E106 Embedded Electronics Le1 Le3 Le4 Le Ex1 Ex P-block Documentation, Seriecom, Pulse sensor,, R, P, series and parallel K1 LAB1 Pulse sensors, Menu program Start of program task Kirchhoffs laws Node analysis Two terminal RR AD Le5 Ex3 K LAB Two terminal, AD, omparator/schmitt Le6 Le8 Ex6 Le13 Ex4 Ex5 Le10 Le7 Le9 Le11 Le1 Ex7 presentation wr. exam K3 LAB3 Transients PWM Phasor ω PWM P AP/ND-sensor K4 LAB4 Step-up, R-oscillator L-osc, D-motor, P PWM LP-filter Trafo + Guest lecturer presentation of program task Trafo, Ethernetcontakt
Phasor - vector ω π f X L ω L 1 X ω
What s inside the circuit? (11.4)
What s inside the circuit? (11.4)
? R A instruments,, only shows the phasors magnitude all the rest you have to figure out yourself!?? We write omplex variables (magnitude and argument) as underlined,, while not underlined variables are the magnitude of the complex numbers. sing phasor charts one can interpret the magnitudes.
Phasor chart??
Phasor chart R R R + 1 3 3 3 4 5 R R 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) 00 V, f 50 Hz, L 0,318 H, R 1 100 Ω, R 50 Ω.
Phasor chart (11.6) 00 V, f 50 Hz, L 0,318 H, R 1 100 Ω, R 50 Ω. X L ω L π 50 0,318 100 Ω
Phasor chart (11.6)
Phasor chart (11.6) hoose LR as reference phase ( horizontal ).
Phasor chart (11.6) The Väl current LR som R has riktfas the ( same horisontell direction ). as LR.
Phasor chart (11.6) The Strömmen Väl current LR som R L 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 Väl two currents LR som L ligger har riktfas R samma and 90 ( efter L riktning horisontell can be LR added och som ). har as LR lika vectors. lång to visare the current som. R eftersom is longer R 1 och than L har R and samma L växelströmsmotstånd. (Pythagorean (X L 100 Ω, theorem R 1 100 applies!). Ω)
Phasor chart (11.6) urrent De Strömmen Väl två strömmarna passes through LR som L ligger har riktfas samma 90 R och the ( efter riktning horisontell L lower kan LR adderas resistor och som ). har vektoriellt R LR lika. The voltage. lång visare till drop som strömmen R R eftersom gets the. same R 1 blir och direction L ggr. har samma längre as. än växelströmsmotstånd. R eller L (enligt (X LR has L 100 pythagoras the length Ω, R 1 sats). 100 R 100, Ω) R has length 50. Because R we get R LR /.
Phasor chart (11.6) Voltage Strömmen De Strömmen Väl två strömmarna can passerar finally LR som L ligger har riktfas samma be 90 R genom och determined ( efter riktning horisontell L den kan nedre LR adderas as the resistorn och som ). har vektoriellt vector Rsum LR lika. lång. visare till of LR and som Spänningsfallet strömmen R. R eftersom. R 1 blir och R L får ggr. har samma samma längre riktning än växelströmsmotstånd. R eller som L. (enligt Phase (X LR har L 100 ϕ pythagoras längden is Ω, the Rangle 1 R sats). 100, 100 between Ω) R har and längden. 50. The current after the Eftersom R blir R LR /. voltage - inductive is the ratio between lenghts of and. character
Mathematica (11.6) 11.6.nb R R LR L
Phasor chart (11.7) Draw the phasor chart for this circuit. At the frequency f applies that X R and X L R/. is a suitable reference phase.
Phasor chart (11.7) Start with as reference phase ( horizontal ).
Phasor chart (11.7) urrent Böra med R has the som same riktfas direction ( horisontel as. ).
Phasor chart (11.7) urrent Strömmen Böra med leads R har som 90 samma riktfas before riktning ( horisontel and som is equally ).. long as R because X R.
Phasor chart (11.7) urrent Strömmen Böra med and R ligger har R som samma are 90 summed riktfas före riktning ( horisontel och to som. är lika ).. stor som R eftersom is times X longer R. than and R (Pythagorean theorem applies).
Phasor chart (11.7) Strömmarna Strömmen 1 leads 90 Böra med before R ligger har och. som samma 90 R riktfas före summeras riktning ( horisontel och som ihop är lika till ).. stor. som R The eftersom är length ggr. Xis längre 1 R. X än L eller R R/ R (enligt R R/ pythagoras sats).
Phasor chart (11.7) Voltage Strömmarna Strömmen 1 ligger 1 90 and Böra med före R ligger har och are. som samma 90 summed R riktfas före summeras to voltage riktning ( horisontel och som ihop är lika till. ).. stor. som R Längden eftersom är ggr. är X längre 1 R. Xän L eller R R/ R (enligt R R/ pythagoras sats).
Phasor chart (11.7) Spänningarna Strömmarna Strömmen 1 ligger 90 Böra med före 1 och R ligger har och. som samma 90 summeras R riktfas före summeras ihop riktning ( horisontel och som ihop till är lika till spänningen ).. stor.. som R One eftersom can see Längden är ggr. är X from längre 1 R. the chart that becomes equal long to Xän L eller R R/ R (enligt R R/ 1. The angle ϕ 0 and thereby and are in pythagoras phase. sats). nductive or capacitive character?
omplex numbers, ω-method omplex OHM s law for R L and. omplex OHM s law for. ] Re[ ] m[ arctan ) arg( ϕ X X L X L X R ω ω ω ω 1 1 L L L L L R R f π ω
ω mpedance (1.)
ω mpedance (1.) One can imagine that the phasor chart shows the complex plane, and then splitting the current phasor in real part and imaginary part:
ω mpedance (1.) 10 0 8,6 + 5 ( cos(30 ) + sin(30 )) (8,6 (8,6 5) 5) 0 189 1100 99 19,1 11,1
ω mpedance (1.) 19,1 11,1 One possible solution is then a series circuit with R and R 1 19,1 Ω X 11,1 ω 1 87 µ F π 50 ( 11,1) apacitor has negative reaktance. 19,1-11,1
ω mpedance (1.) Another possible solution is a parallel circuit with R and one then thinks on as divided in to current composants R and which are perpendicular to each other.
ω mpedance (1.) R' X ' R cos30 sin 30 0 10 0,87 0 10 0,5 5,3 Ω 44 Ω X 1 1 ' ' 7,6 µ F ω' π 50 44
ω mpedance (1.) s there any way to find out which of the two proposed circuits actually contain??
omplex impedance (1.6) Determine the complex impedance AB of this circuit. 1 ω 1 0,05 0 ωl + 0
omplex numbers calculation AB (15 + 0) (10 0) 15 + 0 + 10 0 550 100 5 4 [ Ω] Here the denominator directly was a real number so there no need to multiply with the complex conugate of the denominator, otherwise the calculations had been moore extensive
omplex numbers calculation AB (15 + 0) (10 0) 15 + 0 + 10 0 ( 4i) Online Scientific alculator
"heavy" calculations! (1.9) alculate impedance. alculate current. alculate (current branching). alculate L (voltage divider). 1 ω ωl + 6 1 0,15 8 1.9.nb Mathematica can be to help!
alculate impedance ( 8) ( + 8) R R 8 + 8 1,88 0,47 R + X L + R 1 3 + 6+ (1,88 0,47) 4,88 + 5,53 4,88 + 5,53 7, 38 Ω
alculate current is reference phase, real. 30 (4,88 4,88 + 5,53 (4,88 5,53) 5,53) 146,5 165,9,7 3 4,88 + 5,53,7 + 3 4 A
alculate current R (, 7 3) + 8 R X (, 7 3) ( + 8) 0,86 + 0, 46 8 (+ 8) + 0,86 0, 46 0,98 A
L complex conugate method? L X L X + L R + R 1 30 6+ 6 (1,88 0,47 ) + 3 6 (4,88 5,53) 30 18,3 + 16, 4,88 + 5,53 (4,88 5,53) L 18,3 + 16, 4,4 V
Amount and phase ) arg( ) arg( ) arg( 1 1 1 1 + ) arg( ) arg( ) arg( 1 1 1 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 X L 6 L 30 X + + R 6+ (1,88 0,47 ) + 3 L R 6 30 4,88 + 5,53 30 1 L 4,88 6 + 5,53 6 30 (90 48,6 ) 4,4 41, 4 7,38 4,4 V 90 arctan 5,53 4,88
Derive the complex current (1.7) Set up the complex current (with as reference phase). Note! One does not always have give the answer of the form a+b. 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 (1.7) Set up the complex current (with as reference phase). Note! One does not always have give the answer of the form a+b. 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 + + b d a + c + b d arg ( ) arg( a + b) arg( c + d ) Note!
Derive the complex current (1.7) Set up the complex current (With as the reference phase). ) 1 ( 1 ) 1 ( L R LR L L R L R ω ω ω ω ω ω ω + + + + + LR L L R ω ω ω ) 1 ( + + 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 (1.8) 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 ) ωl ωl
if time permits
Qualifying part on written exam! An A voltage with the frequency f 5,3 khz supplies an inductor L mh in paralell with a resistor R 80 Ω and in series with a capacitor 1 µf. One measures the voltage 3 V. a) alculate b) alculate LR (first LR ). c) alculate d) Scetch the phasor diagram for the circuit ( LR L R ).
Qualifying part on written exam! a) alculate 100 ma 0,1 10 1 10 5,3 3 1/ 3 ) 6 3 reference a π ω ω
Qualifying part on written exam! b) alculate LR (first LR ). 5,11 3,3 3,9 3,3 3,9 ) 39 (33 0,1 39 33 66,6 80 66,6 80 80 66,6 66,6} 10 10 5,3 { ) ( ) ( ) ( ) ( ) 3 3 + + + + + + + + + LR LR LR LR L L R L R LR L R L R L R L R L R L R b π ω ω ω ω ω ω ω ω ω ω
Qualifying part on written exam! c) alculate c) + LR 3+ 3,9 3,3 6,9 + 3,3 6,9 + 3,3 7,65 V
Qualifying part on written exam! d) Scetch the phasor diagram for the circuit ( LR L R ). d) L R R L LR ωl 3,9 3,3 0,049 + 0,059 66,6 0,049 + 0,059 0,076 76 ma LR 3,9 3,3 0,049 + 0,041 R 80 ( 0,049) + 0,041 0,064 64 ma