Is it purely a function of the total bias current? How do I calculate the Class A power available for an F5T for a given supply voltage (in my case ~48V), Bias Current(in my case ~ 3.6A - I will have six output pairs) and a load of 4 ohms?
Thanks,
Denis
Thanks,
Denis
It's down to your bias and load.
As the load will vary in impedance over the frequency spectrum it's difficult to give an exact figure.
But at 3.6A into 4R it should still be in Class A at approx. 100W.
As the load will vary in impedance over the frequency spectrum it's difficult to give an exact figure.
But at 3.6A into 4R it should still be in Class A at approx. 100W.
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F5/F5T is PP, bias level is half the peak Class A output current.
3.6A bias corresponds to +200W peak output power, aka ~100W continuous in 4 ohm. Both Class A.
(a (regular) F5T transits to Class AB the exact same way as an F5. Only difference is that the rail loss is lower, feedback ensures that the output voltage still tracks the input signal)
3.6A bias corresponds to +200W peak output power, aka ~100W continuous in 4 ohm. Both Class A.
(a (regular) F5T transits to Class AB the exact same way as an F5. Only difference is that the rail loss is lower, feedback ensures that the output voltage still tracks the input signal)
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I'd be interested to compare my Aleph 4 (100W Class A) with the F5T.
Anyone with the F5T within 50 miles of B90 (Birmingham), give me a call.
Anyone with the F5T within 50 miles of B90 (Birmingham), give me a call.
calculation for class A power in push-pull is:
(bias*2)*bias*2)*impedance=peak class A power. (as long as the voltage is high enough)
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