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.
(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)
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.