Question

A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _______ \(\Omega\).

Answer 1

Correct Answer: 300

Step 1: Understanding the Concept:
A Zener diode is used as a voltage regulator. To prevent it from overheating and being destroyed, the current flowing through it must not exceed a maximum threshold determined by its power rating. A series resistor is used to drop the excess voltage and limit this current.
Step 2: Key Formula or Approach:
Maximum allowed Zener current: $I_{Z(max)} = \frac{P_{max}}{V_Z}$
Voltage across the series resistor: $V_s = V_{in} - V_Z$
Minimum safe series resistance: $R_{min} = \frac{V_s}{I_{Z(max)}}$
Step 3: Detailed Explanation:
Given parameters:
Zener voltage $V_Z = 10 \text{ V}$
Maximum power dissipation $P_{max} = 0.5 \text{ W}$
Source voltage $V_{in} = 25 \text{ V}$
First, calculate the maximum safe current that the Zener diode can handle.
\[ I_{Z(max)} = \frac{P_{max}}{V_Z} = \frac{0.5 \text{ W}}{10 \text{ V}} = 0.05 \text{ A} \] When connected in the circuit, the Zener diode will lock the voltage across its terminals at $10 \text{ V}$. The remaining voltage from the supply must be dropped across the series resistor $R$.
\[ V_s = V_{in} - V_Z = 25 \text{ V} - 10 \text{ V} = 15 \text{ V} \] To ensure the Zener diode does not blow out under the worst-case scenario (when no load is attached, and all current flows through the Zener), the series resistor must restrict the total circuit current to exactly $I_{Z(max)}$.
\[ R_{min} = \frac{V_s}{I_{Z(max)}} = \frac{15 \text{ V}}{0.05 \text{ A}} \] \[ R_{min} = \frac{15}{0.05} = 300\ \Omega \] Step 4: Final Answer:
The minimum resistance required is $300\ \Omega$.