Chapter 2

Q. 2.SP.31

A Zener diode has the specifications V_Z = 5.2  \text{V}  and  P_{D  \max} = 260  \text{mW}. Assume R_Z = 0.   (a)  Find the maximum allowable current i_Z when the Zener diode is acting as a regulator.   (b)  If a single-loop circuit consists of an ideal 15-V dc source V_S, a variable resistor R, and the described Zener diode, find the range of values of R for which the Zener diode remains in constant reverse breakdown with no danger of failure.


Verified Solution

(a)              i_{Z  \max} = I_Z = \frac{P_{D  \max}}{V_Z} + \frac{2.60  \times  10^{-3}}{5.2} = 50  \text{mA}

(b)    By KVL,
V_S = Ri_Z + V_Z         so that       R = \frac{V_s  –  V_Z}{i_z}

From Section 2.10, we know that regulation is preserved if
R \leq \frac{V_S  –  V_Z}{0.1 I_{Z \max}} = \frac{15  –  5.2}{(0.1)(50  \times  10^{-3})} = 1.96  k\Omega

Overcurrent failure is avoided if
R \geq \frac{V_S  –  V_Z}{I_{Z \max}} = \frac{15  –  5.2}{50  \times  10^{-3}} = 196  \Omega

Thus, we need  196  \Omega \leq R \leq 196  k\Omega.