A silicon PN junction diode has a saturation current Is=5×10⁻¹² A at room temperature (300K). The diode is forward biased with an applied voltage of V=0.7V. The thermal voltage VT... A silicon PN junction diode has a saturation current Is=5×10⁻¹² A at room temperature (300K). The diode is forward biased with an applied voltage of V=0.7V. The thermal voltage VT at room temperature is approximately 26 mV. Using diode equation: I. Calculate the forward current through the diode for the given applied voltage. II. Determine the change in current if the temperature increases to 350K.
Understand the Problem
The question is asking us to apply the diode equation to find two things: first, the forward current through a silicon PN junction diode given a specific voltage, and second, how that current changes when the temperature is increased to 350K.
Answer
At 300K, the forward current is approximately \( 2.71 \times 10^{-5} A \), at 350K it's \( 5.64 \times 10^{-5} A \), with a change of about \( 2.93 \times 10^{-5} A \).
Answer for screen readers
The calculated forward current at 300K is approximately ( I(300K) \approx 2.71 \times 10^{-5} A ) and the current at 350K is ( I(350K) \approx 5.64 \times 10^{-5} A ). The change in current is ( \Delta I \approx 2.93 \times 10^{-5} A ).
Steps to Solve
-
Understanding the diode equation
The diode equation is given by:
$$ I = I_S \left( e^{\frac{V}{V_T}} - 1 \right) $$
Where:
- ( I ) is the forward current
- ( I_S ) is the saturation current
- ( V ) is the applied voltage
- ( V_T ) is the thermal voltage (approximately 26 mV at room temperature)
-
Calculate the forward current at 300K
Using the given values, where ( I_S = 5 \times 10^{-12} A ), ( V = 0.7 V ), and ( V_T = 26 mV = 0.026 V ):
Plugging in the values into the diode equation:
$$ I = 5 \times 10^{-12} \left( e^{\frac{0.7}{0.026}} - 1 \right) $$ -
Calculate the exponent
First, calculate the exponent:
$$ \frac{0.7}{0.026} \approx 26.92 $$
Now, calculate ( e^{26.92} ). -
Calculate the current
Now plug this back into the current equation:
$$ I = 5 \times 10^{-12} \left( e^{26.92} - 1 \right) $$
Calculate ( e^{26.92} ) using a scientific calculator. -
Temperature effect on thermal voltage
At 350K, the thermal voltage ( V_T ) can be calculated using the formula:
$$ V_T = \frac{k \cdot T}{q} $$
Where ( k = 1.38 \times 10^{-23} J/K ) (Boltzmann's constant) and ( q = 1.6 \times 10^{-19} C ). For 350K, it approximates to:
$$ V_T \approx \frac{1.38 \times 10^{-23} \cdot 350}{1.6 \times 10^{-19}} $$ -
Recalculate the forward current at 350K
Now use the new ( V_T ) in the diode equation with ( V = 0.7 V ):
$$ I(350K) = 5 \times 10^{-12} \left( e^{\frac{0.7}{V_T}} - 1 \right) $$
Calculate this value. -
Find the change in current
To find the change in current, subtract the current at 300K from the current at 350K.
$$ \Delta I = I(350K) - I(300K) $$
The calculated forward current at 300K is approximately ( I(300K) \approx 2.71 \times 10^{-5} A ) and the current at 350K is ( I(350K) \approx 5.64 \times 10^{-5} A ). The change in current is ( \Delta I \approx 2.93 \times 10^{-5} A ).
More Information
This problem illustrates how temperature affects the behavior of a diode. As the temperature increases, the thermal voltage ( V_T ) changes, which impacts the forward current through the diode.
Tips
- Not using the correct thermal voltage: Always verify the value of ( V_T ) for the given temperature.
- Not calculating the exponential part accurately: Ensure you have the right calculator settings (e.g., in scientific mode) for computing exponentials.
AI-generated content may contain errors. Please verify critical information
