0
\$\begingroup\$

I'm simulating a class AB power amplifier.

According to the theory that I learned, the voltage at C8(2) should be equal to C10(1.) Why is there a huge gap between them (5.83 vs 1.89?)

The input impedance of this circuit is an approximation with R10 = R11. Is that correct?

enter image description here

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ How can the reader visualize this big gap. I looked at the oscilloscope and didn't recognize the numbers you talk about. I'd be more concerned about the clipping. \$\endgroup\$
    – Andy aka
    May 9, 2023 at 8:16
  • \$\begingroup\$ I highlight 3 voltage at C10(1), C8(2), D2(A). That is not right way to see gap ? \$\endgroup\$ May 9, 2023 at 8:19
  • \$\begingroup\$ If you can see the gap in your image that's fine but, I have no idea what the scales are nor where this gap is that you refer to. \$\endgroup\$
    – Andy aka
    May 9, 2023 at 8:29
  • \$\begingroup\$ @ErnestoG I think that at one time, just one BJT is work, and the other is cut-off. \$\endgroup\$ May 9, 2023 at 8:33
  • \$\begingroup\$ @Southgoodman that's valid at the peaks of a sine wave (and somewhere before them too), but at DC they should be holding a valid DC point. Is 16V the middle of your supply? \$\endgroup\$
    – Designalog
    May 9, 2023 at 8:37

2 Answers 2

1
\$\begingroup\$

I see Q9's Vbe is reverse biased, the emitter is at a higher voltage than the base. It won't work.

You have to put some effort to bias the bases. I have no idea what your VCC value is, but the fact that D1 is reversed biased as well means this isn't properly biased. Perhaps you're not allowing enough current to flow through the diodes?

\$\endgroup\$
9
  • \$\begingroup\$ Vcc is 30V. Oh, D1 is really in trouble. I change R10, and R11 to 50 Ohm, and the two wave is the same. Thank you. But look like finding a 50Ohm is not realistic when I make the PCB. \$\endgroup\$ May 9, 2023 at 8:49
  • \$\begingroup\$ imgur.com/a/GwG2r1P \$\endgroup\$ May 9, 2023 at 8:51
  • \$\begingroup\$ @Southgoodman why don't you make a proper current source instead of a resistor, then? \$\endgroup\$
    – Designalog
    May 9, 2023 at 8:51
  • \$\begingroup\$ Ah, My teacher permits me to use only Vcc = 30V and Ground, and no more voltage or current source. \$\endgroup\$ May 9, 2023 at 8:53
  • 1
    \$\begingroup\$ @Southgoodman there exists 49.9 ohm standard resistors. \$\endgroup\$
    – Designalog
    May 9, 2023 at 8:54
1
\$\begingroup\$

Typical current gain of a TIP41/TIP42 is 75, and you produce the base current for an 8ohm load through 2k resistors at a voltage amplification of 1.

You are missing a driver transistor stage to make this work. A Sziklai pair configuration does not raise the biasing voltage requirements, a Darlington pair adds another diode forward voltage to the biasing.

\$\endgroup\$
5
  • \$\begingroup\$ Can I use Common Collector configuration instead of Darlington Pair ? \$\endgroup\$ May 9, 2023 at 9:32
  • 1
    \$\begingroup\$ @Southgoodman Huh? You already have both transistors of your push-pull pair in Common Collector configuration. And cascading two transistors in Common Collector configuration is a Darlington pair. \$\endgroup\$
    – user107063
    May 9, 2023 at 9:43
  • \$\begingroup\$ @Southgoodman your question has morphed from why doesn't my buffer work to a full design with specifications, apparently. Why don't you change your question into the core of the problem and not starting with "why is my solution failing". \$\endgroup\$
    – Designalog
    May 9, 2023 at 10:04
  • \$\begingroup\$ @ErnestoG "This is not a site for getting your homework done." Posting a whole spec will invariably lead to "show your work so far", and that's exactly what has been done here. Yes, this will likely lead to followup questions instead of getting everything done by someone else at once. This is how it is supposed to work. \$\endgroup\$
    – user107063
    May 9, 2023 at 11:05
  • \$\begingroup\$ @user107063 the original question was very limited. If he extends it to what was needed AND his solution would be much better than showing only his solution and asking why it doesn't work. \$\endgroup\$
    – Designalog
    May 9, 2023 at 11:17

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.