Electricity and Magnetism-Filters and AC circuits

Topics: part of the course? Power and energy Filters 8.022 (E&M) – Lecture 18 RCL circuits: the hardest of the easiest More on complex impedance 1 2 Work wiThe End. 􀀄 􀀄 () () XV t =􀀄 􀀄 R C L X 1i Z = iC Z i= L ω ω ⎧ ⎪⎪⎨ ⎪ ⎪⎩ G. Sciolla – MIT 8.022 – Lecture 17 Last time: AC driven RCLs Simple solution when introducing following rules: th complex V and I Real currents and voltages are just the real part of the Generalization of Ohm’s law to complex V and I: Analyze circuit as if it were DC with only resistors Take the real part of I(t) and V(t) V and I. I tZ where Z is the of componen Z = R mpedancet X: 3 “Analyze as DC with only resistors” I1Z1=V1; I2Z2=V2; V1+V2Æ Zeq=Z1+Z2 V1/Z1=V2/Z2=V/Zeq; V1=V2=V Æ 1/Zeq=1/Z1+1/Z2 Æ Z1 Z2 ~ Z1 Z2 ~ G. Sciolla – MIT 8.022 – Lecture 17 What do I mean with this statement? Impedances in series Same current flowing in each element =V; V=ZI Impedances in parallel Same voltage drop across each element Same rules as resistors in series and parallel! 4 Is the current leading or lagging? () () CV t =􀀄 􀀄 Z R 1/ωC ωL Z V Ι φ=-φΖ G. Sciolla – MIT 8.022 – Lecture 17 Instead of thinking of the problems in terms of complex currents, think in terms of complex impedance! Generalized Ohm’s law: All what we really care about is amplitude of I and relative phase between I and V Trick: let’s choose V real (no law against it!) and draw the complex I, V and Z in the complex plane I tZ I=|V|/|Z| 25 Is current leading or lagging? (2) lImaginary part: ZL “pulls up” by ωC pullωC Æ ZL>ZC Æ φΖ>0 ZL0, I will be lagging V<0, I will be leading V 3G. Sciolla – MIT 8.022 – Lecture 17 7 Power in RCL circuits Given NB: when we say light bulb has a P of 100W we are referriα−β)=cosαcosβαsinβ ()Pt = 0 0 () () ) Vt V t It I t ω ω φ =⎧ ⎨ = −⎩ 0 0 2 0 1 )2 )2| | T T T P V tI tT V t tZ ω ω ω φπ ω ω ω φπ = = − = = − ∫ ∫ ∫ 2 0 2 2 0 0 cos sin2| | VP t t t t dtZ ω ω π πω ω ω φ ω ω φπ ⎡ ⎤ = +⎢ ⎥ ⎣ ⎦∫ ∫ Power delivered in a circuit is The average power over a period T will be ng to

Using the identity: cos(+sinwe obtain: () ( ) V t I t cos cos( () ( ) co s co s( co s cos( V t I t dt dt dt cos cos cos sin dt 8 Power in RCL circuits (2) Since: φ=0 Æ iiÆ φ=0 when φo Æ Introducing: iÆ remembering that 22 20 0 2 0 1 12 2 2| ( sin 02 VP Z t tω π ω π ω ωπ φωω ω ωπ ⎧ =⎪⎪ ⇒ =⎨ ⎪ = ⎪⎩ ∫ ∫ 2 2cos|( RMS VP Z φ ωω = = 0 0and 2 2 V IV I= = cos R Zφ ω = G. Sciolla – MIT 8.022 – Lecture 17 NB: Power depends on relative phase between I and V cosno power disspated in the crcuit no work done! cos= 90when Z is purely imaginary: R needed! RMS (root mean squared) voltage and currents: NB: in the US: outlet voltage s 120 V. This is the RMS voltage: Vmax=170 co s cos ) | co s tdt dt ( ) ) | RMS RI RMS RMS |( ) | 49 Power vs. frequency ω Æ At what ω ω is the max P? 2 2 2 2 2|( 1 V VP R RZ R L C ω ω ω = = ⎛ ⎞+ −⎜ ⎟⎝ ⎠ P(ω) ω 0 1 10L C LC ω ω ωω− = ⇒ = = behavior! 2VP R = G. Sciolla – MIT 8.022 – Lecture 17 NB: Z depends on power dissipated depends on driving frequency! is P is max? What What is the corresponding phase? Zero: the imaginary part due to C and L exactly cancel out! ) | RMS RMS Resonant max RMS 10 ω0 in term of L and C 0 0 1 1L CLC ω ω ω = ⇔ = Z R1/ωC ωL The imaginary part due to C exactly Æ G. Sciolla – MIT 8.022 – Lecture 17 What does ω=ωmean in terms of L and C? Remember: Back to the phasor representation for Z compensates the one due to L Z is purely real! 5G. Sciolla – MIT 8.022 – Lecture 17 11 How good is the resonant system? Width: ∆ω power goes to Pmax/2: ω1 ω2 P(ω) ωω1ω2 2 2 2 2 1 21 V VR L RR CR L C ω ωω ω = ⇒ − = ± ⎛ ⎞+ −⎜ ⎟⎝ ⎠ 1 2 1 1 1 2 2 2 2 2 1 1 0 1 1 0 L RC LCL RC ω ω ω ω ω ωω ω ⎧ − = −⎪ ⎧ + − =⎪ ⎪ ⇒⎨ ⎨ − − =⎪⎪ ⎩− = ⎪⎩ 2 2 1 2 12 2 0 2 4 2 4 2 res RC RLC LRC L L R C Q ω ω ω ω ω ω ω ω ⎧ − ± + = =⎪⎪ ⇒∆ = − = ⇒⎨ ± +⎪ = = = ∆ ⎪⎩ l Definition: width of resonance wrt the height between the points where the and RMS RMS LC RC RC R C LC R C LC unphysica12 Application: FM antenna L=8.22 µH -12 F Ω VRMS=9.13 µV 8 0 0 0 0 1 10 2 i2 x LC MHz ω ωω ν π = = = ⇒ = = L C R G. Sciolla – MIT 8.022 – Lecture 17 Consider the following circuit: C=0.27 pF=0.27x10R=75 The radio signal in the air induces an alternated emf in the antenna: Find frequency of incoming wave for which antenna is in tune Resonance frequency: 6.7 106 YES, FM rado! πν 613 Application: FM antenna (cont) L=8.22 µH -12 F R=75 Ω VRMS=9.13 µV Calculate I∆VRMS 0 0 0 II = (| |2 V V NBZ R = = = L C R C C S R C i1Zil=C iRMSV mVR µ ω = lly llillll ciii⇒ G. Sciolla – MIT 8.022 – Lecture 17 C=0.27 pF=0.27x10RMS across C RMS : at resonance |Z |=R) RMS RMS RM MS Queston: V =0.66 V=I mV whe V =9 V. How 0.66 can ths happen? L and C cancel amost perfectZ can be sma whe C and L are large and Z~rea. NB: arcuts wth good Q value have this feature! 14 Application: FM antenna (cont) Q factor 6R = L MHz ωω ν π ∆∆ = = = 2 i L C R i il ili⇒ 0res LQ R ω ω ω = = = ∆ ii iG. Sciolla – MIT 8.022 – Lecture 17 Calculate width of resonance 9 10 1.4 Q: is this a good antenna? ⇒∆ Can wencrease L? No,t woud change frequency decreasing Rs the souton 73 good but not enough for a radio. How can this be improved? No, since separaton between statonss ~ 0.2 MHz 7MIT 8.022 – Lecture 17 15 Low pass RL filter 0 2 2 2|| V VI Z V VI R L ω ω ω = = ⇒ + = = + + 􀀄 􀀄 􀀄 􀀄 􀀄 2 R 0 R 0 2 2 V il: V 0 R VRV IR R L ω ω ω → →⎧ ⇒ ⇒⎨ →∞ →⎩ = = + G. Sciolla – MIT 8.022 – Lecture 17 RCL circuits have a frequency dependent response: they can act as filters (select only certain frequencies) Example: RL circuit Calculate the complex current Ri L Ri L 0: V low pass fter G. Sciolla – 16 High pass RL filter L Same complex current 0 2 2 2|| V VI Z V VI R L ω ω ω = = ⇒ + = = + + 􀀄 􀀄 􀀄 􀀄 􀀄 0 2 2 2 0 0 L 2 2 2 0 0 L 02 2 2 0: hiil: V 0 V L LVV R L LV LV RR L LV LV VLR L ωω ω ω ω ω ω ω ω ω ω ⎧ →⎪ ⎪⎪ ⇒ ⇒⎨ →∞⎪ ⎪ ⎪ = = + → → → + → → → + ⎩ What if we take the voltage Vacross the inductor? Ri L Ri L gh pass fter LI 817 Low pass RC filter it 0 2 2 2 || 1 V VI iZ R C V VI iR RC C ω ω ω = = ⇒ − = = − + 􀀄 􀀄 􀀄 􀀄 􀀄 0 0 2 2 2 2 2 2 R 0 R: V 01 1C V I VVCV C CRR C ω ω ω ω ω ω → →⎧ ⇒ ⇒⎨ →∞ →⎩ = = = ++ Demo on filters G. Sciolla – MIT 8.022 – Lecture 17 Let’s now study the voltage across a capacitor of a driven RC circuThe complex current is now: 0: V low pass filter 18 High pass RC filter R Same complex current 0 2 2 2 || 1 V VI iZ R C V VI iR RC C ω ω ω = = ⇒ − = = − + 􀀄 􀀄 􀀄 􀀄 􀀄 0 0 2 2 2 0 2 2 R 2 R 0 hiil1 1 r: VR V CRVV RI CR VR C ω ω ω ω ω → →⎧ ⇒ ⇒⎨ →∞ →⎩ = = = ++ G. Sciolla – MIT 8.022 – Lecture 17 What if we take the voltage Vacross the resistor? 0: V gh pass fte 919 Summary and outlook Today: End of RCL circuits Next time: G. Sciolla – MIT 8.022 – Lecture 17 Some tricks to make RCL calculations easier Power dissipated in RCL circuits Antennas and high and low pass filters Back to Maxwell’s equation: The missing ingredient! 10