Question 3.SP.33: The amplifier of Fig. 3-27 uses an Si transistor for which V......
The amplifier of Fig. 3-27 uses an Si transistor for which V_{BEQ} = 0.7 \text{V}. Assuming that the collector-emitter bias does not limit voltage excursion, classify the amplifier according to Table 3-4 if (a) V_B = 1.0 \text{V} and v_S = 0.25 \cos ωt \text{V}, (b) V_B = 1.0 \text{V} and v_S = 0.5 \cos ωt \text{V}, (c) V_B = 0.5 \text{V} and v_S = 0.6 \cos ωt \text{V}, (d) V_B = 0.7 \text{V} and v_S = 0.5 \cos ωt \text{V}.
Table 3-4
| Class | Percentage of Active-Region Signal Excursion |
| A | 100 |
| AB | between 50 and 100 |
| B | 50 |
| C | less than 50 |
As long as v_S + V_B > 0.7 \text{V} , the emitter-base junction is forward-biased; thus classification becomes a matter of determining the portion of the period of v_S over which the above inequality holds.
(a) v_S + V_B ≥ 0.75 \text{V} through the complete cycle; thus the transistor is always in the active region, and the amplifier is of class A.
(b) 0.5 ≤ v_S + V_B ≤ 1.5 \text{V} ; thus the transistor is cut off for a portion of the negative excursion v_S. Since cutoff occurs during less than 180°, the amplifier is of class AB.
(c) -0.1 ≤ v_S + V_B ≤ 1.1 \text{V}, which gives conduction for less than 180° of the period of v_S, for class C operation.
(d) v_S + V_B ≥ 0.7 \text{V} over exactly 180° of the period of v_S, for class B operation.