Question E7B01:
For what portion of the signal cycle does each active element in a push-pull, Class AB amplifier conduct?  

Choices:  
A: More than 180 degrees but less than 360 degrees  
B: Exactly 180 degrees  
C: The entire cycle  
D: Less than 180 degrees    

Correct Answer is  A: More than 180 degrees but less than 360 degrees  

For a push-pull, Class AB amplifier, each device is biased slightly above cutoff so it conducts a little more than half the cycle. With two devices in push-pull, their conduction overlaps to cover the full cycle with reduced crossover distortion.

Correct answer — A: More than 180 degrees but less than 360 degrees
- In Class AB, the conduction angle per device is:
  $$180^\circ < \theta < 360^\circ$$
- The slight overlap eliminates the crossover notch seen in ideal Class B.

Why the other choices are incorrect
- B: Exactly 180 degrees — That is ideal Class B, not AB. Class AB conducts slightly more than $180^\circ$.
- C: The entire cycle — $360^\circ$ conduction is Class A, not AB.
- D: Less than 180 degrees — That is Class C, used for non-linear RF modes (e.g., CW/FM), not for linear amplification like SSB/AM.

Useful rules of thumb
- Conduction angle per device:
  - Class A: $$\theta = 360^\circ$$
  - Class B: $$\theta = 180^\circ$$
  - Class AB: $$180^\circ < \theta < 360^\circ$$
  - Class C: $$\theta < 180^\circ$$
- Push-pull coverage: the pair together covers $$360^\circ$$ of the signal with some overlap in Class AB.
- Bias/current clues:
  - Class A: significant quiescent current, highly linear, least efficient.
  - Class B: near-zero quiescent current, higher efficiency, crossover distortion risk.
  - Class AB: small quiescent current ($I_Q > 0$), linear enough for SSB/AM, efficiency between Class A and B (roughly 50–70% practical).