Question

A Zener of power rating 1W is to be used as a voltage regulator. If Zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7V, what should be the value of ${R_s}$ ​ for safe operation (Figure)?

Answer 1

Hint: The value of the resistance for safe operation should be such that it can handle the fluctuation in the input voltage. The Zener diode will assist in regulating the voltage by limiting the current in the circuit.
Formula used: In this solution, we will use the following formula
$I = \dfrac{P}{V}$ where $I$ is the current in the Zener diode, $P$ is the power rating of the Zener, and $V$ is the voltage across the Zener diode.
- Ohm’s law: $V = IR$ where $V$ is the potential difference, $I$ is the current in the circuit, and $R$ is the resistance

Complete step by step answer:
We’ve been given the Zener diode in the image has a power rating of 1W and breakdown voltage of $5V$. Then the current in the Zener diode will be determined as
$I = \dfrac{1}{5} = 0.2A$
Now the value of the resistance should be such that it can handle the fluctuation in the voltage for the Zener current that we just calculated. So, using Ohm’s law, we can write
$V = IR$
The potential difference that the resistance has to be sustained is $V = 7 - 3 = 4V$ and the current will be equal to the Zener current. So, its resistance has to be
$R = \dfrac{V}{I} = \dfrac{4}{{0.2}}$
Which gives us
$R = 20\Omega $

Note: When being used as a voltage regulator, the Zener diode is connected in reverse bias mode. The current that we obtained in the Zener diode will remain constant over a range of fluctuating input voltages hereby maintaining a steady voltage difference to the load resistance despite fluctuating input voltages. For the Zener diode to regulate the power, there is a minimum Zener current that must be maintained for which the supply voltage must be greater than the breakdown voltage of the Zener diode.