text: # â€؛ gan.1 â€؛ teaching â€؛ fall08 â€؛ lec4.pdf lecture 4: r-l-c circuits and...
  
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  • RLC series circuit: ● What's VR?
  
  ◆ Simplest way to solve for V is to use voltage divider equation in complex notation:
  
  ◆ Using complex notation for the apply voltage Vin = V0cosωt = Real(V0 e jωt ):
  
  ■ We are interested in the both the magnitude of VR and its phase with respect to Vin.
  
  ■ First the magnitude:
  
  K.K. Gan 1
  
  Lecture 4: R-L-C Circuits and Resonant Circuits
  
  RL C V = V
  
  X
  
  in cosω0
  
  XL C
  
  t
  
  VR = VinR
  
  R+ XC + XL
  
  = VinR
  
  R+ 1 jωC
  
  + jωL
  
  VR = V0e
  
  jωt R
  
  R+ j ωL − 1 ωC
  
    
  
    
  
  VR = V0e
  
  jωt R
  
  R+ j ωL − 1 ωC
  
    
  
    
  
  = V0R
  
  R2 + ωL − 1 ωC
  
    
  
     2
  
  K.K. Gan 1 L4: RLC and Resonance Circuits
  
  • K.K. Gan 2
  
  ■ The phase of VR with respect to Vin can be found by writing VR in purely polar notation. ❑ For the denominator we have:
  
  ❑ Define the phase angle φ :
  
  ❑ We can now write for VR in complex form:
  
  ■ Finally, we can write down the solution for V by taking the real part of the above equation:
  
  R+ j ωL − 1 ωC
  
    
  
     = R2 + ωL −
  
  1 ωC
  
    
  
     2 exp j tan−1
  
  ωL − 1 ωC
  
  R
  
    
  
    
  
   
  
   
  
   
  
   
  
  VR = VoR e
  
  jωt
  
  e jφ R2 + ωL − 1 ωC
  
    
  
     2
  
  = VR e j(ωt−φ)
  
  tanφ = Imaginary X Real X
  
  = ωL − 1
  
  ωC R
  
  VR = Real V0R e
  
  j(ωt−φ)
  
  R2 + ωL − 1 ωC
  
    
  
     2
  
  = V0R cos(ωt −φ)
  
  R2 + ωL − 1 ωC
  
    
  
     2
  
  L4: RLC and Resonance Circuits
  
  Depending on L, C, and ω, the phase angle can be positive or negative! In this example, if ωL > 1/ωC, then VR(t) lags Vin(t).
  
  • K.K. Gan 3
  
  ◆ VR
  
  • K.K. Gan 4
  
  1e-1 1e-7
  
  100
  
  V = V cos t0in ω
  
  Vout
  
  K.K. Gan 4
  
  1e-1 1e-7
  
  100
  
  inV = Vcosω t
  
  Vout
  
  φ(- )
  
  L4: RLC and Resonance Circuits
  
  Bode plot of magnitude of VR/Vin vs. frequency
  
  • K.K. Gan 5
  
  ◆ In general VC(t), VR(t), and VL(t) are all out of phase with the applied voltage.
  
  ◆ I(t) and VR(t) are in phase in a series RLC circuit.
  
  ◆ The amplitude of VC, VR, and VL depend on ω.
  
  ◆ The table below summarizes the 3 cases with the following definitions:
  
  ● RLC circuits are resonant circuits ◆ energy in the system "resonates" between the inductor and capacitor ◆ “ideal" capacitors and inductors do not dissipate energy ◆ resistors dissipate energy i.e. resistors do not store energy
  
  Gain Magnitude Phase VR/Vin R/Z -φ VL/Vin ωL/Z π/2 - φ VC/Vin 1/ωCZ -π/2 - φ
  
  K.K. Gan 5
  
  Z = R2 + (ωL −1/ωC)2[ ] 1/2
  
  tanφ = (ωL −1/ωC) /R
  
  L4: RLC and Resonance Circuits
  
  • K.K. Gan 6
  
  ● Resonant Frequency: ◆ At the resonant frequency the imaginary part of the impedance vanishes. ◆ For the series RLC circuit the impedance (Z) is:
  
  ◆ At resonance (series, parallel etc):
  
  ◆ At the resonant frequency the following are true for a series RLC circuit: ■ |VR| is maximum (ideally = Vin) ■ φ = 0 ■
  
  ☞ The circuit acts like a narrow band pass filter.
  
  ● There is an exact analogy between an RLC circuit and a harmonic oscillator (mass attached to spring):
  
  ◆ x ⇔ q (electric charge), L ⇔ m, k ⇔ 1/C ◆ B (coefficient of damping) ⇔ R
  
  K.K. Gan
  
  Z = R+ XL + XC = R+ j(ωL −1/ωC)
  
  Z = R2 + (ωL −1/ωC)2[ ] 1/2
  
  ωL =1/ωC
  
  ωR = 1 LC
  
  VC Vin
  
  = VL Vin
  
  = L
  
  R C (VC or VL can be > Vin!)
  
  m d 2x dt2
  
  +B dx dt
  
  + kx = 0 damped harmonic oscillator
  
  L d 2q dt2
  
  + R dq dt
  
  + q C
  
  = 0 undriven RLC circuit
  
  6 L4: RLC and Resonance Circuits
  
  • K.K. Gan 7
  
  ● Q (quality factor) of a circuit: determines how well the RLC circuit stores energy ◆ Q = 2π (max energy stored)/(energy lost) per cycle ◆ Q is related to sharpness of the resonance peak:
  
  K.K. Gan 7 L4: RLC and Resonance Circuits
  
  • K.K. Gan 8
  
  ◆ The maximum energy stored in the inductor is LI2/2 with I = IMAX. ■ no energy is stored in the capacitor at this instant because I and VC are 900 out of phase.
  
  ■ The energy lost in one cycle:
  
  ■ There is another popular, equivalent expression for Q
  
  ❑ ωU (ωL) is the upper (lower) 3 dB frequency of the resonance curve. ❍ Q is related to sharpness of the resonance peak. ❑ Will skip the derivation here as it involves a bit of algebra. ❍ two crucial points of the derivation:
  
  K.K. Gan 8
  
  power × (time for cycle) = IRMS 2 R× 2π
  
  ωR = 12 Imax
  
  2 R× 2π ωR
  
  Q = 2π LIMax
  
  2
  
  2
  
   
  
   
  
  2π ωR
  
  RIMax 2
  
  2
  
   
  
   
  
  = ωRL R
  
  Q = ωR ωU −ωL
  
  VR Vin
  
  = 1
  
  1+Q2 ω ωR
  
  − ωR ω
  
   
  
   
  
  2
  
  Q ω ωR
  
  − ωR ω
  
   
  
    = ±1
  
  L4: RLC and Resonance Circuits
  
  at the upper and lower 3 dB points
  
  • K.K. Gan 9
  
  ● Q can be measured from the shape of the resonance curve ■ one does not need to know R, L, or C to find Q!
  
  ● Example: Audio filter (band pass filter)
  
  ■ Audio filter is matched to the frequency range of the ear (20-20,000 Hz).
  
  K.K. Gan 9
  
  Q = ωR ωU −ωL
  
  L4: RLC and Resonance Circuits
  
  • K.K. Gan 10
  
  ● Let's design an audio filter using low and high pass RC circuits.
  
  ◆ Ideally, the frequency response is flat over 20-20,000 Hz, and rolls off sharply at frequencies below 20 Hz and above 20,000 Hz. ■ Set 3 dB points as follows: ❑ lower 3 dB point : 20 Hz = 1/2πR1C1
  
  ❑ upper 3 dB point: 2x104 Hz = 1/2πR2C2
  
  ■ If we put these two filters together we don't want the 2nd stage to affect the 1st stage. ❑ can accomplish this by making the impedance of the 2nd (Z2) stage much larger than R1. ❑ Remember R1 is in parallel with Z2.
  
  ■ In order to insure that the second stage does not "load" down the first stage we need: R2 >> R1 since at high frequencies Z2 ⇒ R2
  
  ■ We can now pick and calculate values for the R's and C's in the problem. ❑ Let C1 = 1 µF ⇒ R1 = 1/(20Hz 2πC1) = 8 kΩ ❑ Let R2 > 100R1 ⇒ R2 = 1 MΩ, and C2 = 1/(2x104 Hz 2πR2) = 8 pf
  
  ☞ R1 = 8 kΩ, C1 = 1 µF R2 = 1 MΩ, C2 = 8 pf
  
  K.K. Gan 10
  
  Z1 = R1+1/ jωC1 Z2 = R2 +1/ jωC2
  
  L4: RLC and Resonance Circuits
  
  • ◆ Exact derivation for above filter: ◆ In the above circuit we treated the two RC filters as independent. ◆ Why did this work? ◆ We want to calculate the gain (|Vout/Vin|) of the following circuit:
  
  ■ Working from right to left, we have:
  
  ❑ ZT is the total impedance of the circuit as seen from the input. ❑ Z1 is the parallel impedance of R1 and R2, in series with C2.
  
  ■ Finally we can solve for the gain G = |Vout/Vin|:
  
  K.K. Gan 11 K.K. Gan 11
  
  Vout =VaX2 /(X2 + R2 ) Va =VinZ1 /ZT
  
  Z1 = R1 R2 + X2( ) R1 + R2 + X2
  
  ZT = X1 + Z1
  
  Va = VinR1 R2 + X2( )
  
  X1 R1 + R2 + X2( )+ R1 R2 + X2( )
  
  Vout Vin
  
  = R1X2
  
  X1 R1 + R2 + X2( )+ R1 R2 + X2( )
  
  L4: RLC and Resonance Circuits
  
  • K.K. Gan 12
  
  ■ We can relate this to our previous result by rewriting the above as:
  
  ■ If we now remember the approximation (R1
  
  • K.K. Gan 13
  
  ● Another Example: Calculate |I| and the phase angle between Vin and I for the following circuit:
  
  ■ First calculate |I|. ◆ The total current out of the input source (I) is related to Vin and the total impedance (ZT) of the circuit by Ohm’s law: ◆ The total impedance of the circuit is given by the parallel impedance of the two branches:
  
  ◆ Putting in numerical values for the R's and X's we have:
  
  ◆ We can now find the magnitude of the current:
  
  K.K. Gan 13
  
  I =Vin /ZT
  
  1/ZT =1/Z1+1/Z2 Z1 = R1+ X1 Z2 = R2 + X2
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