IE1206 Embedded Electronics - KTH · 2015-03-19 · IE1206 Embedded Electronics Transients PWM Phasor j ωPWM CCP CAP/IND-sensor Le1 Le3 Le6 Le8 Le2 Ex1 Le9 Ex4 Le7 Written exam William
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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 william@kth.se
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 william@kth.se
Easy to generate a sinusoidal voltage
Our entire power grid works with sinusoidal voltage.
When the loop rotates with constant speed in a magnetic field a sine wave is generated.
So much easier, it can not be …
William Sandqvist william@kth.se
The sine wave – what do you remember?
periodT
timet
y
amplitudetopY ,ˆ
RMSY
2
ˆ12)sin(ˆ)(
YY
TfftYty ==== πωω
William Sandqvist william@kth.se
(11.1) Phase ϕ
)sin(ˆ)( ϕω += tYty
If a sine curve does not begin with 0 the function expression has a phase angle ϕ.
Specify this function mathematically:
)10002sin(6)( ϕπ +⋅⋅⋅= ttu
y
)30(rad52,06
3arcsin)sin(63)0( °==
=⇒⋅== ϕϕu
)52,06283sin(6)( +⋅⋅= ttu
William Sandqvist william@kth.se
Apples and pears?In circuit analyses it is common (eg. in textbooks) to expresses the angle of the sine function mixed in radians ω·t [rad] and in degrees ϕ [°].
This is obviously improper, but practical (!). The user must "convert" phase angle to radians to calculate the sine function value for any given time t.
(You have now been warned …)
)306283sin(6)( °+⋅⋅= ttu
Conversion:x[°]= x[rad] ⋅57,3x[rad]= x[°]⋅0,017
? ?
William Sandqvist william@kth.se
William Sandqvist william@kth.se
Mean and effective value
T
ttu
UttuU
T
T
T
∫∫ === 0
2
0
1med
d)(
0d)(
All pure AC voltages, has the mean value0.
• More interesting is the effective value– root mean square, rms.
William Sandqvist william@kth.se
(11.2) Example. RMS.
V63,11015
1058
1015
105)0)2(2(d)(
3
3
3
3220
2
=⋅⋅⋅=
⋅⋅⋅+−+== −
−
−
−∫T
ttu
U
T
The rms value is what is normally used for an alternating voltage U. 1,63 V effective value gives the same power in a resistor as a 1,63 V pure DC voltage would do.
RMS, effective value
William Sandqvist william@kth.se
Sine wave effective value
2
1
2
1d)(sin2 ==∫ xx
)(sin2 x
2
1
• Effective value is often called RMS ( Root Mean Square ).
sin2 has the mean value ½
Therefore:
2
UU =
RMS, effective value
Ex. 11.3
William Sandqvist william@kth.se
William Sandqvist william@kth.se
Addition of sinusoidal quantities
)sin(ˆ111 ϕω += tAy
)sin(ˆ222 ϕω += tAy
?21 =+ yy
William Sandqvist william@kth.se
Addition of sinusoidal quantitiesWhen we shall apply the circuit laws on AC circuits, we must add the sines. The sum of two sinusoidal quantities of the same frequency is always a new sine of this frequency, but with a newamplitude and a new phase angle.
( Ooops! The result of the rather laborious calculations are shown below).
+++⋅−++=
=+=+⋅=+⋅=
)cos(A)cos(A
)sin(A)sin(Aarctansin)cos(AA2AA
)()()()sin(A)()sin(A)(
2211
22112121
22
21
21222111
ϕϕϕϕωϕϕ
ϕωϕω
t
tytytyttytty
William Sandqvist william@kth.se
Sine wave as a pointer
A sinusoidal voltage or current,
can be represented by a pointer that rotates (counterclockwise) with the angular velocity ω [rad/sec] .
)sin(ˆ)( tYty ⋅⋅= ω
Wikipedia Phasors
William Sandqvist william@kth.se
Simpler with vectors
If you ignore the "revolution" and adds the pointers with vector addition, as they stand at the time t = 0, it then becomes a whole lot easier!
Wikipedia Phasors
http://en.wikipedia.org/wiki/Phasors
William Sandqvist william@kth.se
Pointer with complex numbers
°⋅⋅+°⋅⇔⋅⇔°∠ ° 30sinj1030cos10e103010 30j
A AC voltage 10 V that has the phase 30° is usually written:
10 ∠ 30° ( Phasor)
Once the vector additions require more than the most common geometrical formulas, it is instead preferable to represent pointers with complex numbers.
In electricity one uses j as imaginary unit, as i is already in use for current.
baz j+=
Imaginary axis
Real axis
William Sandqvist william@kth.se
Phasor
A pointer (phasor) can either be viewed as a vector expressed inpolar coordinates, or as a complex number.
It is important to be able to describe alternating current phenomena without necessarily having to require that the audience has a knowledge of complex numbers - hence the vector method.
The complex numbers and jω-method are powerful tools that facilitate the processing of AC problems. They can be generalized to the Fourier transform and Laplace transform, so the electro engineer’s use of complex numbers is extensive.
Sinusoidal alternating quantities can be represented as pointers, phasors.
”amount” ∠∠∠∠ ”phase”
William Sandqvist william@kth.se
peak/effective value - phasor
baz j+=
The phasor lengths corresponds to sine peak values, but since the effective value only is the peak value scaled by 1/√2 so it does not matter if you count with peak values or effective values - as long as you are consistent!
Imaginary axis
Real axis
William Sandqvist william@kth.se
The inductor and capacitor counteracts changes
William Sandqvist william@kth.se
The inductor and capacitor counteracts changes, such as when connecting or disconnecting a source to a circuit.
What if the source then is sinusoidal AC – which is then changing continuously?
?
William Sandqvist william@kth.se
Alternating current through resistor
RR
RRRRRR )sin(ˆ)()()()sin(ˆ)(
IRU
tIRtuRtitutIti
⋅=⋅⋅=⇒⋅=⋅= ωω
A sinusoidal currentiR(t) through a resistor R provides a proportional sinusoidal voltage drop uR(t) according to Ohm's law. The current and voltage are in phase. No energy is stored in the resistor.
Phasors UR and IR become parallel to each other.
RR IRU ⋅=
The phasor may be a peak pointer or effective value pointer as long as you do not mix different types.
• Complex phasor
• Vector phasor
William Sandqvist william@kth.se
Alternating current through inductor
LL
LL
ILU
tILtILtut
tiLtutIti
⋅=
+⋅⋅=⋅⋅=⇒=⋅=
ω
πωωωωω )2
sin(ˆ)cos(ˆ)(d
)(d)()sin(ˆ)( L
LLLL
A sinusoidal current iL(t) through an inductor provides, due to self-induction, a votage drop uL(t) which is 90° before the current. Energy stored in the magnetic field is used to provide this voltage.
When using complex pointers one multiplies ωL with ”j”, this rotates the voltage pointer +90° (in complex plane). The method automatically keeps track of the phase angles!
LLLL jj IXILU ⋅=⋅= ω
• Vector phasor
• Complex phasor
The phasor UL will be ωL·IL and it is 90° before IL. The entity ωL is the ”amount” of the inductor’s AC resistance, reactance XL [Ω].
William Sandqvist william@kth.se
Alternating current through capacitor
)2
sin(ˆ1)cos(ˆ1
)()sin(ˆ)(
d)(1
)()(1
d
d1
d
)(d
CCCCC
CCCC
πωω
ωω
ω −⋅⋅=⋅⋅−=⇒⋅=
⋅=⇒⋅=⋅=⇒= ∫
tIC
tIC
tutIti
ttiC
tutiCt
q
Ct
tu
C
QU
A sinusoidal current iC(t) throug a capacitor will charge it with the ”voltage drop” uC(t) that lags 90°behind the current. Energy is storered in the electric field.
• Vector phasor
Phasor UC is IC/(ωC) and it lags 90°after IC. The entity 1/(ωC) is the ”amount” of the capacitor’s AC resistance, reactance XC [Ω].
CC
1I
CU ⋅=
ω
William Sandqvist william@kth.se
Complex phasor and the sign of reactance
If you use complex phasor you get the -90° phase by dividing (1/ωC) with ”j ”.
CXI
CI
CU CCCC ωωω
11-j
j
1 −=⇒⋅=⋅= • Complex phasor
The method with complex pointer automatically keeps track of the phase angles if we consider the capacitor reactance XCas negative, and hence the inductor reactance XL as positive.
L+C in series
William Sandqvist william@kth.se
Ωj5 Ω− j4 Ω+ j1
Ωj4 Ω− j5Ω− j1
=
=´L
´C
William Sandqvist william@kth.se
William Sandqvist william@kth.se
Reactance frequency dependency
fC
XLX CL
πωω
ω
2
1
=⋅
=⋅=
][ΩLX
][ΩLX
]Hz[f]Hz[f
William Sandqvist william@kth.se
LOG – LOG plot
Often electronics engineers use log-log scale. The inductor and capacitor reactances will then both be "linear" relationshipin such charts.
( ) ][log Ω− scaleX L ( ) ][log Ω− scaleX C
( ) ]Hz[log scalef − ( ) ]Hz[log scalef −
William Sandqvist william@kth.se
William Sandqvist william@kth.se
R L C
In general, our circuits are a mixture of different R L and C. The phase between I and U is then not ±90° but can have any intermediate value. Positive phase means that the inductances dominates over capacitances, we have inductive character IND . Negative phase means that the capacitance dominates over the inductances, we have capacitive character CAP.
The ratio between the voltage U and current I, the AC resistance, is called impedaceZ [Ω]. We then have OHM´s AC law: I
UZ =
CAP
William Sandqvist william@kth.se
Phasor diagrams
In order to calculate the AC resistance, the impedance, Z, of a composite circuit one must add currents and voltages phasors to obtain the total current I and the total voltage U.
I
UZ = The phasor diagram is our "blind stick" in to
the AC World!
William Sandqvist william@kth.se
Ex. Phasor diagram (11.5)
At a certain frequency f the capacitor has the reactance |XC| and the resistor Rhas the same amount (absolute value), R [Ω].
Elementary diagrams for R L and C
Use the elementary diagrams for R and C as building blocks to draw the whole circuit phasor diagram (for this actual frequency f ).
Try it your self …
William Sandqvist william@kth.se
Example. Phasor diagram.
2R ||UI
R
UIIUI 2
RC2C ==⊥
1) U2 reference phase ( = horizontal )
2)
3)
4) CCR 2 IIIII ⋅=+=
5)
21C1
1
22 UURIRIU
IU
⋅=⇒⋅⋅=⋅=
⊥
6)21 UUU +=
William Sandqvist william@kth.se
William Sandqvist william@kth.se
Impedance ZThe circuit AC resistance, impedance Z, one get as the ratio between the length of U and Iphasors. The impedance phase ϕϕϕϕis the angle between U and Iphasors.
The current is before the voltage in phase, so the circuit has a capacitive character, CAP.
( Something else had hardly been to wait since there are no coils in the circuit )
William Sandqvist william@kth.se
William Sandqvist william@kth.se
Complex phasors, jω-method
°−=−==⋅=⋅=
°+==⋅=⋅=°==⋅=
90)jarg()j
1arg(
j
1j
90)jarg(jj
0)arg(
CCCC
LLLL
RR
CCC
IXIU
LLIXIU
RRIU
ωω
ϕω
ωϕωϕ
Complex phasors. OHM’s law for R L and C.
Complex phasors. OHM’s law for Z.
][Im][Im
arctan][Re
][Imarctan)arg(][Re][Re
)arg()arg(arg)arg(
ZIU
R
X
Z
ZZZIU
IUI
UZ
I
UZZIU
⋅=
=
=⋅=
−=
===⋅= ϕ
In fact, there will be four useful relationships!• Re, • Im, • Abs, • Arg
William Sandqvist william@kth.se
Ex. Complex phasors.
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
William Sandqvist william@kth.se
Ex. Complex phasors.
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
William Sandqvist william@kth.se
Ex. Complex phasors.
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
William Sandqvist william@kth.se
Ex. Complex phasors.
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. I
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. I
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
26,12,14,0j2140
j2,14,0)j31(
)j31(
j3-1
4
j)5-(5j10-
20
j1
22
C//R
=+=+=
+=++⋅=
+=
+==
,,I
ZC
U
Z
UI
ω
William Sandqvist william@kth.se
Ex. Complex phasors. U1
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. U1
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
65,12)4(12j412
j412j)-10(j)2,14,0(j
1
221
1
=−+=−=
−=⋅+=⋅=
U
CIU
ω
William Sandqvist william@kth.se
Ex. Complex phasors. U2
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. U2
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
94,848j48
j48j)3(1
j)3(1
j3-1
j120
j)5-(5j10-
j5-520
j1
222
C//R
C//R2
=+=+=
+=++⋅−⋅=
+⋅=
+⋅=
U
ZC
ZUU
ω
Voltage divider:
William Sandqvist william@kth.se
Ex. Complex phasors. IC
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. IC
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
89,08,04,0j8,04,0
j8,04,0j10-
j48
j1
22C
2C
=+=+=
+−=+=⋅=
I
C
UI
ω
William Sandqvist william@kth.se
Ex. Complex phasors. IR
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
William Sandqvist william@kth.se
Ex. Complex phasors. IR
U = 20 V C = 320 µF R = 10 Ω f = 50 Hz
j1010320502j
1
2j
1
j
16
−=⋅⋅⋅
=⋅⋅
= −ππω CfC
j55)j1010(
)j1010(
j1010
)j10(10
j1
j1
R//C −=++⋅
−−⋅=
+
⋅=
CR
CR
Z
ω
ω
89,04,08,0j4,08,0
j4,08,010
j48
22R
2R
=+=+=
+=+==
I
R
UI
William Sandqvist william@kth.se
You get the phasor chart by plotting the points in the complex plane!
William Sandqvist william@kth.se
Vrida diagrammet …When we draw the phasor diagram it was natural to have U2 as reference phase (=horizontal), with the jω-method U was the natural choice of reference phase (=real).
Because it is easy to rotate the chart, so, in practice, we have the freedom of choosing any entity as the reference.
))7,26sin(j)7,26(cos(
7,268
4arctan)j48arg()arg( 2
°−⋅+°−×
°=
=+=U
Multiply the all complex numbers by this factor and rhe rotation will take effect!
William Sandqvist william@kth.se
William Sandqvist william@kth.se
SummarySinusoidal alternating quantities can be represented as pointers, phasors,
”amount” ∠∠∠∠ ”phase”.
A pointer (phasor) can either be seen as a vector expressed in polar coordinates, or as a complex number.
Calculations are usually best done with the complex method, while phasor diagrams are used to visualize and explain alternating current phenomena.
William Sandqvist william@kth.se
Notation
X
x Instant value
X Top value
XX Absolute value, the amount
Complex phasor
William Sandqvist william@kth.se
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