Question

A diode current increases from 1 mA to 10 mA. Dynamic resistance changes by:

• A. 10\( \times \) increase
• B. 10\( \times \) decrease
• C. Same
• D. 100\( \times \) decrease

Hint:

Dynamic resistance of a diode is inversely proportional to current: \( r_d \propto \frac{1}{I} \).
If current increases by any factor \( k \), the dynamic resistance decreases by the same factor \( k \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to calculate the change in the dynamic (or AC) resistance of a semiconductor diode when the forward operating current increases from 1 mA to 10 mA.
Dynamic resistance is the resistance offered by the diode to a small AC signal at a specific DC operating point.

Step 2: Key Formula or Approach:
The current-voltage relationship of a forward-biased diode is given by the Shockley diode equation:
\[ I = I_s \left( e^{\frac{V}{\eta V_T}} - 1 \right) \approx I_s e^{\frac{V}{\eta V_T}} \] The dynamic resistance (\( r_d \)) is defined as the reciprocal of the derivative of current with respect to voltage:
\[ r_d = \frac{dV}{dI} \] Differentiating the diode equation:
\[ \frac{dI}{dV} = \frac{I_s e^{\frac{V}{\eta V_T}}}{\eta V_T} = \frac{I}{\eta V_T} \] Therefore, the dynamic resistance is:
\[ r_d = \frac{\eta V_T}{I} \] Where:
\( \eta \) is the ideality factor of the diode.
\( V_T \) is the thermal voltage (approximately 26 mV at room temperature).
\( I \) is the DC operating current.

Step 3: Detailed Explanation:
Let us analyze the relationship and perform the calculation:

• Proportionality Relationship: From the formula, since \( \eta \) and \( V_T \) are physical constants, the dynamic resistance is inversely proportional to the operating current:
\[ r_d \propto \frac{1}{I} \]
• Ratio Calculation:
- Let the initial current be \( I_1 = 1\text{ mA} \) and the corresponding dynamic resistance be \( r_{d1} \).
- Let the final current be \( I_2 = 10\text{ mA} \) and the corresponding dynamic resistance be \( r_{d2} \).
- Setting up the ratio:
\[ \frac{r_{d2}}{r_{d1}} = \frac{I_1}{I_2} = \frac{1\text{ mA}}{10\text{ mA}} = \frac{1}{10} \] - This simplifies to:
\[ r_{d2} = \frac{r_{d1}}{10} \] - This result shows that the dynamic resistance decreases to one-tenth of its original value, which is a \( 10\times \) decrease.

Step 4: Final Answer:
When the diode current increases from 1 mA to 10 mA, the dynamic resistance experiences a \( 10\times \) decrease.