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Question 3.SP.28: In the circuit of Fig. 3-8(a), RC = 300 Ω, RE = 200 Ω, R1 = ......

In the circuit of Fig. 3-8(a), R_C = 300  Ω,  R_E = 200  Ω,  R_1 = 2  kΩ,  R_2 = 15  kΩ,  V_{CC} = 15  \text{V}, and β =  110 for the Si transistor.    Assume that I_{CQ} ≈ I_{EQ} and V_{CE\text{sat}} ≈ 0.    Find the maximum symmetrical swing in collector current    (a) if an ac base current is injected, and   (b) if V_{CC} is changed to 10 V but all else remains the same.

3.8
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(a)  From (3.5) and (3.7),
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{(2  ×  10^3)(15  ×  10^3)}{17  ×  10^3} = 1.765  kΩ  \quad  \text{and}  \quad  V_{BB} = \frac{2  ×  10^3}{17  ×  10^3}15 = 1.765  \text{V}

so        I_{CQ} ≈ I_{EQ} =  \frac{V_{BB}  –  V_{BEQ}}{R_B/(β  +  1)  +  R_E} = \frac{1.765  –  0.7}{1765/111  +  200} = 4.93  \text{mA}

By KVL around the collector-emitter circuit with I_{CQ} ≈ I_{EQ},
V_{CEQ} = V_{CC}  –  I_{CQ}(R_C + R_E) = 15  –  (4.93 × 10^{-3})(200 + 300) = 12.535  \text{V}

Since V_{CEQ} > V_{CC}/2 = 7.5  \text{V}, cutoff occurs before saturation, and i_C can swing ±4.93  \text{mA} about I_{CQ} and remain in the active region.
(b)        V_{BB} = \frac{R_1}{R_1  +  R_2}V_{CC} = \frac{2  ×  10^3}{17  ×  10^3}10 = 1.1765  \text{V}

so that        I_{CQ} ≈ I_{EQ} = \frac{V_{BB}  –  V_{BEQ}}{R_B/(β  +  1)  +  R_E} = \frac{1.1765  –  0.7}{1765/111  +  200} = 2.206  \text{mA}

and          V_{CEQ} = V_{CC}  –  I_{CQ}(R_C + R_E) = 10  –  (2.206 × 10^{-3})(0.5) = 8.79  \text{V}

Since V_{CEQ} > V_{CC}/2 = 5  \text{V}, cutoff again occurs before saturation, and i_C can swing ±2.206  \text{mA} about I_{CQ} and remain in the active region of operation.    Here, the 33.3 percent reduction in power supply voltage has resulted in a reduction of over 50 percent in symmetrical collector-current swing.

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