Potential drop across forward junction p-n diode is `0.7` V. If a battery of 4 V is applied, calculate the resistance to be put in series, if the maximum current in the circuit is `5 mA`.
Hint: `R = 4 - 07/5 xx 10^(-3) = 660 Omega`

Options

A. `660 Omega`
B. `350 Omega`
C. `475 Omega`
D. `500 Omega`

Answer

To solve the problem of calculating the resistance to be put in series with a forward-biased p-n diode, we can follow these steps:

### Step 1: Understand the Circuit
We have a battery of 4 V connected in series with a p-n diode that has a forward voltage drop of 0.7 V. We need to find the resistance (R) that will limit the current to a maximum of 5 mA.

### Step 2: Calculate the Voltage Across the Resistor
The total voltage provided by the battery is 4 V, and the voltage drop across the diode is 0.7 V. Therefore, the voltage across the resistor (V_R) can be calculated as:
\[ V_R = V_{battery} - V_{diode} = 4V - 0.7V = 3.3V \]

### Step 3: Use Ohm's Law to Find the Resistance
Ohm's Law states that:
\[ V = I \times R \]
Where:
- \( V \) is the voltage across the resistor (3.3 V),
- \( I \) is the current flowing through the circuit (5 mA = 5 × 10^{-3} A),
- \( R \) is the resistance we want to find.

Rearranging Ohm's Law to solve for \( R \):
\[ R = \frac{V_R}{I} = \frac{3.3V}{5 \times 10^{-3} A} \]

### Step 4: Calculate the Resistance
Now, substituting the values into the equation:
\[ R = \frac{3.3}{5 \times 10^{-3}} = \frac{3.3}{0.005} = 660 \, \Omega \]

### Conclusion
The resistance to be put in series with the p-n diode is **660 Ω**.

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