Hi,Do you know how to calculate the power output from an amplifier?
(Vrms squared)/speaker impedance. The amplifier you show uses a +3V supply and output transistors that drive the output capacitor up to about +2.7V with the bootstrapping shown and down to about +0.9V. Then the peak-to-peak voltage swing is 2.7V - 0.9V= 1.8V. The RMS voltage is 1.8V/2.828= 0.64V (almost nothing) then the power in a 4 ohm speaker is (0.64V squared)/ 4 ohms= 0.1W! The power is less as the battery runs down.
The peak output current is (1.8V/2)/4 ohms= 225mA so the transistors shown cannot be used since their maximum allowed current is only 100mA and their datasheet shows that they work poorly below 50mA.
For 5W into 4 ohms you need 4.5V RMS which is a peak-to-peak output swing of 12.7V so the power supply must be about 15V and produce 5W for the output power and about 5W for heating. The 10W at 15V is a power supply current of 667mA.
I simulated your amplifier and corrected it:
You sure you want to get rid of R4?
That looks like it helps filter the DC a little, in both directions ... into and out of the amplifier ... and possibly limit current into the amplifier.
Isnt there a better alternative to the two diode bias scheme? I'd have to look around but i am sure there is a transistor circuit that gets the biasing better than the two diodes. There's a name for it too, cant remember right now.