When a silicon PN junction is in forward bias | vedclass

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Question

(C) In a forward-biased $PN$ junction circuit with a series resistance $R$,the total voltage $V$ of the battery is equal to the sum of the knee voltag

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) In a forward-biased $PN$ junction circuit with a series resistance $R$,the total voltage $V$ of the battery is equal to the sum of the knee voltage $V_k$ of the diode and the voltage drop across the resistor $V_R$.According to Kirchhoff's voltage law:$V = V_k + I 	imes R$Given:Battery voltage $V = 2.6\, V$Knee voltage $V_k = 0.6\, V$Current $I = 5\, mA = 5 	imes 10^{-3}\, A$Substituting the values into the equation:$2.6 = 0.6 + (5 	imes 10^{-3}) 	imes R$$2.0 = 5 	imes 10^{-3} 	imes R$$R = \frac{2.0}{5 	imes 10^{-3}}$$R = 0.4 	imes 10^3 = 400\,\Omega$.Therefore,the value of the series resistance is $400\,\Omega$.

When a silicon $PN$ junction is in forward biased condition with a series resistance,it has a knee voltage of $0.6\, V$. The current flowing through it is $5\, mA$ when the $PN$ junction is connected to a $2.6\, V$ battery. The value of the series resistance is.....$\Omega$.

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