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This is a beginner's theoretical question. I was looking at the figures on this site: http://www.electronics-tutorials.ws/amplifier/amp_6.html and noticed that the so called "conducting angle" of a class AB amplifier is given this way:

enter image description here

It is said to conduct between 180° and 360°. What I don't understand is that the output signal is coming from two transistors, additive. I see that a single trasistor in the pair does not conduct in the full 360° range. But the amplifier as a whole does conduct at any angle right? The transfer function of the amplifier won't be a straight line (obviously), but I don't see where the signal is clipped. Please help me understand.

UPDATE: here is a graph showing two functions:

f(x) = min(-0.6, sin(x)) g(x) = max(0.6, sin(x))

I believe that the response of the whole amp to a sinusoidal input is something like f(x)+g(x)

enter image description here

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  • \$\begingroup\$ Its not explained very well. That graph shows the operating point of just one transistor of the complementary pair operating in class B. \$\endgroup\$
    – Steve G
    Dec 16, 2016 at 11:02
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    \$\begingroup\$ Don`t use these tutorial pages. They contain errors. In the shown picture, the input signal is on the Vce axis. Is Vce really the input for an amplifier? A more correct figure is here:images.google.de/… \$\endgroup\$
    – LvW
    Dec 16, 2016 at 11:05
  • \$\begingroup\$ Okay, that page shows a different graph learnabout-electronics.org/Amplifiers/images/class-AB.gif but it is still clipped. It is also for one transistor only, right? \$\endgroup\$
    – nagylzs
    Dec 16, 2016 at 11:13
  • \$\begingroup\$ Yes - of course. Both transistors are active for different times. Both half waves must be superimposed. Note that the correct graph also shows the "cross-over distortions" caused by the lower part of the Ic=f(Vbe) transfer curve. \$\endgroup\$
    – LvW
    Dec 16, 2016 at 11:52

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Judging from the single diagram you post, run away from those web pages.

It seems they've managed to make a simple concept difficult to understand, and got it wrong in the process. What it appears that diagram is trying to show is how one of the transistors operates. However, what they are showing is class B operation, not class AB.

In class B, each transistor only conducts for exactly half the cycle, as shown. In class AB, there is a little crossover between the two transistors. Exactly in the middle, they are both conducting some. In the diagram you show, point Q should really be moved to the left a little, and the "output signal" biased up a little.

Added

Yes, your new diagram now shows true class AB behavior.

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  • \$\begingroup\$ I'm sorry, I got the wrong diagram from that website. Replacing now. \$\endgroup\$
    – nagylzs
    Dec 16, 2016 at 13:00
  • \$\begingroup\$ Diagram replaced. So it seems that it is indeed between 180 and 360 degrees, but that is just for one transistor. The question is: will the amplifier as a amplify in 360°? (I guess so?) \$\endgroup\$
    – nagylzs
    Dec 16, 2016 at 13:03
  • \$\begingroup\$ Okay, so the answer is: yes, the class AB amp does amplify signals in 360°, but it is not linear. \$\endgroup\$
    – nagylzs
    Dec 16, 2016 at 13:23
  • \$\begingroup\$ @nag: Each individual transistor is non-linear over the whole signal range, but the whole resulting amplifier is intended to be linear. \$\endgroup\$
    – Olin Lathrop
    Dec 16, 2016 at 13:26
  • \$\begingroup\$ Olin Lathrop - "intended to be linear"??? And what about the cross-over effect? What then is the difference to class-A operation? Can you give a gain value for class AB? \$\endgroup\$
    – LvW
    Dec 16, 2016 at 13:29

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