Explain Voltage divider bias or self-bias

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written 5.6 years ago by

teamques10 ★ 65k

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Features:

The resistors $R_1$ and $R_2$ form a potential divider to apply a fixed voltage $V_B$ to the base.
A resistance $R_E$ has been connected in the emitter circuit. This resistor is not present in the fixed bias or collector to the base bias circuit.
Base voltage $$V_B=V_{R2}=\frac{R_2}{R_1+R_2 } \times …

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