ki thuat dien tu bai 19

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• ­­_Vẽ­sơ­đồ­tương­đương­một­chiều­của­mạch,­tìm­V_B,­V_E,­V_BE,­
V_CE.

• _{_Vẽ­sơ­đồ­tương­đương­tương­xoay­chiều­của­mạch­và­tìm­:}

• +Trở­kháng­vào

• _{+Trở­kháng­ra}

• +Hệ­số­khuếch­đại­A_V

• Biết­β_DC=175,β_ac=200,

• R_L=190KΩ,V_CC=15V,R₁=39KΩ,R₂=6,2KΩ,

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•

_Giải:

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• 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)

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• ­r_in=R₁//R₂//R_in(base)

• R_in(base)=(r_e+R_E1).(β_ac+1)­­với­r_e=25.10-3_(V)/I
E

• =9,395(Ω)

• =(9,395+270).(200+1)=56158,395(Ω)

• =>r_in≈4,884(KΩ)

• r_out=R_C//R_L≈1,979(KΩ)

• A_V=V_out/V_in=V_c/V_b=(I_c.(R_C//R_L))/(I_e.(r_e+R_E1))

• =(β_ac/(­β_ac+1))/((­R_C//R_L)/_­.(r_e+R_E1))

• ≈(­R_C//R_L)/_­.(r_e+R_E1)≈7,084(V)

• A_i=I_out/I_in=(V_out/r_out)/(V_in/r_in)=A_V.(r_out.r_in)

• _{­­­≈2,871(V)}

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