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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 • Complex numbers • Linear equations d2I R dI 1 I0 2 L dt LC dt V(t)=V0ejωt V I 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] [volts] Potential difference: V=VA–VB Energy to move unit charge from A to B in electric field B V E ds A B W QE ds A E V Charge Q=e Current: rate of flow of charge dQ I nAve dt No. electrons/unit vol Cross-section area of conductor Drift velocity [amps] Power: rate of change of work dW d dV dQ QV Q V P dt dt dt dt IV [watts] Ohm’s law Voltage difference current Resistor symbols: L I R A V V IR R=Resistance Ω[ohms] L R A 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] 1 g R conductivity [seimens/m] 1 2 V Power dissipation by resistor: P IV I2R R 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 V I0 Rint 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 2f T 50Hz power, audio, radio… RMS values AC Power dissipation VRM S Why 2? P IRM SVRM S V0 2 VRM S R 2 Square root of mean of V(t)2 VRMS 1 T T 0 V t dt 2 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 n I4 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 V n 0 4kΩ R3 Calculate the voltage across R2 Kirchoff’s voltage law: V n 0 -V0+IR1+IR2+IR3=0 +IR1 + –V0 – 1kW I + – 5V=I(1+3+4)kΩ V0 + 3kW – 4kW + – +IR1 5V I 0.625mA 8000W +IR2 VR2=0.625mA×3kΩ=1.9V Series / parallel circuits R1 R2 R3 RT Resistors in series: RTotal=R1+R2+R3… Resistors in parallel R1 R2 R3 1 RT 1 n R n 1 1 1 R1 R 2 R 3 Two parallel resistors: R1R 2 RT R1 R 2 R n n Potential divider R1 V0 R2 V0R 2 R1 R 2 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 mA I 3 2 mA 3 3 V3 R3 I 3 4 V 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 R1 VX + I1 + 9V R2 I3 I2 - R3 - R1=3kΩ R2=2kΩ R3=6kΩ + 2V - 0V 0 0 I 2 I 3 I1 VX (2V ) VX VX 9V R2 R3 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!!! 0 R1 I 2 I 3 I1 I1 VX (2V ) VX VX 9V 0 R2 R3 R1 9V + VX R2 I3 R3 0V 1 1 1 9V 2V VX 3mA 1mA 2mA R 2 R 3 R1 R1 R 2 1 1 1 1 VX 2mA 2 6 3 kW VX 2mA 1kW VX 2V 9V 2V 7 mA 3kW 3 2V 1 I3 mA 6kW 3 I1 I2 2V 2V 2mA 2kW 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 Veq=VOS VL0 VOS Req I SS Req = resistance between terminals when all voltages sources shorted – Warning! This is not always obvious! R2 Vos V0 R1 R2 I1 V0 + R1 I2 R2 Iss RL VL R eq V0 I ss R1 R2 V0 R1R 2 R1 R 2 V0 R1R 2 R1 Norton’s theorem Any linear network of voltage/current sources and resistors I0 Equivalent circuit Req R1 V0 Ieq Rload Ieq Req Rload VL IR IL IL R eq VL=V0–ILR1 Ieq V0 R eq R eqIeq R eqIL VL R eq R1 V0 R V0 R R A R1=3kΩ 9V + R2=2kΩ R3= 6kΩ + 1kΩ 2V A 2V + B B using Thevenin’s theorem: from Prev. Already know VAB =Veq= 2V A I1 9V + R1 3kW R2=2kΩ 2V Iss + I2 I1 9V 3ma 3kW I SS B 2V 1ma 2kW I1 I 2 2ma I2 Req Veq I SS 1kW 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 Example: Superposition R1=3kΩ R2=2kΩ R3=6kΩ IA 9V + R 2R 3 R1 R2 R3 R1 IC IB R3 R2 9V IA 2mA 4.5kW R 2R 3 1.5kW R2 R3 ICR 3 IBR 2 9V IC 0.5mA 4.5kW 1.5 3V 4.5 IB 1.5mA R1 IE IG IF R3 2V + R1=3kΩ R2=2kΩ R3=6kΩ R2 2V + IG IE R3 R1 R1R 3 2kW R1 R 3 IF R1R 3 R1 R 3 4kW R2 2V IF 0.5mA 4kW R2 IGR 3 IER 1 2V 1 IG mA 6 2 1V 4 1 IE mA 3 R1 + 9V I1 R1 + 9V R2 I3 I2 R3 + = IA IC R3 2V R2 R1 IB + IE R2 IG R3 IF + I1=IA+IE I2=IB+IF I3=IC+IG 1 I1 2 mA 3 7 mA 3 I2 1.5 0.5mA 1 I3 0.5 mA 6 1 mA 3 2mA 2V Matching: maximum power transfer Find R to give maximum L power in load V0 Rin Rload 2 VL2 R 1 2 L P V0 2 RL R in R L R L dP 2 R in R L 2R L R in R L V0 0 4 dR L Rin RL 2 R in R L 2R L 0 Rin RL 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
