Bài giảng Electric circuit theor - Chapter V: Active circuits - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng 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 2 Active Circuits 1. Dependent Sources 2. Analysis of Circuits with Dependent Sources 3. The Operational Amplifier: Basic Concepts and Subcircuits 4. Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn 3 Dependent Voltage Source i vc + – +v – Voltage-controlled voltage source (VCVS): vc = µv x Current-controlled voltage source (CCVS): vc = rmix Active Circuits - sites.google.com/site/ncpdhbkhn 4 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 5 Active Circuits 1. Dependent Sources 2. Analysis of Circuits with Dependent Sources a. Branch Current Method b. Node Voltage Method c. Mesh Current Method d. Equivalent Subcircuits 3. The Operational Amplifier: Basic Concepts and Subcircuits 4. Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn 6 Branch Current Method (1) a Ex. 1 – 4Ω i +− + = 4A v 0.5 vx ix i i c 4 0 x + ix 4i− 6 i = 12 + x – 6Ω ic = =×= 12V b ic0.5 v x 0.5 4 ii xx 2 i+− i2 i + 4 = 0 −i + i =− 4 i = 6A →  x x →  x →  x − = − = = 4ix 6 i 12 4ix 6 i 12 i 2 A Active Circuits - sites.google.com/site/ncpdhbkhn 7 Branch Current Method (2) Ex. 2 R2 E + – bi:− i − i = 0 c 2 3 i2 + − = βi1 ci:1 i 3 J 0 a b ARi:− Ri + Ri −= E 0 11 33 22 ic R1 i1 i R3 i= β i 3 c 1 J c βi− i − i = 0  1 2 3 → + − = i1 i 3 J 0  − + −= Ri11 Ri 33 Ri 22 E 0 Active Circuits - sites.google.com/site/ncpdhbkhn 8 Node Voltage Method (1) a Ex. 1 – 4Ω 11  12 i + =+ − 4A v 0.5 vx   va4 i c x 4 6  4 + ix + = = − – 6Ω i ic0.5 v x 0.5(12 v a ) c 12V b 11  12 →+  v =+−4 0.5(12 − v ) 4 6  a 4 a  12−v 12 − (12) − i =a = = 6A  x 4 4 →v = − 12V →  a v −12 i =−a =− = 2 A  6 6 Active Circuits - sites.google.com/site/ncpdhbkhn 9 Node Voltage Method (2)   Ex. 2 R2 E 1+ 1 − 1 =−+ E   va vJi b c + – RR12  R 2 R 2     −1 + 1 + 1 =− E βi1  va  v b i c a b  R2 RR 23  R 2 ic R1 i1 R3 =β = β va ic i 1 R1 J c   111++β −=+ 1 E   va v b J RRR  R R →  121 2 2      −+1β 1 ++ 11 =− E    va  v b  RR21  RR 23  R 2 Active Circuits - sites.google.com/site/ncpdhbkhn 10 Mesh Current Method (1) a Ex. 1 – 4Ω i −+ − = 4A v 0.5 vx 4(im 4) 6( i m i c ) 12 x im + ix + = =×= – 6Ω i ic0.5 v x 0.5 4 ii xx 2 c = − 12V b 2(im 4) → −+[ − −=] 4(im 4) 6 i m 2( i m 4) 12 i= i −=4 10 −= 4 6A →i = 10 →  x m m =−+=− + − = i im i c 10 2(10 4) 2A Active Circuits - sites.google.com/site/ncpdhbkhn 11 Mesh Current Method (2) Ex. 2 R2 E + + += + – Ri1A Ri 2 D Ri 3 B E 0 iD − = βi1 iB i A J a b i− i = i ic B D c R i R 1 1 iA 3 = β iB ic i 1 J c → ++ =−− + + β (RRRi123 )A ERRJRi () 23 21 = − iA i 1 →+++β =+ + (RRR123 RiE 21 )() RRJ 23 Active Circuits - sites.google.com/site/ncpdhbkhn 12 1. Temporarily treat the dependent R2 E source(s) as independent + – source(s). i2 βi1 2. Perform branch current (or nodal a b or mesh) analysis as usual to i R i c R determine the branch current (or 1 1 i3 3 nodal or mesh) equations. J c 3. Express the value of dependent source(s) in terms of the branch currents (or node voltages or mesh currents). 4. Solve the equations. Active Circuits - sites.google.com/site/ncpdhbkhn 13 Active Circuits Ex. 3 a + – + rmix E – J vy b c + – d R1 R4 R2 ix R3 gmvy e Active Circuits - sites.google.com/site/ncpdhbkhn 14 Active Circuits 1. Dependent Sources 2. Analysis of Circuits with Dependent Sources a. Branch Current Method b. Node Voltage Method c. Mesh Current Method d. Equivalent Subcircuits 3. The Operational Amplifier: Basic Concepts and Subcircuits 4. Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn 15 Equivalent Circuits (1) Ex. 1 a a Find i? – 4Ω – 4Ω + i 4A vx 0.5 vx 4A vx 0.5 vx + ix + ix + voc + – – 6Ω – ic ic 12V b 12V b v E= v R = open-circuit eq open-circuit eq b ishort-circuit 1 12 = + − R v4 i + eq 4a 4 c – Eeq i 6Ω i=0.5 v = 0.5(12 − v ) c x a c 1 12 →v =+−4 0.5(12 − v ) →v = − 4V 4a 4 a oc Active Circuits - sites.google.com/site/ncpdhbkhn 16 Equivalent Circuits (2) Ex. 1 a a Find i? – 4Ω – 4Ω i 4A vx 0.5 vx 4A vx 0.5 vx + ix + ix + isc + – – 6Ω ic ic 12V b 12V b v E= v R = open-circuit eq open-circuit eq b ishort-circuit 4+−iii −=→ 0 iii =−+ 4 R x scc sc x c + eq – E 6Ω = eq i ic0.5 v x → = ic 6 v = 12 V c v= v →  x a b = = →i =−+3 6 4 = 1A ix 12 / 4 3A sc Active Circuits - sites.google.com/site/ncpdhbkhn 17 Equivalent Circuits (3) Ex. 1 a Find i? – 4Ω i 4A v 0.5 vx = = − x Eeq v open-circuit 4 V + ix + – 6Ω ic = isc 1A 12V b v −4 R =open-circuit = = −4 Ω b eq i 1 short-circuit R + eq – E i 6Ω −E −( − 4) eq i =eq = = 2 A + − + c Req 6 4 6 Active Circuits - sites.google.com/site/ncpdhbkhn 18 Equivalent Circuits (4) Ex. 2 R2 E E = 16 V; J = 2 A; R1 = 4 Ω; R2 = 6 Ω; + – β = 2; find R ? eq i2 v βi1 = open-circuit a b Req + ishort-circuit ic R1 i1 v − −+ + = oc (vc v b ) ERi22 Ri 11 0 J – = − c voc v b v c → = + − voc R11 i R 22 i E = →= +β − i1 J voc R1 J R 2 J E = =β = β =×+××− = ii2c i 1 J 4 2 6 2 2 16 16V Active Circuits - sites.google.com/site/ncpdhbkhn 19 Equivalent Circuits (5) Ex. 2 R2 E E = 16 V; J = 2 A; R1 = 4 Ω; R2 = 6 Ω; + – β = 2; find Req ? v βi1 = open-circuit a b Req i i short-circuit R i c −+ =→ =− 1 1 isc iJi1sc0 i sc Ji 1   J 1+ 1 =−+ E c   va J i c R1 R 2  R 2 a v i=β i = β a βi R c 1 1 ic 2 R1 R i 1 1 + va 5.09 → = →=i = = 1.27 A – va 5.09V 1 J E R1 4 → =− = isc 2 1.27 0.73A Active Circuits - sites.google.com/site/ncpdhbkhn 20 Equivalent Circuits (6) Ex. 2 R2 E E = 16 V; J = 2 A; R1 = 4 Ω; R2 = 6 Ω; + – β = 2; find Req ? v βi1 = open-circuit a b Req ishort-circuit ic R1 i1 v = 16 V J oc c = isc 0.73A 16 →R = =22 Ω eq 0.73 Active Circuits - sites.google.com/site/ncpdhbkhn 21 Equivalent Circuits (7) Ex. 3 a + – + rmix E – J vy b c + – d R1 R3 R2 ix gmvy e Active Circuits - sites.google.com/site/ncpdhbkhn 22 Active Circuits 1. Dependent Sources 2. Analysis of Circuits with Dependent Sources 3. The Operational Amplifier: Basic Concepts and Subcircuits 4. Analysis of Circuits with Op Amps Active Circuits - sites.google.com/site/ncpdhbkhn 23 The Operational Amplifier (1) + Input– Output vo v+ + µ(v – v ) – + – µ = ∞ v– Active Circuits - sites.google.com/site/ncpdhbkhn 24 The Operational Amplifier (2) J R = + vo v− RJ – vo + =µ −= µ − vo ( vv+ − )( Ev − ) io + µE− RJ – →v− = E 1+ µ µ(E+ RJ ) →v = o + µ J R 1 v v µ = ∞ – o v+ + io = = + v− v + E – →  – = + µ(v+ – v–) vo E RJ E Active Circuits - sites.google.com/site/ncpdhbkhn 25 The Operational Amplifier (3) i+ v+ + vo v– – io i– i+= i − = 0 v+= v − = ∞ Ri = Ro 0 Active Circuits - sites.google.com/site/ncpdhbkhn 26 The Operational Amplifier (4) R R = 1 2 vi R1 i i R vi → = − 2 vo v i ii = − R1 – vo R2 i i + vo io v + io i – v R R  i o v= 1 v →v =1 + 2  v i i+ o o i R1 R 2 R1  R1 R2 Active Circuits - sites.google.com/site/ncpdhbkhn 27 The Operational Amplifier (5) v + i – = vo vo v i v v v1 1+ 2 R R R1 1 2 v1 v v  R R =1 + 2 1 R o vo  R o 2 v2 – R1 R 2  + vo v2 R2 Active Circuits - sites.google.com/site/ncpdhbkhn 28 The Operational Amplifier (6) R1 R2 v1 v R3 – – + vo v2 v+ R4 v− v v− v 1 − = − o R1 R 2   R2 R 4 R 2 v− v+ v + →=+v1 v −=− vkvkv 2 = o   + 2 1 22 11 R1  RR 34 R 1 R3 R 4 v+= v − Active Circuits - sites.google.com/site/ncpdhbkhn 29 Active Circuits 1. Dependent Sources 2. Analysis of Circuits with Dependent Sources 3. The Operational Amplifier: Basic Concepts and Subcircuits 4. Analysis of Circuits with Op Amps 1. Branch Current Method 2. Node Voltage Method 3. Mesh Current Method Active Circuits - sites.google.com/site/ncpdhbkhn 30 Analysis of Circuits with Op Amps i+ v+ + i+= i − = 0 vo v– – = io v+ v − i– io v+ + v+ + µ(v – v ) g (v – v ) – + – vo m + – v µ → ∞ v gm → ∞ – – – ix ix io + + r i βi – m x vo x rm → ∞ β → ∞ – Active Circuits - sites.google.com/site/ncpdhbkhn 31 Branch Current Method (1) Ex. 1 – + E1 = 7 V; E2 = 4 V; E3 = 2 V; R 1 E3 R3 R2 + R = R = R = 2 k Ω; R = 6 k Ω; E1 vo 1 2 3 4 + – + R5 = 3 k Ω; solve the circuit? – – R5 E2 R4 – + R 1 E3 R3 R2 + E1 + vx – + – v+ + – R5 + µ(v – v ) – + – v vo o E2 + µ → ∞ µv – v– R4 x – Active Circuits - sites.google.com/site/ncpdhbkhn 32 Branch Current Method (2) Ex. 1 i1 i3 E1 = 7 V; E2 = 4 V; E3 = 2 V; – + R R1 = R2 = R3 = 2 k Ω; R4 = 6 k Ω; 1 E3 R3 R2 + E1 R5 = 3 k Ω; solve the circuit? + vx – + – i2 – R5 vo + − = E + i1 i 2 i 3 0 2 i µv – R4 4 x − = − R11 i R 22 i E 1 E 2 µ +12 R i+ v + R i = E → = 22x 44 2 i1 → = 2µ + 6 i1 0.5mA + − = R33 i R 54 i vx E 3 µ → ∞ + = µ (R4 Ri 5 ) 4 v x Active Circuits - sites.google.com/site/ncpdhbkhn 33 Node Voltage Method (1) Ex. 1 – + E1 = 7 V; E2 = 4 V; E3 = 2 V; R 1 E3 R3 R2 + R = R = R = 2 k Ω; R = 6 k Ω; E1 vo 1 2 3 4 + – + R5 = 3 k Ω; solve the circuit? – – R5 E2 R4 – + R 1 E3 R3 R2 + E1 + vx – + – v+ + – R5 + µ(v – v ) – + – v vo o E2 + µ → ∞ µv – v– R4 x – Active Circuits - sites.google.com/site/ncpdhbkhn 34 Node Voltage Method (2) Ex. 1 i1 i3 v1 E1 = 7 V; E2 = 4 V; E3 = 2 V; – + R R1 = R2 = R3 = 2 k Ω; R4 = 6 k Ω; 1 E3 R3 R2 + E1 R5 = 3 k Ω; solve the circuit? + vx – + – i2 i5 + − = – i1 i 2 i 3 0 R5 v2 vo + = E2 + i i 0 – 4 5 R i4 µv x − − 4 = E1 v 1 = E2 v 1 i1 ; i2 R1 R2  E− v E − v E− v + v − + 1 1+ 2 1 −3o 1 = 0 E3 vo v 1  i =  R1 R 2 R 3 3  R3 − −  v2 v2 v o v2 v2 v o + = 0 i = ; i =  4 5  R4 R 5 R4 R5 Active Circuits - sites.google.com/site/ncpdhbkhn 35 Node Voltage Method (3) Ex. 1 i1 i3 v1 E1 = 7 V; E2 = 4 V; E3 = 2 V; – + R R1 = R2 = R3 = 2 k Ω; R4 = 6 k Ω; 1 E3 R3 R2 + E1 R5 = 3 k Ω; solve the circuit? + vx – + – i2 i5 – R  E− v E − v E− v + v v 5 1 1+ 2 1 −3o 1 = 0 2 vo  + R R R E2  1 2 3 i µv –  R4 4 x v v− v  2 +2 o = 0   R4 R 5 =µ = µ − vo v x ( vv1 2 ) 2µ + 3 → = 7− 6 v1 3 → = → = = µ + 3 v1 6V i1 0.5mA µ → ∞ 2 Active Circuits - sites.google.com/site/ncpdhbkhn 36 Mesh Current Method (1) Ex. 1 – + E1 = 7 V; E2 = 4 V; E3 = 2 V; R 1 E3 R3 R2 + R = R = R = 2 k Ω; R = 6 k Ω; E1 vo 1 2 3 4 + – + R5 = 3 k Ω; solve the circuit? – – R5 E2 R4 – + R 1 E3 R3 R2 E1 i + x + – ix io – R5 vo βix E2 R4 β → ∞ ic = βix Active Circuits - sites.google.com/site/ncpdhbkhn 37 Mesh Current Method (2) Ex. 1 i1 E1 = 7 V; E2 = 4 V; E3 = 2 V; – + R R1 = R2 = R3 = 2 k Ω; R4 = 6 k Ω; 1 E3 R3 R2 E1 i R = 3 k Ω; solve the circuit? + 5 x i + – D i – i A B R5 vo E2 R4 + −=− ic = βix Ri12A Ri( A i B ) E 12 E Ri(−+ i )( Ri −= i ) E 9β − 80 2BA 4 Bc 2 →i = + − = A 18β − 70 → = Ri3D Ri 5( D i c ) E 3 iA 0.5mA =β = β − β → ∞ → = ic i x( ii BD ) i1 0.5mA Active Circuits - sites.google.com/site/ncpdhbkhn 38