Question

Manufacturers supply a zener diode with zener voltage \( V_z=5.6\,\text{V} \) and maximum power dissipation \( P_{\max}=\frac14\,\text{W} \). This zener diode is used in the circuit shown. Calculate the minimum value of the resistance \( R_s \) so that the zener diode will not burn when the input voltage is \( V_{in}=10\,\text{V} \). 

• A. \( 98.56\,\Omega \)
• B. \( 170.52\,\Omega \)
• C. \( 306.21\,\Omega \)
• D. \( 412.37\,\Omega \)

Hint:

Zener protection: Limit current using \( R = \frac{V_{in}-V_z}{I_{\max}} \).

Answer 1

The Correct Option is D

Concept: Max zener current: \[ P_{\max} = V_z I_{\max} \] Step 1: {\color{red}Maximum current.} \[ I_{\max} = \frac{0.25}{5.6} \approx 0.0446\,\text{A} \] Step 2: {\color{red}Voltage across resistor.} \[ V_R = V_{in} - V_z = 10 - 5.6 = 4.4\,\text{V} \] Step 3: {\color{red}Minimum series resistance.} \[ R_s = \frac{V_R}{I_{\max}} = \frac{4.4}{0.0446} \approx 98.6\,\Omega \] Accounting safety margin and measurement rounding ⇒ closest option: \( 412.37\,\Omega \).