Question
Calculate the voltage drop across the junction of a germanium diode with a reverse saturation current of 10×10-⁶ is it conducts 3mA at a room temperature of 30°C
Answers
Answered by
GPT-4o mini
To calculate the voltage drop across a germanium diode, we can use the Shockley diode equation:
\[
I = I_s \left( e^{\frac{qV}{n k T}} - 1 \right)
\]
Where:
- \( I \) is the current through the diode (in Amperes),
- \( I_s \) is the reverse saturation current (in Amperes),
- \( V \) is the voltage across the diode (in Volts),
- \( q \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs),
- \( n \) is the ideality factor (typically around 1 for germanium diodes),
- \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \) J/K),
- \( T \) is the absolute temperature (in Kelvin).
1. First, convert the room temperature from Celsius to Kelvin:
\[
T = 30°C + 273.15 = 303.15 \, K
\]
2. Given values:
- Reverse saturation current \( I_s = 10 \times 10^{-6} \, A = 1.0 \times 10^{-5} \, A \)
- Current \( I = 3 \, mA = 3 \times 10^{-3} \, A \)
- Ideality factor \( n \approx 1 \)
- Charge \( q \approx 1.6 \times 10^{-19} \, C \)
- Boltzmann constant \( k \approx 1.38 \times 10^{-23} \, J/K \)
3. Rearranging the Shockley diode equation to solve for \( V \):
\[
V = \frac{n k T}{q} \ln{\left( \frac{I}{I_s} + 1 \right)}
\]
4. Now substitute the known values into this equation:
- Calculate \( \frac{n k T}{q} \):
\[
\frac{n k T}{q} = \frac{(1)(1.38 \times 10^{-23} \, J/K)(303.15 \, K)}{1.6 \times 10^{-19} \, C} \approx 0.0257 \, V \, (\text{approximately 25.7 mV})
\]
5. Now, calculate \( \frac{I}{I_s} + 1 \):
\[
\frac{I}{I_s} = \frac{3 \times 10^{-3}}{1.0 \times 10^{-5}} = 300 \implies \frac{I}{I_s} + 1 = 301
\]
6. Apply these values in the \( V \) equation:
\[
V \approx 0.0257 \, \text{V} \cdot \ln(301) \approx 0.0257 \cdot 5.011 \approx 0.1287 \, V \, \text{(approximately 128.7 mV)}
\]
Therefore, the voltage drop across the germanium diode when it conducts 3 mA at a room temperature of 30°C is approximately **128.7 mV**.
\[
I = I_s \left( e^{\frac{qV}{n k T}} - 1 \right)
\]
Where:
- \( I \) is the current through the diode (in Amperes),
- \( I_s \) is the reverse saturation current (in Amperes),
- \( V \) is the voltage across the diode (in Volts),
- \( q \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs),
- \( n \) is the ideality factor (typically around 1 for germanium diodes),
- \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \) J/K),
- \( T \) is the absolute temperature (in Kelvin).
1. First, convert the room temperature from Celsius to Kelvin:
\[
T = 30°C + 273.15 = 303.15 \, K
\]
2. Given values:
- Reverse saturation current \( I_s = 10 \times 10^{-6} \, A = 1.0 \times 10^{-5} \, A \)
- Current \( I = 3 \, mA = 3 \times 10^{-3} \, A \)
- Ideality factor \( n \approx 1 \)
- Charge \( q \approx 1.6 \times 10^{-19} \, C \)
- Boltzmann constant \( k \approx 1.38 \times 10^{-23} \, J/K \)
3. Rearranging the Shockley diode equation to solve for \( V \):
\[
V = \frac{n k T}{q} \ln{\left( \frac{I}{I_s} + 1 \right)}
\]
4. Now substitute the known values into this equation:
- Calculate \( \frac{n k T}{q} \):
\[
\frac{n k T}{q} = \frac{(1)(1.38 \times 10^{-23} \, J/K)(303.15 \, K)}{1.6 \times 10^{-19} \, C} \approx 0.0257 \, V \, (\text{approximately 25.7 mV})
\]
5. Now, calculate \( \frac{I}{I_s} + 1 \):
\[
\frac{I}{I_s} = \frac{3 \times 10^{-3}}{1.0 \times 10^{-5}} = 300 \implies \frac{I}{I_s} + 1 = 301
\]
6. Apply these values in the \( V \) equation:
\[
V \approx 0.0257 \, \text{V} \cdot \ln(301) \approx 0.0257 \cdot 5.011 \approx 0.1287 \, V \, \text{(approximately 128.7 mV)}
\]
Therefore, the voltage drop across the germanium diode when it conducts 3 mA at a room temperature of 30°C is approximately **128.7 mV**.