IE1206 Embedded Electronics2 1 1 sin 2 ( x) dx ... Phasor diagrams In order to calculate the AC resistance, the impedance, Z, of a composite circuit one must add currents and voltages

IE1206 Embedded Electronics

Transients PWM

Phasor j PWM CCP CAP/IND-sensor

Le1

Le3

Le6

Le8

Le2

Ex1

Le9

Ex4 Le7

Written exam

William Sandqvist [email protected]

PIC-block Documentation, Serialcom Pulse sensorsI, U, R, P, series 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


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

…


The sine wave – what do you remember?

periodT

timet

y

amplitudetopY ,ˆ

RMSY

2

ˆ12)sin(ˆ)(

YY

TfftYty


(11.1) Phase

)sin(ˆ)( tYty

If a sine curve does

not begin with 0 the

function expression

has a phase angle .

Specify this function mathematically:

)10002sin(6)( ttu

y

)30(rad52,06

3arcsin)sin(63)0(

u

)52,06283sin(6)( ttu


Apples and pears?

In circuit analyses it is common (eg. in textbooks) to

expresses the angle of the sine function mixed in

radians ·t [rad] and in degrees [°].

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] 57,3

x[rad]= x[]0,017? ?



Mean and effective value

T

ttu

UttuU

T

T

T

0

2

0

1med

d)(

0d)(

All pure AC voltages, has the mean value 0.

More interesting is the effective value – root

mean square, rms.


(11.2) Example. RMS.

V63,11015

1058

1015

105)0)2(2(d)(

3

3

3

322

0

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


Sine wave effective value

2

1

2

1d)(sin 2 xx

)(sin 2 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



Addition of sinusoidal quantities

)sin(ˆ111 tAy

)sin(ˆ222 tAy

?21 yy


Addition of sinusoidal quantities

When 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 new

amplitude 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

2

2

2

1

21222111

t

tytytyttytty


Sine wave as a pointer

A sinusoidal voltage or current,

can be represented by a pointer that

rotates (counterclockwise) with the

angular velocity [rad/sec] .

)sin(ˆ)( tYty

Wikipedia Phasors

http://en.wikipedia.org/wiki/Phasors

http://en.wikipedia.org/wiki/File:Unfasor.gif


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/File:Sumafasores.gif


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


Phasor

A pointer (phasor) can either be viewed as a vector expressed in

polar 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 j-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”


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


The inductor and capacitor

counteracts changes


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?

?


Alternating current through resistor

RR

RRRRRR )sin(ˆ)()()()sin(ˆ)(

IRU

tIRtuRtitutIti

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


Alternating current through inductor

LL

LL

ILU

tILtILtut

tiLtutIti

)

2sin(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 L with ”j”, this rotates the

voltage pointer +90° (in complex plane). The method automatically keeps

track of the phase angles!

LLLL jj IXILU

Vector phasor

Complex phasor

The phasor UL will be L·IL and it is 90° before IL. The entity L is the

”amount” of the inductor’s AC resistance, reactance XL [].


Alternating current through capacitor

)2

sin(ˆ1)cos(ˆ1

)()sin(ˆ)(

d)(1

)()(1

d

d1

d

)(d

CCCCC

CCCC

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/(C) and it lags 90° after IC.

The entity 1/(C) is the ”amount” of the

capacitor’s AC resistance, reactance XC [].

CC

1I

CU


Complex phasor and the sign of reactance

If you use complex phasor you get the

-90° phase by dividing (1/C) with ”j”.

CXI

CI

CU CCCC

11-j

j

1 Complex phasor

The method with complex pointer automatically keeps track

of the phase angles if we consider the capacitor reactance XC

as negative, and hence the inductor reactance XL as positive.

L+C in series


j5 j4 j1

j4 j5

j1

´L

´C



Reactance frequency dependency

f

CXLX CL

2

1

][

LX

][

LX

]Hz[f]Hz[f


LOG – LOG plot

Often electronics engineers use log-log scale. The inductor

and capacitor reactances will then both be "linear" relationship

in such charts.

][log scaleX L ][log scaleXC

]Hz[log scalef ]Hz[log scalef



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 impedace Z []. We then have OHM´s AC law: I

UZ

CAP


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!


Ex. Phasor diagram (11.5)

At a certain frequency f the capacitor has the reactance |XC| and the resistor R

has the same amount (absolute value), R [].

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 …


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



Impedance ZThe circuit AC resistance,

impedance Z, one get as the ratio

between the length of U and I

phasors. The impedance phase

is the angle between U and I

phasors.

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 )



Complex phasors, j-method

90)jarg()j

1arg(

j

1j

90)jarg(jj

0)arg(

CCCC

LLLL

RR

CCC

IXIU

LLIXIU

RRIU

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

In fact, there will be four useful relationships!

Re, Im, Abs, Arg


Ex. Complex phasors.

U = 20 V C = 320 F R = 10 f = 50 Hz



j1010320502j

1

2j

1

j

16

CfC

U = 20 V C = 320 F R = 10 f = 50 Hz



j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC



U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z


Ex. Complex phasors. I

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z


Ex. Complex phasors. I

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

26,12,14,0j2140

j2,14,0)j31(

)j31(

j3-1

4

j)5-(5j10-

20

j

1

22

C//R

,,I

ZC

U

Z

UI


Ex. Complex phasors. U1

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z



U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

65,12)4(12j412

j412j)-10(j)2,14,0(j

1

22

1

1

U

CIU



U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z



U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

94,848j48

j48j)3(1

j)3(1

j3-1

j120

j)5-(5j10-

j5-520

j

1

22

2

C//R

C//R2

U

ZC

ZUU

Voltage divider:


Ex. Complex phasors. IC

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z


Ex. Complex phasors. IC

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

89,08,04,0j8,04,0

j8,04,0j10-

j48

j

1

22

C

2C

I

C

UI


Ex. Complex phasors. IR

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z


Ex. Complex phasors. IR

U = 20 V C = 320 F R = 10 f = 50 Hz

j1010320502j

1

2j

1

j

16

CfC

j55)j1010(

)j1010(

j1010

)j10(10

j

1

j

1

R//C

CR

CR

Z

89,04,08,0j4,08,0

j4,08,010

j48

22

R

2R

I

R

UI


You get the phasor

chart by plotting the

points in the complex

plane!


Rotate diagram …When we draw the phasor

diagram it was natural to have

U2 as reference phase

(=horizontal), with the j-

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 the rotation will take effect!


Mathematica (11.5)

u = 20;

c = 320*10^-6;

r = 10;

f = 50;

w = 2*Pi*f;

xc = 1/(I*w*c);

N[xc]

zrc =

r*xc/(r+xc);

N[zrc]

i = u/(xc+zrc);

N[i]

N[Abs[i]]

u1 = i*xc;

N[u1]

N[Abs[u1]]

u2 =

u*zrc/(xc+zrc);

N[u2]

N[Abs[u2]]

ic = u2/xc;

N[ic]

N[Abs[ic]]

ir = u2/r;

N[ir]

N[Abs[ir]]

Mathematica understands complex

numbers ( I ) and has the functions

Abs[]

Arg[]

(180/Pi)*Arg[] to get degres

11.5.nb


Mathematica (11.5)

11.5.nb

Phasor plot

U I



Summary

Sinusoidal 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.


Notation

X

x Instant value

X Top value

XX Absolute value, the amount, magnitude

Complex phasor


Documents

IE1206 Embedded Electronics2 1 1 sin 2 ( x) dx ... Phasor diagrams In order to calculate the AC resistance, the impedance, Z, of a composite circuit one must add currents and voltages