Question

A zener diode of power rating 2 W 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 _________ \(\Omega\). 

 

Hint:

When designing a Zener regulator circuit for safety, always consider the worst-case scenario. For the Zener diode itself, this means maximum input voltage and minimum load current (often zero). This combination leads to the maximum possible current flowing through the Zener, which must not exceed its power rating limit.

Answer 1

Correct Answer: 20

Step 1: Understanding the Question:
We need to find the value of the series resistance \(R_S\) required for the safe operation of a Zener diode regulator. "Safe operation" implies that the power dissipated by the Zener diode does not exceed its maximum power rating.
Step 2: Key Formula or Approach:
1. The Zener diode maintains a constant voltage \(V_Z\) across it.
2. The maximum current the Zener can safely handle is \( I_{Z,max} = \frac{P_Z}{V_Z} \).
3. The current through the series resistor is \( I_S = \frac{V_{in} - V_Z}{R_S} \).
4. The total current \(I_S\) splits into the Zener current \(I_Z\) and the load current \(I_L\): \( I_S = I_Z + I_L \).
5. The worst-case condition for the Zener (maximum power dissipation) occurs when the input voltage is maximum and the load current is minimum (or zero if no load is specified).
Step 3: Detailed Explanation:
Given values:
- Power rating of Zener diode, \( P_Z = 2 \) W.
- Zener breakdown voltage, \( V_Z = 10 \) V.
- Input voltage range, \( V_{in} = 6 \) V to \( 14 \) V.
First, note that the Zener diode will only regulate when \( V_{in} \ge V_Z \). So, the operational input range is actually from 10 V to 14 V. The 6 V value is below the breakdown voltage.
Calculate the maximum safe Zener current:
\[ I_{Z,max} = \frac{P_Z}{V_Z} = \frac{2 \, \text{W}}{10 \, \text{V}} = 0.2 \, \text{A} \] The Zener current \(I_Z\) will be maximum when the input voltage \(V_{in}\) is maximum and the load current \(I_L\) is minimum. Since no load is specified in the problem, we assume the worst-case scenario where there is no load connected, i.e., \( I_L = 0 \).
In this case, all the current from the source resistor passes through the Zener diode: \( I_S = I_Z \).
This maximum current occurs at the maximum input voltage, \( V_{in,max} = 14 \) V. To ensure safe operation, we must limit this current to be no more than \(I_{Z,max}\).
\[ I_S \le I_{Z,max} \] \[ \frac{V_{in,max} - V_Z}{R_S} \le I_{Z,max} \] To find the minimum resistance \(R_S\) that guarantees safety, we use the equality:
\[ R_S = \frac{V_{in,max} - V_Z}{I_{Z,max}} \] \[ R_S = \frac{14 \, \text{V} - 10 \, \text{V}}{0.2 \, \text{A}} = \frac{4 \, \text{V}}{0.2 \, \text{A}} = 20 \, \Omega \] A resistance of 20 \(\Omega\) will ensure that even under the worst conditions (14 V input, no load), the current through the Zener will not exceed its safe limit.
Step 4: Final Answer:
The value of R\(_S\) for safe operation should be 20 \(\Omega\).