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CP2 Circuit Theory Dr Todd Huffman [email protected] http://www-pnp.physics.ox.ac.uk/~huffman/ Aims of this course: Understand basic circuit components (resistors, capacitors, inductors, voltage and current sources, op-amps) Analyse and design simple linear circuits – + + + Circuit Theory: Synopsis Basics: voltage, current, Ohm’s law… Kirchoff’s laws: mesh currents, node voltages… Thevenin and Norton’s theorem: ideal voltage and current sources… Capacitors: Stored energy, RC and RL transient circuits Inductors: AC theory: complex notation, phasor diagrams, RC, RL, LCR circuits, resonance, bridges… Op amps: ideal operational amplifier circuits… Op-amps are now on the exam syllabus Reading List • Electronics: Circuits, Amplifiers and Gates, D V Bugg, Taylor and Francis Chapters 1-7 • Basic Electronics for Scientists and Engineers, D L Eggleston, CUP Chapters 1,2,6 • Electromagnetism Principles and Applications, Lorrain and Corson, Freeman Chapters 5,16,17,18 • Practical Course Electronics Manual http://www-teaching.physics.ox.ac.uk/practical_course/ElManToc.html Chapters 1-3 • Elementary Linear Circuit Analysis, L S Bobrow, HRW Chapters 1-6 • The Art of Electronics, Horowitz and Hill, CUP Why study circuit theory? • Foundations of electronics: analogue circuits, digital circuits, computing, communications… • Scientific instruments: readout, measurement, data acquisition… • Physics of electrical circuits, electromagnetism, transmission lines, particle accelerators, thunderstorms… • Not just electrical systems, also thermal, pneumatic, hydraulic circuits, control theory Mathematics required • Differential equations 2 d I dt • Complex numbers L dt 1 LC I 0 V(t)=V0ejωt I • Linear equations 2 R dI V Z Z R jX V0–I1R1–(I1–I2)R3 = 0 (I1–I2)R3–I2R2+2 = 0 Covered by Complex Nos & ODEs / Vectors & Matrices lectures Charge, voltage, current Charge: determines strength of electromagnetic force quantised: e=1.62×10-19C [coulombs] Potential difference: V=VA–VB [volts] Energy to move unit charge from A to B in electric field B E V V E ds A B W Q E ds A Charge Q=e Current: rate of flow of charge I dQ dt nAve No. electrons/unit vol Cross-section area of conductor Drift velocity [amps] Power: rate of change of work P dW dt d dt QV Q dV dt V dQ dt IV [watts] Ohm’s law Voltage difference current Resistor symbols: L I R A V V IR R L A R=Resistance Ω[ohms] ρ=Resistivity Ωm Resistivities Silver 1.6×10-8 Ωm Copper 1.7×10-8 Ωm Manganin 42×10-8 Ωm Distilled water 5.0×103 Ωm PTFE ~1019 Ωm Conductance [seimens] g 1 R conductivity [seimens/m] Power dissipation by resistor: P IV I 2R V 2 R 1 Voltage source V0 + – V0 battery cell Rload Ideal voltage source: supplies V0 independent of current Real voltage source: Rint Rload Vload=V0–IRint Constant current source I0 Rload Ideal current source: supplies I0 amps independent of voltage Symbol: or Real current source: I0 Rint Rload I load I 0 V R int AC and DC DC (Direct Current): Time independent + – V=V0 AC (Alternating Current): + - Constant voltage or current Time dependent Periodic I(t)=I0sin(ωt) 2 2 f T 50Hz power, audio, radio… RMS values 2 AC Power dissipation VRM S Why 2 ? P IRMS VRMS VRMS R V0 2 Square root of mean of V(t)2 VRM S 1 T T 0 V 2 t dt Passive Sign Convention Passive devices ONLY - Learn it; Live it; Love it! R=Resistance Ω[ohms] V IR Two seemingly Simple questions: Which way does the current flow, left or right? Voltage has a ‘+’ side and a ‘-’ side (you can see it on a battery) on which side should we put the ‘+’? On the left or the right? Given V=IR, does it matter which sides for V or which direction for I? Kirchoff’s Laws I Kirchoff’s current law: I1 I2 I3 Sum of all currents at a node is zero I1+I2–I3–I4=0 I I4 n 0 (conservation of charge) Here is a cute trick: It does not matter whether you pick “entering” or “leaving” currents as positive. BUT keep the same convention for all currents on one node! II Kirchoff’s voltage law: Around a closed loop the net change of potential is zero (Conservation of Energy) R1 I 1kΩ V0 5V 3kΩ R2 Vn 0 4kΩ R3 Calculate the voltage across R2 Kirchoff’s voltage law: Vn 0 -V0+IR1+IR2+IR3=0 +IR1 + –V0 – 1kW I + – 5V=I(1+3+4)kΩ V0 + 3kW – 4kW + – +IR1 I 5V 8000 W 0 . 625 mA +IR2 VR2=0.625mA×3kΩ=1.9V Series / parallel circuits R1 R2 R3 RT R n Resistors in series: RTotal=R1+R2+R3… Resistors in parallel R1 R2 R3 1 RT R n Two parallel resistors: 1 1 R1 n 1 R2 RT 1 R3 R 1R 2 R1 R2 n Potential divider R1 V0 R2 V 0R 2 R1 R2 USE PASSIVE SIGN CONVENTION!!! Show on blackboard Mesh currents I1 + + 9V + + I1 R2 I3 R3 I2 I2 + R1 R1=3kΩ R2=2kΩ R3=6kΩ 2V First job: Label loop currents in all interior loops Second job: USE PASSIVE SIGN CONVENTION!!! Third job: Apply KCL to elements that share loop currents Define: Currents Entering Node are positive I1–I2–I3 = 0 I3 = I1-I2 Mesh currents Fourth job: I1 + + 9V + + I1 R2 I3 R3 I2 I2 + R1 R1=3kΩ R2=2kΩ R3=6kΩ 2V Apply Kirchoff’s Voltage law around each loop. Last job: USE Ohm’s law and solve equations. Mesh currents R1 -9V+I1R1+I3R3 = 0 –I3R3+I2R2+2V = 0 9V/kΩ = 9I1–6I2 -2V/kΩ = -6I1+8I2 I2=1 mA R2 I3 I2 R3 I2 + I1 + + 9V + + I1 R1=3kΩ R2=2kΩ R3=6kΩ 2V I3 =I1–I2 Solve simultaneous equations I1 5 3 mA V3 R3 I 3 4 V I3 2 3 mA Node voltages R1 I1 9V VX + + R1=3kΩ R2=2kΩ R3=6kΩ R2 I3 I2 R3 - + 2V - 0V Step 1: Choose a ground node! Step 2: Label voltages on all unlabled nodes Step 3: Apply KCL and ohms law using the tricks Node voltages VX R1 + I1 9V + R2 I3 I2 - R3 - R1=3kΩ R2=2kΩ R3=6kΩ + 2V - 0V 0 0 I 2 I 3 I1 V X ( 2V ) R2 VX R3 V X 9V R1 All currents leave all labeled nodes And apply DV/R to each current. Only one equation, Mesh analysis would give two. USE PASSIVE SIGN CONVENTION!!! R1 0 0 I 2 I 3 I1 V X ( 2V ) R2 VX R3 I1 9V V X 9V + R1 VX R2 I3 R3 0V 1 1 1 9V 2V V X 3 mA 1mA 2 mA R3 R1 R1 R2 R2 1 1 1 1 VX 2 mA 6 3 kW 2 VX 1k W I1 I3 2 mA 9V 2V 7 3k W 2V 6k W 1 3 3 mA VX 2 V mA I2 2V 2V 2k W 2mA I2 + 2V Thevenin’s theorem Any linear network of voltage/current sources and resistors Veq Req Equivalent circuit In Practice, to find Veq, Req… RL (open circuit) RL0 (short circuit) IL0 VL0 Veq=VOS R eq V OS I SS Req = resistance between terminals when all voltages sources shorted – Warning! This is not always obvious! I1 V0 + V os V 0 R1 I2 R2 R2 R1 R 2 V0 Iss RL VL R eq I ss V0 R1 R2 R1 R2 V0 R1 R 1R 2 R 1 R 2 Norton’s theorem Any linear network of voltage/current sources and resistors I0 Equivalent circuit Req R1 V0 I eq Rload VL=V0–ILR1 I eq Req I eq I R I L V0 Rload VL R eq R eq I eq R eq I L VL R eq R eq R 1 V0 R V0 R R IL A R1=3kΩ 9V + R2=2kΩ R3= 6kΩ + 1kΩ A 2V + 2V B B using Thevenin’s theorem: from Prev. Already know VAB =Veq= 2V A I1 9V + R2=2kΩ R1 3 k W 2V Iss + I2 I1 9V 3kW I2 3 ma 2V 2 kW 1ma I SS I 1 I 2 2 ma B R eq V eq I SS 1k W A R1=3kΩ 9V R2=2kΩ + R3= 6kΩ + A 2V 2mA 1kΩ B B using Norton’s theorem: Same procedure: Find ISS and VOC IEQ = ISS and REQ = VOC/ISS Superposition R1 + 9V I1 R1 + 9V IA I1=IA+IE R2 I3 I2 R3 + = IC R3 2V Important: label Everything the same directions! R2 R1 IB + I2=IB+IF IE R2 IG R3 I3=IC+IG IF + 2V R2 R1 + 9V IA IC IB R3 R1=3kΩ R2=2kΩ R3=6kΩ Example: Superposition IA 9V R1 + IA 9V 4 . 5k W R1 IC IB R3 R2 2 mA R 2R 3 R2 R3 4 . 5k W R 2R 3 R2 R3 1 . 5k W I C R 3 IBR 2 9 V I C 0 . 5 mA 1 .5 4 .5 3V I B 1 . 5 mA R1 IE IG IF R3 2V + R1=3kΩ R2=2kΩ R3=6kΩ R2 2V + IG IE R3 R1 R 1R 3 R1 R3 2k W R2 IF IF 4k W R1 R3 4k W R2 2V R 1R 3 I G R 3 I ER 1 2 V 0 . 5 mA IG 1 6 mA 2 4 1V IE 1 3 mA R1 + 9V I1 R1 + 9V R2 I3 I2 R3 + = IA IC R3 2V R2 R1 IB + R2 IE IG R3 IF + I1=IA+IE I2=IB+IF I3=IC+IG 1 I1 2 mA 3 I 2 1 . 5 0 . 5 mA 1 I 3 0 . 5 mA 6 7 3 mA 2 mA 1 3 mA 2V Matching: maximum power transfer Find R to give maximum L power in load V0 Rin Rload 2 P dP dR 2 V0 L VL RL R in 2 2 V0 R in RL 1 RL RL 2 R L 2R L R in R L 2 R in RL R in R L 2 R L 0 4 0 R in R L Maximum power transfer when RL=Rin Note – power dissipated half in RL and half in Rin Circuits have Consequences • Problem: – My old speakers are 60W speakers. – Special 2-4-1 deal at ElCheap-0 Acoustics on 120W speakers!! • (“Offer not seen on TV!”) – Do I buy them? • Depends! 4W, 8W, or 16W speakers? • Why does this matter? And Now for Something Completely Different