1. **Problem statement:** Given a transistor amplifier circuit and parameters $V_{CC} = 20$, $R_1 = 100000$, $R_2 = 50000$, $R_E = 1000$, $r_s = 2000$, $R_L = 1000$, $h_{fe} = -100$, $h_{rc} = 1$, and $h_{ic} = 1500$. We need to: a) Draw the AC h-parameter equivalent circuit. b) Calculate: i. Current gain ii. Voltage gain iii. Power gain iv. Input impedance v. Output impedance 2. **Drawing the AC h-parameter equivalent circuit:** - The $h$-parameter model for a transistor uses parameters $h_{fe}$ (current gain), $h_{rc}$ (reverse voltage gain, neglected here), $h_{ic}$ (input impedance), and $h_{oe}$ (output admittance, neglected). - Since $h_{rc}$ is given as 1 but the problem says to neglect $h_{oc}$ (likely output admittance), we focus on using $h_{fe}$, $h_{ic}$ for the small signal equivalent. - The equivalent circuit model consists of: - Input port: voltage $v_{be}$ across input terminals with input impedance $h_{ic} = 1500 \Omega$. - A dependent current source at output: $i_c = h_{fe} i_b$ where $h_{fe} = -100$. - Output port voltage and current linked accordingly, with load $R_L$ and emitter resistor $R_E$ included appropriately. 3. **Assumptions:** - Capacitors $C_1$, $C_2$ are considered short circuits for AC analysis. - Neglect $h_{oc}$ (output admittance) effect. - For $h_{fe}$ negative, use its magnitude $|h_{fe}| = 100$ for gain calculations. 4. **Calculate i. Current gain ($A_i$):** - Current gain is ratio $A_i = \frac{i_{out}}{i_s}$. - The source current $i_s$ flows through $r_s$ and $R_1$ into the input. - The input impedance is $Z_{in} = R_1 || R_2 || h_{ic}$. Calculate parallel resistance: $$R_{in} = \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{h_{ic}} \right)^{-1} = \left( \frac{1}{100000} + \frac{1}{50000} + \frac{1}{1500} \right)^{-1}$$ Calculate each term: $$\frac{1}{100000} = 0.00001, \quad \frac{1}{50000} = 0.00002, \quad \frac{1}{1500} \approx 0.0006667$$ Sum: $$0.00001 + 0.00002 + 0.0006667 \approx 0.0006967$$ Inverting: $$R_{in} \approx \frac{1}{0.0006967} \approx 1435 \ \Omega$$ Total input current: $$i_{in} = \frac{v_s}{r_s + R_{in}}$$ Assuming voltage source $v_s$ and since current gain is ratio output current to $i_s$ (source current), $i_s$ equals $i_{in}$. Current gain through transistor: $$A_i = |h_{fe}| \times \frac{R_L}{R_L + R_E}$$ Since emitter resistor $R_E$ is 1000 and load $R_L$ is 1000: $$\frac{R_L}{R_L + R_E} = \frac{1000}{1000 + 1000} = \frac{1000}{2000} = 0.5$$ So: $$A_i = 100 \times 0.5 = 50$$ 5. **Calculate ii. Voltage gain ($A_v$):** Voltage gain is: $$A_v = \frac{v_{out}}{v_s}$$ Where output voltage is across $R_L$, and input voltage divides over source resistance and input resistance: Voltage at base: $$v_{in} = v_s \times \frac{R_{in}}{r_s + R_{in}} = v_s \times \frac{1435}{2000 + 1435} = v_s \times \frac{1435}{3435} \approx 0.4178 v_s$$ Output voltage is output current times $R_L$: Output current is approximately $$i_{out} = |h_{fe}| i_b \times \frac{R_L}{R_L+R_E} = 50 i_b$$ Input current $i_b = \frac{v_{in}}{R_{in}}$ Thus $$v_{out} = i_{out} R_L = 50 \times \frac{v_{in}}{R_{in}} \times R_L = 50 \times v_{in} \times \frac{R_L}{R_{in}}$$ Plugging values: $$A_v = \frac{v_{out}}{v_s} = 50 \times 0.4178 \times \frac{1000}{1435} = 50 \times 0.4178 \times 0.6969 \approx 14.55$$ 6. **Calculate iii. Power gain ($A_p$):** Power gain defined as: $$A_p = A_v \times A_i$$ Plug in results: $$A_p = 14.55 \times 50 = 727.5$$ 7. **Calculate iv. Input impedance ($Z_{in}$):** Already calculated: $$Z_{in} = R_1 || R_2 || h_{ic} = 1435 \ \Omega$$ 8. **Calculate v. Output impedance ($Z_{out}$):** Output impedance looking into the emitter through $R_E$ and $R_L$: Since emitter is connected to $R_E$ and $R_L$ in series, $$Z_{out} = R_E + R_L = 1000 + 1000 = 2000 \ \Omega$$ **Final answers:** - Current gain $A_i = 50$ - Voltage gain $A_v \approx 14.55$ - Power gain $A_p \approx 727.5$ - Input impedance $Z_{in} \approx 1435 \ \Omega$ - Output impedance $Z_{out} = 2000 \ \Omega$