2
\$\begingroup\$

Thank you all, let me ask about a schematic design of a Class A amplifier schematic on internet. Please explain to me about how much voltage gain is obtained for the amplifier in this design? Is it 10x times or 100x times? How do we calculate the 10-14 watt power?

enter image description here

Please explain step by step.

\$\endgroup\$
2
  • \$\begingroup\$ explain to me about how much voltage amplifier the design Explain what? Are you asking to explain how to do design? Depending on your current skill level that takes years to master! \$\endgroup\$ Commented Apr 16, 2020 at 7:37
  • \$\begingroup\$ are you any recomendation step how to study, i only understand basic amplifier, for advanced i stii confuse \$\endgroup\$
    – Ridwan
    Commented Apr 17, 2020 at 6:38

2 Answers 2

1
\$\begingroup\$

This amplifier must crawl before it can walk. Your idle current is not well defined, thermal runaway is possible. The bottom NPN Darlington is not being driven in any way so it is trying to be a constant source. This means that the max efficiency if all the parts were ideal would be 25% instead of the normal 50% max. Your 680k seems really high so such a high prospective gain means little or no feedback. If you used say a 68k then there would of course be much less gain but lower distortion.

\$\endgroup\$
0
\$\begingroup\$

base voltage is about 3 volts

emitter voltage about 2 volts

emitter current = 2v/150 or 12mA

The reac thus is 2 ohms (transconductance is 0.48 amps/volt)

the gain of the transistor is 1.8K / 2 ohms or about 900x

this circuit also has huge input capacity caused by Miller multiplication of Cob * gain of the 2n3904; this will be about 10,000 picofarads

but this circuit also has global negative feedback, implemented right at the input

\$\endgroup\$
1
  • \$\begingroup\$ thank your for your answer, are you any recomendation website to understand that schematic? \$\endgroup\$
    – Ridwan
    Commented Apr 17, 2020 at 6:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.