A zener diode of power rating
Question:

A zener diode of power rating $2 \mathrm{~W}$ is to be used as

a voltage regulator. If the zener diode has a breakdown of $10 \mathrm{~V}$ and it has to regulate voltage fluctuated between $6 \mathrm{~V}$ and $14 \mathrm{~V}$, the value of $\mathrm{R}_{\mathrm{s}}$ for safe operation should be $\Omega .$

Solution:

When unregulated voltage is $14 \mathrm{~V}$ voltage across zener diode must be $10 \mathrm{~V}$ So potential difference across resistor $\Delta \mathrm{V}_{\mathrm{Rs}}=4 \mathrm{~V}$ and $\mathrm{P}_{\text {zener }}=2 \mathrm{~W}$

$\mathrm{VI}=2$

$\mathrm{I}=\frac{2}{10}=0.2 \mathrm{~A}$

$\Delta \mathrm{V}_{\mathrm{Rs}}=\mathrm{IR}_{\mathrm{s}}$

$4 \times 0.2 \mathrm{R}_{\mathrm{s}} \Rightarrow \mathrm{R}_{\mathrm{s}}=\frac{40}{2}=20 \Omega$

Administrator

Leave a comment

Please enter comment.
Please enter your name.