Memorial University of Newfoundland
Department of Physics and Physical Oceanography
Physics 2055 Laboratory
RC Filter Circuits
Introduction
In an AC circuit, Ohm’s law is amended to take into account the behaviour of capacitors
and inductors, which are characterized by a quantity known as impedance. The impedance,
Z, of a circuit is given by
Z = R + jX
(1)
√
where j = −1 and X is called the reactance. The modulus of Z is written as
|Z| =
q
(R2 + X 2 )
(2)
Ohm’s law then takes the form V = I|Z|.
A Capacitor and Resistor in a Simple Series Circuit
Consider the circuit shown in Fig (1). It can be shown that the ratio of the output voltage
to the input voltage (commonly called gain) is given by
Vout
R
=r
2
Vin
1
R2 + ωC
(3)
C
Vin
R
Vout
Figure 1: A simple RC filter circuit
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1. Construct the circuit using R ∼ 1 kΩ and C ∼ 47 nF (measure these first). Set the
signal generator to output a sine wave with amplitude 5 volts pk- pk. You will use the
oscilloscope to measure the voltage across the resistor. Why can’t you use a digital
meter here?
2. Before you take any readings, quickly increase the frequency from about 100 Hz to
100 kHz and observe how the voltage across the resistor varies with frequency. Is this
a high pass filter or a low pass filter?
3. Now do more careful measurements of voltage as a function of frequency and plot
the gain over the full range of frequencies used. Check that the input signal remains
constant by adjusting the amplitude control of the waveform generator as necessary.
Since the range of frequencies is very large, it is usual to plot this sort of graph on
a semilogarithmic scale, with frequency on the x-axis plotted on a logarithmic scale.
(Igor Pro users should select Modify Axis ... Mode: Log). It is not necessary to fit
your data to the functional form of Eq (3).
4. Swap the positions of the resistor and capacitor and obtain a second set of data, where
Vout is now the voltage measured across the capacitor. In this case the gain is given by
1/ωC
Vout
=r
2
Vin
1
R2 + ωC
(4)
Is this a high pass filter or a low pass filter? Plot this data on the same axes as the
first graph.
5. Explain your observations in terms of a potential divider, taking into account the
predictions of Eqs (3) and (4).
√
6. Where the two curves cross, VR = VC = Vin / 2. At this point the resistance of R
equals the reactance of C, i.e.
1
R = XC =
(5)
ωC
The crossover frequency ν◦ is therefore given by
ν◦ =
1
2πRC
Calculate ν◦ using Eq(6) and compare with the value obtained from your graph.
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(6)
Phase in an RC circuit
For the circuit you have just studied, take a closer look at the oscilloscope traces from vin
and vout . At some intermediate frequency, the two traces will differ in phase.
T
vin
vout
t
Figure 2: Phase shift in filter circuit
Here the output lags the input by time t, corresponding to a phase shift of
φ=−
t
× 360◦
T
The phase depends on frequency, which for this particular filter circuit is given by
tan φ = −ωCR
(7)
For the original circuit, the phase angle by which vout leads vin is given by
tan φ =
1
ωCR
(8)
where vout is the signal measured across the resistor.
1. Use the laboratory computers to plot graphs of phase angle versus frequency for both
circuits over the range 0–100 kHz predicted by Eqns (7) and (8). IgorPro users can
plot these functions by typing on the command line:
Make/O/N=500 wave0 wave1 // make space for 2 functions (500 points each)
SetScale/I x, 0, 10000, wave0 wave1 // plot function from x=0 to x=10000
wave0=<function 1>
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wave1=<function 2>
Display wave0 wave1
(you will need to provide your own functions for function1 and function2)
2. Establish that the phase difference from the oscilloscope traces agrees with your prediction. Do this for a few points only - one at very low and very high frequencies and
somewhere in the middle, and indicate these points on your computer-generated plots.
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