I am reading about output stages of an amplifier and classes of amplifier. The following circuit is named as "Class B" amplifier.

enter image description here

From my understanding, the above circuit should be a class AB amplifier. Because we have a constant biasing, so both transistors would be slightly ON even if the signal is OFF, giving rise to small quiscient current. The crossover distortion is reduced greatly. But I am not sure why in my notes, they regard it as a class B amplifier.

I understand that if we adjust the battery voltage vbatn and vbatp, then would it become a class AB amplifier.

And I am not sure how the following circuit is regarded as Class C source follower: (This below circuit is actually class B from my understanding)

enter image description here

So, could we operate the circuit in class B or class AB or class C? Are class B,AB,C are terms for "Operation"?

f anybody could assist on the above, it would be helpful.

  • \$\begingroup\$ Doing some research goes a long way. \$\endgroup\$ Mar 23, 2018 at 21:56

2 Answers 2


The classes were originally arranged to describe the conduction angle for a single quadrant of a power amplifier.

  • In class-A, active conduction occurs throughout all \$360^\circ\$ of the period.
  • In class-B, active conduction occurs for \$180^\circ\$, or one-half of the period.
  • In class-C, active conduction occurs for \$\lt 180^\circ\$ (but more than \$0^\circ\$.) Anything less than class-B would be class-C.
  • In class-AB, active conduction occurs for more than \$180^\circ\$ but less than \$360^\circ\$ of the period. Anything more than class-B but not fully class-A is called class-AB.

The above pretty much covered all the cases (except for exactly \$0^\circ\$, which is trivial and no one cares about.)

But the above applied to a single quadrant. (Back when, for radio transmitters, a linear amplifier would have ONE really big vacuum tube for the power stage and you often operated it class-C.)

Now consider the case where you can afford the cost of two quadrants and you have an active push-pull drive capability. The above ideas still apply, but they apply ONLY to one of the quadrants. (On the assumption that both quadrants are operated in the same class as the other one.)

So in your first diagram, with sufficient biasing, the upper quadrant is operating in class-AB: meaning, "more than \$180^\circ\$ but less than \$360^\circ\$."

  • Of course, if the biasing is large enough (it never is, because no sane person would actually do it), then it could even be operating in class-A, with both quadrants fully active for the entire period. Or if the biasing were too little, or even reversed, then it could be operating in class-C.

In your lower diagram, the biasing is insufficient to allow either quadrant to operate in class-B, class-AB, or class-A. So the quadrants must be operating in class-C.

Class-AB, as it is often used in the context of push-pull, two-quadrant power output stages is usually meant to imply "slightly more than \$180^\circ\$" for each quadrant. The two-quadrant power output stage is operated this way with transistors to provide a little bit of overlapping coverage during the short transition "shoot-through" angle as the transistor of one quadrant takes over and the other quadrant backs off. This reduces cross-over distortion at the expense of wasted power due to a very short range of shoot-through angles.

It is all about the conduction angle for one quadrant.

  • \$\begingroup\$ Wouldn't the conduction angle always be less than or equal to 90 degree if it is for single quadrant? \$\endgroup\$
    – emnha
    Jan 5 at 20:00
  • \$\begingroup\$ @emnha I probably could have written that better. I meant to say that a single active quadrant's behavior in the output stage, with respect to the input signal (\$360^\circ\$), defines the class of output operation, regardless of the number of active quadrants in the output. A class-B with two active quadrants operating half the time doesn't make it a class-A output. It's still the fact that each quadrant operates class-B. Of course, a different class-B stage may only have one quadrant operating. It's also still class-B. Behavior of the two with respect to the load is different, though. \$\endgroup\$
    – jonk
    Jan 6 at 3:54
  • \$\begingroup\$ thanks for the reply. Is this the quadrant concept that you're refereing to? google.com/… How would you apply that concept of quadrant to this? zpag.net/Electroniques/English/amplifiers/Images/amplif16.jpg \$\endgroup\$
    – emnha
    Jan 7 at 14:31
  • \$\begingroup\$ @emnha It would be much better if you would get a copy of the ARRL handbook and look there. A weak version of that can be found on Wiki's 'Power Amplifier Classes' page, though. \$\endgroup\$
    – jonk
    Jan 7 at 21:51
  • \$\begingroup\$ I have basics about the operation and biasing condition for each. I was just confused about your concept of quadrant. \$\endgroup\$
    – emnha
    Jan 9 at 2:22

Class A = both transistors are ON all the time.

Class AB = both transistors are ON at idle, then up to a certain output current. When output current is higher than a certain limit, one of the transistors turns off.

Class B = either one transistor or the other is ON, but not both. The transistor that is ON is determined by output current polarity.

Class C = either one transistor is ON, or the other is ON, or both are OFF. ie, during part of the cycle they are both OFF and no output current flows. This can be used if the load is a tuned circuit (which is excited by the fundamental and rejects harmonics) or a motor, a solenoid, etc. Using this mode is intentional, the goal is high efficiency. The term "class C" also applies to a single transistor power amplifier if the transistor conducts for less than half (180°) of the cycle.

Note that class B doesn't actually exist. Since transistors threshold voltages vary with process and temperature, in practice you will never be able to adjust the bias voltages accurately enough to get class B. You will either get class AB if the gate bias is a bit high, and some conduction overlap will occurs, or if the gate bias is a bit low you will get class C when both transistors are off during part of the cycle.

Class C increases crossover distortion massively, so it is not recommended if you care about distortion. However, it has an advantage: no current is wasted keeping both transistors conducting at idle. So if the application doesn't care about crossover distortion it is a good choice.

Additionally when one considers a transistor to be "off" is a bit debatable, is it off when current is 1mA? or 1µA? or the leakage current? Or a negligible current relative to the output current?...

Anyway. The last schematic (without voltage sources to bias the gates) depends on your FET's threshold voltage.

For example, it can be class AB if you use JFETs which are ON when Vgs=0V.

If you use MOSFETs which require Vgs of a few volts (Vgsth) to begin turning on, then at idle both FETs are off, and you need to move the input voltage by at least one Vgsth up or down to turn one FET on. So it is class C.

Adding the gate bias voltage sources (as in the first schematic) would make it class AB once the bias is high enough to turn both transistors on at idle (zero output current).

  • 2
    \$\begingroup\$ Class C means something other than what you think. It's not what you get when a class B driver has a little gap. That's just a not-quite-accurate class B. Class C exploits resonance, and therefore usually works over a very narrow frequency range. Class C would not be suitable for a audio amplifier, for example. \$\endgroup\$ Mar 23, 2018 at 20:34
  • \$\begingroup\$ I think class A uses only one transistor. \$\endgroup\$ Mar 23, 2018 at 20:37
  • \$\begingroup\$ @Harry Svensson - Both transistors could be on, but in the linear region-- as one nearly turns off, the other nears full on, so if you look at their currents, both transistors show full sine waves, 180 degrees out of phase. \$\endgroup\$
    – Bort
    Mar 23, 2018 at 21:33
  • 1
    \$\begingroup\$ @HarrySvensson in a push pull class A both transistors can be on at all times, this happens for example in an opamp when the load is a high impedance, and output current is low. \$\endgroup\$
    – bobflux
    Mar 23, 2018 at 21:36
  • \$\begingroup\$ @OlinLathrop Got me ;) I was too lazy to spend the time to write this... \$\endgroup\$
    – bobflux
    Mar 23, 2018 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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