Explain collector to base bias circuit

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written 6.5 years ago by

teamques10 ★ 69k

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

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:

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

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Applying kVL

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

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