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Determine Q Point and draw the dc load line for the following circuit.
1 Answer
written 3.9 years ago by |
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
VB−IBRB−VBE−IERE=0
But, IE=(1+β)IB
VB−VBE−IB[RB+(1+β)RE]=0
IB=VB−VBERB+(1+β)RE
IB=5−0.73750+(1+100)3000
IB=14.01 μA
IC=βIB
IC=100×62.91×10−6
IC=1.4 mA__
IC≈IE
Applying KVL to Collector - Emitter Loop
VCC−ICRC−VCE−ICRE=0
VCE=VCC−IC(RC+RE)
VCE=20−1.4×10−3(2×103+3×103)
VCE=13 V__