- This is most popular method of biasing and stabilization.
- There are two resistance R1 and R2 is connected across supply
voltage V
_{CC}to provide biasing. The emitter resistance R_{E}provides stabilization. **There are two resistances R1 and R2 provides voltage divider in this method therefore it is called as voltage divider method.**- The voltage
drop across resistance R
_{2}forward biases base – emitter junction resulting base current flows. - This will result in zero signal collector current flows.

**Mathematical analysis**

- Let us consider that the current I
_{1}flows through resistance R_{1}. - As the base current I
_{B}is very small, therefore we consider that the current flows through resistance R_{2}is also I_{1}.

__Collector current I___{C}
I

_{1}= V_{CC}/ ( R_{1}+ R_{2})
Voltage drop across resistance R

_{2}= R_{2}I_{1}
V

_{2}= [ V_{CC}/ ( R_{1}+ R_{2}) ] R_{2}
Apply KVL to base circuit

V

_{2}= Voltage drop across Base emitter junction + voltage drop across
resistance R

_{E}
= V

_{BE}+ V_{E}
= V

_{BE}+ I_{E}R_{E}
As base current is very small I

_{E}≈ I_{C}
V

_{2}= V_{BE}+ I_{C }R_{E }................ ( 1 )
I
_{C} = V_{2} – V_{BE} / R_{E.....} ( 2 ) |

- The value of collector current does not depend upon current amplification factor β.
- As the collector current I
_{C}depends upon V_{BE}but V_{2}>> V_{BE}therefore I_{C}is practically does not depend upon V_{BE}. - We can say that the
collector current I
_{C}is independent parameter therefore it provides good stabilization.

__Collector to Emitter voltage__
Applying KVL to collector circuit

V

_{CC}= I_{C}R_{C}+ V_{CE}+ I_{E}R_{E}
AS I

_{E}≈ I_{C}
V

_{CC}= I_{C}R_{C}+ V_{CE}+ I_{C}R_{E}
= I

_{C }( R_{C}+ R_{E}) + V_{CE}
V

_{CE}= V_{CC}– I_{C }( R_{C}+ R_{E})

__Stabilization__- The stabilization is provided by emitter resistance R
_{E}. Let us consider the equation ( 1 )

V

_{2}= V_{BE}+ I_{C }R_{E }- As the collector current increases with increase in the
temperature therefore the voltage drop across resistance R
_{E}also increases. - As the voltage drop across resistance R2 does not depends upon I
_{C}, the voltage across base to emitter ( V_{BE}) decreases. - As the base to
emitter voltage decrease, the base current also decreases and this will result
in collector current I
_{C}restore to its original value.

Voltage drop across R2 = [ V

_{CC}/ ( R1 + R2 ) ] R2
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