Topic: Electronics Engineering
In the circuit given below, assume that $V_{CC}=15 V$; Z1, Z2, Z3 and Z4 are identical zener diodes with breakdown voltage of 5 V; $R1=R4=5~k\Omega$ $R2=R3=10~k\Omega$. Find $V_{0}$ when Q1 is OFF.
- A. 7.5 V
- B. 5 V
- C. 10 V
- D. 8.33 V
Answer: C. 10 V
Explanation: Apply KCL at node B (when Q1 is OFF).
$\frac{15-V_{0}}{R_{1}}=\frac{V_{0}-5}{R_{2}}+\frac{V_{0}-5}{R_{3}}$
Since $R_{2}=R_{3}$, $\frac{15-V_{0}}{5~k}=2(\frac{V_{0}-5}{10~k}) \Rightarrow 15-V_{0}=V_{0}-5 \Rightarrow 2V_{0}=20 \Rightarrow V_{0}=10~V$.