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(C) For forward biasing,the change in voltage is $\Delta V = 2.4 \, V - 2.0 \, V = 0.4 \, V$.The change in current is $\Delta I = 80 \, mA - 60 \, mA

Vedclass · Accepted Answer

$20 \, \Omega, 8 	imes 10^6 \, \Omega$

Vedclass · Answer

(C) For forward biasing,the change in voltage is $\Delta V = 2.4 \, V - 2.0 \, V = 0.4 \, V$.The change in current is $\Delta I = 80 \, mA - 60 \, mA = 20 \, mA = 20 	imes 10^{-3} \, A$.The forward bias resistance is $r_{fb} = \frac{\Delta V}{\Delta I} = \frac{0.4}{20 	imes 10^{-3}} = \frac{0.4}{0.02} = 20 \, \Omega$.For reverse biasing,the change in voltage is $\Delta V = -2 \, V - 0 \, V = -2 \, V$.The change in current is $\Delta I = -0.25 \, \mu A - 0 \, \mu A = -0.25 \, \mu A = -0.25 	imes 10^{-6} \, A$.The reverse bias resistance is $r_{rb} = \frac{|\Delta V|}{|\Delta I|} = \frac{2}{0.25 	imes 10^{-6}} = 8 	imes 10^6 \, \Omega$.Thus,the forward and reverse bias resistances are $20 \, \Omega$ and $8 	imes 10^6 \, \Omega$ respectively.

Condition	$V$	$I$
Forward biasing	$2.0 \, V$	$60 \, mA$
Forward biasing	$2.4 \, V$	$80 \, mA$
Reverse biasing	$0 \, V$	$0 \, \mu A$
Reverse biasing	$-2 \, V$	$-0.25 \, \mu A$

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