
I thought about it and being a rather smallish AB biased amp I reckoned it could be done properly, and commercial amp designers do it all the time
But I wanted the amp to handle heavy loads and low impedance, so I knew I should be very careful
Fore one the heatsinks are oversized, and could handle amps with twice the wattage
But I also made the heatsink actually stick out through the box bottom, not just small holes but really sticking through the bottom plate
And further the exit holes in the top plate was positioned right above the heatsink
This way I created a convection cooling
All the air suched into the box passes through the heatsink bottom
And all the air that leaves the box will also be right at the heatsink
Doing this I enhanced the air shift inside the box
And I wouldnt be surpriced if this is actually more effektive than an outside mounted heatsink
It may be slightly hotter, but more stable constant temperature
Another thing is that heatsinks are really never completely plane, and should fore an optimal result be machined grinded
btw, my heatsinks mostly never gets hot, not even at full blast into lowish impedance
I have always mounted heatsinks outside the box, wouldnt have it any other way
But with my last build I had to consider inside heatsinks due to the layout I wanted to do
I thought about it and being a rather smallish AB biased amp I reckoned it could be done properly, and commercial amp designers do it all the time
But I wanted the amp to handle heavy loads and low impedance, so I knew I should be very careful
Fore one the heatsinks are oversized, and could handle amps with twice the wattage
But I also made the heatsink actually stick out through the box bottom, not just small holes but really sticking through the bottom plate
And further the exit holes in the top plate was positioned right above the heatsink
This way I created a convection cooling
All the air suched into the box passes through the heatsink bottom
And all the air that leaves the box will also be right at the heatsink
Doing this I enhanced the air shift inside the box
And I wouldnt be surpriced if this is actually more effektive than an outside mounted heatsink
It may be slightly hotter, but more stable constant temperature
Another thing is that heatsinks are really never completely plane, and should fore an optimal result be machined grinded
btw, my heatsinks mostly never gets hot, not even at full blast into lowish impedance
Information Source
Good advice can be hard to follow:
One thing that Aavid provides is a plot of resistance as a function of Pd. Attached is one from a model specified at a resistance of .44. It is the lowest in a very extensive product line.
Note well that over a broad range of Pd figures, it never reaches .44!
That notwithstanding, the Aavid web site has so many extrusion profiles it will likely have something close to whatever profile another mfg. might offer, wherever resistance as a function of Pd is not given. Then it would be possible to get close.
(If I were to construct a catalog search for heatsink profiles, it would start with Pd, an ideal sink temperature rise above ambient, and an upper limit for temperature rise above ambient.)
Good advice can be hard to follow:
One thing that Aavid provides is a plot of resistance as a function of Pd. Attached is one from a model specified at a resistance of .44. It is the lowest in a very extensive product line.
Note well that over a broad range of Pd figures, it never reaches .44!
That notwithstanding, the Aavid web site has so many extrusion profiles it will likely have something close to whatever profile another mfg. might offer, wherever resistance as a function of Pd is not given. Then it would be possible to get close.
(If I were to construct a catalog search for heatsink profiles, it would start with Pd, an ideal sink temperature rise above ambient, and an upper limit for temperature rise above ambient.)
Attachments
Hi Toolr,
look again, that graph is for low power dissipation values.
Take one pair of coordinates:- 95W and 0.5C/W.
That gives a deltaT of 47.5Cdegrees.
Most heatsinks are rated for a deltaT of 70 to 80Cgdrees. It is also rated for the whole of the heat contact face isothermally at heatsink temperature.
Both of these parameters are difficult to achieve in practice.
Most manufacturers will give data on how to re-rate for temperature heated face size, device location, elevation, air speed, etc.
But, back to that 0.44C/W an assume this is rated for deltaT=70degC.
The dissipated power applied evenly over the whole surface will be 70 / 0.44 = 159W.
But any group of devices dissipating a total of 159W will be running with a Tc >= Ta + deltaT + Rth c-s * [Pd/device].
Let's put some real numbers to this.
4devices with Rth c-s = 0.6C/W and ambient air passing the heatsink = 30degC will run at Tc>=123.85degC.
This is obviously running very hot and that's why that graph shows data for low power dissipation.
look again, that graph is for low power dissipation values.
Take one pair of coordinates:- 95W and 0.5C/W.
That gives a deltaT of 47.5Cdegrees.
Most heatsinks are rated for a deltaT of 70 to 80Cgdrees. It is also rated for the whole of the heat contact face isothermally at heatsink temperature.
Both of these parameters are difficult to achieve in practice.
Most manufacturers will give data on how to re-rate for temperature heated face size, device location, elevation, air speed, etc.
But, back to that 0.44C/W an assume this is rated for deltaT=70degC.
The dissipated power applied evenly over the whole surface will be 70 / 0.44 = 159W.
But any group of devices dissipating a total of 159W will be running with a Tc >= Ta + deltaT + Rth c-s * [Pd/device].
Let's put some real numbers to this.
4devices with Rth c-s = 0.6C/W and ambient air passing the heatsink = 30degC will run at Tc>=123.85degC.
This is obviously running very hot and that's why that graph shows data for low power dissipation.
Clarification
AndrewT,
The tenor of my post was a caveat, so I think we are more on the same page than your response indicates in my reading of it.
You write:
My point was that .5 is greater than .44, obviously; that is, ignore the .44 figure!
Let's add some more numbers to the mix. On the assumption that you have a sink with a resistance of .44, you might think an attempt to dissipate 20 Watts will give you almost 9 degrees C above ambient, when in fact at that Pd you will have a resistance of .75, and a rise above ambient of 15 degrees!
Again, as pacificblue pointed out, you need the figure for YOUR Pd. This can be hard to obtain. Plots such as the one above can help you do that.
But what about off the chart?
I think the assumption that .44 C/W is rated for a deltaT=70degC is risky; I would bet deltaT is higher still for the .44 figure, way higher. Still I agree entirely that we are dealing here with "running very hot," as you put it, and whether the plot covers "low power" I leave to future design(er)s. Frankly I think .44 is absurd; I suppose I should have said that outright. I can't see why anyone would publish such a figure at all. That figure sounds like a heat sink of this type at 200C located in a sub zero freezer.
BTW it is part # 62335.
AndrewT,
The tenor of my post was a caveat, so I think we are more on the same page than your response indicates in my reading of it.
You write:
My point was that .5 is greater than .44, obviously; that is, ignore the .44 figure!
Let's add some more numbers to the mix. On the assumption that you have a sink with a resistance of .44, you might think an attempt to dissipate 20 Watts will give you almost 9 degrees C above ambient, when in fact at that Pd you will have a resistance of .75, and a rise above ambient of 15 degrees!
Again, as pacificblue pointed out, you need the figure for YOUR Pd. This can be hard to obtain. Plots such as the one above can help you do that.
But what about off the chart?
I think the assumption that .44 C/W is rated for a deltaT=70degC is risky; I would bet deltaT is higher still for the .44 figure, way higher. Still I agree entirely that we are dealing here with "running very hot," as you put it, and whether the plot covers "low power" I leave to future design(er)s. Frankly I think .44 is absurd; I suppose I should have said that outright. I can't see why anyone would publish such a figure at all. That figure sounds like a heat sink of this type at 200C located in a sub zero freezer.
BTW it is part # 62335.
the reason the manufacturers quote that figure is that they have all agreed on a similar deltaT of around 70 to 80Cdegrees. We as builders are supposed to be aware of the standard and what we need to do to make use of the standard information.
We as users MUST de-rate the standard Rth s-a to suit what we require and to suit the operating conditions.
I habitually multiply Rth s-a by 1.33 as a first guess at dissipation ability. Find out deltaT and then adjust Rth s-a more accurately once I have an idea at the operating temperatures.
This is part of the reason I recommend our beginners to double the dissipation requirement that National suggest in their design tables.
If National's table says use 4C/W for Ta=25degC then consider your design using around 2C/W.
We as users MUST de-rate the standard Rth s-a to suit what we require and to suit the operating conditions.
I habitually multiply Rth s-a by 1.33 as a first guess at dissipation ability. Find out deltaT and then adjust Rth s-a more accurately once I have an idea at the operating temperatures.
This is part of the reason I recommend our beginners to double the dissipation requirement that National suggest in their design tables.
If National's table says use 4C/W for Ta=25degC then consider your design using around 2C/W.
A model proposal
Point taken; here's another pass at the issue.
In an article on heatsink construction, the model
50/sqrt A
is proposed (http://w1.859.telia.com/~u85920178/begin/heat-0.htm).
This give very conservative figures...in the case before us thus far, we can use the given surface area to obtain a resistance of .91, a figure that corresponds to dissipating less than 10 watts and a delta T of under 10 deg. C.
By altering the constant in direct proportion, we can get different figures. Instead of 50, for around 20 watts/delta T 15 deg. C use 40, for around 60watts/delta T 35, use 33.
I wish I could get a function in there that would really predict what will happen from just wattage!
Of course 1.33 and 2 times are easier!
AndrewT said:
Point taken; here's another pass at the issue.
In an article on heatsink construction, the model
50/sqrt A
is proposed (http://w1.859.telia.com/~u85920178/begin/heat-0.htm).
This give very conservative figures...in the case before us thus far, we can use the given surface area to obtain a resistance of .91, a figure that corresponds to dissipating less than 10 watts and a delta T of under 10 deg. C.
By altering the constant in direct proportion, we can get different figures. Instead of 50, for around 20 watts/delta T 15 deg. C use 40, for around 60watts/delta T 35, use 33.
I wish I could get a function in there that would really predict what will happen from just wattage!
Of course 1.33 and 2 times are easier!
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.