Outline
• We have moved to room 210. (Wed Oct 8
only we will use room 170)
• RC circuits
– Response to sine waves, cont’d
– High/low pass filters
recap
• Vin(t) = A cos(ω t)
• Assume that V1 = B cos(ω t + ϕ)
• Then
– ϕ = – tan–1 (RCω),
Vin
– B = A/[1+(RCω) 2 ] 1/2.
R
• RCL circuit(s) (L = inductance or coil)
Ohm’s Law of capacitor, revisited
• I = C dV/dt for sinusoidal signal
(V = V0 sinω t) is
• I = Cω V0 cosω t.
• Remember that for a resistor,
• I = (1/R) V.
• So ignoring the difference between sine and
cosine, RC = 1/Cω.
Can we express this “perpendicular”
idea by a imaginary number?
• Like RC = 1/iωC?
• Yes! This idea of resistance is called
impedance. We use Z instead of R to
signify that this is not the same as normal
resistance, and use j instead of i for the
imaginary number to avoid confusion with
“currents.” So, ZC = 1/jωC?
C
Then this circuit is a voltage
divider!
• V1 = Vin (1/Cω)/[R+(1/Cω)]
= Vin /[1+RCω]
Vin
• It should really be
Vin/[1+(RCω) 2] 1/2.
• The difference is like adding
two distances, which are
perpendicular to each other,
linearly (wrong) vs. properly.
R
RC = 1/C ω
More formal approach
• Consider that the input is Vin(t) = Aejωt =
A[cosω t + j sinω t], with an understanding
that the imaginary part should be ignored.
• Similarly, the solution is expressed like
V1(t) = Bej(ωt+ϕ) with the same understanding.
• The current through C is I(t) = C jωBej(ωt+ϕ) =
C jω V1(t). Differentiation = multiplication!
• This implies that the “resistance” of C is
1/jCω and dependent on the freq. of signal, ω.
1
Now this IS a voltage divider!
• V1(t) = [1/jCω]/{R + [1/jCω]} Vin (t)
= 1/{jRCω + 1} Aejωt .
• Really, we are talking about the real-part of
this expression.
• What is the magnitude of 1/{jRCω + 1}?
• What is the “phase” of 1/{jRCω + 1}?
• If you re-write 1/{jRCω + 1} = rejϕ what are
r and ϕ?
Low-pass filter
• In the previous circuit,
– ω→0
• ϕ = – tan–1(RCω) → 0, and B = A. i.e. no change in
signal
– ω → ∞ (RCω » 1)
• ϕ → – π/2 (–90°) and B « A
• Low freq. signal passes w/o modification
whereas high freq. signal is reduced.
Summary of RC
High-pass filter
• When one swaps R and C, the circuit becomes
high-pass filter.
• When RCω = 1, how much is the signal reduced?
• This is integrator
(t « RC)
• This is low-pass
filter (RCω «1)
Vin
R
C
– 1/√2 both in low- and high-pass filters.
• In the unit of dB, this is 20 log √2 reduction,
which is approx. 3, so this freq. is called 3dB
point.
• By the way, why do we have 20 in front of log to
get dB? [d ≡ 1/10!]
RCL circuits
•
•
•
•
What’s “L?”
Ohm’s Law of L?
LC circuit
RLC circuit(?)
• Differentiator
(t » RC)
• High-pass filter
(RCω »1)
Vin
C
R
V2
Physics of L
• What do you remember (from freshman
physics classes) about coil?
– Current, I, in a coil produce B field
– Change in B field (or flux) in a coil produce
EMF (voltage)
V = dΦ/dt, where Φ is flux:
→
→
Φ = ∫S B • d a (We will not worry too much
about the sign of the voltage here.
– Combining these two gives you, V ∝ dI/dt or V
= L dI/dt, where L is called “self inductance.”
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HW: estimate inductance
• Due next Monday, Sept. 15
• You want to make a 0.1 mH inductance (to filter
high-freq. noise for your sensitive circuit) using a
toroid of 5 mm outer radius and 2.5 mm inner
radius. The thickness of the toroid is 2 mm. How
many turns do you have to wind a wire to get the
desired inductance? The ferrite core material has
magnetic susceptibility of 100. Assume that the
cross section of the toroid is rectangle in shape (to
make computation a bit easier).
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