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, we need to calculate the resistance that should be placed in series with a forward-biased PN junction diode when a 4V battery is applied, and the maximum current in the circuit is 5mA.

### Step-by-Step Solution:

1. **Identify the Voltage Drop Across the Diode:**
The potential drop across the forward junction of the PN diode is given as \( V_D = 0.7 \, V \).

2. **Determine the Total Voltage from the Battery:**
The total voltage provided by the battery is \( V_{battery} = 4 \, V \).

3. **Calculate the Voltage Across the Resistor:**
According to Kirchhoff's Voltage Law (KVL), the sum of the voltage drops in the circuit must equal the total voltage supplied by the battery. Therefore, the voltage across the resistor \( V_R \) can be calculated as:
\[ V_R = V_{battery} - V_D = 4 \, V - 0.7 \, V = 3.3 \, V \]

4. **Determine the Maximum Current:**
The maximum current in the circuit is given as \( I = 5 \, mA = 5 \times 10^{-3} \, A \).

5. **Use Ohm's Law to Calculate Resistance:**
Ohm's Law states that \( V = IR \). Rearranging this gives us:
\[ R = \frac{V_R}{I} \]
Substituting the values we have:
\[ R = \frac{3.3 \, V}{5 \times 10^{-3} \, A} = \frac{3.3}{0.005} = 660 \, \Omega \]

6. **Conclusion:**
The resistance that should be put in series with the diode is \( R = 660 \, \Omega \).

### Final Answer:
The resistance to be put in series is \( 660 \, \Omega \).

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