Draw the dc load line for above circuit.

585views

written 3.9 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

DevarenjiniP • 3.2k

electronic devices and circuits

ADD COMMENT EDIT

1 Answer

6views

written 3.9 years ago by

teamques10 ★ 69k

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

When IC=0

VCB=20

When VCB=0

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

Create a free account to keep reading this post.

and 3 others joined a min ago.

ADD COMMENT EDIT

Please log in to add an answer.