In a PN-junction diode,the reverse saturation | vedclass

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Question

(C) The formula for the forward current $I$ in a $PN$-junction diode is given by $I = I_s (e^{\frac{eV}{k_B T}} - 1)$.Given:Reverse saturation current

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$20.376 	imes {10^{-3}} \ A$

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(C) The formula for the forward current $I$ in a $PN$-junction diode is given by $I = I_s (e^{\frac{eV}{k_B T}} - 1)$.Given:Reverse saturation current $I_s = {10^{-5}} \ A$Voltage $V = 0.2 \ V$Temperature $T = 27^\circ C = 27 + 273 = 300 \ K$Boltzmann constant $k_B = 1.4 	imes {10^{-23}} \ J/K$Elementary charge $e = 1.6 	imes {10^{-19}} \ C$Exponent value $\frac{eV}{k_B T} = \frac{1.6 	imes {10^{-19}} 	imes 0.2}{1.4 	imes {10^{-23}} 	imes 300} = \frac{0.32 	imes {10^{-19}}}{4.2 	imes {10^{-21}}} = \frac{32}{4.2} \approx 7.62$.Substituting the values into the equation:$I = {10^{-5}} 	imes (e^{7.62} - 1)$$I = {10^{-5}} 	imes (2038.6 - 1)$$I = {10^{-5}} 	imes 2037.6$$I = 20.376 	imes {10^{-3}} \ A$.

In a $PN$-junction diode,the reverse saturation current is ${10^{-5}} \ A$ at $27^\circ C$. Calculate the forward current for a voltage of $0.2 \ V$. Given: $[\exp(7.62) = 2038.6, \ k_B = 1.4 \times {10^{-23}} \ J/K, \ e = 1.6 \times {10^{-19}} \ C]$.

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