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  $196 \Omega \leq R \leq 196 k\Omega$.