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29th December 2012, 09:20 AM  #11  
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Quote:
accordind to your calculation. the First watt F5 will give 6.7W class A at 8ohm.(1.3A bias) acctualy it gives 27W Last edited by AudioSan; 29th December 2012 at 09:26 AM. 

29th December 2012, 09:31 AM  #12 
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3.6 amps is the bias and is the same thing as the peak current (in class A) and the rms current is 1.414 times lower. Do we talk about the same thing here?
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29th December 2012, 09:35 AM  #13 
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peak current is TWICE the bias

29th December 2012, 09:40 AM  #14 
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from F5 manual:
For this sort of circuit, a 1.3 amp bias means that the amplifier will operate Class A to 2.6 amps of output current. To understand this, imagine a condition where Q3 and Q4 are idling at 1.3 amps, so that all the current is going from the V+ voltage rail to the V voltage rail, and none is going through the loudspeaker. The power of 2.6 amps into 8 ohms is I^2 * R, or 2.6 * 2.6 * 8 = 54 watts. This is the peak value, and the nature of an undistorted sine wave is that the peak wattage is twice the average, so this circuit would operate 27 watts average Class A into 8 ohms. At currents above 2.6 amps one of the transistors will shut off, leaving the other to continue to increase beyond the 2.6 amps in what is known as Class AB. 
29th December 2012, 09:46 AM  #15 
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So 1.3A is giving you 27W Class A from the F4.
With the Aleph 4, 2.5A bias gives you a true 100W RMS Class A. Last edited by KatieandDad; 29th December 2012 at 10:00 AM. 
29th December 2012, 09:57 AM  #16 
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Join Date: Feb 2009

katie. class A power have nothing to do with your rail voltage.
it is, (2.5*2)*(2.5*2)*8/2=100W class A power at 8ohm. 
29th December 2012, 02:22 PM  #17 
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check this thread how nelson calculates the class A power..
Leaving Class A? Martin Colloms vs Nelson Pass 
29th December 2012, 02:35 PM  #18 
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Location: USA

I would build it with a tiered power supply.
With a ±20V supply for the low rail the idle would only be 144W, that could be done with only two or three pair of IRFP240/9240.
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29th December 2012, 03:11 PM  #19  
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Quote:
414w peak is 207w average. 60V is about right for 200W/8ohms. Air forced cooling Mandatory. Last edited by bobodioulasso; 29th December 2012 at 03:20 PM. 

29th December 2012, 07:03 PM  #20 
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Join Date: Aug 2012

I would be using the IGBT heatsink and each heatsink will have 4 Hexfets dimensions
heatsink width 5 inch length 6 inch depth is also 6 inches... wouldnt that be sufficient? how do I calculate how much each hexfet is dessipating the heat? 
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