Inductor (電感) : L
unit: Henryy (H)
( )
1
di(t)
V(t) = L
dt
V(t): voltage
i(t): current
L: inductance
-3
m = 10
-6
μ = 10
2
3
4
5
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Relayy ((繼電器))
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Relayy ((繼電器))
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Relayy ((繼電器))
光華商場附近，捷運站的火警系統。
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An electric
A
l i current i flowing
fl i aroundd a circuit
i i produces
d
a
magnetic field and hence a magnetic flux Φ through the
circuit.
i i
The ratio
Th
i off the
h magnetic
i flux
fl to the
h current is
i called
ll d
the inductance, or more accurately self-inductance of
the
h circuit.
i i
The term
Th
t
was coined
i d by
b Oliver
Oli
Heaviside
H i id in
i February
F b
1886. It is customary to use the symbol L for
i d t
inductance,
possibly
ibl in
i honour
h
off the
th physicist
h i it
Heinrich Lenz.
From: www.wikipedia.org
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The quantitative
Th
i i definition
d fi i i off the
h inductance
i d
in
i SI units
i
(webers per ampere) is
L=Φ / i
In honour of Joseph Henry, the unit of inductance has
been given the name henry (H): 1H = 1Wb/A.
In the above definition, the magnetic flux Φ is that
caused by the current flowing through the circuit
concerned.
From: www.wikipedia.org
11
The equation relating inductance and flux linkages can
be rearranged as follows:
Φ = Li
Taking the time derivative of both sides of the equation
yields:
dΦ
di
dL
= L +i
dt
dt
dt
From: www.wikipedia.org
12
In most physical cases
cases, the inductance is constant with
time and so
dΦ
di
=L
dt
dt
By Faraday's Law of Induction we have:
dΦ
= −ε = vL
dt
where ε is the Electromotive force (emf) and v is the
iinduced
d d voltage.
l
Note that
h the
h emff is
i opposite
i to the
h
induced voltage.
From: www.wikipedia.org
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Inductors in series
V1
V2
Vn
L1 L2
Ln
i1 = i2 = . . . = in
E
Le
E
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di3
din
di1
di2
E = V1 + V2 + V3 + ... + Vn = L1
+ L2
+ L3
+ ... + Ln
dt
dt
dt
dt
i1 = i2 = i3 = ... = in = i
di
di
E = Leq = ( L1 + L2 + L3 + ... + Ln )
dt
dt
∴ Leq = L1 + L2 + L3 + ... + Ln
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Inductors in parallel
p
V1 V2 V3
Vn
Le
E
E
L1 L2 L3
Ln
i1 i2 i3
in
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di3
din
di1
di2
E = V1 = V2 = V3 = ... = Vn = L1
= L2
= L3
= ... = Ln
dt
dt
dt
dt
i = i1 + i2 + i3 + ... + in
din
di di1 di2 di3
=
+
+
+ ... +
==>
dt dt dt dt
dt
E
E E E
E
= + + + ... +
Leq L1 L2 L3
Ln
1
1 1 1
1
∴
= + + + ... +
Le L1 L2 L3
Ln
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Example 1: Direct Current (DC) and charging
VR
+
_
R
+
V
L
VL
_
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Example 1: Direct Current (DC) and charging
iR
VR
+
0 = V – VR – VL
_
R
iL
iR = iL = i(t)
+
V
L
_
VL
•
i
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Example 1: Direct Current (DC) and charging
iR
VR
+
0 = V – VR – VL
_
R
iL
iR = iL = i(t)
+
V
L
•
iR
VL
_
VR = iR R
di(t)
VL(t) = L
dt
= L i’(t)
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VR + VL = V
i R + L i’ = V
Li’ + R i = V
R
V
i’ +
i =
L
L
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VR + VL = V
y’ + ay = b
-at
y(t) = c1e + c2
i R + L i’ = V
Li’ + R i = V
R
V
i’ +
i =
L
L
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VR + VL = V
y’ + ay = b
-at
y(t) = c1e + c2
i R + L i’ = V
Li’ + R i = V
R
V
i’ +
i =
L
L
If i(t=0) = 0
V
-t/τ
Then i(t) =
Then,
(1 - e )
R
where τ = L/R
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Q
Question:
i
If R = 2000, L = 6, V = 5,
What is τ?
A
Answer:
τ = L/R = 6/2000 = 0.003
0 003 sec
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If i(t) =
V
(1 - e -t/τ ) , where τ = L/R
R
What is VL(t)?
di(t)
Answer: VL(t) = L
dt
=L
V
1
-t//τ
(
)e
R τ
= V e
-t//τ
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solution : VL (t ) = Ve
−
t
τ
V
, i (t ) =
R
t
− ⎞
⎛
τ
1
−
e
⎜
⎟ , where τ = L / R
⎝
⎠
V
τ
V
R
τ
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clf;
clear;
close
l
all;
ll
v = 5;
r = 2000;
l = 6;
tau = l / r;
time = ones(1000,1);
voltage = time;
current = time;
figure(1)
subplot(2,1,1);
plot(time/tau,voltage/v,'color','r','LineWidth',2);
l t(ti /t
lt / ' l ' ' ' 'Li Width' 2)
ylim([-0.2 1.2])
title('voltage');
subplot(2,1,2);
plot(time/tau,current/(v/r),'color','g','LineWidth',2);
ylim([-00.22 1.2])
ylim([
1 2])
title('current');
for i = 1:1:1000
time(i)
( ) = i*tau/100;;
current(i) = v/r*(1-exp(-time(i)/tau));
voltage(i) = v*(exp(-time(i)/tau));
end
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http://en.wikipedia.org/wiki/Oliver_Heaviside
http://en.wikipedia.org/wiki/Joseph_Henry
http://en.wikipedia.org/wiki/Heinrich_Lenz
http://maps.google.com
golden bridge, San Francisco, CA, USA
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Example 2: Alternating Current (AC)
A
iL
E = Vm sin((ωt))
+
~
L
_
VL
i
30
Example 2: Alternating Current (AC)
i
A
iL
E = Vm sin((ωt))
+
~
L
_
VL
•
i
vL(t) = E = Vm sin(ωt)
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vL(t) = E = Vm sin(ωt)
vL(t) = L i’(t)
1
1
i’(t) = vL =
[ Vm sin(ω t) ]
L
L
- Vm
Vm
i(t) =
cos(ω t) =
sin(ω t - π )
ωL
ωL
2
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vL(t) = E = Vm sin(ωt)
Vm
iL(t) =
sin(ω t - π ) = Im sin(ω t - π )
ωL
2
2
here Im = Vm / (ωL)
The inductive reactance is defined as: XL
Vm
XL =
=
Im
ωL
1
= (2πf) L
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Vm
π
vL(t) = E = Vm sin(ωt) , iL(t) =
sin(ω t - )
ωL
2
Question:
E = 10 sin( 200π t)
L=2mH
Please calculate: i(t) and XL
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Vm
π
iL(t) =
sin(ω t )
ωL
2
π
= 10 / (200π × 0.002) sin(ω t - )
= 7.96 sin(ω t - π )
2
2
XL = (2πf) L = 200π × 0.002
0 002
= 1.26
1 26 Ω
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π
VL(t) = E = Vm sin(ωt) , iL(t) = Im sin(ω t - )
2
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clc
l
clear
close all
vm = 10;
w = 200*pi;
l = 0.002;
0 002;
cycle = 2;
dt = cycle*(2*pi)/w;
figure(1)
fi
(1)
plot(time*w,voltage,'color','r','LineWidth',2);
hold on;
plot(time*w current 'color'
plot(time*w,current,
color ,'g'
g ,'LineWidth'
LineWidth ,2);
2);
legend('v(t)', 'i(t)');
set(gca,'XTick',0:pi:4*pi)
set(gca 'XTickLabel'
set(gca,
XTickLabel ,{{'0'
0 ,'pi'
pi ,'2pi'
2pi ,'3pi'
3pi ,'4pi'})
4pi })
xlim([0 4*pi])
ylim([-12 12])
time = ones(1000,1);
voltage = time;
current = time;;
for i = 1:1:1000
time(i) = i*dt/1000;
voltage(i) = (vm)*sin(w*time(i));
current(i) = 0 - (vm/w/l)*cos(w*time(i));
end
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VL(t) = Vm sin(ωt)
Vm
π
iL(t) =
sin(ω t - )
ωL
2
Vm
=
sin(ω t + θ )
ωL
在電感器上，電流波形落後電壓波形 π/2。
(相位) = + θ = - π/2
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Transformer (變壓器)
39
40
41
Transformer (變壓器)
From: www.wikipedia.org
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43
The transformer
Th
t
f
is
i based
b d on two
t principles:
i i l
(1)an electric current can produce a magnetic field
(2) changing
(2)a
h i magnetic
ti field
fi ld within
ithi a coil
il off wire
i induces
i d
a voltage across the ends of the coil
By changing the current in the primary coil, it changes
th strength
the
t
th off its
it magnetic
ti field;
fi ld since
i
the
th changing
h i
magnetic field extends into the secondary coil, a
voltage
lt
is
i induced
i d d across the
th secondary.
d
From: www.wikipedia.org
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I d ti law
Induction
l
The voltage
Th
lt
induced
i d d across the
th secondary
d
coil
il may be
b
calculated from Faraday's law of induction, which states
th t
that:
dΦ
VS = N S
dt
where VS is the instantaneous voltage, NS is the number of
t
turns
in
i the
th secondary
d
coil
il andd Φ equals
l the
th magnetic
ti
flux through one turn of the coil.
From: www.wikipedia.org
45
Since the
Si
th same magnetic
ti flux
fl passes through
th
h both
b th the
th
primary and secondary coils in an ideal transformer,
th instantaneous
the
i t t
voltage
lt
across the
th primary
i
winding
i di
equals
dΦ
dΦ
VP = N P
dt
Taking
T
ki the
th ratio
ti off the
th two
t equations
ti
for
f VS andd VP gives
i
the basic equation for stepping up or stepping down the
voltage
lt
VP N P
=
VS N S
From: www.wikipedia.org
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Ideally,
Id
ll th
the transformer
t
f
is
i perfectly
f tl efficient;
ffi i t all
ll the
th
incoming energy is transformed from the primary circuit
t the
to
th magnetic
ti field
fi ld andd thence
th
to
t the
th secondary
d
circuit.
i it
The incoming
Th
i
i electric
l t i power mustt equall the
th outgoing
t i
power. Pin coming = IPVP = Pout going = ISVS
giving
i i the
th ideal
id l transformer
t
f
equation
ti
VP N P I S
=
=
VS N S I P
From: www.wikipedia.org
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