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Determine Q Point and draw the dc load line for the following circuit.
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Assuming β=100

RB=R1||R2

RB=15×103×5×10315×103+5×103

RB=3.75 KΩ

VTH=VB=VCC×R2R1+R2

VB=20×5×10315×103+5×103

VB=5 V

Applying KVL to the Base - Emitter Loop

VBIBRBVBEIERE=0

But, IE=(1+β)IB

VBVBEIB[RB+(1+β)RE]=0

IB=VBVBERB+(1+β)RE

IB=50.73750+(1+100)3000

IB=14.01 μA

IC=βIB

IC=100×62.91×106

IC=1.4 mA__

ICIE

Applying KVL to Collector - Emitter Loop

VCCICRCVCEICRE=0

VCE=VCCIC(RC+RE)

VCE=201.4×103(2×103+3×103)

VCE=13 V__

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