Απάντηση: Re: Απάντηση: Υπάρχει αντιστοιχία Α τάξης και Α-Β ;
http://sound.westhost.com/class-a.htm
http://sound.westhost.com/project10.htm
Determining The Optimum Current
The ideal operating current for a Class-A amp will be about 110% of the peak speaker current. If the loudspeaker system has a nominal impedance of 8 Ohms (the design impedance for this amp), then with a +/- 22V supply the maximum (theoretical) speaker current is
I = V / R = 22 / 8 = 2.75A
In my original calculations, I actually decided on a quiescent current of 2.6A - this is really Ok, because the above calculation does not consider the losses in the output stage. In practice, it is likely that up to 3 Volts will be lost in the output circuit, based on the losses in the output devices, emitter resistors and driver transistors.
This now gives a maximum voltage of 19V peak (2.375A @ 8 Ohms). Applying the 110% fudge factor gives an operating current of 2.6125A, or 2.6 close enough. If these peaks are met in practice, this gives an output power of 22.5W into 8 Ohms.
Note that the current in the -ve supply rail remains constant, but that in the +ve supply rail will vary from the normal steady state current (same as the -ve supply). At signal extremes, the current will double (upper transistors turned on), or will drop to almost zero for negative peaks. This is common for single-ended Class-A amplifiers, although you will not see it in the text for most designs. This can complicate the design of the power supply.
Χοντρα χοντρα, η ισχυς σε ταξη Α ισουται (ειναι λιγο μικροτερη στην πραγματικοτητα) με την αντισταση φορτιου επι το τετραγωνο του ρευματος ηρεμιας .
ΟΙΚΟΝΟΜΟΥ ΤΑΣΟΣ said:
http://sound.westhost.com/class-a.htm
http://sound.westhost.com/project10.htm
Determining The Optimum Current
The ideal operating current for a Class-A amp will be about 110% of the peak speaker current. If the loudspeaker system has a nominal impedance of 8 Ohms (the design impedance for this amp), then with a +/- 22V supply the maximum (theoretical) speaker current is
I = V / R = 22 / 8 = 2.75A
In my original calculations, I actually decided on a quiescent current of 2.6A - this is really Ok, because the above calculation does not consider the losses in the output stage. In practice, it is likely that up to 3 Volts will be lost in the output circuit, based on the losses in the output devices, emitter resistors and driver transistors.
This now gives a maximum voltage of 19V peak (2.375A @ 8 Ohms). Applying the 110% fudge factor gives an operating current of 2.6125A, or 2.6 close enough. If these peaks are met in practice, this gives an output power of 22.5W into 8 Ohms.
Note that the current in the -ve supply rail remains constant, but that in the +ve supply rail will vary from the normal steady state current (same as the -ve supply). At signal extremes, the current will double (upper transistors turned on), or will drop to almost zero for negative peaks. This is common for single-ended Class-A amplifiers, although you will not see it in the text for most designs. This can complicate the design of the power supply.
Χοντρα χοντρα, η ισχυς σε ταξη Α ισουται (ειναι λιγο μικροτερη στην πραγματικοτητα) με την αντισταση φορτιου επι το τετραγωνο του ρευματος ηρεμιας .
