E1206 Embedded Electronics Le1 Le3 Le4 Le2 Ex1 Ex2 PC-block Documentation, Seriecom, Pulse sensor,, R, P, series and parallel KC1 LAB1 Pulse sensors, Menu program Start of program task Kirchhoffs laws Node analysis Two ports R2R AD Le5 Ex3 KC2 LAB2 Two port, AD, Comparator/Schmitt Le6 Le8 Ex6 Le13 Ex4 Ex5 Le10 Le7 Le9 Le11 Le12 Ex7 presentation wr. exam KC3 LAB3 Transients PWM Step-up, RC-oscillator Phasor jω PWM CCP CAP/ND-sensor KC4 LAB4 LC-osc, DC-motor, CCP PWM LP-filter Trafo + Guest lecturer presentation of program task Trafo, Ethernetcontakt
Phasor - vector ω 2π f X L ω L 1 X C ω C
What s inside the circuit? (11.4)
What s inside the circuit? (11.4)
Phasor chart??
Phasor chart 2 2 2 C R C R C R + 1 3 3 3 4 5 2 2 2 2 R R C R Now all values are known! The two voltages have 90 phase angle. Pythagorean theorem applies!??
Phasor chart 1 R 3 3V 5V
Phasor chart (11.6) 200 V, f 50 Hz, L 0,318 H, R 1 100 Ω, R 2 50 Ω.
Phasor chart (11.6) 200 V, f 50 Hz, L 0,318 H, R 1 100 Ω, R 2 50 Ω. X L ω L 2π 50 0,318 100 Ω
Phasor chart (11.6)
Phasor chart (11.6) Choose LR as reference phase ( horizontal ).
Phasor chart (11.6) The Välj current LR som R has riktfas the ( same horisontell direction ). as LR.
Phasor chart (11.6) The Strömmen Välj current LR som RL har lags riktfas samma 90 ( behind riktning horisontell LR som and ). has LR. an equally long pointer as R because R 1 and L has the same impedance. ( X L 100 Ω, R 1 100 Ω )
Phasor chart (11.6) The Strömmen Välj two currents LR som L ligger har riktfas R samma and 90 ( efter L riktning horisontell can be added LR och som ). har as LR lika vectors. lång to visare the current som. R eftersom is 2 longer R 1 och than Lhar R and samma L växelströmsmotstånd. (Pythagorean (X L 100 Ω, theorem R 1 100 applies!). Ω)
Phasor chart (11.6) Current De Strömmen Välj två strömmarna passes through LR som L ligger har riktfas samma 90 R och the ( efter riktning horisontell L lower kan adderas resistor LR och som ). har vektoriellt R LR lika. 2. The voltage lång visare till drop som strömmen R2 R eftersom gets the. same R 1 blir och direction 2 L ggr. har samma längre as. än växelströmsmotstånd. R eller L (enligt (X LR has L 100 pythagoras the Ω, length R 1 sats). 100 R 100, Ω) R2 has length 50. Because R 2 we get R2 LR / 2.
Phasor chart (11.6) Voltage Strömmen De Strömmen Välj två strömmarna can passerar finally LR som L ligger har riktfas samma 90 R genom be och determined ( efter riktning horisontell L den kan nedre adderas as the resistorn LR och som ). har vektoriellt vector Rsum LR lika. lång 2. visare till of LR and som Spänningsfallet strömmen R2. R eftersom. R 1 blir och R2 2 L får ggr. har samma samma längre riktning än växelströmsmotstånd. R eller som. L (enligt Phase (X LR har L 100 ϕpythagoras längden is Ω, the Rangle 1 R sats). 100, 100 between Ω) R2 har and längden. 50. The current after the Eftersom R 2 blir R2 LR / 2. voltage - inductive is the ratio between lenghts of and. character
Phasor chart (11.7) Draw the phasor chart for this circuit. At the frequency f applies that X C R and X L R/2. 2 is a suitable reference phase.
Phasor chart (11.7) Start with 2 as reference phase ( horizontal ).
Phasor chart (11.7) Current Börja med R has 2 the som same riktfas direction ( horisontel as 2. ).
Phasor chart (11.7) Current Strömmen Börja med C leads R 2 har som 90 samma riktfas before riktning ( 2 horisontel and som is equally 2 ).. long as R because X C R.
Phasor chart (11.7) Current Strömmen Börja med C and C R ligger 2 har som R samma are 90 summed riktfas före riktning ( 2 horisontel och to som. är lika 2 ).. stor som R eftersom is 2 times X C longer R. than C and R (Pythagorean theorem applies).
Phasor chart (11.7) Strömmarna Strömmen 1 leads 90 Börja med before C R ligger C 2 har och. som samma 90 riktfas R före summeras riktning ( 2 horisontel och som ihop är lika till 2 ).. stor. som R The eftersom är length 2 ggr. Xis C längre 1 R. X än L C 2 eller R R/2 R (enligt R R/ pythagoras 2 sats).
Phasor chart (11.7) Voltage Strömmarna Strömmen 1 ligger 90 1 and Börja med före 2 C R ligger C 2 har och. are som samma 90 summed riktfas R före summeras to voltage riktning ( 2 horisontel och som ihop är lika till. 2 ).. stor. som R Längden eftersom är 2 ggr. är X C längre 1 R. Xän L 2 C eller R R/2 R (enligt R R/ 2pythagoras sats).
Phasor chart (11.7) Spänningarna Strömmarna Strömmen 1 ligger 90 Börja med före 1 och C R ligger C 2 har och. som samma 90 2 summeras riktfas R före summeras ihop riktning ( 2 horisontel och som ihop till är lika till spänningen 2 ).. stor.. som R One eftersom can see Längden är 2 ggr. är X from C längre 1 R. the chart that becomes equal long to Xän L 2 C eller R R/2 R (enligt R R/ 2 1. The angle ϕ 0 and thereby and are in phase. pythagoras sats). nductive or capacitive character?
Complex numbers, jω-method C X C X L X L X R ω ω ω ω 1 j 1 j j j C C C C C L L L L L R R Complex OHM s law for R L and C. Complex OHM s law for. ] Re[ ] m[ arctan ) arg( ϕ f π ω 2
jω mpedance (12.2)
jω mpedance (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:
jω mpedance (12.2) 10 220 8,6 + 5j ( cos(30 ) + j sin(30 )) (8,6 (8,6 5j) 5j) 220 1892 1100j 99 19,1 11,1j
jω mpedance (12.2) 19,1 11,1j One possible solution is then a series circuit with R and C R C 1 19,1 Ω X C 11,1 ωc 1 287 µ F 2π 50 ( 11,1) Capacitor has negative reaktance. 19,1-11,1j
jω mpedance (12.2) Another possible solution is a parallel circuit with R and C one then thinks on as divided in to current composants R and C which are perpendicular to each other.
jω mpedance (12.2) R' X ' C R C cos30 sin 30 220 10 0,87 220 10 0,5 25,3 Ω 44 Ω X C 1 1 ' C' 72,6 µ F ωc' 2π 50 44
jω mpedance (12.2) s there any way to find out which of the two proposed circuits actually contain??
Complex impedance (12.6) Determine the complex impedance AB of this circuit.
Complex numbers calculation AB (15 + j20) (10 j20) 15 + j20 + 10 j20 550 j100 25 22 j4 [ 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 AB (15 + j20) (10 j20) 15 + j20 + 10 j20 ( 22 4i) Online Scientific Calculator
"heavy" calculations! (12.9) Calculate impedance. Calculate current. Calculate C (current branching). Calculate L (voltage divider).
Calculate impedance 2 ( 8j) (2 + 8j) R R C C 2 8j 2 + 8j 1,88 0,47 j 1 j R + X L + R C 3 + 6j+ (1,88 0,47 j) 4,88 + 5,53j 2 2 4,88 + 5,53 7, 38 Ω
Calculate current is reference phase, real. 30 (4,88 4,88 + 5,53j (4,88 5,53j) 5,53j) 146,5 165,9 j 2,7 3j 2 2 4,88 + 5,53 2,7 2 + 3 2 4 A
Calculate current C R 2 (2, 7 3j) + j 2 8j 2 C R 2 X C (2, 7 3j) 2 (2 + 8j) 0,86 + 0, 46j 2 8j (2 + 8j) + C 2 2 0,86 0, 46 0,98 A
L complex conjugate method? L jx L jx + L R C + R 1 30 6j+ 6j (1,88 0,47 j) + 3 6j (4,88 5,53j) 30 18,3 + 16,2j 4,88 + 5,53j (4,88 5,53j) L 18,3 2 + 16,2 2 24,4 V
Amount and phase ) arg( ) arg( ) arg( 2 1 2 1 2 1 2 1 + ) arg( ) arg( ) arg( 2 1 2 1 2 1 2 1
L amount phase method? Amount phase method, the polar form, often gives simpler calculations, but nowdays most math programs and pocket calculators handles complex numbers directly jx L 6j L 30 jx + + R 6j+ (1,88 0,47 j) + 3 L R C 1 6j 6 90 30 30 4,88 + 5,53j 2 2 5,53 4,88 + 5,53 arctan 4,88 6 30 (90 48,6 ) 24,4 41, 4 7,38 24,4 V L
Derive the complex current (12.7) Set up the complex current (with 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 (12.7) Set up the complex current (with 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. a c + + jb jd a + c + jb jd arg ( ) arg( a + jb) arg( c + jd ) Note!
Derive the complex current (12.7) Set up the complex current (With as the reference phase). ) 1 j( j j 1 j j ) j 1 ( C L R LR C L C L R L C R ω ω ω ω ω ω ω + + + + + LR C L C L R ω ω ω j ) 1 j( + + That it is which is the reference phase can be seen that we let the voltage be a real number! Sufficiently simplified!
Derive the complex current (12.8) Set up the complex current (With as the reference phase). Now it will be easier! The voltage is located directly across the parallel branch with the inductance L. (We need not concern ourselves with R and C) j jωl ωl