# 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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