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IE1206 Embedded Electronics
Transients PWM
Phasor jω PWM CCP CAP/IND-sensor
Le1
Le3
Le6
Le8
Le2
Ex1
Le9
Ex4 Le7
Written exam
William Sandqvist [email protected]
PIC-block Documentation, Seriecom Pulse sensorsI, U, R, P, serial and parallel
Ex2
Ex5
Kirchhoffs laws Node analysis Two-terminals R2R AD
Trafo, Ethernet contactLe13
Pulse sensors, Menu program
Le4
KC1 LAB1
KC3 LAB3
KC4 LAB4
Ex3Le5 KC2 LAB2 Two-terminals, AD, Comparator/Schmitt
Step-up, RC-oscillator
Le10Ex6 LC-osc, DC-motor, CCP PWM
LP-filter TrafoLe12 Ex7 Display
Le11
•••• Start of programing task
•••• Display of programing task
William Sandqvist [email protected]
Transformer
William Sandqvist [email protected]
Voltage ratio
2
1
2
1
2211 d
d
d
d
N
N
U
U
tNU
tNU
=
Φ=Φ=
N1 : N2
William Sandqvist [email protected]
Ideal transformer I0 = 0
N1⋅I0 = N1⋅ I1 – N2⋅ I2
Magnetisig current I0 ≈ 0 is smallcompared to the work currents I1 and I2. The transformer itself has a high inductance.
William Sandqvist [email protected]
Current ratio
2
1
2
1
1
2
2211
0021 )0,(
N
N
U
U
I
I
IUIU
IPPP
=≈
⇒⋅=⋅==
N1 : N2
William Sandqvist [email protected]
Eddy current losses
Eddy currents – currents inside the iron core is prevented with lacquered ( = isolation ) sheet metal.
William Sandqvist [email protected]
E I -core
• EI-core is very economical to manufacture!
William Sandqvist [email protected]
E I -core
William Sandqvist [email protected]
Toroid
Toroid core has a low leakage field – so it will not disturb nearby electronics!
How do one wind such a transformer?
William Sandqvist [email protected]
Automatic Winding of toroidal core
William Sandqvist [email protected]
William Sandqvist [email protected]
Transformer (15.4)
William Sandqvist [email protected]
Transformer (15.4)
2
1
1
2
2
1
2
1
=
⇒=
I
I
U
U
William Sandqvist [email protected]
Transformer (15.4)
2
1
1
2
2
1
2
1
=
⇒=
I
I
U
U
8102,010010 1111 =⋅−=⇒=−⋅− UUIR
William Sandqvist [email protected]
Transformer (15.4)
2
1
1
2
2
1
2
1
=
⇒=
I
I
U
U
8102,010010 1111 =⋅−=⇒=−⋅− UUIR
42
8
2
112 ==⋅= UU
William Sandqvist [email protected]
Transformatorn (15.4)
2
1
1
2
2
1
2
1
=
⇒=
I
I
U
U
8102,010010 1111 =⋅−=⇒=−⋅− UUIR
42
8
2
112 ==⋅= UU 4,01
212 =⋅= II
William Sandqvist [email protected]
Transformer (15.4)
2
1
1
2
2
1
2
1
=
⇒=
I
I
U
U
8102,010010 1111 =⋅−=⇒=−⋅− UUIR
42
8
2
112 ==⋅= UU 4,01
212 =⋅= II
Ω=== 104,0
4
2
22 I
UR
William Sandqvist [email protected]
William Sandqvist [email protected]
2
2
2
121
2
2
2
2
1
21
2
22
1
1
1
2
22
1
11
RN
NR
I
U
N
N
IN
N
UN
N
I
U
I
UR
I
UR
⋅
=
⋅
====
←
RN
N ⋅
2
2
1
Transforming impedances
William Sandqvist [email protected]
2
2
2
121
2
2
2
2
1
21
2
22
1
1
1
2
22
1
11
RN
NR
I
U
N
N
IN
N
UN
N
I
U
I
UR
I
UR
⋅
=
⋅
====
←
RN
N ⋅
2
2
1
Transforming impedances
William Sandqvist [email protected]
Ex. Transforming impedances
A transformer has the voltage ratio 240V/120V.
We have two capacitors 1µF and 16 µF. How should one connect to get 5 µF ?
William Sandqvist [email protected]
Ex. Transforming impedances
)4/(
12
1
1
221
2
CCZ
CZ
ωω
ω
=⋅=
⇒=
←
A transformer has the voltage ratio 240V/120V.
We have two capacitors 1µF and 16 µF. How should one connect to get 5 µF ?
William Sandqvist [email protected]
Ex. Transforming impedances
)4/(
12
1
1
221
2
CCZ
CZ
ωω
ω
=⋅=
⇒=
←
16 Fµ4 Fµ
A transformer has the voltage ratio 240V/120V.
We have two capacitors 1µF and 16 µF. How should one connect to get 5 µF ?
William Sandqvist [email protected]
William Sandqvist [email protected]
Series and parallel connection of inductors
(Ex. 15.6) Assuming that none of the coils parts magnetic lines of force with each other but are completely independent components, they can be treated series and parallel inductors just as if they were resistors.
H36
4444
4
64444
4
ERS =+
+⋅+
⋅
+⋅+
=L
William Sandqvist [email protected]
Series and parallel connection of inductors?We have previously studied serial and parallel coils as if they were completely independent components that do not share magnetic lines with each other.
We are now treating coils with interconnected flow
??
Inductive coupling
William Sandqvist [email protected]
d
du r i
t
ϕ= ⋅ +
A portion of the flow in the coil 1 is interconnected with flow from the coil 2.
11 1 1 1 1 1 2
d
du r i i L i M
t
ϕ ϕ= ⋅ + = ⋅ + ⋅In same way: 2
2 2 2 2 2 2 1
d
du r i i L i M
t
ϕ ϕ= ⋅ + = ⋅ + ⋅
Induction
Inductive coupling
William Sandqvist [email protected]
1 21 1 1 1
2 12 2 2 2
d d
d dd d
d d
i iu r i L M
t ti i
u r i L Mt t
= ⋅ + +
= ⋅ + +
1 1 1 1 1 2
2 2 2 2 2 1
j j
j j
U r I L I MI
U r I L I MI
ω ωω ω
= ⋅ + += ⋅ + +
jω-method:
± M is called mutual inductance
21LL
Mk =
Coupling factor:
The coupling factor indicates how much of the flow a coil has in common with another coil
An ideal transformer has coupling factor k = 1 (100%)
Series with mutual inductance
William Sandqvist [email protected]
Series connection has the same current
212111 LLL IMjILjU ωω ±= 121222 LLL IMjILjU ωω ±=
1 2 1 2 12 21
1 2( )L L L LI I I U U U M M M
U I j L M L Mω= = = + = = ⇒
= ⋅ ± + ±
1 2( 2 )U
j L L MI
ω= + ±
11121 LL IUML 22212 LL IUML
Derive:
Series with mutual inductance
William Sandqvist [email protected]
MLLLTOT 221 ++= MLLLTOT 221 −+=
Series connection has the same current I1 = I2 =I
M can can contribute or counter act to the flow, this gives ±sign. Therefore, coil winding polarity is usually indicated by adot convention in schematics.
M-dot M-dot M-dot M-dot
”Dot” convention
William Sandqvist [email protected]
An increasing current in to a dot results in induced voltages with directions that would give increasing currents out of other dots.
”Dot” convention
William Sandqvist [email protected]
An increasing current in to a dot results in induced voltages with directions that would give increasing currents out of other dots.
In parallel with mutual inductance
William Sandqvist [email protected]
MLL
MLLLTOT 221
221
−+−⋅=
MLL
MLLLTOT 221
221
++−⋅=
TOTL TOTL
Parallel connected coils Antiparal conected coils
Ex. 15.7 Series connection
William Sandqvist [email protected]
51 =L 102 =L 153 =L
212 =M 323 =M
113 =M
[H]
Ex. 15.7 Series connection
William Sandqvist [email protected]
51 =L 102 =L 153 =L
212 =M 323 =M
113 =M
[H]
LTOT =
L1 + M12 – M13 +
L2 + M12 – M23 +
L3 – M23 – M13 =
= 5 +2 –1 + 10 + 2 – 3 + 15 –3 –1 = 26 [H]
William Sandqvist [email protected]
Measuring the mutual inductance?
William Sandqvist [email protected]
+TOTL −TOTL
MLLLTOT 221 ++=+ MLLLTOT 221 −+=−
Measuring the mutual inductance?
William Sandqvist [email protected]
+TOTL −TOTL
MLLLTOT 221 ++=+ MLLLTOT 221 −+=−
4−+ −= TOTTOT LLM
Variometer (to an antique radio)
William Sandqvist [email protected]
)(
221αfM
MLLLTOT=
±+=
William Sandqvist [email protected]
A bad actuator can become a good sensor
William Sandqvist [email protected]
1906
The industry's "rugged" position sensor
William Sandqvist [email protected]
Differential transformer
William Sandqvist [email protected]
LVDT Linear Variable Differential Transformer
The secondary coils are connected in series but with opposite polarity – when the core is in the middle U = 0.
core
primary coil
secondary coilsecondary coil
LVDT design
William Sandqvist [email protected]
William Sandqvist [email protected]
The output voltage is relatively high – it makes this a popular sensor …
LVDT principle
LVDT probe
Output signal changes phase180°°°° exactly when the core pass the middle point.
William Sandqvist [email protected]
A XOR-gate kan indicate this change.
A LVDT probe can keep track on that the thicknes is correct.
Guess application?
It is important to ensure that the ATM does not distribute "double" bills …
William Sandqvist [email protected]
William Sandqvist [email protected]
Periodic differential transformer
Renywell Spherical encoder
LVDT-principel witin a core, and then keep track on how many cores that have passed.
A similar sensor?
William Sandqvist [email protected]