Question E7B04:
What is the operating point of a Class A common emitter amplifier?  

Choices:  
A: Approximately halfway between saturation and cutoff  
B: Approximately halfway between the emitter voltage and the base voltage  
C: At a point where the bias resistor equals the load resistor  
D: At a point where the load line intersects the zero bias current curve    

Correct Answer is  A: Approximately halfway between saturation and cutoff  

Correct answer: A — Approximately halfway between saturation and cutoff

Why this is correct
- A Class A common-emitter amplifier conducts for the entire cycle (360°). To get maximum undistorted output swing, you bias the transistor at the midpoint of its DC load line—about halfway between cutoff ($I_C \approx 0$) and saturation ($V_{CE} \approx V_{CE(sat)}$).
- This midpoint gives roughly equal headroom for positive and negative signal swings, minimizing clipping and distortion.

Useful picture in equations (resistive collector load):
- DC load line:
  $$V_{CE} + I_C R_C = V_{CC}$$
- Midpoint (idealized) Q-point:
  $$V_{CE(Q)} \approx \frac{V_{CC}}{2}, \quad I_{C(Q)} \approx \frac{V_{CC}}{2R_C}$$
- Saturation and cutoff references:
  $$V_{CE(sat)} \approx 0.2\ \text{V (silicon BJT)}, \quad I_C \approx 0\ \text{at cutoff}$$

Why the other choices are incorrect
- B: “Halfway between the emitter voltage and the base voltage”
  - The CE operating point is defined on the collector characteristics (by $I_C$ and $V_{CE}$), not by averaging base and emitter voltages. Also, $V_{BE}$ is roughly fixed near 0.6–0.7 V for silicon; “halfway” between base and emitter has no design meaning.

- C: “Bias resistor equals the load resistor”
  - No general rule. The Q-point depends on $V_{CC}$, $R_C$, the bias network, and transistor parameters. Making a base-bias resistor equal to $R_C$ does not place the Q-point at the midpoint.

- D: “Where the load line intersects the zero bias current curve”
  - The zero base current curve is the cutoff condition. Intersecting there biases the transistor off, which is not Class A operation.

Rule of thumb
- For a CE Class A stage with resistor load, set:
  $$V_C \approx \frac{V_{CC}}{2}, \quad I_C \approx \frac{V_{CC}}{2R_C}$$
- With emitter degeneration:
  $$V_E \approx I_E R_E, \quad V_B \approx V_E + 0.7\ \text{V}$$
Bias near the midpoint to achieve maximum symmetrical undistorted output swing.