Explain collector to base bias circuit

5views

written 6.5 years ago by

teamques10 ★ 69k

enter image description here

$V_{cc}-(I_c+I_B ) R_c-I_B R_B-V_{BE}=0$

But

$I_c=β_dc I_B \\ ∴V_{cc}=(R_B+R_C ) I_B+β_{dc} I_B R_c+V_{BE} \\ ∴V_{cc}=[R_B+R_C+β_{dc} R_c ] I_B+V_{BE} \\ I_{BQ}=\frac{V_{cc}-V_{BE}}{[RB+(1+β_{dc} ) R_c ]} \\ ∴V_{cc}=(R_B+R_c ) I_B+I_c R_c+V_{BE}$

Collector current:

$I_{CQ}=β_{dc} I_{BQ} \\ I_{BQ}=\frac{β_{dc} [V_{cc}-V_{BE}]}{[R_B+(1+β_{dc}) R_c ]}$

enter image description here

Applying kVL

$V_{cc}=(I_c+I_B ) R_c+V_{CEQ} \\ ∴V_{CEQ}=V_{cc}-(I_C+I_B ) R_c$

Create a free account to keep reading this post.

and 5 others joined a min ago.

ADD COMMENT EDIT