Resonant Circuits - Lecture Notes - Circuits II | EE 221, Study notes of Electrical and Electronics Engineering

Download Resonant Circuits - Lecture Notes - Circuits II | EE 221 and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! Lecture 15: Resonant Circuits Vijay Singh∗ April 6, 2003 Abstract Resonance in RLC circuits. 1 Resonance in RLC circuits V1 I R L C VR VC VL + + + + − − − − Figure 1: Series RLC circuit. V1 G L I + − IG IC IL C Figure 2: Parallel RLC circuit. Z(jω) — Input impedance of the series RLC circuit. Y (jω) — Input admittance of the parallel RLC circuit. Z(jω) = R + jωL + 1 jωC (1) Y (jω) = G + jωC + 1 jωL (2) Z(jω) = real when ωL = 1 ωC (3) Y (jω) = real when ωL = 1 ωC (4) ∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky, Lexington, KY, USA. E-mail: vsingh@engr.uky.edu. Document prepared by Ramprasad Potluri using LATEX. Figures created using Metagraf. 1 Equations (3) and (4) are satisfied when ω = ω0 = 1√ LC ∆ = Resonant Frequency (5) At ω = ω0, Z = R, and Y = G (6) |Z| R ω0 (ωL− 1ωC ) ω Figure 3: Frequency response of series RLC circuit. |Y | G ω0 (ωC − 1ωL ) ω Figure 4: Frequency response of parallel RLC circuit. 2 Phasor diagrams 2.1 Series RLC circuit 1. For ω < ω0, ωL < 1 ωC , |jωLI| < ∣∣∣ IjωC ∣∣∣, |VL| < |VC | (Figure 5). 2. For ω > ω0, ωL > 1 ωC , |jωLI| > ∣∣∣ IjωC ∣∣∣, |VL| > |VC | (Figure 6). 3. At ω = ω0, ωL = 1 ωC , |jωLI| = ∣∣∣ IjωC ∣∣∣, |VL| = |VC | (Figure 7). Note that |VC | can be much greater than |V1| [p.443 (text)]. 2.2 Parallel RLC circuit IC = jωV1, and IL = V1 jωL . 1. At ω = ω0, ωC = 1 ωL , |IC | = |IL|. IG and V1 are in phase. I and V1 are in phase (Figure 8). 2. At ω < ω0, |jωCV1| < ∣∣∣ V1jωL ∣∣∣, |IC | < |IL|. I lags V1 (Figure 9). 3. At ω > ω0, |jωCV1| > ∣∣∣ V1jωL ∣∣∣, |IC | > |IL|. I leads V1 (Figure 10). 2 2.5 Computation of ∆ω Finding ωL0 and ωH0: We know that at ω = ωL0 and ω = ωH0, |M(ω)| = 1√2 . Thus, using Equation (23), we get:   1 1 + Q2 ( ω ω0 − ω0 ω )2   1/2 = 1√ 2 (16) =⇒ 1 + Q2 ( ω ω0 − ω0 ω )2 = 2 =⇒ Q ( ω ω0 − ω0 ω )2 = ±1 at ω = ωL0, ω = ωH0 =⇒ Q ( ω2 − ω20 ω0ω ) = ±1 (17) So, Qω2 −Qω20 ± ωω0 = 0 or, ω2 ± ωω0 Q − ω20 = 0 ω = ± ω0 2Q ± √( ω0 Q )2 + 4ω20 = ω0 2Q ± ω0 √( 1 2Q )2 + 1 (18) Taking the positive values, we get: ωH0 = ω0  + 1 2Q + 1 2 √( 1 2Q )2 + 1   (19) Taking the negative values, we get: ωL0 = ω0  − 1 2Q + √( 1 2Q )2 + 1   (20) BW = ∆ω = ωH0 − ωL0 = ω0 Q (21) Mutliplying equations (19) and (20), we get: ωH0 × ωL0 = ω20 [ − 1 4Q2 + ( 1 4Q2 + 1 )] (22) = ω20 =⇒ ω0 = √ωL0ωH0 (23) 5 For the parallel RLC circuit of Figure 2: IS = IG + IC + IL = VS [ G + jωC + 1 jωL ] = VS [ G + j ( ωC − 1 ωL )] When the circuit is in resonance: IS = GVS 2.6 Bode plot of a series RLC circuit From Figure 1, we know: Y (jω) ∆ = Input admittance = 1 R + jωL + 1 jωC = jωC 1 + (jω)2LC + jωRC (24) Denominator of Equation (24) reminds us of the standard form of a quadratic pole: (jωτ)2 + 2ξτjω + 1 (25) where, τ = 1/ω0. Expression (25) can also be written as: (jω)2 ω20 + 2ξ ω0 jω + 1 (26) Comparing Expression (26) with the denominator of Equation (24), we get: ω20 = 1 LC (27) 2ξ ω0 = RC (28) From Equations (27) and (28), we get: ξ = R 2 Cω0 = R 2 C 1√ LC = R 2 √ C L (29) We also know that Q = ω0L R = 1√ LC L R = 1 R √ L C (30) From equations (29) and (30), we have: Q = ξ 2 (31) Thus, a quadratic pole can be realized from an RLC circuit. In many cases, we would like to avoid the inductor altogether. 6