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Chapter 11

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)

Step-by-Step

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.