In a p-n junction diode the reverse saturation current is `10 muA`. What will be the forward current for a voltager of 0.2 V ?

Answer

To find the forward current for a p-n junction diode when a voltage of 0.2 V is applied, we can use the diode equation:

\[ I = I_0 (e^{\frac{qV}{kT}} - 1) \]

Where:
- \( I \) is the forward current,
- \( I_0 \) is the reverse saturation current,
- \( V \) is the applied voltage,
- \( q \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) C),
- \( k \) is Boltzmann's constant (approximately \( 1.38 \times 10^{-23} \) J/K),
- \( T \) is the absolute temperature in Kelvin (assumed to be around 300 K for room temperature).

Given:
- \( I_0 = 10 \, \mu A = 10 \times 10^{-6} \, A \)
- \( V = 0.2 \, V \)

### Step 1: Calculate the value of \( \frac{qV}{kT} \)
Using the values:
- \( q \approx 1.6 \times 10^{-19} \, C \)
- \( k \approx 1.38 \times 10^{-23} \, J/K \)
- \( T \approx 300 \, K \)

First, calculate \( kT \):
\[ kT = (1.38 \times 10^{-23} \, J/K)(300 \, K) \approx 4.14 \times 10^{-21} \, J \]

Now calculate \( \frac{qV}{kT} \):
\[ \frac{qV}{kT} = \frac{(1.6 \times 10^{-19} \, C)(0.2 \, V)}{4.14 \times 10^{-21} \, J} \]

Calculating the numerator:
\[ 1.6 \times 10^{-19} \times 0.2 = 3.2 \times 10^{-20} \, C \cdot V \]

Now, divide by \( kT \):
\[ \frac{3.2 \times 10^{-20}}{4.14 \times 10^{-21}} \approx 7.72 \]

### Step 2: Substitute into the diode equation
Now substitute this value back into the diode equation:
\[ I = I_0 (e^{\frac{qV}{kT}} - 1) \]
Since \( e^{\frac{qV}{kT}} \) is much larger than 1, we can approximate:
\[ I \approx I_0 e^{\frac{qV}{kT}} \]

Substituting \( I_0 = 10 \times 10^{-6} \, A \):
\[ I \approx (10 \times 10^{-6}) e^{7.72} \]

### Step 3: Calculate \( e^{7.72} \)
Using a calculator:
\[ e^{7.72} \approx 2245.3 \]

### Step 4: Calculate the forward current
Now calculate \( I \):
\[ I \approx (10 \times 10^{-6}) \times 2245.3 \]
\[ I \approx 22.453 \, mA \]

Thus, the forward current for a voltage of 0.2 V is approximately:
\[ I \approx 22.5 \, mA \]

### Final Answer:
The forward current for a voltage of 0.2 V is approximately **22.5 mA**.

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