Question: Question 2 [25 marks total] a) The voltage-divider-biased (VDB) amplifier is shown in Figure 4 with the following parameters: v_g = 50mV; V_CC = 10V; R_G = 500 \Omega; R_1 = 10k\Omega; R_2 = 2.2k\Omega; R_C = 3.6k\Omega; R_E = 1k\Omega; R_L = 10k\Omega; \beta = 200; i) Find the Q-point; (5 marks) ii) Find the voltage gain; (4 marks) iii) Find the output voltage, v_out. (3 marks)

1. **Problem Statement:** We have a voltage-divider-biased (VDB) BJT amplifier with given parameters: $$v_g = 50\text{mV} = 0.05\text{V}, V_{CC} = 10\text{V}, R_G = 500\Omega, R_1 = 10k\Omega, R_2 = 2.2k\Omega, R_C = 3.6k\Omega, R_E = 1k\Omega, R_L = 10k\Omega, \beta = 200.$$ We need to find: - i) The Q-point (quiescent point) which includes the collector current $I_C$ and collector-emitter voltage $V_{CE}$. - ii) The voltage gain $A_v$. - iii) The output voltage $v_{out}$. --- 2. **Step i) Find the Q-point:** - First, find the base voltage $V_B$ using the voltage divider formed by $R_1$ and $R_2$: $$V_B = V_{CC} \times \frac{R_2}{R_1 + R_2} = 10 \times \frac{2200}{10000 + 2200} = 10 \times \frac{2200}{12200} = 1.8033\text{ V}.$$ - The base-emitter voltage drop $V_{BE}$ for silicon transistor is approximately $0.7\text{ V}$. - Therefore, emitter voltage $V_E$ is: $$V_E = V_B - V_{BE} = 1.8033 - 0.7 = 1.1033\text{ V}.$$ - The emitter current $I_E$ is: $$I_E = \frac{V_E}{R_E} = \frac{1.1033}{1000} = 1.1033 \text{ mA}.$$ - Since $I_C \approx I_E$ for large $\beta$, we take: $$I_C \approx 1.1033 \text{ mA}.$$ - Next, find collector voltage $V_C$: $$V_C = V_{CC} - I_C R_C = 10 - (1.1033 \times 10^{-3}) \times 3600 = 10 - 3.9719 = 6.0281\text{ V}.$$ - Finally, collector-emitter voltage $V_{CE}$: $$V_{CE} = V_C - V_E = 6.0281 - 1.1033 = 4.9248\text{ V}.$$ **Q-point:** $$I_C = 1.1033 \text{ mA}, \quad V_{CE} = 4.9248 \text{ V}.$$ --- 3. **Step ii) Find the voltage gain $A_v$:** - The small-signal emitter resistance $r_e$ is approximately: $$r_e = \frac{25\text{ mV}}{I_E} = \frac{0.025}{1.1033 \times 10^{-3}} = 22.66 \Omega.$$ - Since the emitter resistor $R_E$ is bypassed by a capacitor, the AC emitter resistance is just $r_e$. - The voltage gain formula for a common-emitter amplifier with emitter bypass capacitor is: $$A_v = - \frac{R_C \parallel R_L}{r_e}.$$ - Calculate $R_C \parallel R_L$: $$R_C \parallel R_L = \frac{R_C R_L}{R_C + R_L} = \frac{3600 \times 10000}{3600 + 10000} = \frac{36000000}{13600} = 2647.06 \Omega.$$ - Therefore, $$A_v = - \frac{2647.06}{22.66} = -116.87.$$ --- 4. **Step iii) Find the output voltage $v_{out}$:** - The input AC voltage is $v_g = 50\text{ mV} = 0.05\text{ V}$. - The voltage gain magnitude is $|A_v| = 116.87$. - Output voltage amplitude: $$v_{out} = A_v \times v_g = -116.87 \times 0.05 = -5.8435 \text{ V}.$$ - The negative sign indicates a phase inversion. --- **Final answers:** - Q-point: $$I_C = 1.1033 \text{ mA}, V_{CE} = 4.9248 \text{ V}.$$ - Voltage gain: $$A_v = -116.87.$$ - Output voltage: $$v_{out} = -5.8435 \text{ V}.$$