Lecture Electric circuit theory - Active circuits presents the following content: Dependent sources, analysis of circuits with dependent sources, the operational amplifier: Basic concepts and subcircuits, analysis of circuits with Op Amps.
Nguyễn Công Phương Electric Circuit Theory Active Circuits Contents I Basic Elements Of Electrical Circuits II Basic Laws III Electrical Circuit Analysis IV Circuit Theorems V Active Circuits VI Capacitor And Inductor VII First Order Circuits VIII.Second Order Circuits IX Sinusoidal Steady State Analysis X AC Power Analysis XI Three-phase Circuits XII Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV Two-port Networks Active Circuits - sites.google.com/site/ncpdhbkhn Active Circuits Dependent Sources Analysis of Circuits with Dependent Sources The Operational Amplifier: Basic Concepts and Subcircuits Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn Dependent Voltage Source i vc + – + v – Voltage-controlled voltage source (VCVS): vc = µvx Current-controlled voltage source (CCVS): vc = rmix Active Circuits - sites.google.com/site/ncpdhbkhn Dependent Current Source i + ic v – Voltage-controlled current source (VCCS): ic = gmvx Current-controlled current source (CCCS): ic = βix Active Circuits - sites.google.com/site/ncpdhbkhn Active Circuits Dependent Sources Analysis of Circuits with Dependent Sources a b c d Branch Current Method Node Voltage Method Mesh Current Method Equivalent Subcircuits The Operational Amplifier: Basic Concepts and Subcircuits Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn Branch Current Method (1) a Ex – 4Ω ix + i − ic + = 4A vx + ix + i − 2ix + = → 4ix − 6i = 12 – ic = 0.5v x = 0.5 × 4ix = 2ix 0.5vx 6Ω ic ix + 4ix − 6i = 12 i 12V b −ix + i = −4 ix = A → → 4ix − 6i = 12 i = A Active Circuits - sites.google.com/site/ncpdhbkhn Branch Current Method (2) R2 Ex +– b : ic − i2 − i3 = c : i1 + i3 − J = A : R1i1 − R3i3 + R2 i2 − E = ic = β i1 E i2 βi1 a R1 ic i1 i3 b R3 J c β i1 − i2 − i3 = → i1 + i3 − J = R i − R i + R i − E = 3 2 11 Active Circuits - sites.google.com/site/ncpdhbkhn Node Voltage Method (1) a Ex vx i 0.5vx 6Ω ic ix + – 4A + 12 1 1 + va = + − ic 4 6 ic = 0.5vx = 0.5(12 − va ) – 4Ω 12V b 12 1 1 → + va = + − 0.5(12 − va ) 4 6 12 − va 12 − ( −12) = 6A ix = = → va = −12 V → i = − va = − −12 = A 6 Active Circuits - sites.google.com/site/ncpdhbkhn Node Voltage Method (2) 1 va R1 E +– βi1 a R1 v ic = β i1 = β a R1 1 + +β R1 R2 → − + β R2 R2 Ex 1 E + v − v = J − i + a b c R R R R2 1 E − v + + v = i − b c a R R R R2 ic i1 b R3 J c − E vb = J + R2 R2 1 E v + + v = − b a R1 R R R2 Active Circuits - sites.google.com/site/ncpdhbkhn 10 The Operational Amplifier (1) + Output – Input vo v+ + – v– µ(v+ – v–) µ=∞ Active Circuits - sites.google.com/site/ncpdhbkhn 24 The Operational Amplifier (2) J vo = v− + RJ R vo – + vo = µ (v+ − v− ) = µ ( E − v− ) – E J R vo v– v+ – Active Circuits - sites.google.com/site/ncpdhbkhn – E + + v− = v+ = E → vo = E + RJ + µ E − RJ → v− = 1+ µ µ ( E + RJ ) → vo = 1+ µ µ =∞ io io µ(v+ – v–) 25 The Operational Amplifier (3) i+ v+ + vo – v– i– io i+ = i− = v+ = v− Ri = ∞ Ro = Active Circuits - sites.google.com/site/ncpdhbkhn 26 The Operational Amplifier (4) vi = R1ii R2 → vo = − vi R1 R2 vi ii – vo + vo = − R2ii R1 io + io vo – R2 R1 vi = vo → vo = 1 + vi R1 + R2 R1 vi ii R1 Active Circuits - sites.google.com/site/ncpdhbkhn R2 27 The Operational Amplifier (5) + vi – vo = vi v1 v2 + R1 R2 v1 R1 v1 R1 R2 v2 v2 R2 Active Circuits - sites.google.com/site/ncpdhbkhn Ro – vo + v1 v2 vo = + Ro R1 R2 vo 28 The Operational Amplifier (6) R1 R2 v1 R3 v1 − v− v− − vo = R1 R2 v2 − v+ v+ = R3 R4 – vo + v2 v– v+ R4 R2 R4 R2 → vo = + 1 v2 − v1 = k2v2 − k1v1 R1 R1 R3 + R4 v+ = v− Active Circuits - sites.google.com/site/ncpdhbkhn 29 Active Circuits Dependent Sources Analysis of Circuits with Dependent Sources The Operational Amplifier: Basic Concepts and Subcircuits Analysis of Circuits with Op Amps Branch Current Method Node Voltage Method Mesh Current Method Active Circuits - sites.google.com/site/ncpdhbkhn 30 Analysis of Circuits with Op Amps i+ v+ + – v– vo io i– + v+ + – v– µ(v+ – v–) v o µ →∞ – ix + + rmix rm → ∞ i+ = i− = v+ = v− io v+ gm(v+ – v–) gm → ∞ v– ix io βix β→∞ vo – Active Circuits - sites.google.com/site/ncpdhbkhn 31 – Branch Current Method (1) Ex R1 E1 + + R5 vo µvx + – R4 + v+ – + – E2 R3 v– Active Circuits - sites.google.com/site/ncpdhbkhn – + vx – E1 R4 + R2 vo R5 – – R1 R3 – + E2 –+ E3 R2 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? –+ E3 µ(v+ – v–) v o µ →∞ – 32 Branch Current Method (2) i1 – R1i1 − R2i2 = E1 − E2 R2i2 + vx + R4i4 = E2 R3i3 + R5i4 − vx = E3 R5 R4 i4 vo µvx + E2 vx – i2 R3 + E1 + i1 + i2 − i3 = R2 –+ E3 – R1 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? i3 – Ex µ + 12 → i1 = 2µ + → i1 = 0.5 mA µ →∞ ( R4 + R5 )i4 = µ vx Active Circuits - sites.google.com/site/ncpdhbkhn 33 Node Voltage Method (1) Ex R1 E1 + + R5 vo µvx + – R4 + v+ – + – E2 R3 v– Active Circuits - sites.google.com/site/ncpdhbkhn – + vx – E1 R4 + R2 vo R5 – – R1 R3 – + E2 –+ E3 R2 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? –+ E3 µ(v+ – v–) v o µ →∞ – 34 Node Voltage Method (2) E1 vx R5 v2 – R4 i4 i5 vo µvx + E2 – i2 R3 + R2 –+ E3 + i1 + i2 − i3 = i4 + i5 = E1 − v1 E2 − v1 i1 = ; i2 = R1 R2 E3 − vo + v1 i3 = R3 v2 − vo v2 i4 = ; i5 = R4 R5 R1 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? i3 v1 – i1 – Ex E1 − v1 E2 − v1 E3 − vo + v1 =0 R + R − R3 v2 + v2 − vo = R4 R5 Active Circuits - sites.google.com/site/ncpdhbkhn 35 Node Voltage Method (3) E1 vx R5 v2 R4 i4 i5 vo µvx + E2 – i2 R3 + R2 –+ E3 + – E1 − v1 E2 − v1 E3 − vo + v1 =0 R + R − R3 v2 + v2 − vo = R4 R5 vo = µ vx = µ (v1 − v2 ) R1 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? i3 v1 – i1 – Ex 2µ + 7−6 → v1 = → v = V → i = = 0.5 mA µ +3 1 µ→∞ Active Circuits - sites.google.com/site/ncpdhbkhn 36 Mesh Current Method (1) Ex R1 E1 vo – + R5 – – E2 R3 + R2 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? –+ E3 R4 R1 –+ E3 R2 E1 ix + + ix R5 – – E2 R3 vo R4 io βix β→∞ ic = βix Active Circuits - sites.google.com/site/ncpdhbkhn 37 Mesh Current Method (2) Ex i1 R1 R2 E1 + iA – – E2 R3iD + R5 (iD − ic ) = E3 ic = β ix = β (iB − iD ) ix iB R3 iD R5 vo R4 R1i A + R2 (i A − iB ) = E1 − E2 R2 (iB − i A ) + R4 (iB − ic ) = E2 –+ E3 + E1 = V; E2 = V; E3 = V; R1 = R2 = R3 = kΩ; R4 = kΩ; R5 = kΩ; solve the circuit? ic = βix → iA = 9β − 80 18β − 70 → iA = 0.5 mA β →∞ → i1 = 0.5 mA Active Circuits - sites.google.com/site/ncpdhbkhn 38 ... J c Active Circuits - sites.google.com/site/ncpdhbkhn 13 Active Circuits Ex a rmix + E – R1 +– J c b R2 ix R3 + vy – d R4 gmvy e Active Circuits - sites.google.com/site/ncpdhbkhn 14 Active Circuits. .. XI Three-phase Circuits XII Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV Two-port Networks Active Circuits - sites.google.com/site/ncpdhbkhn Active Circuits. .. Elements Of Electrical Circuits II Basic Laws III Electrical Circuit Analysis IV Circuit Theorems V Active Circuits VI Capacitor And Inductor VII First Order Circuits VIII.Second Order Circuits