Banned
Joined 2002
JasonL said:
I don't doubt that the heatsink may be sufficient for you but 300 W power from a heat sink is much. My QRO amp (see my homepage) becomes hotter than hell if I torture if at full power. Therefore I have overheat protection. The amp has a little too low cooling ability but for homeuse it's no problem.
some like it hot
djnigma,
The power dissipation of anything is the product of voltage across it times the current through it (the real part of the product, that is, if they are out of phase).
For a class B or AB amp (minimum bias current in output stage) the theoretical efficiency is about 75% or so. As a rule of thumb this means that for every watt dissipated by the speaker the amp will dissipate 0.3W.
First calculate the highest average speaker power. If the maximum voltage amplitude across the speaker is Vm and the speaker load is R, the average power will be Vm^2/(2xR). In your design you can probably only swing 65V across the speaker because of the GS drop across the output FETs. So the average power will be 530W.
If the average speaker power is 530W then if the amp is 70% efficient the amp will dissipate 530*0.3=160W as heat. Since the amp has two output devices each will dissipate 80W. These figures are approximate.
I would normally choose a heatsink so that its temperature stays below 70C. To work out the heatsink temperature, take it's rating in degC/W, multiply by the device dissipation and add room temperature. For example, say you have one heatsink per channel. It must dissipate 160W and room temperature is 25C. The heatsink needs to have a temperature rise less than (70-25)/160=0.3deg/W or the equivalent with forced air cooling.
djnigma,
The power dissipation of anything is the product of voltage across it times the current through it (the real part of the product, that is, if they are out of phase).
For a class B or AB amp (minimum bias current in output stage) the theoretical efficiency is about 75% or so. As a rule of thumb this means that for every watt dissipated by the speaker the amp will dissipate 0.3W.
First calculate the highest average speaker power. If the maximum voltage amplitude across the speaker is Vm and the speaker load is R, the average power will be Vm^2/(2xR). In your design you can probably only swing 65V across the speaker because of the GS drop across the output FETs. So the average power will be 530W.
If the average speaker power is 530W then if the amp is 70% efficient the amp will dissipate 530*0.3=160W as heat. Since the amp has two output devices each will dissipate 80W. These figures are approximate.
I would normally choose a heatsink so that its temperature stays below 70C. To work out the heatsink temperature, take it's rating in degC/W, multiply by the device dissipation and add room temperature. For example, say you have one heatsink per channel. It must dissipate 160W and room temperature is 25C. The heatsink needs to have a temperature rise less than (70-25)/160=0.3deg/W or the equivalent with forced air cooling.
Banned
Joined 2002
i dont use a fan and the board is from anthony the amps runs cool itself the fan was just there because i was using it for some thing and then i put my amp there to take a pic.. the amp runs fine and strong. im thinking about buying another board to build 2 n-channels or a pair of av800's
J'
J'
Re: some like it hot
traderbam, you're not quite right. The maximium of heat do you not get at full power!...if the load is purely resistive.
Example, 60 V DC, perfect output stage, 8 ohm load
Max power dissipation (60^2)/(4*8) = 112 W (at 1/3 of full power)
Power dissipation at full power
(60^2)/(2*8) * (1-(pi/4)) = 48,3 W
traderbam said:
traderbam, you're not quite right. The maximium of heat do you not get at full power!...if the load is purely resistive.
Example, 60 V DC, perfect output stage, 8 ohm load
Max power dissipation (60^2)/(4*8) = 112 W (at 1/3 of full power)
Power dissipation at full power
(60^2)/(2*8) * (1-(pi/4)) = 48,3 W
some like it hotter
Peranders,
You are quite right. The worst case amp power doe not correspond with maximum speaker power. D'oh!
Ok, so you forced me to spend the morning remembering how to do integral calculus. And I have come up with the following. It isn't quite the same result as yours but it is close. Tell me what you think:
Let Vs=psu voltage (single rail of a split rail supply)
Vp = peak amplitude across the speaker
Pamp = average power dissipated by the amplifier
Pspkr = average power dissipated by the speaker
R = resistive speaker load
Pspkr = Vp^2/(2.R)
Pamp = 2.(Vp/R).[ (Vs/pi) - (Vp/4) ]
The maximum Pamp occurs when Vp = 2.Vs/pi
At this point Pamp(max) = 2.Vs^2/(R.pi^2) and the amplifier efficiency is 50%
Or Pamp(max) = (2/R).(Vs/pi)^2 which is approximately Vs^2/(5.R)
Example 1
=======
Taking the previous example, Vs=70, Vp=65, R=4
Pamp = 2.(65/4).(70/3.14 - 65/4) = 196W and Pspkr=528W, efficiency is 73%
but maximum Pamp occurs when Vp = 140/pi = 45V
Pamp(max) = (2/4).(70/3.14)^2 = 248W and Pspkr=248W, efficiency is 50%
So each output transistor needs to dissipate 124W average
(not 80W as I concluded before).
Example 2
=======
Vs=60, R=8, perfect o/p stage
Pamp(max) = (2/8).(60/3.14)^2 = 91W
Each transistor needs to dissipate 45.5W average.
Peranders,
You are quite right. The worst case amp power doe not correspond with maximum speaker power. D'oh!
Ok, so you forced me to spend the morning remembering how to do integral calculus. And I have come up with the following. It isn't quite the same result as yours but it is close. Tell me what you think:
Let Vs=psu voltage (single rail of a split rail supply)
Vp = peak amplitude across the speaker
Pamp = average power dissipated by the amplifier
Pspkr = average power dissipated by the speaker
R = resistive speaker load
Pspkr = Vp^2/(2.R)
Pamp = 2.(Vp/R).[ (Vs/pi) - (Vp/4) ]
The maximum Pamp occurs when Vp = 2.Vs/pi
At this point Pamp(max) = 2.Vs^2/(R.pi^2) and the amplifier efficiency is 50%
Or Pamp(max) = (2/R).(Vs/pi)^2 which is approximately Vs^2/(5.R)
Example 1
=======
Taking the previous example, Vs=70, Vp=65, R=4
Pamp = 2.(65/4).(70/3.14 - 65/4) = 196W and Pspkr=528W, efficiency is 73%
but maximum Pamp occurs when Vp = 140/pi = 45V
Pamp(max) = (2/4).(70/3.14)^2 = 248W and Pspkr=248W, efficiency is 50%
So each output transistor needs to dissipate 124W average
(not 80W as I concluded before).Example 2
=======
Vs=60, R=8, perfect o/p stage
Pamp(max) = (2/8).(60/3.14)^2 = 91W
Each transistor needs to dissipate 45.5W average.
Re[03]: some like it hot
Per-Anders,
Can you clarify a couple points in the formulas for me please?
1) Is the constant "4" of the "4*8" for "Max power dissipation" to factor for 4 pairs or 4 output devices?
2) Is the constant "4" of the "pi/4" for "Power dissipation at full power" to factor for 4 pairs or 4 output devices?
3) Is the constant "2" of "2*8" for "Power dissipation at full power" to factor for 2 pairs or 2 output devices?
I been imersed in a pile of formulas and those with a number of different opinions or formulas so I can properly calculate and properly add safety margins where I want to a mosfet power amp design. So skip what losses and values to use for losses, I am having a darn time finding the math for ideal no losses calculations before I add in losses. Any insight to the above qestions I have asked will hopefully reduce my confusion and help me sort out the forest for trees!
Regards,
John L. Males
Willowdale, Ontario
Canada
13 January 2005 21:22
peranders said:
Per-Anders,
Can you clarify a couple points in the formulas for me please?
1) Is the constant "4" of the "4*8" for "Max power dissipation" to factor for 4 pairs or 4 output devices?
2) Is the constant "4" of the "pi/4" for "Power dissipation at full power" to factor for 4 pairs or 4 output devices?
3) Is the constant "2" of "2*8" for "Power dissipation at full power" to factor for 2 pairs or 2 output devices?
I been imersed in a pile of formulas and those with a number of different opinions or formulas so I can properly calculate and properly add safety margins where I want to a mosfet power amp design. So skip what losses and values to use for losses, I am having a darn time finding the math for ideal no losses calculations before I add in losses. Any insight to the above qestions I have asked will hopefully reduce my confusion and help me sort out the forest for trees!
Regards,
John L. Males
Willowdale, Ontario
Canada
13 January 2005 21:22
Check page 7
http://www.web-ee.com/primers/files/DesignSem4.pdf
Search words for Google: class AB power dissipation formula
http://www.google.se/search?q=class+AB+power+dissipation+formula&hl=sv&lr=&start=0&sa=N
http://www.web-ee.com/primers/files/DesignSem4.pdf
Search words for Google: class AB power dissipation formula
http://www.google.se/search?q=class+AB+power+dissipation+formula&hl=sv&lr=&start=0&sa=N
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