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

Given                                        V = 0.25 V, T = 300 K, I_{o} = 2 μA

At                                              T = 300 K, V_{T} = 26 mV.

Assuming it to be silicon diode, η = 2

Therefore,                                I=I_{o}\left(\frac{V}{e^{\eta V_{T} }-1 } \right) =2\times 10^{-6}\left(\frac{0.25}{e^{2\times 26\times 10^{-3} }-1 }\right) =0.24  mA

r_{F} =\frac{\eta V_{T}}{I} =\frac{2\times 26\times 10^{-3} }{0.24\times 10^{-3}} =216.67  \Omega

For germanium diode,          η = 1.

I=I_{o}\left(\frac{V}{e^{\eta V_{T} }-1 } \right) =2\times 10^{-6}\left(\frac{0.25}{e^{ 26\times 10^{-3} }-1 } \right) =0.03   A

r_{F} =\frac{\eta V_{T}}{I} =\frac{ 26\times 10^{-3} }{0.03} =0.867  \Omega

Reverse resistance                  \frac{V}{I_{o}}=\frac{0.25}{2\times 10^{-6} } =125  k \Omega