Transcript WATKINS - Chabot College

Engineering 43
Chp 6.2
Inductors
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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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
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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
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 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
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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
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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
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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
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 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
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 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
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 Time Integrate Power to
Find the Energy (Work)
t2
d 1 2 
wL (t1 , t 2 )    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
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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 103 A
 A
m

10
 s 
2 103 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
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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 *103 (20 *10 3 ) 2
w(2,4)  2.0 µJ
Engineering-43: Engineering Circuit Analysis
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
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
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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
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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
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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
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 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 103  A 
v  10 10 [ H ] 
2 103  s 
3
Engineering-43: Engineering Circuit Analysis
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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
m0
v  0V
v (t )  L
m
di
(t )
dt
0  10
( A / s)
43
v  50mV
v  5 103 ( H )  20( A / s); 0  t  1ms  100mV
Engineering-43: Engineering Circuit Analysis
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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
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 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
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300mA
240 mA
C  100 F  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 103  A 
v  100 10 H 
20 106  S 
6
s
L  100  H  10%
di
dt
 Use v (t )  L (t )
Engineering-43: Engineering Circuit Analysis
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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
it   2te 4t
t0
t 0
 Find: v(t), tmax for imax, tmin for vmin
Engineering-43: Engineering Circuit Analysis
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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
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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
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-06-2_Inductors.ppt
200
150
100
50
0
-50
0
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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
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0.3
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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
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0.3
0.4
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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