ENGR-44_Lec-06-2_Inductors.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Engineering 43 Chp 6.2 Inductors

ENGR-44_Lec-06-2_Inductors.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

ENGR-44_Lec-06-2_Inductors.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis The Inductor  Second of the Energy-Storage Devices  Basic Physical Model:  Details of Physical Operation Described in PHYS 4B or ENGR45  Note the Use of the PASSIVE Sign Convention Ckt Symbol

ENGR-44_Lec-06-2_Inductors.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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

ENGR-44_Lec-06-2_Inductors.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductance Defined  From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law  Where the Constant of Proportionality, L, is called the INDUCTANCE  L is Measured in Units of “Henrys”, H 1H = 1 Vs/Amp  Inductors STORE electromagnetic energy  They May Supply Stored Energy Back To The Circuit, But They CANNOT CREATE Energy  For a Linear Inductor The Flux Is Proportional To The Current Thru it

ENGR-44_Lec-06-2_Inductors.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductance Sign Convention  Inductors Cannot Create Energy; They are PASSIVE Devices  All Passive Devices Must Obey the Passive Sign Convention

ENGR-44_Lec-06-2_Inductors.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Circuit Operation  Recall the Circuit Representation  Separating the Variables and Integrating Yields the INTEGRAL form  Previously Defined the Differential Form of the Induction Law  In a development Similar to that used with caps, Integrate −  to t 0 for an Alternative integral Law

ENGR-44_Lec-06-2_Inductors.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor Model Implications  From the Differential Law Observe That if i L is not Continuous, di L /dt → , and v L must also →   This is NOT physically Possible  Thus i L must be continuous  Consider Now the Alternative Integral law  If i L is constant, say i L (t 0 ), then The Integral MUST be ZERO, and hence v L MUST be ZERO This is DC Steady-State Inductor Behavior –v L = 0 at DC –i.e; the Inductor looks like a SHORT CIRCUIT to DC Potentials

ENGR-44_Lec-06-2_Inductors.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor: Power and Energy  From the Definition of Instantaneous Power  Time Integrate Power to Find the Energy (Work)  Subbing for the Voltage by the Differential Law  Units Analysis J = H x A 2  Again By the Chain Rule for Math Differentiation  Energy Stored on Time Interval Energy Stored on an Interval Can be POSITIVE or NEGATIVE

ENGR-44_Lec-06-2_Inductors.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Inductor: P & W cont.  In the Interval Energy Eqn Let at time t 1  Then To Arrive At The Stored Energy at a later given time, t  Thus Observe That the Stored Energy is ALWAYS Positive In ABSOLUTE Terms as i L is SQUARED ABSOLUTE-POSITIVE-ONLY Energy-Storage is Characteristic of a PASSIVE ELEMENT 0)(  tLi L )( 2 1 )( 2 t tw L 

ENGR-44_Lec-06-2_Inductors.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Given The i L Current WaveForm, Find v L for L = 10 mH  The Derivative of a Line is its SLOPE, m  The Differential Reln  Then the Slopes  And the v L Voltage

ENGR-44_Lec-06-2_Inductors.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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  Running the No.s

ENGR-44_Lec-06-2_Inductors.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Student Exercise  Let’s Turn on the Lights for 5-7 minutes to allow YOU to solve an INTEGRAL Reln Problem  Given The Voltage Wave Form Across L  Then Find i L For L = 0.1 H i(0) = 2A t > 0

ENGR-44_Lec-06-2_Inductors.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Given The Voltage Wave Form Across L, Find i L if L = 0.1 H i(0) = 2A  The PieceWise Function  The Integral Reln  A Line Followed by A Constant; Plotting

ENGR-44_Lec-06-2_Inductors.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example - Energy  The Current Characteristic  The Initial Stored Energy  The Energy Eqn  The “Total Stored Energy” Energy Stored on Interval Can be POS or NEG  Energy Stored between 0-2  → Consistent with Previous Calculation

ENGR-44_Lec-06-2_Inductors.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Example  Notice That the ABSOLUTE Energy Stored At Any Given Time Is Non Negative (by sin 2 ) i.e., The Inductor Is A PASSIVE Element-  Find The Voltage Across And The Energy Stored (As Function Of Time)  For The Energy Stored

ENGR-44_Lec-06-2_Inductors.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis L = 10 mH; Find the Voltage

ENGR-44_Lec-06-2_Inductors.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis L = 5 mH; Find the Voltage

ENGR-44_Lec-06-2_Inductors.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis C & L Specifications  Capacitors Standard Range  1 pF to  50 mF (50000 µF) Working (DC) Voltage Range  Vdc Standard Tolerances –± 5% –± 10% –± 20%  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

ENGR-44_Lec-06-2_Inductors.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Cap Spec Sensitivity  Given The Voltage Waveform Find the Current Variation Due to Cap Spec Tolerance  Use VOLTAGE WAVEFORM dt dv Cti C  )( 240 mA

ENGR-44_Lec-06-2_Inductors.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Ind Spec Sensitivity  Given The Current Waveform Find the Voltage Variation Due to L Spec-Tolerance  Use CURRENT WAVEFORM

ENGR-44_Lec-06-2_Inductors.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work  Let’s Work This Problem  Find: v(t), t max for i max, t min for v min

ENGR-44_Lec-06-2_Inductors.ppt 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis By MATLAB % 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] = 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] = -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

ENGR-44_Lec-06-2_Inductors.ppt 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis By MATLAB % 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] = 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] = -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

ENGR-44_Lec-06-2_Inductors.ppt 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

ENGR-44_Lec-06-2_Inductors.ppt 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

ENGR-44_Lec-06-2_Inductors.ppt 27 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis