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CP2 Circuit Theory Dr Todd Huffman t.huffman1@physics.ox.ac.uk 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 + – + +
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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: 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 Stored energy, RC and RL transient circuits
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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
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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
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Mathematics required Differential equations Complex numbers Linear equations V(t)=V 0 e jωt V 0 –I 1 R 1 –(I 1 –I 2 )R 3 = 0 (I 1 –I 2 )R 3 –I 2 R 2 +2 = 0 Covered by Complex Nos & ODEs / Vectors & Matrices lectures
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Charge, voltage, current Potential difference: V=V A –V B Energy to move unit charge from A to B in electric field Charge: determines strength of electromagnetic force quantised: e=1.62×10 -19 C [coulombs] [volts]
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Power: rate of change of work No. electrons/unit vol Cross-section area of conductor Drift velocity Charge Q=e Current: rate of flow of charge [amps] [watts]
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Ohm’s law Voltage difference current I L A V R=Resistance Ω[ohms] ρ=Resistivity Ωm Resistor symbols: R
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Resistivities Silver1.6×10 -8 Ωm Copper1.7×10 -8 Ωm Manganin42×10 -8 Ωm Distilled water5.0×10 3 Ωm PTFE~10 19 Ωm Conductance [seimens] conductivity [seimens/m] Power dissipation by resistor:
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Voltage source V0V0 Ideal voltage source: supplies V 0 independent of current V0V0 R int Real voltage source: V load =V 0 –IR int R load +–+– battery cell
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Constant current source Ideal current source: supplies I 0 amps independent of voltage Real current source: R load I0I0 R int I0I0 Symbol: or
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AC and DC DC (Direct Current): Constant voltage or current Time independent V=V 0 AC (Alternating Current): Time dependent Periodic I(t)=I 0 sin(ωt) 50Hz power, audio, radio… +–+– + -
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RMS values AC Power dissipation Why ? Square root of mean of V(t) 2
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Passive Sign Convention Passive devices ONLY - Learn it; Live it; Love it! R=Resistance Ω[ohms] Which way does the current flow, left or right? Two seemingly Simple questions: 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?
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Kirchoff’s Laws I Kirchoff’s current law: Sum of all currents at a node is zero I1I1 I3I3 I2I2 I4I4 I 1 +I 2 –I 3 –I 4 =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!
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II Kirchoff’s voltage law:Around a closed loop the net change of potential is zero (Conservation of Energy) V0V0 I R1R1 R2R2 R3R3 5V 3kΩ 4kΩ Calculate the voltage across R 2 1kΩ
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Kirchoff’s voltage law: V0V0 I 1k 3k 4k -V 0 +IR 1 +IR 2 +IR 3 =0 + + + + – – – – –V 0 +IR 1 +IR 2 +IR 1 5V=I(1+3+4)kΩ V R2 =0.625mA×3kΩ=1.9V
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Series / parallel circuits R1R1 R2R2 R3R3 Resistors in series: R Total =R 1 +R 2 +R 3 … R1R1 R2R2 R3R3 Resistors in parallel Two parallel resistors:
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Potential divider R1R1 R2R2 V0V0 USE PASSIVE SIGN CONVENTION!!! Show on blackboard
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Mesh currents 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 I1I1 I2I2 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ Second job: USE PASSIVE SIGN CONVENTION!!! First job: Label loop currents in all interior loops + + + Third job: Apply KCL to elements that share loop currents Define: Currents Entering Node are positive I 1 –I 2 –I 3 = 0 I 3 = I 1 -I 2
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Mesh currents 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 I1I1 I2I2 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ Last job: USE Ohm’s law and solve equations. Fourth job: Apply Kirchoff’s Voltage law around each loop. + + +
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Mesh currents 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 I1I1 I2I2 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ -9V+I 1 R 1 +I 3 R 3 = 0 I 3 =I 1 –I 2 –I 3 R 3 +I 2 R 2 +2V = 0 9V/k Ω = 9I 1 –6I 2 -2V/kΩ = -6I 1 +8I 2 I 2 =1 mA + + + Solve simultaneous equations
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Node voltages 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ VXVX 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
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Node voltages 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ VXVX 0V USE PASSIVE SIGN CONVENTION!!! - - + - Only one equation, Mesh analysis would give two. All currents leave all labeled nodes And apply V/R to each current.
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2V 9V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 VXVX 0V
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Thevenin’s theorem V eq R eq Any linear network of voltage/current sources and resistors Equivalent circuit
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In Practice, to find V eq, R eq … R L (open circuit) I L 0 V eq =V OS R L 0 (short circuit) V L 0 V0V0 + R1R1 RLRL I1I1 I ss R2R2 I2I2 VLVL R2R2 I2I2 R eq = resistance between terminals when all voltages sources shorted – Warning! This is not always obvious!
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Norton’s theorem R eq Any linear network of voltage/current sources and resistors Equivalent circuit I0I0
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V0V0 R1R1 R eq R load V L =V 0 –I L R 1 V0V0 R R
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using Thevenin’s theorem: from Prev. Already know V AB =V eq = 2V 2V + + R 1 =3kΩR 2 =2kΩ R 3 = 6kΩ A B 9V 2V + + A B R 2 =2kΩ + A B 2V 1kΩ I ss I1I1 I2I2
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using Norton’s theorem: 2V + + R 1 =3kΩR 2 =2kΩ R 3 = 6kΩ A B 9V 1kΩ 2mA A B Same procedure: Find I SS and V OC I EQ = I SS and R EQ = V OC /I SS
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Superposition 9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 9V + R1R1 R2R2 R3R3 IAIA IBIB ICIC 2V + R1R1 R2R2 R3R3 IEIE IFIF IGIG = + I 1 =I A +I E I 2 =I B +I F I 3 =I C +I G Important: label Everything the same directions!
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9V + R1R1 R2R2 R3R3 IAIA IBIB ICIC + R1R1 R2R2 R3R3 IAIA IBIB ICIC R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ Example: Superposition
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2V + R1R1 R2R2 R3R3 IEIE IFIF IGIG + R2R2 R3R3 IFIF IEIE IGIG R1R1 R 1 =3kΩ R 2 =2kΩ R 3 =6kΩ
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9V 2V + + R1R1 R2R2 R3R3 I1I1 I2I2 I3I3 9V + R1R1 R2R2 R3R3 IAIA IBIB ICIC 2V + R1R1 R2R2 R3R3 IEIE IFIF IGIG = + I 1 =I A +I E I 2 =I B +I F I 3 =I C +I G
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Matching: maximum power transfer V0V0 R in R load Find R L to give maximum power in load Maximum power transfer when R L =R in Note – power dissipated half in R L and half in R in
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Circuits have Consequences Problem: –My old speakers are 60W speakers. –Special 2-4-1 deal at El- Cheap-0 Acoustics on 120W speakers!! (“Offer not seen on TV!”) –Do I buy them? Depends! 4 , 8 , or 16 speakers? Why does this matter?
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And Now for Something Completely Different
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