In the given circuit, the voltage across load | vedclass

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Question

(A) The circuit contains a $15 \, V$ $DC$ source, a Germanium $(Ge)$ diode, a Silicon $(Si)$ diode, a $1.5 \, k\Omega$ resistor, and a load resistor $

Vedclass · Accepted Answer

$8.75$

Vedclass · Answer

(A) The circuit contains a $15 \, V$ $DC$ source, a Germanium $(Ge)$ diode, a Silicon $(Si)$ diode, a $1.5 \, k\Omega$ resistor, and a load resistor $R_L = 2.5 \, k\Omega$ in series.The barrier potential for a $Ge$ diode is $V_{Ge} = 0.3 \, V$ and for a $Si$ diode is $V_{Si} = 0.7 \, V$.Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$15 - V_{Ge} - V_{Si} - i(1.5 \, k\Omega) - i(2.5 \, k\Omega) = 0$$15 - 0.3 - 0.7 = i(1.5 + 2.5) \, k\Omega$$14 = i(4 \, k\Omega)$$i = \frac{14}{4} \, mA = 3.5 \, mA$The voltage across the load resistance $R_L$ is:$V_L = i 	imes R_L = 3.5 \, mA 	imes 2.5 \, k\Omega = 8.75 \, V$

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	$-2 \, V$	$-0.25 \, \mu A$

In the given circuit, the voltage across load resistance $(R_L)$ is: (in $V$)

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