Question

A zener diode of breakdown voltage 10 V is connected to an external voltage of 15 V and a resistance R in series. If power of zener diode is 0.4 W. Find value of unknown resistance R :

• A. 125 $\Omega$
• B. 105 $\Omega$
• C. 130 $\Omega$
• D. 115 $\Omega$

Hint:

In Zener diode circuits, always identify if there is a load resistor. If there is no load resistor, the Zener current is the total circuit current. The voltage across the series resistor is always $(V_{in} - V_{Z})$ as long as the input voltage is greater than the breakdown voltage.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A Zener diode is used as a voltage regulator. It is connected in reverse bias with a series resistor $R$ and a DC voltage source. We are given the Zener breakdown voltage, the source voltage, and the power dissipated by the Zener diode. There is no load resistor mentioned, so all the current passing through $R$ also passes through the Zener diode.
Step 2: Key Formula or Approach:
1. Power dissipated by Zener diode: $P_{Z} = V_{Z} \times I_{Z}$.
2. Current through the Zener diode: $I_{Z} = \frac{P_{Z}}{V_{Z}}$.
3. Since the Zener diode and the resistor R are in series, the current through the resistor $I_{R}$ is equal to $I_{Z}$.
4. Voltage drop across the resistor: $V_{R} = V_{in} - V_{Z}$.
5. Using Ohm's Law, $R = \frac{V_{R}}{I_{R}}$.
Step 3: Detailed Explanation:
Given values:
Breakdown voltage of Zener, $V_{Z} = 10$ V
Input voltage, $V_{in} = 15$ V
Power of Zener diode, $P_{Z} = 0.4$ W
First, find the current flowing through the Zener diode ($I_{Z}$):
$I_{Z} = \frac{P_{Z}}{V_{Z}} = \frac{0.4 \text{ W}}{10 \text{ V}} = 0.04 \text{ A}$
Since there is no external load connected in parallel to the Zener, the entire current from the source flows through the series resistor $R$ and then through the Zener diode.
So, the current through the resistor is $I = I_{Z} = 0.04 \text{ A}$.
The voltage drop across the resistor $R$ is the difference between the source voltage and the Zener voltage:
$V_{R} = V_{in} - V_{Z} = 15 \text{ V} - 10 \text{ V} = 5 \text{ V}$.
Now, applying Ohm's law to the resistor $R$:
$R = \frac{V_{R}}{I} = \frac{5 \text{ V}}{0.04 \text{ A}} = \frac{500}{4} \Omega = 125 \Omega$.
Step 4: Final Answer:
The value of the unknown resistance R is 125 $\Omega$.