Calculate the dynamic forward and reverse resistance of a PN junction diode when the applied voltage is 0.25 V at T = 300 K given Io = 2 μA.

Question

Calculate the dynamic forward and reverse resistance of a PN junction diode when the applied voltage is 0.25 V at T = 300 K given [latex]I_{o}[/latex] = 2 μA.

Accepted Answer

Given V = 0.25 V, T = 300 K, [latex]I_{o}[/latex] = 2 μA

At T = 300 K, [latex]V_{T}[/latex] = 26 mV.

Assuming it to be silicon diode, η = 2

Therefore, [latex]I=I_{o}\left(\frac{V}{e^{\eta V_{T} }-1 } ight) =2 imes 10^{-6}\left(\frac{0.25}{e^{2 imes 26 imes 10^{-3} }-1 } ight) =0.24 mA[/latex]

[latex]r_{F} =\frac{\eta V_{T}}{I} =\frac{2 imes 26 imes 10^{-3} }{0.24 imes 10^{-3}} =216.67 \Omega [/latex]

For germanium diode, η = 1.

[latex]I=I_{o}\left(\frac{V}{e^{\eta V_{T} }-1 } ight) =2 imes 10^{-6}\left(\frac{0.25}{e^{ 26 imes 10^{-3} }-1 } ight) =0.03 A[/latex]

[latex]r_{F} =\frac{\eta V_{T}}{I} =\frac{ 26 imes 10^{-3} }{0.03} =0.867 \Omega [/latex]

Reverse resistance [latex]\frac{V}{I_{o}}=\frac{0.25}{2 imes 10^{-6} } =125 k \Omega [/latex]

Question 4.18: Calculate the dynamic forward and reverse resistance of a PN...

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