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The current flowing through a certain P-n junction at room temperature when reverse-biased (bias voltage being very large in comparison to VT) is 0.15μA. Determine the current (in μA) flowing through the diode when applied voltage is 0.12V. (VT = 26 mV)
Correct answer is '15'. Can you explain this answer?
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The current flowing through a certain P-n junction at room temperature...
Large reverse bias current - (I0)= 0.15 x 10-6 A
V = 0.12V & VT = 26 mV = 0.026 V.
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The current flowing through a certain P-n junction at room temperature...
The current flowing through a certain P-n junction at room temperature when reverse-biased can be approximated by the reverse saturation current, IS, of the junction.
The reverse saturation current is given by the equation:
IS = IS0 * e^(V / VT)
where IS0 is the reverse saturation current at 0V bias, V is the bias voltage, and VT is the thermal voltage.
In this case, the bias voltage is very large compared to VT, so e^(V / VT) can be approximated to a very large value.
Therefore, if the current flowing through the junction when reverse-biased is 0.15A, it can be concluded that the reverse saturation current IS0 is very small, or close to zero.
This could be due to factors such as low doping concentration in the junction or high-quality materials used in the fabrication of the junction.
The reverse saturation current is given by the equation:
IS = IS0 * e^(V / VT)
where IS0 is the reverse saturation current at 0V bias, V is the bias voltage, and VT is the thermal voltage.
In this case, the bias voltage is very large compared to VT, so e^(V / VT) can be approximated to a very large value.
Therefore, if the current flowing through the junction when reverse-biased is 0.15A, it can be concluded that the reverse saturation current IS0 is very small, or close to zero.
This could be due to factors such as low doping concentration in the junction or high-quality materials used in the fabrication of the junction.
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The current flowing through a certain P-n junction at room temperature when reverse-biased (bias voltage being very large in comparison to VT) is 0.15μA. Determine the current (in μA) flowing through the diode when applied voltage is 0.12V. (VT = 26 mV)Correct answer is '15'. Can you explain this answer?
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