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T┴n hi?u, h? th?ng v┐ ph?n gi?i m?ch 12

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Linh Tr?n L┷

The document discusses an amplifier circuit analysis problem. It identifies the circuit as having: 1) Negative feedback. 2) A shunt-series feedback configuration. 3) To calculate A1, an equivalent circuit model is drawn using a simple BJT model without hre and hoe. 4) The expression for A2 is derived as Re2/(Rf + Rg + Re2).

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-H? v┐ t┷n: Tr?n L┷ Duy Linh. -MSSV: 18200032. - L?p : 18DTV1_L1. - Tu?n 12. B┐i t?p c?ng c? bu?i 12. Tr━nh b┐y: 1. Cho m?ch sau: a. X│c ??nh lo?i h?i ti?p? Gi?i th┴ch? b. X│c ??nh d?ng h?i ti?p? Gi?i th┴ch? c. ?? t┴nh A1, v? m?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). d. T┴nh A2? Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co T┏m t?t b┐i l┐m: 1. a. Lo?i h?i ti?p l┐ h?i ti?p ?m. b. Ki?u h?i ti?p : shunt C series. c. H━nh v?. d. ?2 = ?? ?0 = ??2 ??+??+??2
a. Lo?i h?i ti?p l┐ h?i ti?p ?m. b. Ki?u h?i ti?p : shunt C series. c. ?? t┴nh A1, v? m?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). V?i: + Series -> D┛ng ?i?n b?ng 0 ( ?? h?) + Shunt -> ?i?n th? b?ng 0 (n?i ??t) + Ph?n c?c AC, t┴n hi?u nh? : . T? n?i t?t. . Vcc n?i ??t. Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co + - - - - ＾p￣ ＾c￣ series shunt
M?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). d. T┴nh A2? ?? = ?0 ??2 ?? + ?? + ??2 => ?2 = ?? ?0 = ??2 ?? + ?? + ??2 Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co RF B1 C2 R1 RC2 R2 hie1 hfe2.ib2 E1 hie2 B2 C1 ib1 RF ib2 R4 RG Vo Vi hfe1.ib1 E2 RG RC1 R3 Vo R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co E2 B1 i0 if

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T┴n hi?u, h? th?ng v┐ ph?n gi?i m?ch 12

  • 1. -H? v┐ t┷n: Tr?n L┷ Duy Linh. -MSSV: 18200032. - L?p : 18DTV1_L1. - Tu?n 12. B┐i t?p c?ng c? bu?i 12. Tr━nh b┐y: 1. Cho m?ch sau: a. X│c ??nh lo?i h?i ti?p? Gi?i th┴ch? b. X│c ??nh d?ng h?i ti?p? Gi?i th┴ch? c. ?? t┴nh A1, v? m?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). d. T┴nh A2? Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co T┏m t?t b┐i l┐m: 1. a. Lo?i h?i ti?p l┐ h?i ti?p ?m. b. Ki?u h?i ti?p : shunt C series. c. H━nh v?. d. ?2 = ?? ?0 = ??2 ??+??+??2
  • 2. a. Lo?i h?i ti?p l┐ h?i ti?p ?m. b. Ki?u h?i ti?p : shunt C series. c. ?? t┴nh A1, v? m?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). V?i: + Series -> D┛ng ?i?n b?ng 0 ( ?? h?) + Shunt -> ?i?n th? b?ng 0 (n?i ??t) + Ph?n c?c AC, t┴n hi?u nh? : . T? n?i t?t. . Vcc n?i ??t. Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co + - - - - ＾p￣ ＾c￣ series shunt
  • 3. M?ch t??ng ???ng, v?i BJT d┫ng m? h━nh ??n gi?n (b? qua hre, hoe). d. T┴nh A2? ?? = ?0 ??2 ?? + ?? + ??2 => ?2 = ?? ?0 = ??2 ?? + ?? + ??2 Vo Vi R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co RF B1 C2 R1 RC2 R2 hie1 hfe2.ib2 E1 hie2 B2 C1 ib1 RF ib2 R4 RG Vo Vi hfe1.ib1 E2 RG RC1 R3 Vo R1 Rc1 Ct R2 Re2 VCC Re1 Q2A 3 1 2 Rg Rf Rb1 Ri Q1A 3 1 2 Rc2 Ci Co E2 B1 i0 if