200W class A psu considerations with IRF240

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Max class A power into 4r with 3.6A bias is 50W and 100W into 8r. 120V x 3.6A = 432W idle power which isn't really possible passively within a realistic size. To achieve 200W into 8 you will need 5A bias and 65V rails, so you will be dissipating 650W at idle.
 
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.
 
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