according to nelsons pow calculations 3.6A Bias
( 3.6 x 2 )^2*8 = 414.72W peak and for continuous 293W in 8 ohms
What sort of calculations are these? 3.6 amps will give you 50 Watts into 8 ohms if we talk class A.
It's pretty simple:
Irms = 3.6/1.414 = 2.546 A
P = I^2*R= 2.546*2.546*8 = 51.84 watts
What sort of calculations are these? 3.6 amps will give you 50 Watts into 8 ohms if we talk class A.
It's pretty simple:
Irms = 3.6/1.414 = 2.546 A
P = I^2*R= 2.546*2.546*8 = 51.84 watts
Sorry yes I forgot about to convert to rms I only got up 30mins ago!
http://www.firstwatt.com/pdf/art_leave_classa.pdf
Most of his commercial amps are hybrids.
I see you've got the F5, I've got the Aleph 4, both of those are true Class A. The F5 about 25W, the Aleph 4 about 100W.
As far as I can recall, the A4 runs at +/- 48V with 2.5A bias.
Most of his commercial amps are hybrids.
I see you've got the F5, I've got the Aleph 4, both of those are true Class A. The F5 about 25W, the Aleph 4 about 100W.
As far as I can recall, the A4 runs at +/- 48V with 2.5A bias.
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What sort of calculations are these? 3.6 amps will give you 50 Watts into 8 ohms if we talk class A.
It's pretty simple:
Irms = 3.6/1.414 = 2.546 A
P = I^2*R= 2.546*2.546*8 = 51.84 watts
no. the formula will be: (3.6x2)x(3.6x2)x8/2=207W
accordind to your calculation. the First watt F5 will give 6.7W class A at 8ohm.(1.3A bias) acctualy it gives 27W
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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.
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.
check this thread how nelson calculates the class A power..
http://www.diyaudio.com/forums/pass-labs/156058-leaving-class-martin-colloms-vs-nelson-pass.html
http://www.diyaudio.com/forums/pass-labs/156058-leaving-class-martin-colloms-vs-nelson-pass.html
according to nelsons pow calculations 3.6A Bias
( 3.6 x 2 )^2*8 = 414.72W peak and for continuous 293W in 8 ohms
414w peak is 207w average. 60V is about right for 200W/8ohms. Air forced cooling Mandatory.
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