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)
Last edited:
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
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)
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.