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For a NPN transistor in CE mode voltage divider bias configuration determine VC and VB-. Given VCC=+20V, VEE=-20V, R1=8.2 KkΩ, R2=2.2KΩ, RC=2.7 KΩ, RE=1.8 KΩ, C1=C2=10 7mu;F and β=120.
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The Base Voltage of a BJT transistor in Voltage divider bias configuration is given by

Vb=Vth=Vcc(R2R1+R2)

Vb=202.28.2+2.2 

     =202.210.4

     =4.320V

The value of Threshold Resistor is parallel combination of R1 and R2.that is

Rth=R1R2(R1+R2)

       =8.2×2.28.2+2.2

       =1.734Ω

Ic=βIb

   =β(VthVbe)Rth+(β+1)RE)

   =120(4.32(4.32+20)(1.734+(121×1800)

   =120(4.3224.32)(1.734+217800)

   =2400217801.734

   =11mA

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