Question

A zener diode of power rating 2W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of $R_s$ for safe operation should be _______ $\Omega$

Answer 1

20

Answer 2

The maximum power dissipated by the Zener diode is $P_{z\_max} = 2$W, and its breakdown voltage is $V_z = 10$V. The maximum Zener current is $I_{z\_max} = \frac{P_{z\_max}}{V_z} = \frac{2W}{10V} = 0.2A$. The input voltage fluctuates between $V_{in\_min} = 6$V and $V_{in\_max} = 14$V. For safe operation, the Zener diode must be in breakdown, so the regulated voltage is $V_z = 10$V. The Zener current is given by $I_z = \frac{V_{in} - V_z}{R_s} - I_L$. To ensure safe operation, $I_z \le I_{z\_max}$. The Zener current is maximum when $V_{in}$ is maximum and $I_L$ is minimum. Assuming the minimum load current $I_{L\_min} = 0$, at $V_{in\_max} = 14$V: $I_{z\_max} = \frac{14V - 10V}{R_s} - 0$ $0.2A = \frac{4V}{R_s}$ $R_s = \frac{4V}{0.2A} = 20 \Omega$. Thus, $R_s$ must be at least $20 \Omega$ for safe operation.