8 February 2020

Biasing Method : Voltage Divider Method


voltage-divider-bias-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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