Let's solve the problem step by step: --- ## **A. Derive an expression for the current transfer ratio** ### **Step 1: Analyze the circuit** This is a current mirror circuit using two identical NPN BJTs (Tr1 and Tr2). - \( I_{REF} \) is the reference current through \( R_3 \) and Tr1. - Tr2 mirrors the current, providing output at \( V_O \). ### **Step 2: Write the KVL for Tr1 and Tr2** Since both transistors are identical and \( V_{BE} \) is the same for both, - The voltage across \( R_1 \) and \( R_2 \) will set the emitter currents. - Assume base currents are negligible due to high \( \beta \). #### For **Tr1**: \[ I_{E1} = I_{C1} + I_{B1} \approx I_{C1} \quad (\text{since } \beta \text{ is very high}) \] \[ I_{C1} = \frac{V_{R1}}{R_1} = \frac{V_{E1}}{R_1} \] #### For **Tr2**: \[ I_{C2} = \frac{V_{R2}}{R_2} = \frac{V_{E2}}{R_2} \] Both emitters are at the same voltage (since bases are tied together and emitters are referenced through resistors). ### **Step 3: Current Transfer Ratio** The current mirror ratio is set by the resistor values in the emitters: \[ \frac{I_{C2}}{I_{C1}} = \frac{V_{E2}/R_2}{V_{E1}/R_1} \] But \( V_{E1} = V_{E2} \) (since both emitters are at same potential), so: \[ \frac{I_{C2}}{I_{C1}} = \frac{R_1}{R_2} \] ### **Step 4: Expression for Output Current** \[ I_{OUT} = I_{C2} = I_{C1} \cdot \frac{R_1}{R_2} \] --- ## **B. Given values, find \( R_3 \) and the output resistance** **Given:** - \( R_1 = 1k\Omega \) - \( R_2 = 2k\Omega \) - \( I_{REF} = 2\,mA \) - \( r_\pi = 25k\Omega \) ### **Step 1: Find \( R_3 \)** Apply KVL at the loop from +8V to ground through \( R_3 \), Tr1, and \( R_1 \): \[ 8V = I_{REF} R_3 + V_{BE} + I_{C1} R_1 \] Since \( I_{REF} \approx I_{C1} \) (base current is negligible), \[ 8V = I_{REF} R_3 + 0.7V + I_{REF} R_1 \] \[ I_{REF} (R_3 + R_1) = 8V - 0.7V \] \[ R_3 + R_1 = \frac{7.3V}{2mA} = 3650 \Omega \] \[ R_3 = 3650 - 1000 = 2650\Omega = 2.65k\Omega \] ### **Step 2: Output Resistance** Ignoring finite output resistance (as per question), output resistance is **ideally infinite**. But if you include transistor output resistance (\( r_o \)), for a current mirror: \[ r_{out} = r_o = \frac{V_A}{I_C} \] Given \( V_A = 100V \), \( I_C = 1mA \) (since \( I_{C2} = I_{REF} \cdot \frac{R_1}{R_2} = 2mA \times \frac{1}{2} = 1mA \)) \[ r_{out} = \frac{100V}{1mA} = 100k\Omega \] --- ## **Summary of Answers** ### **A. Expression for current transfer ratio:** \[ \boxed{\frac{I_{C2}}{I_{C1}} = \frac{R_1}{R_2}} \] ### **B.** \[ \boxed{R_3 = 2.65\,k\Omega} \] \[ \boxed{\text{Output resistance} = 100\,k\Omega} \] --- **Let me know if you'd like further clarification or details!**

The circuit is a current mirror using two identical BJTs. The reference current \(I_{REF}\) flows through resistor \(R_3\) and transistor Tr1, setting its emitter current. Since the transistors are matched with high \(\beta\), the base-emitter voltage (\(V_{BE}\)) is approximately equal, causing Tr2 to mirror the current based on the emitter resistor ratio. The current transfer ratio is determined by the ratio of emitter resistors: \(\frac{I_{OUT}}{I_{REF}} = \frac{R_1}{R_2}\). By selecting resistor values and considering transistor parameters, the circuit provides a stable current output proportional to the reference current.

The circuit is a simple current mirror employing two matched BJTs. The reference current \(I_{REF}\) flows through \(R_3\) and Tr1, establishing its emitter current. Due to transistor matching and high \(\beta\), both BJTs have nearly equal \(V_{BE}\), causing Tr2 to mirror the current based on the ratio of emitter resistors \(R_1\) and \(R_2\). The current transfer ratio is thus \(\frac{I_{OUT}}{I_{REF}} = \frac{R_1}{R_2}\). By choosing resistor values and considering transistor parameters like \(V_{BE}\), the circuit ensures a stable current output proportional to the reference current, useful in biasing and analog circuit applications.