Question

In the given circuit diagram, $I_L = 5$ mA and the Zener voltage is $V_Z = 5$ V. If $i_z = 4 i_L$, find the value of $R_s$ (in $\Omega$).

Hint:

In Zener regulator problems, always apply current division: $I_s = I_L + I_Z$ and remember that the Zener voltage remains constant across the load.

Answer 1

Correct Answer: 800

Given data: $I_L = 5\,\text{mA}$, $V_Z = 5\,\text{V}$, $i_z = 4i_L$.
To find: $R_s$.

First, calculate $i_z$:
$i_z = 4 \times i_L = 4 \times 5\,\text{mA} = 20\,\text{mA}$.

Total current through $R_s$ ($I_t$) is:
$I_t = i_z + i_L = 20\,\text{mA} + 5\,\text{mA} = 25\,\text{mA}$.

Apply voltage drop across $R_s$:
Source voltage $= 25\,\text{V}$
Zener voltage $= 5\,\text{V}$
Voltage across $R_s = 25\,\text{V} - 5\,\text{V} = 20\,\text{V}$.

Ohm's Law ($V = IR$) for $R_s$:
$R_s = \frac{\text{Voltage across } R_s}{I_t} = \frac{20\,\text{V}}{25\,\text{mA}} = \frac{20\,\text{V}}{0.025\,\text{A}} = 800\,\Omega$.

The calculated $R_s$ is $800\,\Omega$, which falls in the expected range [800, 800].