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Question 3.SP.27: In the circuit of Fig. 3-10(a), the transistor is a Si devic......

In the circuit of Fig. 3-10(a), the transistor is a Si device, R_E = 200  Ω,  R_2 = 10R_1 = 10  kΩ,  R_L = R_C = 2  kΩ,  β = 100, and V_{CC} = 15  \text{V}. Assume that C_C and C_E are very large, that V_{CE\text{sat}} ≈ 0, and that i_C = 0 at cutoff.    Find  (aI_{CQ},   (bV_{CEQ},  (c) the slope of the ac load line,  (d) the slope of the dc load line, and  (e) the peak value of undistorted i_L.

3.10
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(a)   Equations (3.5) and (3.7), give
I_{CQ} ≈ I_{EQ} = \frac{V_{BB}  –  V_{BEQ}}{R_B/(β  +  1)  +  R_E}          (3.7)
R_B = \frac{R_1R_2}{R_1  +  R_2} \quad V_{BB} = \frac{R_1}{R_1  +  R_2}V_{CC}          (3.5)

R_B = \frac{(1  ×  10^3)(1  ×  10^3)}{11  ×  10^3} = 909  Ω  \quad  \text{and}  \quad  V_{BB} = \frac{1  ×  10^3}{11  ×  10^3}15 = 1.364  \text{V}

so        I_{CQ} ≈ \frac{V_{BB}  –  V_{BEQ}}{R_B/(β  +  1)  +  R_E} = \frac{1.364  –  0.7}{(909/101)  +  200} = 3.177  \text{mA}

(b)   KVL around the collector-emitter circuit, with I_{CQ} ≈ I_{EQ}, gives
V_{CEQ} = V_{CC}  –  I_{CQ}(R_E + R_c) = 15  –  (3.177 × 10^{-3})(2.2 × 10^3) = 8.01  \text{V}

(c)        \text{Slope} = \frac{1}{R_{ac}} = \frac{1}{R_C} + \frac{1}{R_L} = 2 \frac{1}{2  ×  10^3} = 1  \text{mS}

(d)        \text{Slope} = \frac{1}{R_{dc}} = \frac{1}{R_C  +  R_E} = \frac{1}{2.2  ×  10^3} = 0.454  \text{mS}

(e)  From (3.14), the ac load line intersects the v_{CE} axis at
v_{CE  \max} = V_{CEQ} + I_{CQ}R_{ac} = 8.01 + (3.177 × 10^{-3})(1 × 10^3) = 11.187  \text{V}

Since v_{CE  \max} < 2 V_{CEQ}, cutoff occurs before saturation and thus sets V_{cem}. With the large capacitors appearing as ac shorts,
i_L = \frac{v_L}{R_L} = \frac{v_{ce}}{R_L}

or, in terms of peak values,
I_{Lm} = \frac{V_{cem}}{R_L} = \frac{v_{CE  \max}  –  V_{CEQ}}{R_L} = \frac{11.187  –  8.01}{2  ×  10^3} = 1.588  \text{mA}

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