Question 3.8: The signal source switch of Fig. 3-9(a) is closed, and the t......

The signal source switch of Fig. 3-9(a) is closed, and the transistor base current becomes

i_B = I_{BQ} + i_b = 40 + 20 \sin ωt    μ\text{A}

The collector characteristics of the transistor are those displayed in Fig. 3-9(b).   If V_{CC} = 12  \text{V} and R_{dc} = 1  kΩ, graphically determine (a) I_{CQ} and V_{CEQ}, (b) i_c and v_{ce}, and (c) h_{FE}(= β) at the Q point.

3.9
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(a)   The dc load line has ordinate intercept V_{CC}/R_{dc} = 12  \text{mA} and abscissa intercept V_{CC} = 12  \text{V} and is constructed on Fig. 3-9(b).    The Q point is the intersection of the load line with the characteristic curve i_B = I_{BQ} = 40  μ\text{A}.    The collector quiescent quantities may be read from the axes as I_{CQ} = 4.9  \text{mA} and V_{CEQ} = 7.2  \text{V}.

(b)   A time scale is constructed perpendicular to the load line at the Q point, and a scaled sketch of i_b = 20 \sin ωt  μ\text{A} is drawn [see Fig. 3-9(b)] and translated through the load line to sketches of i_c and v_{ce}. As i_b swings ±20  μ\text{A} along the load line from points a to b, the ac components of collector current and voltage take on the values
i_c = 2.25 \sin ωt  \text{mA} \quad \text{and} \quad  v_{ce} = -2.37 \sin ωt  \text{V}
The negative sign on v_{ce} signifies a 180° phase shift.

(c)   From (3.2) with I_{CEO} = 0 [the i_B = 0 curve coincides with the v_{CE} axis in Fig. 3-9(b)],
β(≡ h_{FE}) ≡ \frac{α}{1  –  α} ≡ \frac{I_C  –  I_{CEO}}{I_B}        (3.2)
h_{FE} = \frac{I_{CQ}}{I_{BQ}} = \frac{4.9  ×  10^{-3}}{40  ×  10^{-6}} = 122.5

It is clear that amplifiers can be biased for operation at any point along the dc load line.    Table 3-4 shows the various classes of amplifiers, based on the percentage of the signal cycle over which they operate in the linear or active region.

Table 3-4

Class Percentage of Active-Region
Signal Excursion
A 100
AB between 50 and 100
B 50
C less than 50

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