lecture03_06_25_2010..
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EE40 Lecture 3 Josh Hug 6/25/2010 “Users [are] reporting a drop in signal strength when the phone is held.” -BBC “If you ever experience this on your iPhone 4, avoid gripping it in the lower left corner in a way that covers both sides of the black strip in the metal band, or simply use one of many available cases.”-Apple EE40 Summer 2010 Hug 1 Logistical Notes • Office Hours – Room reservation has been put in, but no word from the people yet. I’ve got someone looking into it • HW1 due today at 5 PM in the box in 240 Cory • No i>Clicker today [couldn’t get hardware today] EE40 Summer 2010 Hug 2 Nodal Analysis: Example R1 R5 R3 V 1 R2 I1 R4 V2 Using the basic method • 5 unknowns • 2 KCL equations • 3 KVL equations EE40 Summer 2010 Hug 3 Nodal Analysis: Example R1 a c R3 V R2 1 I1 R5 V2 R4 b One equation, one unknown With Node Voltage: ππ − π2 ππ ππ + π1 + −πΌ1 + =0 π 5 π 4 π 1 πΊ5 ππ − π2 + πΊ4 ππ − πΌ1 + πΊ1 ππ + π1 = 0 ππ πΊ5 + πΊ4 + πΊ1 = πΊ5 π2 + πΌ1 − πΊ1 π1 EE40 Summer 2010 Hug 4 Nodal Analysis: Example R1 a c R3 V R2 1 I1 R5 V2 R4 b ππ πΊ5 + πΊ4 + πΊ1 = πΊ5 π2 + πΌ1 − πΊ1 π1 πΊ5 π2 + πΌ1 − πΊ1 π1 ππ = πΊ5 + πΊ4 + πΊ1 π 4 πΌ1 π 1 π 5 − π 5 π1 + π 1 π2 ππ = π 4 π 5 + π 1 π 4 + π 5 EE40 Summer 2010 It’s fine to leave your answer in terms of conductances on HW and exams Hug 5 Dependent Sources • In practice, we’ll want to use controllable sources – Called “dependent sources” since their output is dependent on something external to the source itself vs +_ independent vs(x) +_ dependent • In theory, a dependent source could be a function of anything in the universe – Intensity of light incident on the source – Number of fish within 3 miles EE40 Summer 2010 Hug 6 Dependent Sources • Since we’re building electrical circuits, dependent sources have been developed which are functions of other electrical quantities 10Ω 5V 40Ω 100Ω 3π40 πΌ100 EE40 Summer 2010 3π40 3 × 4π = = = 0.12π΄ 100Ω 100Ω Hug 7 Dependent Sources • Dependent sources allow us to decouple the controller from the controlled – Acceleration of the engine affect by gas pedal – Gas pedal not affected by engine acceleration • This is in contrast to our circuits so far where everything is connected EE40 Summer 2010 Hug 8 Dependent Sources With Feedback • Dependent sources can be coupled to their controller • This is useful for when the controller needs feedback from the thing being controlled • Can be a little tricky to analyze EE40 Summer 2010 Hug 9 Node Voltage With Dependent Source ππ − 100 ππ + − 4π1 = 0 100 20 ππ − 100 π1 = − 100 • There are two ways to proceed: • Direct substitution (almost always better) • Indirect substitution (tough part of the reading) 100Ω ππ 100V EE40 Summer 2010 4π1 20Ω Hug 10 Direct Substitution Method for Dependent Sources ππ − 100 ππ + − 4π1 = 0 100 20 ππ − 100 π1 = − 100 ππ − 100 ππ ππ − 100 + +4 =0 100 20 100 ππ − 100 + 5ππ + 4ππ − 400 = 0 10ππ = 500 ππ = 50 EE40 Summer 2010 Hug 11 Node Voltage With Dependent Source ππ − 100 ππ + − 4π1 = 0 100 20 ππ − 100 π1 = − 100 • There are two ways to proceed: • Direct substitution (almost always better) • Indirect substitution (tough part of the reading) 100Ω ππ ? 100V EE40 Summer 2010 π=π π ? 4π1 20Ω πΌ = π π = π(π π ) ? Hug 12 Indirect Substitution Method for Dependent Sources ππ − 100 ππ + − 4π πΌ= 00 1 = 100 20 100Ω ππ Node voltage vs. dummy source 100 + 100πΌ ππ = 6 4πI 1 100V 20Ω Node voltage vs. controlling current Real source vs. dummy source πΌ = π1 4 100 − ππ π1 = 100 100 + 100πΌ πΌ 100 − 6 = 4 100 πΌ = 2π΄ EE40 Summer 2010 Hug 13 Summary So Far • Dependent sources model controllable electrical sources • Node voltage can be used with dependent sources – Controlling input can be seemingly difficult to deal ππ −100 ππ with, e.g. + − 4π1 = 0 100 20 – Direct Substitution: Replace controlling input with expression in terms of • The node voltages you’re trying to solve for • Known currents if controlling input driven by current source – Indirect Substitution: Treat dependent source as a new variable and solve (better in rare cases) EE40 Summer 2010 Hug 14 Useful Resistive Circuits • Wheatstone Bridge – Used for measuring unknown resistances • Strain Gauge – Used for measuring weight EE40 Summer 2010 Hug 15 Wheatstone Bridge • Named for Charles Wheatstone • Used for measuring resistance of an unknown resistor R R2 1 • Parts: – Known resistors R1 and R2 – Adjustable resistor R3 – Unknown resistance Rx + V – R3 Rx • Basic concept: – If π 1 π 3 = π 2 , π π₯ then no current will flow in the middle branch Invented 1833 by Samuel Christie Hit the charts when remixed by Wheatstone in 1843 EE40 Summer 2010 Hug 16 Finding the value of Rx • Adjust R3 until there is no current in the detector Then, Rx = R2 R1 R3 Derivation: KCL => i1 = i3 and i2 = ix R1 + V – R2 KVL => i3R3 = ixRx and i1R1 = i2R2 i1R3 = i2Rx R3 Rx R3 R1 EE40 Summer 2010 = Rx R2 Hug 17 Strain Gauge Intuition • Resistance is a function of wire length and area • Weight stretches a wire, changing its shape • Can theoretically get weight of a load by seeing how resistance varies when a load is added EE40 Summer 2010 Hug 18 Resistivity • Wire resistance: π = π π π΄ • π is resistivity, measured in Ω·m • Can think of as how tightly molecular lattice holds on to electrons π Material Copper 1.68x10-8 Aluminum 2.82x10-8 Nichrome 1.1x10-6 Glass 1010 EE40 Summer 2010 Hug 19 Wire Gauge Gauge Diameter [mm] Area [mm2] 10 2.58 5.26 14 1.62 2.08 16 1.29 1.31 Resistance of 30m, 16 gauge extension cord? 30m −8 1.68 × 10 Ωπ 1.31 × 10−6 π2 0.384Ω If carrying 10 amps, how much power dissipated? π = ππΌ = πΌ2 π = 38.4 π EE40 Summer 2010 Hug 20 Basic Principle • Pull on resistor: • L=L0+ΔL • A=A0-ΔA • V=LA [constant] • Length wins the battle to control resistance • R=R0+ΔR ΔπΏ Define strain π = πΏ0 Δπ = πΊπΉ × π π 0 EE40 Summer 2010 π π =π π΄ πΏ0 + βπΏ π = π 0 + βπ = π π΄0 − βπ΄ Where the Gauge Factor relates to how length/area change. GF~2 Hug 21 Using Strain to Measure Weight Δπ = πΊπΉ × π π 0 • We’ll attempt to design a circuit which can measure ε – Convert ε Δπ π 0 into a voltage Δπ π 0 Our Circuit Vx(ε) • We’ll leave it to the mechanical engineers to map ε back to load Microcontroller EE40 Summer 2010 Hug 22 Using Strain to Measure Weight ε Δπ π 0 Our Circuit Vx(ε) Microcontroller vx ADC vx_to_delRr0(vx) delRr0 weight EE40 Summer 2010 eps2weight(eps) calc_eps(delRr0) eps Hug 23 One Possible Design • Here, Rx is a variable resistor, where Rx is dependent on strain • As strain varies, so will vx EE40 Summer 2010 Hug 24 One Possible Design vx is what μController sees −π 0 ππ + π 0 ππ₯ + π πππ ππ₯ Δπ = ππ Δπ 1 π= × π€πππβπ‘ = π(π) π 0 πΊπΉ μController calculates ε from known quantities Vx, Vs, R0, Rref, GF, and then weight from ε EE40 Summer 2010 Hug 25 Better Circuits • This will work, but: – ππ₯ ≠ 0 for π = 0 – As source voltage varies (e.g. battery gets old), calibration changes – “zero drift” EE40 Summer 2010 Hug 26 Improvement #1: Half Bridge Works, but requires balanced sources EE40 Summer 2010 Hug 27 Improvement #2: Full Bridge Want π£0 Use voltage divider π 0 π£π = π£π π 0 + Δπ + π 0 π 0 π£π π£π = π£π = π 0 + π 0 2 Works, but requires resistors with value equal to R0 EE40 Summer 2010 Hug 28 Strain Gauge Summary • We can map strain (weight) to resistance • Simplest design (voltage divider) works, but is subject to zero-drift • More complex circuits give different design tradeoffs – Wheatstone-bridge provides arguably the best design • We will explore these tradeoffs in lab on Wednesday EE40 Summer 2010 Hug 29 Useful Resistive Circuit Summary • The Wheatstone bridge (and other designs) provide us with a way to measure an unknown resistance • There are resistors which vary with many useful parameters, e.g. – Incident light – Temperature – Strain • And then… there are always toasters EE40 Summer 2010 Hug 30 Back to Circuit Analysis • Next we’ll discuss a few more circuit analysis concepts – Superposition – Equivalent Resistance • Deeper explanation of equivalent resistance • For circuits with dependent sources – Thevenin/Norton Equivalent Circuits – Simulation EE40 Summer 2010 Hug 31 Superposition • Principle of Superposition: • In any linear circuit containing multiple independent sources, the current or voltage at any point in the network may be calculated as the algebraic sum of the individual contributions of each source acting alone. • A linear circuit is one constructed only of linear elements (linear resistors, and linear capacitors and inductors, linear dependent sources) and independent sources. • Linear means I-V characteristic of all parts are straight when plotted EE40 Summer 2010 Hug 32 Superposition Procedure: 1. Determine contribution due to one independent source • Set all other sources to 0: • Replace independent voltage source by short circuit • independent current source by open circuit 2. Repeat for each independent source 3. Sum individual contributions to obtain desired voltage or current EE40 Summer 2010 Hug 33 Easy Example • Find Vo 12 W 4V +– + 4 W Vo 4A – 12 W 4V +– + 4 W Vo – Voltage Divider: -1V EE40 Summer 2010 Hug 34 Easy Example • Find Vo 12 W 4V +– 4A + 4 W Vo – 12 W + 4A 4 W Vo – Current Divider: -(3A*4Ω)=-12V EE40 Summer 2010 Hug 35 Easy Example • Find Vo 12 W 4A 4V +– + 4 W Vo – V0=-12V-1V=-13V Due to current source EE40 Summer 2010 Due to voltage source Hug 36 Hard Example EE40 Summer 2010 Hug 37 Example Equivalent resistance: 3+2+(3||2) = 5+(3*2)/(3+2)=6.2V Current is 10A 50V loss through top 5Ω, leaving 12V across v0 EE40 Summer 2010 Hug 38 Example VT Vo ππ ππ ππ − ππ + + 31 + =0 3 2 2 ππ − ππ ππ −31 + + =0 2 3 EE40 Summer 2010 This will work… But algebra is easier if we pick a better ground Hug 39 Example VT VB ππ΅ ππ΅ ππ΅ − ππ + + =0 2 3 3 2 ππ΅ = ππ 7 ππ ππ − ππ΅ −31 + + =0 2 3 ππ΅ = 12π EE40 Summer 2010 ππ = −12π Hug 40 Example Vo=-12V EE40 Summer 2010 Vo=12V ππ = −12π + 12π = 0π Hug 41 Note on Dependent Sources • You can use superposition in circuits with dependent sources • However, DON’T remove the dependent sources! Just leave them there. EE40 Summer 2010 Hug 42 Equivalent Resistance Review • If you add a source to any two terminals in a purely resistive circuit – The added source will “see” the resistive circuit as a single resistor 10Ω 1A 25V EE40 Summer 2010 10Ω 10Ω 10Ω 1A 25Ω 25V Hug 43 Alternate Viewpoint 10Ω 10Ω 10Ω 10Ω We can think of the circuit above as a two terminal circuit element with an I-V characteristic Slope: 1/25Ω v i EE40 Summer 2010 Hug 44 Equivalent Resistance • Let’s consider the IV characteristic of the following circuit: αV2 + V2 - R1 R2 I V π2 − π π2 + + πΌπ2 = 0 π 1 π 2 EE40 Summer 2010 Hug 45 Equivalent Resistance αV2) + V2 - R1 R2 π2 − π π2 + + πΌπ2 = 0 π 1 π 2 π 2 π π− π 1 + π 2 + πΌπ 1 π 2 πΌ= π 1 EE40 Summer 2010 I V π 2 π π2 = π 1 + π 2 + πΌπ 1 π 2 πΌ 1 + πΌπ 2 = π π 1 + π 2 + πΌπ 1 π 2 Hug 46 Equivalent Resistance αV2 + V2 - I R1 R2 V v πΌ 1 + πΌπ 2 = π π 1 + π 2 + πΌπ 1 π 2 = 1/Req i Slope: 1+πΌπ 2 π 1 +π 2 +πΌπ 1 π 2 This circuit just acts like a resistor! EE40 Summer 2010 Hug 47 Equivalent Resistance Summary So Far • Purely resistive networks have an I-V characteristic that looks just like their equivalent resistance • Purely resistive networks which also include dependent sources also act like resistors • Let’s see what happens with a circuit with an independent source 5Ω 10V EE40 Summer 2010 I V Hug 48 Equivalent Resistance Summary So Far 5Ω 10V I V π − 10 π πΌ= = −2 5 5 Doesn’t match our basic I-V characteristics… good! v i EE40 Summer 2010 Hug 49 Interestingly… 10Ω 20V I 10Ω V π − 20 π πΌ= + 10 10 π = −2 5 v i EE40 Summer 2010 Hug 50 Equivalent Circuits 10Ω 20V 10Ω Has the exact same I-V characteristic as: 5Ω 10V EE40 Summer 2010 Hug 51 Thevenin Equivalents • We saw before that we can replace a network of resistors (and dependent sources) with a single equivalent resistance • Now, we have that we can replace any circuit we can build so far with a single voltage source and resistor – Not proven, but it’s true, trust me • This two element network is known as a Thevenin equivalent • Generalization of the idea of equivalent resistance Again: Discovered twice, named after the second guy! EE40 Summer 2010 Hug 52 Why is this useful? • Can swap out elements and not have to resolve a big circuit again • Captures the fundamental operation of the circuit as a whole (at chosen two terminals only!) EE40 Summer 2010 Hug 53 Thevenin Algorithm for Independently Sourced Circuits • What you’re ultimately doing is finding the I-V characteristic of the circuit • You can do this by attaching a made up V, and calculating I as on slides 49 and 50 – Often called a “Test Voltage” • This is equivalent to: – Finding the “open circuit” voltage – Finding the “short circuit” current EE40 Summer 2010 Hug 54 Calculating a Thévenin Equivalent 1. Calculate the open-circuit voltage, voc a network of sources and resistors + voc – b 2. Calculate the short-circuit current, isc • Note that isc is in the direction of the open-circuit voltage drop across the terminals a,b ! a network of sources and resistors isc RTh b EE40 Summer 2010 VTh ο½ voc voc ο½ isc Hug 55 Example – On the Board 8A 2Ω 12Ω 12V 6Ω Find the Thevenin Equivalent circuit, by finding 1. VOC 2. ISC EE40 Summer 2010 Hug 56 Finding Thevenin Resistance Directly • If there are no dependent sources in the circuit, we can find the Thevenin Resistance directly • Algorithm is easy: – Set all independent sources to zero • Voltage source becomes short circuit • Current source becomes open circuit – Leave dependent sources intact – Find equivalent resistance between terminals of interest EE40 Summer 2010 Hug 57 Norton Equivalent Circuit • Any network of voltage sources, current sources, and resistors can also be replaced by an equivalent circuit consisting of an independent current source in parallel with a resistor without affecting the operation of the rest of the circuit. Norton equivalent circuit a network of sources and resistors iL vL RL ≡ iN RN – b EE40 Summer 2010 + a + iL vL RL – b Hug 58 Example – On the Board 8A 2Ω 12Ω 12V 6Ω Find the Thevenin Equivalent resistance directly EE40 Summer 2010 Hug 59 Finding IN and RN • We can derive the Norton equivalent circuit from a Thévenin equivalent circuit simply by making a source transformation: RTh – + vTh a a + iL vL RL iN RN – b vTh iL ο½ RTh ο« RL + iL vL RL – b RN iL ο½ iN RN ο« RL voc vTh RN ο½ RTh ο½ ; iN ο½ ο½ isc isc RTh EE40 Summer 2010 Hug 60 Circuit Simulation • Automated equation solvers use our algorithm: – Choose a ground node – Assign node voltage labels to all nodes – Write out a system of N-1 linear equations – Solve (using standard linear algebra techniques) • One pretty handy tool is falstad.com’s circuit simulator • Let’s try a live demo EE40 Summer 2010 Hug 61 Extra Slides EE40 Summer 2010 Hug 62 Delta-Wye Conversion EE40 Summer 2010 Hug 63 Equivalent Resistances R2 R1 V + R3 ο R4 R5 Are there any circuit elements in parallel? series? No Are there any circuit elements in parallel? No What do we do? 1. Be clever and find I-V characteristic directly 2. Apply weirder transformation rules than series or parallel EE40 Summer 2010 Hug 64 Y-Delta Conversion • These two resistive circuits are equivalent for voltages and currents external to the Y and D circuits. Internally, the voltages and currents are different. Rc a a b b R1 Rb R2 Ra R3 c c R1 = RbRc Ra + Rb + Rc EE40 Summer 2010 R2 = RaRc Ra + Rb + Rc R3 = RaRb Ra + Rb + Rc Hug 65 D-Y and Y-D Conversion Formulas Delta-to-Wye conversion Wye-to-Delta conversion Rc a R1 = R2 = b RbRc Ra = Ra + Rb + Rc Rb RaRc Ra c Rb = Ra + Rb + Rc a R3 = RaRb R1R2 + R2R3 + R3R1 R1 R1R2 + R2R3 + R3R1 R2 b R1 R2 Ra + Rb + Rc R3 Rc = R1R2 + R2R3 + R3R1 R3 c EE40 Summer 2010 Hug 66 Circuit Simplification Example Find the equivalent resistance Rab: 2W 2W a a 18W 12W 6W ≡ 9W 4W b 9W Rb=18 Ra=12 Rc=6 EE40 Summer 2010 R3 = RaRb Ra + Rb + Rc 4W b =6 R2=2 3 Hug 67 General Versions of Thevenin Slides EE40 Summer 2010 Hug 68 Equivalent Resistance Summary So Far • Purely resistive networks have an I-V characteristic that look just like their equivalent resistance • Networks which include dependent sources also act like resistors • Let’s see what happens with a circuit with a source R1 I Vs EE40 Summer 2010 V Hug 69 Equivalent Resistance Summary So Far R1 Vs I V π − ππ πΌ= π 1 Doesn’t match our basic I-V characteristics… good! v i EE40 Summer 2010 Hug 70 Interestingly… R1 Vs I R2 V π − ππ π πΌ= + π 1 π 2 v i EE40 Summer 2010 Hug 71 R1 Vs R2 Has the exact same I-V characteristic as: Req Vs EE40 Summer 2010 Hug 72 I 2W 1W + 30 V 4W Vo – EE40 Summer 2010 Hug 73
