ki thuat dien tu bai 19

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• ­­_Vẽ­sơ­đồ­tương­đương­một­chiều­của­mạch,­tìm­VB,­VE,­VBE,­ VCE.

• _Vẽ­sơ­đồ­tương­đương­tương­xoay­chiều­của­mạch­và­tìm­:

• +Trở­kháng­vào

• +Trở­kháng­ra

• +Hệ­số­khuếch­đại­AV

• Biết­βDC=175,βac=200,

• RL=190KΩ,VCC=15V,R1=39KΩ,R2=6,2KΩ,

• Giải:

• VBE=0,7­(V)­vì­JE­phân­cực­thuận

• RIN(Base)=­βDC­.­(RE1+RE2)=175.(270+240) • ­­­­­­­­­­­­­=89250(Ω)>10R2=62.10­­­­3

• =>VB=(R2.VCC)/(R1+R2)=(6,2.15)/(39+6,2) • ­­­­­­­­­=2,057(V)

• VE=VB-0,7=1,357(V)

• =>IE=VE/(RE1+RE2)=1,357/(270+240) • ­­­­­­­­=2,661.10-3(A)

• IC≈IE=2,661.10-3(A)

• =>VCE=VCC-IC.RC-IE(RE1+RE2)

• =15-2,661.10-3.103-2,661.10-3(270+240)

• ­rin=R1//R2//Rin(base)

• Rin(base)=(re+RE1).(βac+1)­­với­re=25.10-3(V)/I E

• =9,395(Ω)

• =(9,395+270).(200+1)=56158,395(Ω)

• =>rin≈4,884(KΩ)

• rout=RC//RL≈1,979(KΩ)

• AV=Vout/Vin=Vc/Vb=(Ic.(RC//RL))/(Ie.(re+RE1))

• =(βac/(­βac+1))/((­RC//RL)/­.(re+RE1))

• ≈(­RC//RL)/­.(re+RE1)≈7,084(V)

• Ai=Iout/Iin=(Vout/rout)/(Vin/rin)=AV.(rout.rin)

• ­­­≈2,871(V)