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• Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Lecture 5 Inductors and transformers • Inductor – Hydraulic analogy – Duality with capacitor – Charging and discharging • Transformer – Physics – Fundamental equations – Dot convention – Comparison to voltage divider Robert R. McLeod, University of Colorado http://hilaroad.com/camp/projects/magnet.html 50 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics The Inductor Packages http://www.electronicsandyou.com/ electronics-basics/inductors.html Physics http://www.orble.com/userimages/user2207_1162388671.JPG Symbol Air core / generic Ferrite core Laminated iron core Robert R. McLeod, University of Colorado 51 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Inductor hydraulics • • • The hydraulic analogy to an inductor is a frictionless waterwheel with mass. Kinetic energy is stored in the wheel. The finite mass makes the wheel resist changes in current. Pump is turned on at t=0. Wheel is initially stationary , that is, I(0)=0. At a later time… …the wheel will be turning with a finite current. If the pump belt was removed, current would continue to flow. Robert R. McLeod, University of Colorado https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/ 52 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Inductor+resistor hydraulics “charging” phase Valve (switch) is open Wheel is initially stationary , that is, IINDUCTOR(0)=0. Pump is turned on at t=0. So all current initially goes through resistor according to Ohm’s law. Pump is constant current At a later time… Valve (switch) is open The wheel is turning, offering very little resistance. Thus a large current will go through the wheel. The same amount is still going through R (by Ohm’s law) but this is now a small fraction of the total current. Robert R. McLeod, University of Colorado https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/ 53 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Inductor+resistor hydraulics “discharge” phase Valve (switch) is closed at t=0 Wheel is initially moving, that is, IINDUCTOR(0)>0. By conservation of current, this current must run through resistor, dissipating energy. Valve (switch) is closed At a later time… The kinetic energy in the wheel is gone and it is nearly stationary. Approximately no current is flowing. Robert R. McLeod, University of Colorado https://ece.uwaterloo.ca/~dwharder/Analogy/Inductors/ 54 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics RL circuit 0.2 ms 1 KHz Vsource I VL + VL I - • • • • • Energy is stored in the magnetic field of the inductor. Power is finite, so the energy does not change instantaneously. The magnetic field is directly related to the current. Thus inductors resist changes in the current. Inductors act like short at DC, open at infinite frequency dI VL = L dt Robert R. McLeod, University of Colorado L τ L = = GL = 0.1[ms] R 55 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics RL charge/discharge Discharge • • Charge • At t=0, the voltage source is replaced by a short. What is the current versus time? • Boundary conditions At t=0, a 12 V voltage source replaces the short. What is the current versus time? Boundary conditions I 0 = VSource R = 1.2 [A ] I0 = 0 I (t = ∞) = VSource R = 1.2 [A ] I (t = ∞) = 0 Time constant Time constant Discharging exponential function Charging exponential function I (t ) = I 0e − t τ L I (t ) = I 0 1 − e − t τ L ( 1.2 1.2 1.0 1.0 0.8 0.8 I [A] I [A] τ L = GL = 1 mH 10 Ω = 0.1 ms τ L = GL = 1 mH 10 Ω = 0.1 ms 0.6 0.4 ) 0.6 0.4 0.2 0.2 0.1 0.2 t [mS] 0.3 0.4 Robert R. McLeod, University of Colorado 0.5 0.1 0.2 0.3 t [mS] 0.4 0.5 56 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics The Transformer Packages http:// www.audiophilejournal.co m/what-is-a-switchingpower-supply/ http://www.ehow.com/ how_7842346_buildsimple-electricaltransformer-school.html http:// en.wikipedia.org/ wiki/Transformer Physics http://www.witricity.com/pages/technology.html Symbol Robert R. McLeod, University of Colorado 57 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics Hydraulic analogy Primary NP gear teeth Secondary NS gear teeth • Two paddle-wheels connected by ideal gears with different #s of teeth. • Rotation ratio of the two shafts is inverse to gear ratio: ω P NS = ω S NP IP NS = IS NP T N • Torque ratio is proportional to gear ratio: P = P TS N S V N • Voltage (pressure) is proportional to torque, so: P = P VS N S ⎛ NS ⎞ ⎛ NP ⎞ VS = I SVS = PS • Is power conserved? PP = I PVP = ⎜ I S ⎝ N P ⎟⎠ ⎜⎝ N S ⎟⎠ • Current is proportional to rotation rate, so • Note gear connection could be flipped, reversing the secondary direction. http://en.wikipedia.org/wiki/Talk%3AHydraulic_analogy Robert R. McLeod, University of Colorado http://pptcrafter.wordpress.com/2013/01/17/drawing-and-animating-gears-in-powerpoint/ 58 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics Physics of the transformer VP - Magnetic flux (primary side) A Φ = NP µ0 I P 2R + Φ IP NS Magnetic flux (secondary side) A Φ = NS µ0 I S 2R IS - Load + VS Robert R. McLeod, University of Colorado Faraday’s Law (primary side) dΦ VP = N P dt NP (1) (2) (3) Faraday’s Law (secondary side) dΦ VS = N S dt (4) 59 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics Engineering equations of transformer Divide equation 2 by equation 4 VP N P = VS N S • VS is proportional to VP • NP/NS = “turns ratio” • E.g. Turns ratio of 10:1 would step primary voltage down by factor of 10. VS = VP/10 Divide equation 1 by equation 3, rearrange IP NS = IS NP • IS is proportional to IP • Current increases if voltage decreases • E.g. Turns ratio of 10:1 would step primary current up by factor of 10. IS = 10 IP Multiply these two equations VP I P PP [W ] N P N S = = =1 VS I S PS [W ] N S N P Robert R. McLeod, University of Colorado Power is conserved. 60 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics Dot convention If either winding is reversed, the secondary current and voltage are reversed. http://en.wikipedia.org/wiki/Transformer • Dots indicates + voltage terminals. • Primary current in, secondary current out dot. Robert R. McLeod, University of Colorado http://en.wikipedia.org/wiki/Dot_convention 61 • Lecture 5: Transformers ECEN 1400 Introduction to Analog and Digital Electronics Compare voltage divider to transformer + IP RP VP + R2 + IS + RS - RS VP VS VS - VS = IS IP RS VP RP + RS IS = IP RS PS = VS I S = VP I P RP + RS • VS lower than VP by R ratio • IS same as IP • Voltage can only decrease • Power lower by R ratio • Circuits share ground • Works at all frequencies • Cheap, small Robert R. McLeod, University of Colorado - - VS = 1 VP NP NS N IS = P IP NS PS = VS I S = VP I P • VS lower than VP by N ratio • IS greater than IP by N ratio • Voltage can increase or decr. • Power conserved • Circuits isolated • Works only for AC • Expensive, heavy 62 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Quiz 5.1 Q: A constant (“DC”) voltage is placed across an inductor on the left and a very high frequency time-varying (“AC”) voltage is placed across the inductor on the right. Large current will flow in which cases? A: No in the DC and no in the AC B: Yes in the DC and no in the AC C: No in the DC and yes in the AC D: Yes in the DC and yes in the AC E: It depends on the AC frequency. The inductor is a short at DC and an open at high frequency. Robert R. McLeod, University of Colorado 63 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Quiz 5.2 Q: At t=0, a switch is closed and the uncharged inductor L begins to charge. What is the current through the inductor at t=1 second? L / R = 1ms A: 0 A One second is 1000x the L/R time B: 1 mA constant, so the inductor will be fully charged and operating like a short. C: 10 mA The voltage across R is thus 10 V. I = V / R = 10 mA D: 100 mA E: I have no idea how to find this. Robert R. McLeod, University of Colorado 64 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Quiz 5.3 Q: A voltage V1 = 9 sin(omega t) is applied to the primary (left) side of this transformer. What is V2 on the secondary side? A: V2 = 3 sin(omega t) B: V2 = 9 sin(omega t) N C: V2 = 27 sin(omega t) V = N V = 3V D: V2 = 9 sin(3 omega t) E: V2 = 9 sin(omega / 3 t) S S P P P Robert R. McLeod, University of Colorado 65 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Bonus quiz 5.1 Q: A function generator is hooked to the series RL circuit above. The peak current through the inductor depends on A: The peak voltage of the source B: The frequency of the source C: The order of R and L in series The system is linear, so the D: A and B but not C • inductor current is to the source E: A and C but not B proportional voltage. • • Robert R. McLeod, University of Colorado The current through the inductor will drop when the frequency is > 1/(L/R) The order makes no difference 66 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Bonus quiz 5.2 Q: The current through an inductor of inductance L=2 H is I(t) = 1 + cos(500 t). The voltage on the inductor is: A: B: C: D: E: V(t)= -1000 sin(500 t) V(t)=2t + (1/250) sin(500 t) V(t)=1/2 + (1/2) cos(500 t) V(t)=2/(1 + cos(500 t)) V(t)=2 + 2 cos(500 t) VL = L dI = 2[− 500 sin (500t )] = −1000 sin (500t ) dt Robert R. McLeod, University of Colorado 67 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Bonus quiz 5.3 Q: As the frequency of the function generator is increased from zero, the peak voltage across the inductor stays near zero, then begins to increase. The frequency where this begins to occur is about A: 5 mHz L / R = 5 ms B: 200 Hz 1 C: 500 Hz f = = 200 Hz T D: 2 KHz E: 200 MHz Robert R. McLeod, University of Colorado 68 • Lecture 5: Inductors ECEN 1400 Introduction to Analog and Digital Electronics Bonus quiz 5.4 I (t ) = I 0e − t L R t − 1 I 0 = I 0e L R 4 L ⎛1⎞ t = − ln⎜ ⎟ R ⎝4⎠ Q: A function generator is hooked to the series RL circuit above. The function generator is set to a square wave which alternates between V0 and 0V (high/ low) with a period >> RL. If a transition from V0 to 0V occurs at t=0, when does the current through the inductor drop to ¼ of its value at t=0? Robert R. McLeod, University of Colorado L ⎛1⎞ A : t = − ln⎜ ⎟ R ⎝4⎠ B:t = − L ⎛3⎞ ln⎜ ⎟ R ⎝4⎠ C:t = − R ⎛1⎞ ln⎜ ⎟ L ⎝4⎠ D:t = − R ⎛3⎞ ln⎜ ⎟ L ⎝4⎠ E:t = e − 1 4 69