Analog Circuits
Introduction:
Resistors, capacitors and inductors are commonly used components in analog circuits. In
various combinations, they can be used to build frequency filters and oscillators. In this
lab, you will learn about, build, and characterize high-pass and low-pass frequency filters,
and resonant circuits.
First, we need a bit of background:
A resistor is a component that dissipates electrical power when a current passes through
it. They are commonly made of carbon films or very thin wire. The voltage drop
induced by passing a current, I, through a resistance, R, is given by:
V = IR = R
!
dq
dt
(1)
A capacitor stores charge. The simplest capacitor consists of two parallel metal plates
separated by a thin insulator. When a constant voltage is applied across a capacitor, a net
positive charge forms on the surface of one metal plate, next to the insulator, and a
negative charge forms on the other. The capacitance of a parallel plate capacitor is given
by:
C=
"r"o A
d
(2)
where εr is the relative permittivity of the insulator, A is the area of (one of) the plates,
and d is the thickness of the insulator separating the plates. The charge stored on a
capacitor is given by:
!
q = CV
(3)
The current that flows into (or out of) a capacitor is therefore:
!
I=
assuming constant C
dq
dV
=C
dt
dt
(4) ,
An inductor is a coil of wire (like an electro-magnet). Energy is stored in the magnetic
field created by current!flowing through the coil. Ideally, the coil (wire) will have no
resistance, and therefore no voltage will drop across the coil when the current is constant.
There will, however, be a voltage drop when the current is changing (think Faraday’s
law/Lenz’s law). The voltage across an inductor (inductance L) is given by:
dI
d 2q
V =L =L 2
dt
dt
!
(5)
In all of the analysis below, we will use complex exponentials to represent sinusoidal
functions: e i" = cos " + isin " Believe it or not, this is much easier than using trig
functions and identities!
Low pass RC filter:
!
Consider the following circuit:
We know that the current through all points in the circuit must be the same, and that the
sum of the voltage drops across the resistor and capacitor must equal the applied voltage,
Vin:
Vin = VR + VC = IR + Vout
dV
dV
but IR = IC = C c = C out
dt
dt
dV
Vin = RC out + Vout
(6)
dt
We will consider the steady-state solutions when the applied voltage (and therefore the
output voltage) is a general sinusoidal function:
!
Vin = Ae i"t ,
Vout = Be i("t +# )
The amplitudes, A and B, are positive real numbers, and the possibility of a difference in
phase between Vin and Vout is considered by including an arbitrary phase shift, φ. We
!
want to find the output amplitude,
relative to the input (B/A), and the relative phase of the
output. Substituting into eqn (6), and solving for B/A:
B
e"i#
e"i[# +arctan($RC )]
=
=
A 1+ i$RC
1+ ($RC) 2
but A and B are both positive real number (and therefore so is B/A), so the angle in the
complex exponential must be equal to 0:
!
%
B
1
=
'
A
1+ ("RC) 2 &
# = $arctan("RC)'(
low pass RC filter (7)
Thus, we have the amplitude and phase of the output voltage of the low pass filter as a
function of frequency. Note that B/A=1 and φ=0 when ω=0, and B/A→0 and φ→-π/2 as
ω→∞. When!ωRC=1, (B/A)2=1/2. This is known as the corner frequency, or halfpower frequency, ωc=1/RC.
High Pass RC Filter:
This is the same circuit as the low pass filter, but we have defined the output voltage as
being the voltage drop across the resistor instead of the capacitor. By using the same
approach, we find that:
*
B
"RC
=
,
2
A
1+ ("RC) ,
+ high pass RC filter (8)
$ 1 ',
# = arctan&
)
% "RC ( ,-
Note that B/A=0 and phi=+π/2 when ω=0, and B/A→1 and φ→0 as ω→∞. Again, the
corner frequency, or half-power frequency, is given by ωc=1/RC.
!
Resonant LCR circuits:
Consider the following circuit:
Again, we know that the sum of the voltage drops across the three components must
equal the applied voltage. Using equations 1, 3 and 5:
L
d 2q
dq 1
+ R + q = Vin (t)
2
dt
dt C
(9)
First, let’s take a closer look at equation 9. It looks remarkably similar to the differential
equation for a driven mechanical oscillator with damping:
!
m
d2x
dx
+c
+ kx = Fapplied (t)
2
dt
dt
(10)
!
Important: We can understand the behaviour of the electrical LCR circuit as an
analogue of a mechanical oscillator by making the following substitutions:
Mechanical
Applied Force, F(t)
Displacement, x(t)
dx
Velocity,
dt
d2x
Acceleration, 2
dt
Mass
(inertia),
m
!
!
Electrical
Applied Voltage, V(t)
Charge, q(t)
dq
Current,
dt
d 2q
Rate of change of current, 2
dt
Inductance,
L
!
!
Spring (potential energy), k
1
1
,
capacitance C
Resistance, R
Damping, c
!
In the mechanical case, energy is converted between
potential energy in the spring,
kx 2
2
(e.g. when the spring is maximally compressed or extended and the velocity is 0) to
mv 2
kinetic energy,
(e.g. when the velocity is maximal and the spring is neither
2
!
compressed nor extended, x=x0) during every oscillation. Energy is dissipated by the
dashpot, c (like a shock absorber).
!
q2
In the electrical case, energy is converted between potential energy in the capacitor,
,
2C
LI 2
and kinetic energy in the inductor,
. Energy is dissipated in the resistor, R.
2
!
Again, to find the frequency response of the system, we will apply a general sinusoidal
input signal and assume the charge, q(t) will also be sinusoidal in time with some
! to the driving voltage,
arbitrary phase shift relative
Vin (t) = Ae i"t ,
q(t) = Be i("t +# )
Substituting into equation 9 we get
!
1
"# 2 LBe i(#t +$ ) + i#RBe i(#t +$ ) + Be i(#t +$ ) = Ae i#t
C
%
1(
Be i(#t +$ )' "# 2 L + i#R + * = Ae i#t
&
C)
or
B
=
A
e"i$
(11)
1
" # 2 L + i#R
C
Again, since both B and A are real numbers, the RHS of 11 must be real, so we can
immediately solve for the phase angle, φ, as a function of frequency:
!
&1
)
#( $ % 2 L + i%R+
'C
*
tan(" ) =
&1
)
,( $ % 2 L + i%R+
'C
*
&
)
( %R +
" = arctan(
+
1
( $ % 2L +
'C
*
In our experiment, we will not be actually measuring the charge, but we will be
measuring the voltage across one of the components. We can measure the voltage across
any of L, C, or R. Let’s assume we measure the voltage across the capacitor. Remember,
!
q=CV.
q(t)
C
1
B
Vc (t) = " Vin (t) " e i# so,
C
A
Vc (t)
1
=
2
Vin (t) 1$ % LC + i%RC
Vc (t) =
(12)
As in the mechanical oscillator, there are different response regimes, depending on the
magnitude of the!damping:
The magnitude of Vc will be a maximum when the denominator of equation 12 is a
minimum. Imagine that the damping, R, is very small. The denominator of 12 will be a
minimum when
1" # 0 2 LC = 0
1
#0 =
LC
" 0 is known as the resonant, or characteristic frequency of the circuit.
!
!
The maximum response, at resonance
is
" Vc (t) %
1
$
' =
2
#Vin (t) &max 1( ) 0 LC + i) o RC
=
1
(i L
=
i) o RC R C
where the factor –i indicates that Vc is π/2 out of phase with Vin.
!
Note that in the case that R=0 (no damping), the response is infinite at " 0 . This is
obviously not physical, as there is always some resistance in the circuit that keeps the
response finite. The response as a function of frequency is a peak. The width of the peak
is determined by the damping – the higher the damping, the wider the peak, and lower the
!
height of the peak.
This is called the under damped regime. The three regimes of operation, and the criteria
defining them are:
Under damped
R<2
L
C
Critically damped
R=2
L
C
R>2
L
C
!
Over damped
!
Resonant circuits (or mechanical oscillators) are often characterized by a dimensionless
“quality factor”, or “Q”, where
!
1 L
"quality factor"= Q =
R C
For under damping, Q>1/2, for critical damping Q=1/2, and for over damping, Q<1/2.
The quality factor has many other interpretations. For instance, in an under damped
oscillator, the quality!factor is equal to the resonant frequency divided by the ½-power
frequency width of the resonance peak (see figure). Note: power varies as V2
Q=
!
1 L "0
=
R C #"
Experiment:
Resonant LCR Circuit
Construct the LCR circuit using the fixed inductor provided. Construct such that channel
2 on the oscilloscope measures the voltage across the capacitor, remembering to connect
both negative connections to the same point in the circuit.
Select R=10Ω and C=0.01µF
Measure the amplitude and phase response (relative to the input) for frequencies up to 30
kHz. This time, the frequencies should be spaced equally on a linear scale. Make sure
you take enough data points to accurately reproduce the resonance peak!
Plot the amplitude and phase response on linear scale graphs. Determine the resonance
frequency, and use this to get an estimate of the inductance. What is the phase at
resonance? What is the quality factor of this circuit? What value of R to you get, if you
calculate it from the quality factor? How does this compare with the expected value?
Set the resistance decade box to 0 Ω. Measure the resonant frequency as a function of
capacitance for C=0.01 µF to 0.1 µF.
Plot C as a function of ω02 and use the slope of this function to determine the inductance.
Select C=0.01 µF. For each value of R between R=0 and R=100Ω, in steps of 20Ω,
measure the half-power frequencies at which the amplitude is equal to 1 2 times the
maximum amplitude (at ω0). The difference between these two frequencies is Δω, the
“full-width, half max” (FWHM) of the resonance peak. Plot Δω as a function of R, and
using the definition of quality factor in the introduction, estimate the value of the inductor
!
from the slope. What is the meaning of the non-zero intercept? Where is the “extra”
resistance coming from?
Low Pass Filter
Construct the low pass RC network using the decade resistance and capacitance boxes.
Use the function generator (set to sine wave; not square or triangle). Use the
oscilloscope to measure both the input waveform (on channel 1) and the output, Vc (on
channel 2). Make sure that the negative (black) connections from oscilloscope
channels 1 and 2 go to the same point in the circuit! (why?)
Select R=1kΩ, C=0.1 µF
Measure the phase and amplitude response from 30 Hz to 30 kHz. Choose your
frequencies so that, when plotted on a logarithmic scale, they will be approximately
equally spaced (think about this…)
Plot the amplitude response in dB (20 log(|Vout|/|Vin|)) vs frequency (in Hz) with
frequency on a log scale.
Plot the phase, in radians, vs frequency (in Hz) with the phase on a linear scale, and
frequency on a log scale.
Identify the frequency at which the power has decreased by a factor of 2 (ie,
(|Vout|/|Vin|)2=0.5). This is called the 3dB or corner frequency (why?), ωc, and determine
the slope of the response above the corner frequency in dB/decade (1 decade is one order
of magnitude in frequency). What is the phase shift at the corner frequency?
Keeping the capacitance constant at 0.1 µF, vary the resistance from 100 to 1000 Ω in
100 Ω steps. Measure the 3dB frequency at each resistance. Plot 1/ωc as a function of R
to verify the expected behaviour. Determine the value of C from the slope, and compare
with the expected value. What is the significance of the intercept (if any)?
Keeping the resistance constant at 100 Ω, vary the capacitance from 0.1 to 1 µF and
measure the 3dB frequency as a function of C. Plot and analyze as above.
High Pass Filter
Construct the high pass RC network using the decade resistance and capacitance boxes.
Use the function generator (set to sine wave; not square or triangle). Use the
oscilloscope to measure both the input waveform (on channel 1) and the output, Vr (on
channel 2). Make sure that the negative (black) connections from oscilloscope
channels 1 and 2 go to the same point in the circuit!
Select R=1kΩ, C=0.01 µF
Measure the phase and amplitude response from 30 Hz to 30 kHz. Choose your
frequencies so that, when plotted on a logarithmic scale, they will be approximately
equally spaced.
Identify the 3dB or corner frequency, ωc, and determine the slope of the response below
the corner frequency in dB/decade. What is the phase shift at the corner frequency?
Plot the amplitude response in dB (20 log (|Vout|/|Vin|)) vs frequency (in Hz) with
frequency on a log scale.
Plot the phase, in radians, vs frequency (in Hz) with the phase on a linear scale, and
frequency on a log scale.
Keeping the capacitance constant at 1 µF, vary the resistance from 100 to 1000 Ω in 100
Ω steps. Measure the 3dB frequency at each resistance. Plot 1/ωc as a function of R to
verify the expected behaviour. Determine the value of C from the slope, and compare
with the expected value. What is the significance of the intercept (if any)?
Keeping the resistance constant at 1000 Ω, vary the capacitance from 0.1 to 1 µF and
measure the 3dB frequency as a function of C. Plot and analyze as above.