IE1206 Embedded Electronicskth.s3-website-eu-west-1.amazonaws.com/ie1206/slides/eng/Ex_pha… · Phasor j ωPWM CCP CAP/IND-sensor Le1 Le3 Le6 Le8 Le2 Ex1 Le9 Ex4 Le7 wr. exam William

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 [email protected]

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

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Kirchhoffs laws Node analysis Two ports R2R AD

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Pulse sensors, Menu program

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KC1 LAB1

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KC4 LAB4

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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


Phasor - vector

I

UZ =

LXL ⋅= ωC

XC ⋅=

ω1

fπω 2=



What’s inside the circuit? (11.4)


What’s inside the circuit? (11.4)



Phasor chart

?

?


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!

?

?


Phasor chart

U = 5V

= 3V

3=RU

1=I



Phasor chart (11.6)

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


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


Phasor chart (11.6)


Phasor chart (11.6)

Choose ULR as reference phase ( = horizontal ).


Phasor chart (11.6)

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


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 )


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!).


Phasor chart (11.6)


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.


Phasor chart (11.6)


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



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.


Phasor chart (11.7)

Start with U2 as reference phase ( = horizontal ).


Phasor chart (11.7)

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


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.


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).


Phasor chart (11.7)


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


Phasor chart (11.7)




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.


Phasor chart (11.7)




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?



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



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:



( )j1,111,19

99

j11001892

)j56,8(

)j56,8(

j56,8

220

)30sin(j)30cos(10

220

−=−=−−⋅

+=

=°⋅+°⋅

==I

UZ



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



• 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.



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



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

?



Complex impedance (12.6)

Determine the complex impedance ZAB of this circuit.


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…


Complex numbers calculation

=−++−⋅+=

20j1020j15

)20j10()20j15(ABZ

Online Scientific Calculator

)i422( −=



"heavy" calculations! (12.9)

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


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


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


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 = + =


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


)arg()arg()arg( 21

212121

ZZZ

ZZZZZZZZ

+=⋅=⋅=⋅=

)arg()arg()arg( 21

2

1

2

1

2

1

ZZZ

Z

Z

Z

ZZ

Z

ZZ

−=

===


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



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.


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.



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!




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).


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IE1206 Embedded Electronicskth.s3-website-eu-west-1.amazonaws.com/ie1206/slides/eng/Ex_pha… · Phasor j ωPWM CCP CAP/IND-sensor Le1 Le3 Le6 Le8 Le2 Ex1 Le9 Ex4 Le7 wr. exam William