calculating heatsink for LM4780

rho · 19th April 2004, 07:28 PM

Hi,

I'm currently designing my fist chip-amp. The schematic shoudn't be too hard.
The thing I am wondering about is the size of the heatsink I'll need for the amp. In pictures I see realy small heatsinks for this type of amps, but when calculating it for the amp I'm designing I get a rather big type of heatsink.
The Vcc will be 68V. The secundary voltage of the transformer is 25V. This would result in (25V x 1.414)-1.4V=33.95V This for 1 rail. That results in 68Vdc.
Now the datasheet teels me that the maximum powerdissipation is equal to Vcc²/(2piRl). Rl = 8 ohm
So PDmax= 92W/ch. The LM4780 has 2 channels, so for the one device this would result in a PDmaxtot = 184W
This is where I don't follow. They advise to look at the gaphs to check the maximum power dissipation for the desired amp. In the graph I can see that there is more heat generated at 30W output/ch then at full power. But the PDmax will only be about 68W at that point. That's about 1/3th of what the calculation shows.

Can anyone help me out with this? It makes quite a big difference in heatsink-size.

maybe a link to the datasheet will help:LM4780

bigparsnip · 19th April 2004, 08:11 PM

you should be using 34V for your Vcc value her rather than the 68V you are currently trying with. This should give about 24W per channel (you have to remember there are two seperate chanels in this one chip, so you have to account for the power from both of them as you have done to give about 50W total) which is a little lower than you have at the moment.

rho · 20th April 2004, 07:16 PM

The datasheets states specificly that you should use Vcc as the rail to rail voltage.
But I have found the mistake: used following formula
Vcc²/(2piRl) but that should be Vcc²/(2pi²Rl).

This way the results fdrom the calculations get very close to the graphs in the datasheet.

19th April 2004, 07:28 PM	#1
rho diyAudio Member Join Date: Sep 2003 Location: Peer, Belgium	calculating heatsink for LM4780 Hi, I'm currently designing my fist chip-amp. The schematic shoudn't be too hard. The thing I am wondering about is the size of the heatsink I'll need for the amp. In pictures I see realy small heatsinks for this type of amps, but when calculating it for the amp I'm designing I get a rather big type of heatsink. The Vcc will be 68V. The secundary voltage of the transformer is 25V. This would result in (25V x 1.414)-1.4V=33.95V This for 1 rail. That results in 68Vdc. Now the datasheet teels me that the maximum powerdissipation is equal to Vcc²/(2piRl). Rl = 8 ohm So PDmax= 92W/ch. The LM4780 has 2 channels, so for the one device this would result in a PDmaxtot = 184W This is where I don't follow. They advise to look at the gaphs to check the maximum power dissipation for the desired amp. In the graph I can see that there is more heat generated at 30W output/ch then at full power. But the PDmax will only be about 68W at that point. That's about 1/3th of what the calculation shows. Can anyone help me out with this? It makes quite a big difference in heatsink-size. maybe a link to the datasheet will help:LM4780

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19th April 2004, 08:11 PM	#2
bigparsnip diyAudio Member Join Date: Apr 2003 Location: Cambridge	you should be using 34V for your Vcc value her rather than the 68V you are currently trying with. This should give about 24W per channel (you have to remember there are two seperate chanels in this one chip, so you have to account for the power from both of them as you have done to give about 50W total) which is a little lower than you have at the moment.

20th April 2004, 07:16 PM	#3
rho diyAudio Member Join Date: Sep 2003 Location: Peer, Belgium	The datasheets states specificly that you should use Vcc as the rail to rail voltage. But I have found the mistake: used following formula Vcc²/(2piRl) but that should be Vcc²/(2pi²Rl). This way the results fdrom the calculations get very close to the graphs in the datasheet.

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