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.
Step-by-Step
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.