Consider a simple pn-junction diode. Suppose this diode carries a current of 50 mA with a forward bias voltage of 200 mV at room temperature (293 K). What is the current when a reverse bias of 200 mV is applied? Strategy We could use Equation (11.9) to solve for I0 in the forward bias case and then

Question

Consider a simple pn-junction diode. Suppose this diode carries a current of 50 mA with a forward bias voltage of 200 mV at room temperature (293 K). What is the current when a reverse bias of 200 mV is applied?

Strategy We could use Equation (11.9) to solve for [latex]I_{0}[/latex] in the forward bias case and then use that [latex]I_{0}[/latex] to find the current in reverse bias. But it is simpler to use the fact that the forward bias current [latex]I_{f}[/latex] and the reverse bias current [latex]I_{r}[/latex] are related by

[latex]\frac{I_{r}}{I_{f}}=\frac{I_{0}\left(e^{e V_{r} / k T}-1\right)}{I_{0}\left(e^{e V_{f} / k T}-1\right)}[/latex]

[latex]I=I_{0}\left(e^{e V / k T}-1\right)[/latex] (11.9)

Accepted Answer

Using the current ratio along with the numerical values [latex]I_{f}=50 mA , V_{f}=+200 mV , ext { and } V_{r}=-200 mV[/latex], we find that

[latex]\frac{e V_{f}}{k T}=\frac{\left(1.602 imes 10^{-19} C ight)(0.200 V )}{\left(1.38 imes 10^{-23} J / K ight)(293 K )}=7.924[/latex]

Therefore

[latex]I_{r}=I_{f} \frac{e^{e V_{r} / k T}-1}{e^{e V_{f} / k T}-1}=(50 mA ) \frac{e^{-7.924}-1}{e^{7.924}-1}=-0.018 mA[/latex]

or [latex]I_{r}=-18 \mu A[/latex]

This is an extremely small current, showing clearly the “oneway” property of the diode.

Question 11.4: Consider a simple pn-junction diode. Suppose this diode carr...

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