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(A) For a silicon diode,the knee voltage or barrier potential is approximately $0.7\, V$.The bulk resistance $(r_f)$ is calculated using the excess vo

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$5\,\Omega, 10\, M\Omega$

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(A) For a silicon diode,the knee voltage or barrier potential is approximately $0.7\, V$.The bulk resistance $(r_f)$ is calculated using the excess voltage across the diode:$r_f = \frac{V_f - V_k}{I_f} = \frac{1.2\, V - 0.7\, V}{100 	imes 10^{-3}\, A} = \frac{0.5\, V}{0.1\, A} = 5\,\Omega$.The reverse resistance $(r_r)$ is calculated using Ohm's law for the reverse bias condition:$r_r = \frac{V_r}{I_r} = \frac{10\, V}{1 	imes 10^{-6}\, A} = 10 	imes 10^6\, \Omega = 10\, M\Omega$.Thus,the bulk resistance is $5\,\Omega$ and the reverse resistance is $10\, M\Omega$.

$A$ silicon diode has a forward voltage drop of $1.2\, V$ for a forward $DC$ current of $100\, mA$. It has a reverse current of $1\times10^{-6}\, A$ for a reverse voltage of $10\, V$. The bulk and reverse resistances of the diode are:

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