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Engineering 43 Chp 6.2 Inductors Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-43: Engineering Circuit Analysis 1 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Capacitance & Inductance Introduce Two Energy STORING Devices • Capacitors • Inductors Outline • Capacitors – Store energy in their ELECTRIC field (electrostatic energy) – Model as circuit element • Inductors – Store energy in their MAGNETIC field – Model as circuit element • Capacitor And Inductor Combinations – Series/parallel combinations of elements • RC OP-AMP Circuits – Electronic Integration & Differentiation Engineering-43: Engineering Circuit Analysis 2 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt The Inductor Second of the Energy-Storage Devices Basic Physical Model: Ckt Symbol Details of Physical Operation Described in PHYS 4B or ENGR45 Engineering-43: Engineering Circuit Analysis 3 Note the Use of the PASSIVE Sign Convention Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Physical Inductor Inductors are Typically Fabricated by Winding Around a Magnetic (e.g., Iron) Core a LOW Resistance Wire • Applying to the Terminals a TIME VARYING Current Results in a “Back EMF” voltage at the connection terminals Some Real Inductors Engineering-43: Engineering Circuit Analysis 4 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductance Defined From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law d vL Where the Constant of Proportionality, L, is called the INDUCTANCE L is Measured in Units of “Henrys”, H • 1H = 1 V•s/Amp Inductors STORE electromagnetic energy For a Linear Inductor The Flux Is Proportional To They May Supply Stored The Current Thru it Energy Back To The diL Circuit, But They LiL vL L CANNOT CREATE dt Energy dt Engineering-43: Engineering Circuit Analysis 5 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductance Sign Convention Inductors Cannot Create Energy; They are PASSIVE Devices All Passive Devices Must Obey the Passive Sign Convention Engineering-43: Engineering Circuit Analysis 6 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductor Circuit Operation Recall the Circuit Representation Separating the Variables and Integrating Yields the INTEGRAL form t 1 iL (t ) vL ( x)dx L Previously Defined the Differential Form of the Induction Law diL vL L dt Engineering-43: Engineering Circuit Analysis 7 In a development Similar to that used with caps, Integrate − to t0 for an Alternative integral Law t 1 iL (t ) iL (t0 ) vL ( x)dx; t t0 L t0 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductor Model Implications From the Differential Law Observe That if iL is not Continuous, diL/dt → , and vL must also → This is NOT physically Possible Thus iL must be continuous diL vL dt Engineering-43: Engineering Circuit Analysis 8 Consider Now the Alternative Integral law t 1 iL (t ) iL (t0 ) vL ( x)dx; t t0 L t0 If iL is constant, say iL(t0), then The Integral MUST be ZERO, and hence vL MUST be ZERO • This is DC Steady-State Inductor Behavior – vL = 0 at DC – i.e; the Inductor looks like a SHORT CIRCUIT to DC Potentials Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductor: Power and Energy From the Definition of Instantaneous Power pL (t ) vL (t )iL (t ) Subbing for the Voltage by the Differential Law diL p L (t ) L (t )iL (t ) dt Again By the Chain Rule for Math Differentiation d d di dt di dt d 1 2 pL t LiL (t ) dt 2 Engineering-43: Engineering Circuit Analysis 9 Time Integrate Power to Find the Energy (Work) t2 d 1 2 wL (t1 , t2 ) LiL ( x) dx dx 2 t1 Units Analysis • J = H x A2 Energy Stored on Time Interval w(t1 , t 2 ) 1 2 1 LiL (t 2 ) LiL2 (t1 ) 2 2 • Energy Stored on an Interval Can be POSITIVE or NEGATIVE Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Inductor: P & W cont. In the Interval Energy Eqn Let at time t1 1 2 LiL (t1 ) 0 2 Then To Arrive At The Stored Energy at a later given time, t 1 2 w(t ) LiL (t ) 2 Thus Observe That the Stored Energy is ALWAYS Positive In ABSOLUTE Terms as iL is SQUARED • ABSOLUTE-POSITIVE-ONLY Energy-Storage is Characteristic of a PASSIVE ELEMENT Engineering-43: Engineering Circuit Analysis 10 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Example Given The iL Current WaveForm, Find vL for L = 10 mH The Derivative of a Line is its SLOPE, m Then the Slopes 10( A / s) 0 t 2ms di 10( A / s) 2 t 4ms dt 0 elsewhere 20 103 A A m 10 s 2 103 s A m 10 s And the vL Voltage di (t ) 10( A / s) 3 vL (t ) 100 10 V 100mV dt L 10 10 3 H The Differential Reln diL vL L dt Engineering-43: Engineering Circuit Analysis 11 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Example Power & Energy The Energy Stored between 2 & 4 mS The Value Is Negative Because The Inductor SUPPLIES Energy PREVIOUSLY STORED with a Magnitude of 2 μJ The Energy Eqn 1 2 1 2 w(2,4) LiL (4) LiL (2) 2 2 Running the No.s w(2,4) 0 0.5 *10 *103 (20 *103 ) 2 w(2,4) 2.0 µJ Engineering-43: Engineering Circuit Analysis 12 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Student Exercise Given The Voltage Wave Form Across L v (V ) 2 Let’s Turn on the Lights for 5-7 minutes to allow YOU to solve an INTEGRAL Reln Problem 2A 2 t (s) 2H Then Find iL For • L = 0.1 H • i(0) = 2A • t>0 Engineering-43: Engineering Circuit Analysis 13 t 1 i(t ) i(0) v( x)dx L0 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Numerical Example Given The Voltage Wave Form Across L , Find iL if • L = 0.1 H • i(0) = 2A The PieceWise Function t v(t ) 2 v( x)dx 2t; 0 t 2 0 L 0.1H i (t ) 2 20t ; 0 t 2s v(t ) 0; t 2 i (t ) i (2s); t 2s v (V ) A Line Followed by A Constant; Plotting 2 2 t (s) 42 i (A) The Integral Reln t 1 i(t ) i(0) v( x)dx L0 Engineering-43: Engineering Circuit Analysis 14 2 2 t (s) Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Numerical Example - Energy The Current Characteristic 42 The Initial Stored Energy w(0) 0.5 * 0.1[ H ](2 A) 2 0.2 J The “Total Stored Energy” i (A) w() 0.5 * 0.1[ H ] * (42 A) 2 88.2 J Energy Stored between 0-2 2 2 t (s) The Energy Eqn w(t1 , t 2 ) 1 2 1 LiL (t 2 ) LiL2 (t1 ) 2 2 • Energy Stored on Interval Can be POS or NEG Engineering-43: Engineering Circuit Analysis 15 w(2,0) 1 2 1 LiL (2) LiL2 (0) 2 2 w(0,2) 0.5 * 0.1* (42) 2 0.5 * 0.1* (2) 2 w(0,2) 88 J → Consistent with Previous Calculation Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Example w L (t ) Find The Voltage Across And The Energy Stored (As Function Of Time) v (t ) For The Energy Stored Engineering-43: Engineering Circuit Analysis 16 Notice That the ABSOLUTE Energy Stored At Any Given Time Is Non Negative (by sin2) • i.e., The Inductor Is A PASSIVE Element- Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt L = 10 mH; Find the Voltage v 100mV v (t ) L 20 103 A v 10 10 [ H ] 2 103 s 3 Engineering-43: Engineering Circuit Analysis 17 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt di (t ) dt L = 5 mH; Find the Voltage m 20mA 1ms m v 100mV 10 20 ( A / s) 2 1 v 50mV m0 v 0V v (t ) L m di (t ) dt 0 10 ( A / s) 43 v 50mV v 5 103 ( H ) 20( A / s); 0 t 1ms 100mV Engineering-43: Engineering Circuit Analysis 18 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt C & L Specifications Capacitors • Standard Range 1 pF to 50 mF (50000 µF) • Working (DC) Voltage Range 6.3-500 Vdc • Standard Tolerances – ± 5% – ± 10% – ± 20% Engineering-43: Engineering Circuit Analysis 19 Inductors • Standard Range 1 nH to 100 mH • Current Rating 10mA-1A • Standard Tolerances – ± 5% – ± 10% • As Wire Coils, Inductors also have a RESISTANCE Specification Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Cap Spec Sensitivity VOLTAGE WAVEFORM Nominal current 300mA 100 10 6 F Given The Voltage Waveform Find the Current Variation Due to Cap Spec Tolerance Use i (t ) C dvC dt Engineering-43: Engineering Circuit Analysis 20 300mA 240 mA C 100F 20% (3) 3 V 600mA 3 2 s Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt Ind Spec Sensitivity Given The Current Waveform Find the Voltage Variation Due to L Spec-Tolerance CURRENT WAVEFORM 200 103 A v 100 10 H 20 106 S 6 s L 100 H 10% di dt Use v (t ) L (t ) Engineering-43: Engineering Circuit Analysis 21 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt WhiteBoard Work Let’s Work This Problem L 50 H i t 0 it 2te 4t t0 t 0 Find: v(t), tmax for imax, tmin for vmin Engineering-43: Engineering Circuit Analysis 22 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid Engineering-43: Engineering Circuit Analysis 23 By MATLAB Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid By MATLAB Engineering-43: Engineering Circuit Analysis 24 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt 200 150 100 50 0 -50 0 0.1 0.2 0.3 Engineering-43: Engineering Circuit Analysis 25 0.4 0.5 0.6 0.7 0.8 0.9 Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt 1 Irwin Prob 5.26: I(t) & v(t) 200 i(t) (mA) 175 Current (mA) 150 125 100 75 50 25 0 -0.1 0.0 0.1 0.2 file = Engr44_prob_5-26_Fall03..xls Engineering-43: Engineering Circuit Analysis 26 0.3 0.4 0.5 0.6 0.7 0.8 0.9 time (s) Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt 1.0 Irwin Prob 5.26: I(t) & v(t) 200 120 i(t) (mA) v(t) (µV) 100 150 80 125 60 100 40 75 20 50 0 25 -20 0 -0.1 -40 0.0 0.1 0.2 file = Engr44_prob_5-26_Fall03.xls Engineering-43: Engineering Circuit Analysis 27 0.3 0.4 0.5 0.6 0.7 0.8 0.9 time (s) Bruce Mayer, PE [email protected] • ENGR-44_Lec-06-2_Inductors.ppt 1.0 Electrical Potential (µV) Current (mA) 175