Audioguru again
 Joined Oct 21, 2019
 6,783
The 10k collector resistor of the CE transistor makes a voltage divider with the 188 ohm base resistors of the power transistors.
It is the 188 ohm resistors (not the CE transistor) that turn on the output transistors but even 188 ohms does not provide enough current.
For a 10W sinewave into 4 ohms the peak current is simply calculated to be 2.25A and the peak output voltage is simply 2.25A x 4 ohms= 9V. If you buy hundreds of output transistors and measure them, then you might find some with a typical current gain of 60 so the peak base current will ne 2.25A/60= 37.5mA. Then you can add ordinary little transistors to drive each output transistor.
Use darlington output transistors instead like this old amplifier:
It is the 188 ohm resistors (not the CE transistor) that turn on the output transistors but even 188 ohms does not provide enough current.
For a 10W sinewave into 4 ohms the peak current is simply calculated to be 2.25A and the peak output voltage is simply 2.25A x 4 ohms= 9V. If you buy hundreds of output transistors and measure them, then you might find some with a typical current gain of 60 so the peak base current will ne 2.25A/60= 37.5mA. Then you can add ordinary little transistors to drive each output transistor.
Use darlington output transistors instead like this old amplifier:
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