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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
3.27
Step-by-Step
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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.

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