The I-V characteristics of a semiconductor di | vedclass

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Question

(A) $1$. For the reverse bias condition at $V = -8 	ext{ V}$,the current is $I = -1 	ext{ }\mu	ext{A} = -1 	imes 10^{-6} 	ext{ A}$.Resistance $R

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$8 	ext{ M}\Omega, 20 	ext{ }\Omega$

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(A) $1$. For the reverse bias condition at $V = -8 	ext{ V}$,the current is $I = -1 	ext{ }\mu	ext{A} = -1 	imes 10^{-6} 	ext{ A}$.Resistance $R = \frac{|V|}{|I|} = \frac{8 	ext{ V}}{1 	imes 10^{-6} 	ext{ A}} = 8 	imes 10^{6} 	ext{ }\Omega = 8 	ext{ M}\Omega$.$2$. For the forward bias condition at $I_{d} = 10 	ext{ mA}$,we calculate the dynamic resistance $R_{d} = \frac{\Delta V}{\Delta I}$.From the graph,we take the points around $I = 10 	ext{ mA}$ (which corresponds to $V = 0.7 	ext{ V}$ approximately,but we can use the given points $0.6 	ext{ V}$ at $5 	ext{ mA}$ and $0.8 	ext{ V}$ at $15 	ext{ mA}$):$R_{d} = \frac{0.8 	ext{ V} - 0.6 	ext{ V}}{(15 	ext{ mA} - 5 	ext{ mA})} = \frac{0.2 	ext{ V}}{10 	imes 10^{-3} 	ext{ A}} = \frac{0.2}{0.01} 	ext{ }\Omega = 20 	ext{ }\Omega$.Thus,the resistances are $8 	ext{ M}\Omega$ and $20 	ext{ }\Omega$.

The $I-V$ characteristics of a semiconductor diode are shown in the figure. The resistance of the diode at $V = -8 \text{ V}$ and $I_{d} = 10 \text{ mA}$ are respectively:

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