IE1206 Embedded Electronicskth.s3-website-eu-west-1.amazonaws.com/ie1206/slides/eng/Ex_pha… ·...

IE1206 Embedded Electronics

Transients PWM

Phasor jω PWM CCP CAP/IND-sensor

Le1

Le3

Le6

Le8

Le2

Ex1

Le9

Ex4 Le7

wr. exam

William Sandqvist william@kth.se

PIC-block Documentation, Seriecom, Pulse sensorI, U, R, P, series and parallel

Ex2

Ex5

Kirchhoffs laws Node analysis Two ports R2R AD

Trafo, EthernetcontaktLe13

Pulse sensors, Menu program

Le4

KC1 LAB1

KC3 LAB3

KC4 LAB4

Ex3Le5 KC2 LAB2 Two port, AD, Comparator/Schmitt

Step-up, RC-oscillator

Le10Ex6 LC-osc, DC-motor, CCP PWM

LP-filter Trafo + Guest lecturerLe12 Ex7 presentation

Le11

•••• Start of program task

•••• presentation of program task

William Sandqvist william@kth.se

Phasor - vector

I

UZ =

LXL ⋅= ωC

XC ⋅=

ω1

fπω 2=

William Sandqvist william@kth.se

William Sandqvist william@kth.se

What’s inside the circuit? (11.4)

William Sandqvist william@kth.se

What’s inside the circuit? (11.4)

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Phasor chart

?

?

William Sandqvist william@kth.se

Phasor chart

222CR

CR

CR

UUU

UU

III

+=

⊥==

13

3345 2222 ====−=−=

R

UIUUU R

CR

Now all values are known!

The two voltages have 90°phase angle. Pythagorean theorem applies!

?

?

William Sandqvist william@kth.se

Phasor chart

U = 5V

= 3V

3=RU

1=I

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Phasor chart (11.6)

U = 200 V, f = 50 Hz, L = 0,318 H, R1 = 100 Ω, R2 = 50 Ω.

William Sandqvist william@kth.se

Phasor chart (11.6)

U = 200 V, f = 50 Hz, L = 0,318 H, R1 = 100 Ω, R2 = 50 Ω.

|XL| = ω⋅L = 2π⋅50⋅0,318 = 100 Ω

William Sandqvist william@kth.se

Phasor chart (11.6)

William Sandqvist william@kth.se

Phasor chart (11.6)

Choose ULR as reference phase ( = horizontal ).

William Sandqvist william@kth.se

Phasor chart (11.6)

Välj ULR som riktfas ( = horisontell ).The current IR has the same direction as ULR.

William Sandqvist william@kth.se

Phasor chart (11.6)

Välj ULR som riktfas ( = horisontell ).Strömmen IR har samma riktning som ULR.The current IL lags 90° behind ULR and has an equally long pointer as IR because R1 and L has the same impedance. ( |XL| = 100 Ω, R1 = 100 Ω )

William Sandqvist william@kth.se

Phasor chart (11.6)

Välj ULR som riktfas ( = horisontell ).Strömmen IR har samma riktning som ULR.Strömmen IL ligger 90° efter ULR och har lika lång visare som IR eftersom R1 och L har samma växelströmsmotstånd. (XL = 100 Ω, R1 = 100 Ω)

The two currents IR and IL can be added as vectors to the current I. I is √2 longer than IR and IL(Pythagorean theorem applies!).

William Sandqvist william@kth.se

Phasor chart (11.6)

Välj ULR som riktfas ( = horisontell ).Strömmen IR har samma riktning som ULR.Strömmen IL ligger 90° efter ULR och har lika lång visare som IR eftersom R1 och L har samma växelströmsmotstånd. (XL = 100 Ω, R1 = 100 Ω)

De två strömmarna IR och IL kan adderas vektoriellt till strömmen I. I blir √2 ggr. längre än IR eller IL(enligt pythagoras sats).

Current I passes through the lower resistor R2. The voltage drop UR2 gets the same direction as I. ULR has the length IR⋅100, UR2 has length I⋅50. Because I = IR⋅√2 we get UR2 = ULR / √2.

William Sandqvist william@kth.se

Phasor chart (11.6)

Välj ULR som riktfas ( = horisontell ).Strömmen IR har samma riktning som ULR.Strömmen IL ligger 90° efter ULR och har lika lång visare som IR eftersom R1 och L har samma växelströmsmotstånd. (XL = 100 Ω, R1 = 100 Ω)

De två strömmarna IR och IL kan adderas vektoriellt till strömmen I. I blir √2 ggr. längre än IR eller IL(enligt pythagoras sats).

Strömmen I passerar genom den nedre resistorn R2. Spänningsfallet UR2 får samma riktning som I. ULR har längden IR⋅100, UR2 har längden I⋅50. Eftersom I = IR⋅√2 blir UR2 = ULR / √2.

Voltage U can finally be determined as the vector sum of ULR and UR2.

• Phaseϕ is the angle between U and I.

• Z is the ratio between lenghts of U and I.

The current after the voltage - inductive character

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Phasor chart (11.7)

Draw the phasor chart for this circuit. At the frequency fapplies that |XC| = Rand |XL| = R/2.

U2 is a suitable reference phase.

William Sandqvist william@kth.se

Phasor chart (11.7)

Start with U2 as reference phase ( = horizontal ).

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Current IR has the same direction as U2.

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Strömmen IR har samma riktning som U2.Current IC leads 90° before U2 and is equally long as IRbecause XC = R.

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Strömmen IR har samma riktning som U2.Strömmen IC ligger 90° före U2 och är lika stor som IReftersom XC = R.

Current IC and IR are summed to I.

I is √2 times longer than IC and IR (Pythagorean theorem applies).

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Strömmen IR har samma riktning som U2.Strömmen IC ligger 90° före U2 och är lika stor som IReftersom XC = R.

Strömmarna IC och IR summeras ihop till I.

I är √2 ggr. längre än IC eller IR (enligt pythagoras sats).

U1 leads 90° before I.

The length is U1 = I⋅XL = √2⋅IR⋅R/2 = IR⋅R/ √2

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Strömmen IR har samma riktning som U2.Strömmen IC ligger 90° före U2 och är lika stor som IReftersom XC = R.

Strömmarna IC och IR summeras ihop till I.

I är √2 ggr. längre än IC eller IR (enligt pythagoras sats).

U1 ligger 90° före I.

Längden är U1 = I⋅XL = √2⋅IR⋅R/2 = IR⋅R/ √2

Voltage U1 and U2 are summed to voltage U.

William Sandqvist william@kth.se

Phasor chart (11.7)

Börja med U2 som riktfas ( = horisontel ).Strömmen IR har samma riktning som U2.Strömmen IC ligger 90° före U2 och är lika stor som IReftersom XC = R.

Strömmarna IC och IR summeras ihop till I.

I är √2 ggr. längre än IC eller IR (enligt pythagoras sats).

U1 ligger 90° före I.

Längden är U1 = I⋅XL = √2⋅IR⋅R/2 = IR⋅R/ √2

Spänningarna U1 och U2 summeras ihop till spänningen U.

One can see from the chart that U becomes equal long to U1.The angle ϕ = 0 and thereby U and I are in phase.

Inductive or capacitive character?

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Complex numbers, jω-method

CX

CIXIU

LXLIXIU

RIU

ωω

ωω1

j

1j

jj

CCCCC

LLLLL

RR

−=⋅=⋅=

=⋅=⋅=⋅=

Complex OHM’s law for R L and C.

Complex OHM’s law for Z.

===⋅=

][Re

][Imarctan)arg(

Z

ZZ

I

UZZIU ϕ

f⋅= πω 2

William Sandqvist william@kth.se

William Sandqvist william@kth.se

jω Impedance (12.2)

William Sandqvist william@kth.se

jω Impedance (12.2)

One can imagine that the phasor chart shows the complex plane, and then splitting the current phasor in real part and imaginary part:

William Sandqvist william@kth.se

jω Impedance (12.2)

( )j1,111,19

99

j11001892

)j56,8(

)j56,8(

j56,8

220

)30sin(j)30cos(10

220

−=−=−−⋅

+=

=°⋅+°⋅

==I

UZ

William Sandqvist william@kth.se

jω Impedance (12.2)

j1,111,19 −==I

UZ

19,1-11,1jF287

)1,11(502

1

1,111

1,19

µπ

ω

=−⋅⋅

−=

−=−=Ω=

C

CXR C

Capacitor has negative reaktance.

• One possible solution is then a series circuit with Rand C

William Sandqvist william@kth.se

jω Impedance (12.2)

• Another possible solution is a parallel circuit with R’ and C’one then thinks on I as divided in to current composants IR

and IC which are perpendicular to each other.

William Sandqvist william@kth.se

jω Impedance (12.2)

Ω=⋅

=°

==

Ω=⋅

=°

==

445,010

220

30sin|'|

3,2587,010

220

30cos'

CC

R

I

U

I

UX

I

U

I

UR

F6,7244502

1'

'

1' µ

πω=

⋅⋅=⇒= C

CX C

William Sandqvist william@kth.se

jω Impedance (12.2)

Is there any way to find out which of the two proposed circuits Z actually contain?

?

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Complex impedance (12.6)

Determine the complex impedance ZAB of this circuit.

William Sandqvist william@kth.se

Complex numbers calculation

=−++−⋅+=

20j1020j15

)20j10()20j15(ABZ

=−=25

100j550

][4j22 Ω−=

Here the denominator directly was a real number so there no need to multiply with the complex conjugate of the denominator, otherwise the calculations had been moore extensive…

William Sandqvist william@kth.se

Complex numbers calculation

=−++−⋅+=

20j1020j15

)20j10()20j15(ABZ

Online Scientific Calculator

)i422( −=

William Sandqvist william@kth.se

William Sandqvist william@kth.se

"heavy" calculations! (12.9)

• Calculate impedance Z.• Calculate currentI.• CalculateIC (current branching).• CalculateUL (voltage divider).

William Sandqvist william@kth.se

Calculate impedance Z

j82

)j8(2|| −

−⋅=CRZ =++⋅

−−⋅=

j82

)j82(

j82

)j8(2||CRZ

j47,088,1 −=

j53,588,4j)47,088,1(j63 +=−++=

Ω=+= 38,753,588,4 22Z

=++= CRL ZXRZ ||1 j

William Sandqvist william@kth.se

Calculate current I

j53,588,4

30

+==

Z

UI

U is reference phase, real.

=−−⋅

+==

)j53,588,4(

)j53,588,4(

j53,588,4

30

Z

UI

j37,253,588,4

j9,1655,14622

−=+−=

A437,2 22 =+=I

William Sandqvist william@kth.se

Calculate current IC

2

2

2(2,7 3j)

j 2 8jC

C

RI I

R X= = − ⋅ =

+ −

(2,7 3j) 2 (2 8j)0,86 0,46j

2 8j (2 8j)

− ⋅ += ⋅ = +− +

2 20,86 0,46 0,98 ACI = + =

William Sandqvist william@kth.se

UL complex conjugate method?

=++

=1||j

j

RZX

XUU

CRL

LL

=+−+

=3)j47,088,1(j6

j630

V4,242,163,18 22 =+=LU

j2,163,18j)53,588,4(

)j53,588,4(

j53,588,4

j630 +=

−−⋅

+=

Amount and phase

William Sandqvist william@kth.se

)arg()arg()arg( 21

212121

ZZZ

ZZZZZZZZ

+=⋅=⋅=⋅=

)arg()arg()arg( 21

2

1

2

1

2

1

ZZZ

Z

Z

Z

ZZ

Z

ZZ

−=

===

William Sandqvist william@kth.se

UL amount ∠ phase method?

=++

=1||j

j

RZX

XUU

CRL

LL =

+−+ 3)j47,088,1(j6

j630

V4,24=LU

Amount ∠ phase method, the polar form, often gives simpler calculations, but nowdays most math programs and pocket calculators handles complex numbers directly …

=

∠+

°∠=+

=

88,453,5

arctan53,588,4

90630

j53,588,4

j630

22

°∠=°−°∠= 4,414,24)6,4890(38,7

630

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Derive the complex current I (12.7)Set up the complex current I (with U as reference phase).

Note! One does not always have give the answer of the form a+jb. The same information, but with less effort, one gets if the answer is expressed as a ratio of complex numbers. Amount and phase can if necessary be taken from the numerator and the denominator directly.

William Sandqvist william@kth.se

Derive the complex current I (12.7)

( ) ( ) ( )dcbaIdc

baI

dc

baI jargjargarg

j

j

j

j +−+=++

=++=

Note!

Set up the complex current I (with U as reference phase).

Note! One does not always have give the answer of the form a+jb. The same information, but with less effort, one gets if the answer is expressed as a ratio of complex numbers. Amount and phase can if necessary be taken from the numerator and the denominator directly.

William Sandqvist william@kth.se

Derive the complex current I (12.7)

Set up the complex current I (With Uas the reference phase).

)1

j(

j

j1

j

j)j1

(

CLR

LRC

L

CLR

LC

RZ

ωω

ω

ωω

ωω

−+

+=

++

⋅+=

LRC

LC

LRU

Z

UI

ωω

ω

j

)1

j(

+

−+==

That it is U which is the reference phase can be seen that we let the voltage be a real number!Sufficiently simplified!

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Derive the complex current I (12.8)

L

U

L

UI

ω−=

ω= j

j

Now it will be easier! The voltage U is located directly across the parallel branch with the inductance L. (We need not concern ourselves with Rand C)

Set up the complex current I (With Uas the reference phase).

William Sandqvist william@kth.se