Question

In a forward biased semiconductor potential changes from 0.9V to 0.6 V. Find current if resistance is $1\Omega$

• A. 300 mA
• B. 200 mA
• C. 450 mA

Hint:

When dealing with active semiconductor devices operating in non-linear regions, always utilize the dynamic resistance ($\frac{\Delta V}{\Delta I}$) rather than the static resistance ($\frac{V}{I}$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a forward-biased semiconductor diode, the voltage-current relationship is intrinsically non-linear.
However, for small localized variations, the change in potential is linearly related to the change in current through a parameter called dynamic resistance.
Step 2: Key Formula or Approach:
The dynamic resistance $r_d$ of a semiconductor diode is explicitly defined by the formula $r_d = \frac{\Delta V}{\Delta I}$.
We can easily rearrange this relationship to solve for the change in current as $\Delta I = \frac{\Delta V}{r_d}$.
Step 3: Detailed Explanation:
First, compute the magnitude of the change in the applied potential difference:
\[ \Delta V = |0.9\text{ V} - 0.6\text{ V}| = 0.3\text{ V} \] The given dynamic resistance for this specific semiconductor region is $r_d = 1\ \Omega$.
Using the rearranged formula to calculate the corresponding current change:
\[ \Delta I = \frac{0.3\text{ V}}{1\ \Omega} = 0.3\text{ A} \] To formally convert the calculated current from Amperes to milliamperes (mA), multiply by 1000:
\[ \Delta I = 0.3 \times 1000\text{ mA} = 300\text{ mA} \] Step 4: Final Answer:
The resulting change in current is $300\text{ mA}$.