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Draw the dc load line for above circuit.

written 3.6 years ago by teamques10 ★ 68k

modified 2.7 years ago by DevarenjiniP • 3.2k

electronic devices and circuits

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1 Answer

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written 3.6 years ago by teamques10 ★ 68k

$-4\times 10^3I_C+20-V_{CB}=0\\ I_C=\dfrac{-V_{CB}}{4\times10^3}+\dfrac{20}{4\times10^3}\cdots\cdots(1)$

When IC=0

VCB=20

When VCB=0

$I_C=\dfrac{20}{4}\times 10^{-3}=5mA$

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