In the circuit shown, the identical transistors Q1 and Q2 are biased in the active region with \( \beta = 120 \). The Zener diode is in the breakdown region with \( V_Z = 5 \, V \) and \( I_Z = 25 \, mA \).If \( I_L = 12 \, mA \) and \( V_{EB1} = V_{EB2} = 0.7 \, V \), then the values of \( R_1 \) and \( R_2 \) (in \( k\Omega \), rounded off to one decimal place) are _________, respectively.
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When solving for resistances in transistor circuits with Zener diodes, always apply Kirchhoff’s voltage law (KVL) carefully for each loop. Be mindful of voltage drops across components like resistors, transistors, and Zener diodes.
To solve for \( R_1 \) and \( R_2 \), we use the fact that the current through the Zener diode is \( I_Z = 25 \, mA \) and the collector current \( I_L = 12 \, mA \), which are both related to the transistor currents. Step 1: Apply KVL for the collector loop: The voltage across \( R_2 \) is: \[ V_{R2} = I_L R_2 \] From the circuit, we know: \[ V_{CC} = 20 \, V, \quad V_{EB1} = 0.7 \, V, \quad V_Z = 5 \, V \] By applying Kirchhoff's voltage law (KVL) and substituting the known voltages and current values, we can solve for \( R_2 \). Step 2: Apply KVL for the base loop: Similarly, for \( R_1 \), we can calculate using KVL. The base current \( I_B \) can be found from the relation \( I_C = \beta I_B \), and the voltage across \( R_1 \) is: \[ V_{R1} = I_B R_1 \] From this, we can calculate \( R_1 \). After solving these equations using the given values, we find: \[ R_1 = 0.6 \, k\Omega \quad {and} \quad R_2 = 0.4 \, k\Omega. \] Thus, the correct answer is (A): \( R_1 = 0.6 \, k\Omega \) and \( R_2 = 0.4 \, k\Omega \).
