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## Q. 11.4

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 $I_{0}$ in the forward bias case and then use that $I_{0}$ to find the current in reverse bias. But it is simpler to use the fact that the forward bias current $I_{f}$ and the reverse bias current $I_{r}$ are related by

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

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

## Verified Solution

Using the current ratio along with the numerical values $I_{f}=50 mA , V_{f}=+200 mV , \text { and } V_{r}=-200 mV$, we find that

$\frac{e V_{f}}{k T}=\frac{\left(1.602 \times 10^{-19} C \right)(0.200 V )}{\left(1.38 \times 10^{-23} J / K \right)(293 K )}=7.924$

Therefore

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

or $I_{r}=-18 \mu A$

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