# Class AB amplifier conduction angle

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:

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)

• Its not explained very well. That graph shows the operating point of just one transistor of the complementary pair operating in class B. Dec 16, 2016 at 11:02
• 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/…
– LvW
Dec 16, 2016 at 11:05
• 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? Dec 16, 2016 at 11:13
• 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.
– LvW
Dec 16, 2016 at 11:52

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.