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

IE1206 Embedded Electronics PIC-block Documentation, Seriecom, Pulse sensor Le1 Le2 I , U , R , P , series and parallel Le3 Ex1 KC1 LAB1 Pulse sensors, Menu program • Start of program task Kirchhoffs laws Node analysis Two ports R2R AD Ex2 Le4 Two port, AD, Comparator/Schmitt Ex3 Le5 KC2 LAB2 Transients PWM Le6 Ex4 Le7 KC3 LAB3 Step-up, RC-oscillator Phasor j ω PWM CCP CAP/IND-sensor Ex5 Le8 Le9 LC-osc, DC-motor, CCP PWM KC4 LAB4 Le11 Ex6 Le10 LP-filter Trafo + Guest lecturer presentation Le12 Ex7 • presentation of program task Trafo, Ethernet contact Le13 wr. exam William Sandqvist william@kth.se

Complex phasors, j ω -method • Complex OHM’s law for R L and C . = ⋅ U I R R R = ⋅ = ⋅ ω ω = π ⋅ j j U I X I L 2 f L L L L 1 = ⋅ = ⋅ j U I X I C C ω C C j C • Complex OHM’s law for Z .   Im [ ] U Z   = ⋅ = ϕ = = arg( ) arctan U I Z Z Z    Re [ ]  I Z 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! Z U Z = ⇒ ω = = 2 2 2 ( ) U U H + + 2 1 Z Z U Z Z 1 2 1 1 2 William Sandqvist william@kth.se

LP HP BP BS LP lowpass HP highpass BP bandpass BS bandstop BP and BS filters can be seen as different combination of LP and HP filters. William Sandqvist william@kth.se

William Sandqvist william@kth.se

Transfer function (14.2) I C Set up an expression of I C = f ( U , ω , R , C ). a) b) Set up the transfer function I C / 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? William Sandqvist william@kth.se

Transfer function (14.2) Answer a) 1 ⋅ R ω ω j j C R C = ⋅ = || R C ω + ω 1 j 1 j C RC + R ω j C U = = ⋅ ω C j I U C C C 1 ω j C William Sandqvist william@kth.se

Transfer function (14.2) ω + R 1 j RC + ω 1 1 j RC R = ⋅ = ⇒ U U U + ω + ω + C 1 j R RC 1 j 1 RC + R + ω 1 j RC R ω j C = I U + ω C 2 j RC William Sandqvist william@kth.se

Transfer function (14.2) Answer b) I C / U ω ω   ω   j I C I C I RC = =   = ° −   C C C arg 90 arctan   ω +   + ω   2 j 2 U RC U 2 U 4 ( ) RC     2 I   =   C arg arctan   ω     U RC William Sandqvist william@kth.se

Transfer function (14.2) Answer c) LP HP BP BS? ω j I C C = + ω 2 j U RC ⋅ 0 1 I { } j I { } ω = = = ω = ∞ = C C 0 0 + ⋅ 2 0 U j U R ⇒ HP I C William Sandqvist william@kth.se

Transfer function (14.2) Answer d) Break frequency? At the break frequency the numerator real part and imaginary part are equal. ω j 1 2 I C = ω = ⇒ = ⋅ C 2 RC f + ω π G 2 j 2 U RC RC 2 2 j ω j 1 I C I R R = = ⇒ = = C C + ω + ⋅ + 2 j 2 j 2 U RC U 2 2 2 R 2 2 William Sandqvist william@kth.se

William Sandqvist william@kth.se

Phasor - vector = ω ⋅ 1 X L L = ω ω = π X C 2 f ⋅ C U Z = I 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 U 1 and it’s output voltage is U 2 . At a some frequency the reactance of the inductor is X L = 2R. Draw the phasor chart of this circuit with I 1 , U 1 and U 2 at this frequency. Use I 1 as reference phase ( = horizontal). William Sandqvist william@kth.se

Phasor chart for voltage divider (11.8) RI 1 3 RI 1 U 2 U 1 2 RI 1 I 1 William Sandqvist william@kth.se

j ω -calculation of the divided voltage + ω + ω 2 2 ( ) j R L U R L U = = 2 2 + ω + ω 4 j U R L U 2 2 16 ( ) R L 1 1 = ω = ⇒ 2 X L R L + 2 2 ( 2 ) 5 1 R R U = = = 2 + 2 2 2 U 20 16 ( 2 ) R R 1 William Sandqvist william@kth.se

Here are some more ”filters” if time permits! William Sandqvist william@kth.se

Filter RLR (14.7) The figure shows a simple filter with two R and one L . a) Derive the filter complex transfer function U 2 / U 1 . b) At what angle frequency ω X will the amount function = be | | / | | 1 / 2 U U 2 1 Give an expresson for this frequency ω X with R L . 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?   1 U U U U   = ω ⇒ = ω = ω ≈ ⇒ = = 2 2 2 2 ) ? ) ( , ) ? ) 0 ? arg ? a b R L c   X X   U U U U 2 1 1 1 1   U U ω ≈ ∞ ⇒ =   = 2 2 ) ? arg ? d     U U 1 1 William Sandqvist william@kth.se

Filter RLR (14.7) + ω R j L ⋅ ω + ω + ω U 1 R j L R R j L R j L = = = = = 2 a ) R || L ⋅ ω ⋅ ω + ω + ω + ω + ω R j L 1 j L R j L j L 2 R j L U R j L + + 1 1 R + ω + ω + ω R j L R j L R j L + ω + ω 2 2 ( ) U 1 R L 1 R j L = = = + ω = + ω 2 2 2 2 2 ) 2 2 ( ) ) 4 ( ) b R L R L + ω + ω U R j 2 L 2 2 2 2 ( 2 ) R L 1 R = ω ⇒ ω X = 2 2 R 2 ( L ) 2 L   + ω + 0 U U R j L R   ω → = ⇒ = = ° 2 2 c ) 0 1 1 arg 0   + ω + 2 0   R j L R U U 1 1 R + jL   + ω + ω 0 1 R j L jL U U ⇒ → ∞ = ⇒ =   ω = ° 2 2 d ) 0 , 5 arg 0   + ω + R 2 0 2 2   R j L j L U U + 2 j L 1 1 ω William Sandqvist william@kth.se

William Sandqvist william@kth.se

Filter LCR if time … (14.8) The figure shows a simple filter with L C and R . a) Derive the filter transfer function U 2 / U 1 . b) At what angular frequency ω x will the denominator be purely imaginary? Give an expression of this frequency ω x with R L and C . c) What value has the amount function at this angular frequency, ω x ? d) What value has the phase function at this angular frequency, ω x ? e) Give an expression of the transfer function between I R / U 1 ( Note! You already have the transferfunction U 2 / U 1 from a )   ω ω ω ω ( ) ( ) ( ) ( ) U U U I   = ω = = = = 2 2 X 2 X R ) ? ) ( , , ) ? ) ? ) arg ? ) ? a b R L C c d e   ω ω ω ω X ( ) ( )  ( )  ( ) U U U U 1 1 1 1 X X William Sandqvist william@kth.se

Filter LCR if time … (14.8) 1 ⋅ R ω ω j C R j C = ⋅ = ) ) || a b R C ω + ω 1 1 j C j RC + R ω j C R + ω + ω 1 1 U j RC j RC R = ⋅ = = 2 + ω ω + ω + R 1 ( 1 ) U j RC j L j RC R ω + 1 j L + ω 1 j RC   R U 1 = = ⇒ ω = ω = 2 2   0 RE RLC R − ω + ω 2 ( )   R RLC j L U LC 1   1 U R R U R C = = ω = = = = 2   2 ) c R − ω + ω 2   ( ) U R RLC j L U L LC L L + 1 1 0 j C C    R    U   = = − ° 2 ) arg   arg 90 d L     j U 1  C    1 1 I U I U = = ⇒ = ⋅ = R 2 R 2 ) ? e I R − ω + ω 2 ( ) U R U U R R RLC j L 1 1 1 William Sandqvist william@kth.se

William Sandqvist william@kth.se

Voltage ratio N 1 : N 2 Φ Φ d d = = U N U N 1 1 2 2 d d t t U N = 1 1 U N 2 2 William Sandqvist william@kth.se

Current ratio N 1 : N 2 = = ( , 0 ) P P P I 1 2 0 0 ⋅ = ⋅ ⇒ U I U I 1 1 2 2 I U N ≈ = 2 1 1 I U N 1 2 2 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 N 1 U 1 I 1 N 2 U 2 I 2 600 225 V 200 9 A ? ? Calculate the two values that are missing. I 1 and U 2 . William Sandqvist william@kth.se

Two values are missing ! (15. 1 ) For a transformer the following data was given: Primary Secondary N 1 U 1 I 1 N 2 U 2 I 2 600 225 V 200 9 A ? ? 75V 3A Calculate the two values that are missing. I 1 and U 2 . n = N 1 / N 2 = 600/200 = 3 1 225 1 9 = = = = = = 75 3 U n U I n I 2 1 1 2 3 3 William Sandqvist william@kth.se

Two values are missing ? (15. 2 ) For a transformer the following data was given: Primary Secondary . N 1 U 1 I 1 N 2 U 2 I 2 230 V 2A 150 12 A ? ? Calculate the two values that are missing. N 1 and U 2 . William Sandqvist william@kth.se