IE1206 Embedded Electronics PIC-block Documentation, Seriecom, Pulse - - PowerPoint PPT Presentation

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 sensor I, U, R, P, series and parallel

Ex2 Ex5

Kirchhoffs laws Node analysis Two ports R2R AD Trafo, Ethernet contact

Le13

Pulse sensors, Menu program

Le4

KC1 LAB1 KC3 LAB3 KC4 LAB4

Ex3 Le5

KC2 LAB2

Two port, AD, Comparator/Schmitt Step-up, RC-oscillator

Le10 Ex6

LC-osc, DC-motor, CCP PWM

LP-filter Trafo + Guest lecturer

Le12 Ex7

presentation

Le11

• Start of program task
• presentation of program task

William Sandqvist william@kth.se

Complex phasors, jω-method

C I X I U L I X I U R I U ω ω j 1 j j j

C C C C L L L L R R

⋅ = ⋅ = ⋅ = ⋅ = ⋅ =

• Complex OHM’s law for R L and C.
• Complex OHM’s law for Z.

        = = = ⋅ = ] [ Re ] [ Im arctan ) arg( Z Z Z I U Z Z I U ϕ

f ⋅ = π ω 2

William Sandqvist william@kth.se

Voltage divider, Transfer function

Simple filters are often designed as a voltage dividers.A filter transfer function, H(ω) or H(f), is the ratio between output voltage and input

• voltage. This ratio we get directly from the voltage divider formula!

2 1 2 1 2 2 1 2 1 2

) ( Z Z Z U U H Z Z Z U U + = = ⇒ + = ω

William Sandqvist william@kth.se

LP HP BP BS

BP and BS filters can be seen as different combination of LP and HP filters.

LP lowpass HP highpass BP bandpass BS bandstop

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Transfer function (14.2)

a) Set up an expression of IC = f(U, ω, R, C). b) Set up the transfer function IC/U the amount function and the phase function. c) What filter type is the transfer function, LP HP BP BS ? d) What break frequency has the transfer function?

C

I

William Sandqvist william@kth.se

Transfer function (14.2)

C U C U I RC R C C C R C R C R

C

ω ω ω ω ω ω ω j j 1 j 1 j j j 1 j 1 ||

C C

⋅ = = + = ⋅ + ⋅ = Answer a)

William Sandqvist william@kth.se

Transfer function (14.2)

RC C U I RC U R RC R RC RC R R RC R U U

C

ω ω ω ω ω ω ω j 2 j 1 j 1 1 j 1 j 1 j 1 j 1

C

+ = ⇒ + + = + + ⋅ + + + =

William Sandqvist william@kth.se

Transfer function (14.2)

      =               − ° =         + = + = RC U I RC U I RC C U I RC C U I

C C C C

ω ω ω ω ω ω 2 arctan arg 2 arctan 90 arg ) ( 4 j 2 j

2

Answer b) IC/U

William Sandqvist william@kth.se

Transfer function (14.2)

{ } { }

R U I j j U I

C C

1 2 = ∞ = = ⋅ + ⋅ = = ω ω Answer c) LP HP BP BS?

C

I

RC C U I C ω ω j 2 j + = HP ⇒

William Sandqvist william@kth.se

Transfer function (14.2)

RC f RC RC C U I

G C

2 2 1 2 j 2 j ⋅ = ⇒ = + = π ω ω ω Answer d) Break frequency? At the break frequency the numerator real part and imaginary part are equal. 2 1 2 2 2 2 j 2 R 2 j j 2 j

2 2

⋅ = + = ⇒ + = + = R R U I RC C U I

C C

ω ω

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Phasor - vector

I U Z = L X L ⋅ = ω C X C ⋅ = ω 1 f π ω 2 =

William Sandqvist william@kth.se

Phasor chart for voltage divider (11.8)

The figure shows a voltage divider. It is connected to an AC voltage source U1 and it’s output voltage is U2. At a some frequency the reactance of the inductor is XL = 2R. Draw the phasor chart of this circuit with I1, U1 and U2 at this frequency. Use I1 as reference phase ( = horizontal).

William Sandqvist william@kth.se

Phasor chart for voltage divider (11.8)

1

I 2 RI 1 RI 1 3 RI 1 U 1 U 2

William Sandqvist william@kth.se

jω-calculation of the divided voltage

2 1 20 5 ) 2 ( 16 ) 2 ( 2 ) ( 16 ) ( j 4 j

2 2 2 2 1 2 2 2 2 2 1 2 1 2

= = + + = ⇒ = = + + = + + = R R R R U U R L X L R L R U U L R L R U U

L

ω ω ω ω ω

William Sandqvist william@kth.se

Here are some more ”filters” if time permits!

Filter RLR (14.7)

William Sandqvist william@kth.se

The figure shows a simple filter with two R and one L. a) Derive the filter complex transfer function U2/U1. b) At what angle frequency ωX will the amount function be Give an expresson for this frequency ωX with R L.

2 / 1 | | / | |

1 2

= U U

c) What value has the amount of the transfer function at very low frequencys, ω≈0? What value has the phase function at very low frequencys? d) What value has the amount of the transfer function at very high frequencys, ω≈∞? What value has the phase function at very high frequencys?

? arg ? ) ? arg ? ) ? ) , ( 2 1 ) ? )

1 2 1 2 1 2 1 2 1 2 1 2

=         = ⇒ ∞ ≈ =         = ⇒ ≈ = = ⇒ = U U U U d U U U U c L R U U b U U a

X X

ω ω ω ω

Filter RLR (14.7)

William Sandqvist william@kth.se

L j R L j R L j R L j L j R L j R L j R L j R L j L j R L j R R R U U L j R L j R L R a ω ω ω ω ω ω ω ω ω ω ω ω ω 2 1 1 1 || )

1 2

+ + = + + + + + = + ⋅ + = + ⋅ + = + ⋅ = 2 ) ( 2 ) ( 4 ) ) ( 2 2 2 1 ) 2 ( ) ( 2 1 2 )

2 2 2 2 2 2 2 2 2 2 1 2

L R L R L R L R L R L R L j R L j R U U b

X =

⇒ = + = + = + + = + + = ω ω ω ω ω ω ω ω ° =         = ⇒ = + + → + + arg 1 1 2 )

1 2 1 2

U U U U R R L j R L j R c ω ω ω ° =         = ⇒ = + + ∞ → + + ⇒ + + arg 5 , 2 1 2 2 2 )

1 2 1 2

U U U U L j jL L j R jL R L j R L j R d ω ω ω ω ω

William Sandqvist william@kth.se

Filter LCR if time … (14.8)

William Sandqvist william@kth.se

The figure shows a simple filter with L C and R. a) Derive the filter transfer function U2/U1. b) At what angular frequency ωx will the denominator be purely imaginary? Give an expression of this frequency ωx with R L and C. c) What value has the amount function at this angular frequency, ωx? d) What value has the phase function at this angular frequency, ωx ? e) Give an expression of the transfer function between IR/U1 ( Note! You already have the transferfunction U2/U1 from a )

? ) ( ) ( ) ? ) ( ) ( arg ) ? ) ( ) ( ) ? ) , , ( ) ? ) ( ) ( )

1 1 2 1 2 1 2

= =         = = = ω ω ω ω ω ω ω ω ω U I e U U d U U c C L R b U U a

R X X X X X

Filter LCR if time … (14.8)

William Sandqvist william@kth.se LC R RLC U U RE L j RLC R R R RC j L j R RC j RC j RC j R L j RC j R U U RC j R C j C j C j R C j R C R b a 1 ) ( ) 1 ( 1 1 1 1 1 1 1 || ) )

2 1 2 2 1 2

= = ⇒ =       + − = = + + = + + ⋅ + + + = + = ⋅ + ⋅ = ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ° − =             =       = = + =       = = + − = 90 arg arg ) 1 ) ( )

1 2 1 2 2 1 2

C L j R U U d L C R C L R U U C L j R LC L j RLC R R U U c ω ω ω

L j RLC R R U U U I R U I U I e

R R R

ω ω + − = ⋅ = ⇒ = = ) ( 1 1 ? )

2 1 2 1 2 1

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Voltage ratio

2 1 2 1 2 2 1 1

d d d d N N U U t N U t N U = Φ = Φ =

N1 : N2

William Sandqvist william@kth.se

Current ratio

2 1 2 1 1 2 2 2 1 1 2 1

) , ( N N U U I I I U I U I P P P = ≈ ⇒ ⋅ = ⋅ = =

N1 : N2

William Sandqvist william@kth.se

William Sandqvist william@kth.se

Two values are missing? (15.1)

For a transformer the following data was given: Primary Secondary N1 U1 I1 N2 U2 I2 600 225 V

?

200

?

9 A Calculate the two values that are missing. I1 and U2.

William Sandqvist william@kth.se

Two values are missing! (15.1)

Primary Secondary N1 U1 I1 N2 U2 I2 600 225 V

?

200

?

9 A

n = N1/N2 = 600/200 = 3

I n I

1 2

1 9 3 3 = = =

3A

U n U

2 1

1 225 3 75 = = =

75V

For a transformer the following data was given: Calculate the two values that are missing. I1 and U2.

William Sandqvist william@kth.se

Two values are missing? (15.2)

Primary Secondary N1 U1 I1 N2 U2 I2

?

230 V 2A 150

?

12 A Calculate the two values that are missing. N1 and U2.

.

For a transformer the following data was given:

William Sandqvist william@kth.se

Two values are missing! (15.2)

Primary Secondary N1 U1 I1 N2 U2 I2

?

230 V 2A 150

?

12 A

.

n = I2/I1 = 12/2 = 6 N1 = N2⋅n = 150⋅6 = 900

900

U2 = U1/n =230/6 = 38,3 V

38V

For a transformer the following data was given: Calculate the two values that are missing. N1 and U2.

William Sandqvist william@kth.se

Two values are missing? (15.3)

Primary Secondary N1 U1 I1 N2 U2 I2 600 225 V

? ?

127 V 9 A Calculate the two values that are missing. I1 and N2. For a transformer the following data was given:

William Sandqvist william@kth.se

Two values are missing! (15.3)

Primary Secondary N1 U1 I1 N2 U2 I2 600 225 V

? ?

127 V 9 A

U U N N N U U N

1 2 1 2 2 2 1 1

225 127 1 77 600 127 225 339 = = = ⇒ = = ⋅ = ,

339

I N N I

1 2 1 2

339 600 9 5 08 = = = , A

5A

For a transformer the following data was given: Calculate the two values that are missing. I1 and N2.

William Sandqvist william@kth.se

Inductive coupling

William Sandqvist william@kth.se

± M is called mutal inductance

2 1L

L M k =

The coupling factor indicates how much of its flow a coil has in common with another coil? An ideal transformer has the coupling factor k = 1 (100%)

M L L M L L LTOT 2

2 1 2 2 1

− + − ⋅ = M L L M L L LTOT 2

2 1 2 2 1

+ + − ⋅ =

• Parallel connected coils
• Anti parallel connected coils

M L L LTOT 2

2 1

+ + = M L L LTOT 2

2 1

− + =

• Series connected coils
• Anti series connected coils

William Sandqvist william@kth.se

Mutal inductance (15.8)

Three inductors L1 = 12, L2 = 6, L3 =5 [H] are series

• connected. When inductors are close to each other the

placement on the circuit board can be important. In the figure to the left a) will inductors to have a portion of the magnetic lines in common. They then have the mutual inductances M12 = 3, M23 = 1, M13 = 1 [H]. In the figure to the right b) the inductors are mounted three dimensional so that there are no shared power magnetic lines. a) Calculate the total inductance for the arrangement in figure a). LTOT = ? b) Calculate the total inductance for the arrangement in figure b). LTOT = ?

William Sandqvist william@kth.se

Mutal inductance (15.8)

[ ] [ ]

H 23 5 6 12 ) b H 16 1 1 5 2 3 6 1 3 12 ) a

3 2 1 13 23 3 23 12 2 13 12 1

= + + = + + = = + − + − − + + − = = + − + − − + + − = L L L L M M L M M L M M L L

TOT TOT

William Sandqvist william@kth.se