Assume that the Zener diode has a constant reverse break down voltage for a current range starting from a minimum required Zener current I_Zmin = 2 mA to its maximum allowable current. The input voltage V_I varies from 7 V to 30 V. The value of R will be :

Concept:

To maintain a constant output voltage using a Zener diode, a series resistor R is used to limit the current through the diode and the load. The resistor R must be chosen such that the Zener current remains between its minimum and maximum values across the input voltage range.

Given:

Zener voltage, \( V_Z = 5\,V \)

Minimum Zener current, \( I_{Z_{min}} = 2\,mA \)

Load resistance, \( R_L = 1\,k\Omega \)

Calculation:

To calculate value of R, use worst-case (maximum voltage) to ensure Zener diode doesn’t exceed max current.

Load current, \( I_L = \frac{V_Z}{R_L} = \frac{5}{1000} = 5\,mA \)

Total current at \( V_i = 30\,V \) is: \( I = I_L + I_{Z_{max}} \)

Assume max allowable current through Zener is unknown, so we ensure Zener operates within range at minimum current.

At minimum voltage, \( V_i = 7\,V \), current through R should be enough to provide both \( I_L = 5\,mA \) and \( I_{Z_{min}} = 2\,mA \)

Total current required: \( I = 7\,mA \)

\( R = \frac{V_i - V_Z}{I} = \frac{7 - 5}{0.007} = \frac{2}{0.007} = 285.7\,\Omega \)

To ensure Zener diode doesn't exceed its maximum current at \( V_i = 30\,V \):

\( I = \frac{30 - 5}{R} = \frac{25}{R} \)

Let Zener maximum current \( I_Z \) be the remaining after supplying the load current (5 mA):

\( I_Z = \frac{25}{R} - 5 \)

Now try values from options. For R = 186 Ω:

\( I = \frac{25}{186} \approx 0.1344\,A \Rightarrow 134.4\,mA \)

\( I_Z = 134.4 - 5 = 129.4\,mA \) → Safe if Zener is rated for it.

Correct Answer: 1) 186 Ω