## 08/02/2020

### Biasing Method : Voltage Divider Method

Mathematical analysis
• Let us consider that the current I1 flows through resistance R1
• As the base current IB is very small, therefore we consider that the current flows through resistance R2 is also I1.
Collector current IC
I1 = VCC / ( R1 + R2 )
Voltage drop across resistance R2 = R2I1
V2 = [ VCC / ( R1 + R2 ) ] R2
Apply KVL to base circuit
V2 = Voltage drop across Base emitter junction + voltage drop across
resistance RE
= VBE + VE
= VBE + IERE
As base current is very small IE ≈ IC
V2 = VBE + IC RE ................ ( 1 )
 IC = V2 – VBE / RE..... ( 2 )

Collector to Emitter voltage
Applying KVL to collector circuit
VCC = ICRC + VCE + IERE
AS IE ≈ IC
VCC = ICRC + VCE + ICRE
= IC ( RC + RE ) + VCE
VCE = VCC – IC ( RC + RE )
Stabilization
V2 = VBE + IC RE
• As the collector current increases with increase in the temperature therefore the voltage drop across resistance RE also increases.
• As the voltage drop across resistance R2 does not depends upon IC, the voltage across base to emitter ( VBE ) decreases.
• As the base to emitter voltage decrease, the base current also decreases and this will result in collector current IC restore to its original value.
Voltage drop across R2 = [ VCC / ( R1 + R2 ) ] R2
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