Hello!
I am planning on assembling a chipamp based on the lm4780
So far i have the pcb, the transformer (+-20V ), i have ordered the IC itself and i can get the various thingies easily (rectifiers, caps, cables, fuses, resistors)
the thing i am stuck with is the heatsink.
something tells me it has to be... quite substantial...
now, i have a heatsink that was used for a ~100W amplifier
this one:
The amp itself is getting the boot, so i am using its heatsink.
If i use it in vertical instead of horizontal orientation i think it can be a bit cooler. I mean, literally cooler, not "dude this is sooo coool".
instead of this:
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________
________
________
________
________
put it in this orientation:
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Also, i thought of an improvement: if i do several horizontal cuts, will the C/W ratio be improved?
I am planning on assembling a chipamp based on the lm4780
So far i have the pcb, the transformer (+-20V ), i have ordered the IC itself and i can get the various thingies easily (rectifiers, caps, cables, fuses, resistors)
the thing i am stuck with is the heatsink.
something tells me it has to be... quite substantial...
now, i have a heatsink that was used for a ~100W amplifier
this one:
An externally hosted image should be here but it was not working when we last tested it.
The amp itself is getting the boot, so i am using its heatsink.
If i use it in vertical instead of horizontal orientation i think it can be a bit cooler. I mean, literally cooler, not "dude this is sooo coool".
instead of this:
________
________
________
________
________
________
put it in this orientation:
||||| |||||
Also, i thought of an improvement: if i do several horizontal cuts, will the C/W ratio be improved?
The size of the heatsink you need depends on several variables, like the max. power the IC is expected to dissipate and the ambient temperature.
Total power dissipation depends on the load you're planning to drive and the power supply voltage.
On page 11 (12 according to the pdf-reader) of the datasheet, there are some graphs of total dissipated power vs output power per channel and PSU voltage for 4,6 and 8 ohm loads. Look up the graph that corresponds most to your setup.
You can also calculate PDmax with equation No. 2 on page 16 and multiply the result by 2.
Next go to page 17 (or 18) of the datasheet and read the section called "Determining the correct heat sink". In it you will find the equation for determining the size of the heatsink (No. 4). This may seem jibberish, but you know most variables already, they're given in the datasheet:
Tjmax = 150 degr. C;
Tamb = typically 25 degr. C, but given hot Greek summers, you might want to calculate with a slightly higher value for the ambient temp;
Pdmax = the value you calculated or looked up in the graph on page 11;
Өjc = typically 0.8 degr. C/W
Өcs = typically 0.2 degr. C/W with the use of thermal compound;
Өsa = this is value you need to calculate. The result will be in degr. C/W and is the thermal resistance of the heatsink to the air. You'll need a heatsink of this value or less.
E.g. with a PSU of +/- 30 V and speakers of 8 ohms, PDmax is 45.6 W (calculated, graph shows slightly higher). Lets assume a Tamb of 25 degr. C. That gives:
Өsa = [(150-25)-45.6*(0.8+0.2)]/45.6=
[125-45.6]/45.6=1.74 degr. C/W.
The heatsink in this setup needs to have a thermal resistance of 1.74 degr. C/W or less.
Look up the thermal resistance of your heatsink. Or if you don't know the brand and type of the heatsink, take measurements and look up the thermal resistance of a heatsink that resembles your's as close as possible. This will give you an estimate of the thermal resistance of your heatsink.
Actually, you should always use a heatsink in vertical orientation since the natural convection of warm air is upwards, not sideways. This is especially important when there's no forced airflow.
No, it will not be improved since you're taking away surface area. A smaller surface leads to a higher thermal resistance. Horizontal slits might actually interfere with the natural convection.
I just realized that if you assume Өjc+Өcs is typically 0.8 + 0.2=1 degr. C/W, Өja (which is Өjs+Өcs+Өsa) becomes Өsa +1.
The use of equation No.3, but with Өja substituted with Өsa+1 gives PDmax = (Tjmax - Tamb)/(Өsa+1).
This can be rewritten to Өsa = [(Tjmax-Tamb)/PDmax] - 1 to give the easiest calculation of the needed heatsink.
In above example:
Өsa = [(150-25)/45.6] -1 = 1.74 degr. C/W
The use of equation No.3, but with Өja substituted with Өsa+1 gives PDmax = (Tjmax - Tamb)/(Өsa+1).
This can be rewritten to Өsa = [(Tjmax-Tamb)/PDmax] - 1 to give the easiest calculation of the needed heatsink.
In above example:
Өsa = [(150-25)/45.6] -1 = 1.74 degr. C/W
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Rod Elliot did a nice article which will help you ascertain the thermal characteristics of your heatsink.
DIY Heatsink
All you need to do is calculate the surface area of your heatsink. If in doubt UNDERESTIMATE.
DIY Heatsink
All you need to do is calculate the surface area of your heatsink. If in doubt UNDERESTIMATE.
Then I don't understand the reason for you question in post #1. All you need to do now is find out the thermal resistance of your heatsink.
Two channels. The formula for given for PDmax is for one amp (channel). Since there are two in one package, I multiplied the result by two to get 45.6 W.
Graph 200586a8 (PSU = +/-30 V and RL = 8 ohms) shows the max. total dissipation peaks at 49 W at around 22.5 W per channel power output. At lower or higher power ouputs, the dissipation in the IC is less.
If you're using 4 ohm speakers, I'm not surprised you come up with a rather low required thermal resistance (huge heatsink). The graphs don't show anything over +/- 25 V PSU with 4 ohm speakers. This is probably why.
Depends on how much material you remove while making the slits. But even if the area is increased, it won't be by much and an upset airflow might actually make things worse.
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Lo-Fi manufacturers base their heatsink calculations on about 70% of maximum dissipation. This is perfectly valid as most people rarely use their systems at full volume.
This is fine until you decide to run the system at full volume.
For safety and longetivity - use the worse case calculations. For compactness you can use fans at full volume - no-one is going to hear a fan if the system is cranking out 100dB of music.
This is fine until you decide to run the system at full volume.
For safety and longetivity - use the worse case calculations. For compactness you can use fans at full volume - no-one is going to hear a fan if the system is cranking out 100dB of music.
Based on 20 VAC transformer with full wave rectification and 4 ohm speakers, I too get an Өsa of 0.7 degr. C/W.
+-20 V tranny, which means ~+-28 V dc, let's say +-30 V, maximum dissipation is 50 W and occurs between 20 W to 25 W of output power. Per channel, so total power dissipation is 100 W.
a more accurate Tamb is 30 C (I live in Greece. Right now it's 35 C)
so
θsa = [(150-30)/100] -1 = 0.2 C/W ... yeah... and that is without mica between heatsink and the IC
this doesn't apply here, because the lm4780 dissipates less power at full volume.
the fan might not be audible itself, but parasitics might creep in and get amplified. I am trying to build a high quality amplifier, I am not using any fan. Consider it a personality complex if you so wish.
The formula for PDmax is for one amp and needs to be multiplied by two to get the right answer for the IC. The graph, however, is total power dissipation and doesn't need to be multiplied by two.
i do not understand the graphs...
i would have felt stupid for not understanding such simple graphs... i would but as a physicist i must say the graphs are not at all clear.
Best guess so far is... for 40 W output per channel (that is, 40 W going to each of two speakers) the amp is dissipating 90 W as heat
I do not think it can dissipate only 45 W as heat when outputting 80 W to the speakers.
i would have felt stupid for not understanding such simple graphs... i would but as a physicist i must say the graphs are not at all clear.
Best guess so far is... for 40 W output per channel (that is, 40 W going to each of two speakers) the amp is dissipating 90 W as heat
I do not think it can dissipate only 45 W as heat when outputting 80 W to the speakers.
Not all power is dissipated in the IC, some actually goes to the speakers...
it can suck 40 W from the power supply and provide 45 W at the output? i think there is some misunderstanding somewhere, you, me, the datasheet, i dunno.
I guess "total power dissipation" is "total power the IC uses minus what goes to both speakers"
Ehh, no. Total power dissipation is what the IC needs to operate.
In the table for 8 Ohm speakers, the total power dissipation on +/- 30 V PSUs peaks at 22.5 W per channel output at about 49 W. So, the power drawn from the PSU = 2* 22.5 + 49 = 94 W. But ONLY the 49 W is heating up the IC, so you have to calculate the heatsink for those 49 W.
PDmax formula gives 22.8 W per channel, so a total of 45.6 W for the whole IC, very close to the 49 W in the graph.
In the table for 8 Ohm speakers, the total power dissipation on +/- 30 V PSUs peaks at 22.5 W per channel output at about 49 W. So, the power drawn from the PSU = 2* 22.5 + 49 = 94 W. But ONLY the 49 W is heating up the IC, so you have to calculate the heatsink for those 49 W.
PDmax formula gives 22.8 W per channel, so a total of 45.6 W for the whole IC, very close to the 49 W in the graph.
read what you wrote again, i don't think it makes sense.
if TPD is what the IC needs, then it wouldn't be the same as the heat dissipated, nor would you go and add the output power to that same figure.
so far the only reasonable conclusion i can draw that does not violate the laws of physics that we know and that is semi-consistent with what the pdf says is that
power consumption = total power dissipated + output power*2
now, whether TPD and OP need to be multiplied by two, it's anyone's guess. unless i get a definitive answer from a NI or TI engineer, i can't be sure for nothing. the datasheet is a fuckin mess
if TPD is what the IC needs, then it wouldn't be the same as the heat dissipated, nor would you go and add the output power to that same figure.
so far the only reasonable conclusion i can draw that does not violate the laws of physics that we know and that is semi-consistent with what the pdf says is that
power consumption = total power dissipated + output power*2
now, whether TPD and OP need to be multiplied by two, it's anyone's guess. unless i get a definitive answer from a NI or TI engineer, i can't be sure for nothing. the datasheet is a fuckin mess
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