# 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.

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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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