Electromagnetism
Physics 15b
Lecture #17
AC Circuit
Purcell 8.1–8.5
What We Did Last Time
LC circuit is a simple harmonic oscillator

L ~ mass, C ~ spring
ω=
RLC circuit is a damped oscillator

Behavior depends on R

Weak/strong/critical damping
+Q
C
LC -Q
1
I
L
0
AC voltage source
V = V0 cos ω t
P=
V0I0
= VrmsIrms
2
V
1
Today’s Goals
Introduce Impedance
Extension of resistance into AC circuits
R, L, and C (and combinations) can be treated together
  Catch: must do this in complex numbers


Use impedance to analyze simple AC circuits
1. Frequency filters

Combine R, L, or C to selectively remove (= filter out) signals with
high or low frequencies
2. Resonance circuit
Combine R + L + C to enhance signals near a particular frequency
  Quality Factor Q

Impedance
Impedance is a generalization of resistance

Drive “something” with an AC voltage source
  Could be an R, C, L or a combination
Z
I(t)
How does the current relate to the voltage?

For R, the answer is simple:
V0 cos ω t = R ⋅ I0 cos ω t

V0 = R ⋅ I0
V(t)
It would be nice if L and C were as simple as V0 = Z ⋅ I0
  Not possible as long as we keep using cosωt and sinωt
To simplify, we introduce a complex notation
V(t) = Re(V0e iω t ) = V0 cos ω t
V(t)
Omit “Re” as implicitly being there
2
Impedance of C
Q(t) = CV(t) = CV0e iω t
Let’s study C first
+Q
C
I(t) =
−Q
I

dQ(t)
= CV0 iω e iω t
dt
The (complex) current can be written as
I(t) = I0e iω t where I0 = iω CV0

V
If we take the real part
I(t) = −CV0ω sin ω t
Define the impedance of a capacitor by
Z(ω ) =
V0
I0
=
1
iω C
V0 = Z ⋅ I0
This is a frequency-dependent
imaginary number
Impedance of L
For an L connected to the same source
L
V(t) = V0e iω t = L
I
V
dI(t)
dt
V0
V
e iω t dt = 0 e iω t
∫
L
iω L
V0
  Therefore I =
0
iω L
V0
  Take the real part I(t) =
sin ω t
Lω
I(t) =
The impedance of an inductor is
V
Another frequency-dependent
Z = 0 = iω L
imaginary number
I0
3
Magnitude of Impedance
Impedance of L and C depends on the
frequency
Z
V0 = Z(ω )I0
I
|Z | V
Z L = iω L
ZR = R
ZC =
ω
Impedance varies depending
on the frequency ω
1
iω C
Phase of Impedance
Capacitor C
I(t) = Re(iω CV0e iω t )
⎛ V0 iω t ⎞
e ⎟
⎝ iω L
⎠
Inductor L I(t) = Re ⎜
=
= −ω CV0 sin ω t

Voltage lags the current by 90º
V(t)

V0
sin ω t
ωL
Voltage leads the current by 90º
I(t)
V(t)
I(t)
Both amplitudes and phases are expressed in the impedance Z
1
π
π
ZC =
→ arg(ZC ) = −
ZL = iω L → arg(ZL ) =
iω C
2
2
4
Combining Impedance
Total impedance of a network of resistors, capacitors, and
inductors can be calculated by adding up ZR, ZC, and ZL

Same rules as resistor addition — just in complex numbers
R
C
L
Z=
Z = R + iω L
1
R
1
R
=
+ iω C 1+ iω RC
R
Impedance summarizes the network’s AC characteristics
iω t
V
V(t) V0e
Z=
=
= 0
iω t
I(t) I0e
I0
Magnitude |Z| is the ratio between peak voltage and peak current
  Complex phase arg(Z) is the phase difference between voltage and
current

Low Pass Filter
Series R + C is driven by an AC voltage source
1
Total impedance is ZRC = R +
iω C
  Current is

V0e iω t
V0
I(t) =
=
e iω t
ZRC
R + 1 iω C
R
Vin = V0eiωt
C
Vout
The “output” voltage around C is
Vout (t) = ZCI(t) =


V0
V0
1
e iω t =
e iω t
iω C R + 1 iω C
iω RC + 1
At low frequency (ωRC << 1), Vout is close to Vin
At high frequency (ωRC >> 1), Vout approaches zero
This network passes low freq. and blocks high freq.
= a low-pass filter
5
Low Pass Filter
The frequency response of a simple low-pass filter is
Vout
Vin

=
1
1
=
iω RC + 1
(ω RC)2 + 1
In log-log plot, it’s two straight
lines joined by a round corner
The cut-off frequency is
ω=

1
RC
1
2π RC
= 12 Vin
f=
This is where Vout
1
0.1
0.01
0.001
0.01 0.1
1
10
100 ω RC
Above the cut-off, Vout drops inversely proportionally to ω
Such filter is called a first-order filter
  EE people call it a “−6 dB/octave” or “−20 dB/decade” filter

Other Filters
Low-pass and high-pass filters can be built with RC or RL


RC
low-pass
CR
high-pass
ω cutoff =
1
RC
LR
low-pass
RL
high-pass
ω cutoff =
R
L
All are first-order, ±6 dB/octave filters
Steeper filters are usually made with active electronics
  Nowadays often with digital signal processing
6
Series RLC Circuit
An AC voltage source drives R + L + C
iω t
  Current is common I = I e
R
0

L
C
VC
VL
Total voltage is
V = V0e iω t = VR +VL +VC
VR
Use impedance
⎛
1 ⎞ iω t
V0e iω t = ⎜ R + iω L +
Ie
iω C ⎟⎠ 0
⎝
Im
iωL
Total impedance
Ztotal
Remember: addition of impedance works
in the same way as resistance
  It just has to be done in the complex
plane
1/iωC

Re
R
Resonance
Im
Total impedance changes with ω
  ZL = iωL grows, ZC = 1/iωC shrinks
Magnitude is
Ztotal

⎛
1
1 ⎞
= R + iω L +
= R2 + ⎜ωL −
iω C
ω C ⎟⎠
⎝
iωL
Ztotal
2
R
Re
i/iωC
Minimum when ZL and ZC cancel each other
min Ztotal = R when iω L +
1
=0
iω C
ω0 =
1
Resonance
LC frequency
At the resonance


Impedance is minimum  Current is maximum
Impedance is real  Current and voltage are in phase
7
Resonance
Im
At the resonance
Impedance is minimum
 Current is maximum
  Impedance is real
 Current and voltage are in phase

iωL
Ztotal
R
Re
1/iωC
R = 2.5Ω
L = 5 µH
C = 2 nF
5Ω
I0
1
1
=
=
V0
Z
R 2 + ω L − ω1C
(
10Ω
)
2
Current peaks at the resonance
9
8
10
11x106 rad/s
Peak is higher and narrower for
smaller R
Phase
Phase of impedance tells us if V and I
are synchronized
⎛ ω L − ω1C ⎞
arg Z = arctan ⎜
⎟
⎝ R ⎠
Im
iωL
Ztotal
( )

It changes from –90º to +90º as ω crosses
the resonance
R = 2.5Ω
9
Re
1/iωC
Current leads voltage below the
resonance
5Ω
8
R
10Ω
10
11x106 rad/s
L = 5 µH
C = 2 nF
Current lags behind voltage
above the resonance
Phase shift is more abrupt for
smaller R
8
Quality Factor
How steep (or narrow) is the resonance?

Look at the frequencies where Im(Z) and Re(Z) are the same size
  That’s where I0 is 1/√2 times the peak value
ωL −

ω2 =
1
R2
R2 R4
+ 2± 3 + 4
LC 2L
L C 4L
Suppose the damping is weak, i.e., R  2 L C
ω2 ≈

1
=R
ωC
1 R
±
LC L
1
R
= ω 02 ± ω 0
LC
L
⎛
R ⎞
ω ≈ ω 0 ⎜ 1±
2ω 0L ⎟⎠
⎝
Difference between the two solutions ω1 and ω2 (ω1 > ω2) is
ω1 − ω 2
R
1
=
≡
Quality factor of the resonance circuit
ω0
ω 0L Q
The larger the Q, the narrower and steeper the resonance
What is Q?
Q is the relative width (in frequency) of the resonance

Ex: FM stations use 87.5–108 MHz signals separated by 200 kHz
 The radio’s tuning circuit must have Q > Qmin =
108 × 10 6 Hz
= 504
200 × 103 Hz
Think of the same RLC circuit as a weak-damped oscillator
− α t ± iω t
  The solution was Q(t) = e
e
R
1
R2
1 2L
=
α R
  In that time, it oscillates by

The decay time is τ =
ωτ ≈ ω 0
α=
2L
,ω=
LC
−
4L2
≈ ω0
2L
= 2Q radian
R
Q is (proportional to) the number
of oscillations it makes before dying
9
Summary
Impedance of R, C, and L
ZR =

VR
=R
IR
ZC =
VC
IC
=
1
iω C
ZL =
VL
= iω L
IL
Generally a frequency-dependent complex number
Low- and high-pass filters

Cut-off frequency ω cutoff =
1
R
or ω cutoff =
RC
L
Resonance of an RLC circuit ω 0 =
1
LC
Current amplitude peaks
  Phase between voltage and current changes by 180º

Quality factor Q
Q = ω0
= narrowness (steepness) of the resonance
  = how many “rings” a weakly damped oscillators make

L
R
10