Question

The resistance of p-n junction diode in forward bias is 25 ohm. How much voltage in forward bias be changed so that the change in current would be 2 mA?

Hint:

For diode problems, always be careful to distinguish between static resistance (\(R = V/I\)) and dynamic resistance (\(R = \Delta V / \Delta I\)). Dynamic resistance is relevant for small signal AC analysis, as in this question. Also, remember to convert units like mA and k\(\Omega\) to base SI units (A and \(\Omega\)) before calculation.

Answer 1

Step 1: Understanding the Concept:
The resistance of a p-n junction diode in forward bias mentioned here is the dynamic or AC resistance (\(R_f\)). It is not the static resistance (\(V/I\)), but rather the ratio of a small change in voltage across the diode to the resulting small change in the current flowing through it.

Step 2: Key Formula or Approach:
The dynamic forward resistance (\(R_f\)) is defined by the relation: \[ R_f = \frac{\Delta V_f}{\Delta I_f} \] where \(\Delta V_f\) is the change in forward bias voltage and \(\Delta I_f\) is the corresponding change in forward current. We need to find \(\Delta V_f\).

Step 3: Detailed Explanation:
We are given the following values: \begin{itemize} \item Forward bias resistance, \(R_f = 25 \, \Omega\). \item Change in current, \(\Delta I_f = 2 \, \text{mA}\). We must convert this to Amperes: \[ \Delta I_f = 2 \times 10^{-3} \, \text{A} \] \end{itemize} Rearranging the formula to solve for the change in voltage \(\Delta V_f\): \[ \Delta V_f = R_f \times \Delta I_f \] Substituting the given values: \[ \Delta V_f = (25 \, \Omega) \times (2 \times 10^{-3} \, \text{A}) \] \[ \Delta V_f = 50 \times 10^{-3} \, \text{V} \] \[ \Delta V_f = 0.05 \, \text{V} \]

Step 4: Final Answer:
The forward bias voltage should be changed by 0.05 Volts.