Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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Simple DC and AC Circuits: Using the oscilloscope and function
generator.
Objectives:
To become familiar with building circuits using the bread-board prototyping kit,
to learn some of the uses of the oscilloscope and function generator. We also
will study some characteristics of simple dc and ac circuits made with resistors
and capacitors and attempt to verify Thevenin’s theorem for a simple example.
Equipment required:
1. Oscilloscope (Tektronix TDS210, dual channel 60 MHz digital scope)
2. Function Generator (BK Precision 4011A which outputs sine, triangle
square wave (with variable offset) and logic level signals up to a frequency
of 5 MHz.
3. Breadboard kit (PB-503)
4. Digital Multimeter (Fluke 187)
5. Resistors: 1kΩ (2), 4.7 kΩ, 10 kΩ (3)
6. Capacitors: 1 µ F (1) (pink), 0.01 µ F (1) (green)
7. quick-clips (2)
8. BNC-to-banana cable (1)
Exercise 1: The voltage divider
Construct the simple voltage divider shown below. Apply an input dc voltage
of + 15 V (provided by the breadboard PB-503 power supply). Measure the
output voltage as shown, and again after you connect another 10 kΩ resistor
across the output. Why does the measured output voltage change?
Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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Now connect a sine wave signal voltage (at a frequency of, say, 1 kHz) to
the input (remove the dc voltage first!) from the function generator. Use the
oscilloscope to adjust the peak-to-peak voltage amplitude on the input to be +
15 V. Measure the output (peak-to-peak) voltage for the two cases shown above.
You can try it with a square or triangle wave too but the result should be the
same.
The voltage you measure without any load resistor is called the “open circuit”
output voltage. If you short the output (connect it directly to ground, the output
will be zero (of course), however, you can measure the “short circuit current” by
connecting one of the multimeters as an ammeter in series with the connection
to ground. Do this for the dc case (it is ok here because there is a 10kΩ
resistor between the +15 V voltage source and your short to ground,
usually it is unwise to short the output of a circuit to ground as it
generally destroys the circuit).
Now you have the open circuit voltage and the short circuit current of this
voltage divider circuit. Do they agree with what you would calculate?
Thevenin’s theorem says that you may replace any part of a linear circuit
with its Thevenin equivalent. What is the Thevenin equivalent of this circuit?
Build the Thevenin equivalent and verify that it gives the same open circuit
output voltage and short circuit current as the circuit above.
Exercise 2: The response of STC (single time constant) circuits built
using a resistor and a capacitor
1 Simple integrator. Connect the circuit below, apply a 3 V. 100 Hz
square wave to the input and observe the output waveform on the oscilloscope.
Display both the input and the output using the two channels of the
scope (trigger on the input channel). Adjust the time base to display
approximately one cycle of the square wave. Measure the “rise time” of
the output voltage. (This is the time it takes to go from 10 % to 90 % of
the maximum output voltage, assuming that the output voltage reaches
a steady value during the up half of the cycle. If it doesn’t decrease the
frequency of the square wave till it does.) How does the rise time of the
output compare to the rise time of the input? Why are they different?
Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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Let’s find, and solve, the differential equation describing the circuit’s response to the step in voltage occurring at t=0 (the rising edge of the square
wave). To begin with we note that the current through resistor R at time
t = 0+ is, by Ohm’s Law,
i(t) =
Vstep − vo (t)
R
where Vstep is the amplitude of the square wave. This current flows into
the capacitor C (neglecting any load caused by the oscilloscope). Thus,
the voltage on the capacitor, which is also the output voltage, increases
as charge accumulates on the capacitor due to the current flowing into it.
Z
1 t
Q(t)
i(t)dt
= V0 +
vo (t) = V0 +
C
C 0
where V0 is the voltage on the capacitor at time t = 0 (which we may
take to be zero). Differentiating both sides of this equation with respect
to time and combining with our earlier equation gives
i(t) = C
dvo (t)
Vstep − vo (t)
=
dt
R
Substituting V 0 (t) = Vstep − vo (t) gives the differential equation
dV 0 (t)
1 0
=−
V (t)
dt
RC
which has solution
V 0 (t) = V 0 (0)e−t/RC
or, substituting back for vo (t),
vo (t) = Vstep − [Vstep − v(0)]e−t/RC
Thus, the output voltage starts at 0, at t = 0+ and increases exponentially
to the value Vstep with a time constant of RC.
This is an example of a “single time constant” circuit, with time constant
τ = R · C. How does the calculated time constant compare to the time
required for the output voltage to climb from 0 to 1/e = 0.63 times its
final value? If there is a significant difference (greater than 10%, what do
you think is the source of this discrepancy?
Now change the frequency of the square wave and observe the changes in
the output waveform. At high frequencies (say above 5 kHz) the output
should resemble a triangle or sawtooth wave with roughly linear rise and
fall slopes. At such frequencies, and higher, this circuit acts as a crude
integrator since the output voltage as a function of time is proportional
to the time integral of the input voltage.
Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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2 Simple differentiator. Connect the circuit below, apply the 3 V 100
Hz square wave to the input and observe the output waveform on the
oscilloscope.
Observe the changes in the output waveform as you change the frequency
of the input square wave. This circuit faithfully (more or less) passes the
initial step voltage at the input, but the output voltage following the step
eventually decays to 0 (if the period of the square wave is long enough).
Using a smaller capacitor will reduce the time constant, RC, and cause
the output voltage to decay to zero more rapidly. In the limit as RC → 0
this circuit appears to “differentiate” the input step voltage, i.e., it passes
a transient whenever the voltage suddenly changes but the output is zero
at other times.
What is the differential equation describing the output voltage as a function of the input step voltage at t=0?
3 The STC Low Pass Filter. Reconnect the circuit from example 1 above
only this time use a sine wave as the input voltage. For a fixed input
voltage level (say Vpeak = 5V), measure the output voltage as a function
of frequency. Take about 12 measurements for frequencies spanning a
range from about 0.1 × ν0 = 1/(2 · π · RC) to about 10 × ν0 = 1/(2 · π · RC).
This Low Pass Filter is nothing but a voltage divider where we use complex impedances rather than pure resistances. The output voltage of this
voltage divider is the voltage across the capacitor. An ideal capacitor has
an impedance Z which is purely imaginary
ZC = −
j
ωC
√
where ω is the angular frequency in radians/s (or s−1 ), and j = −1. To
calculate the output vo (ω) which arises from an input voltage vi (ω) we
need to calculate
T (ω) =
j
− ωC
vo
=
j
vi
R − ωC
Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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We need to put this into standard form where the denominator is real.
We accomplish this by multiplying the numerator and denominator by
the complex conjugate of the denominator. Thus,
T (ω)
=
=
=
R+
R+
j
ωC
j
ωC
2
·
−j
ωC
R−
j
ωC
R
1
) − j ωC
( ωC
1
1 + ( ωCR
)2
1 − j( ωω0 )
1 + ( ωω0 )
where ω0 = 1/RC. Fill in the missing steps in this derivation.
The magnitude of this complex function of frequency is obtained by taking
the square root of the product of the function with its complex conjugate
p
|T (ω)| =
T (ω)T ∗ (ω)
1
= q
1 + ( ωω0 )2
The phase response is
φ(ω) = − tan−1 (
=(T (ω))
ω
) = − tan−1 ( )
<(T (ω))
ω0
where =(T (ω)) is the imaginary part of the complex function T (ω) and
<(T (ω)) is its real part.
Plot the calculated magnitude and phase response (as 20 log |T (ω)| and
φ(ω)) vs log(ω), recalling that ω = 2π · ν is the angular frequency in
radians/s. Plot the experimental measurements of the magnitude response
on the same graph for comparison.
At the frequency ν3dB = ω3dB /2π = 1/(2πRC) the magnitude of the
output voltage should be 3 dB below that of the input voltage. Check this
experimentally.
How do your data, and those of some of the others in your class, compare
with the theoretical curve? Can you comment on any variability there
may be?
4 The STC High Pass Filter. Now reconnect the circuit from example
2 using a sine wave as the input. For a fixed input voltage measure the
output voltage as a function of frequency spanning a range from 0.1 to 10
times the 3dB frequency ν3dB = ω3dB /2π = 1/(2πRC).
Calculate the magnitude response function and the phase for the output of
this circuit and plot them as you did for part 3. Compare your measured
and theoretical values (and those of others in your class).
Physics 3100 Electronics, Fall 2009, Lab 1. Simple DC and AC Circuits
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You will see that the 3 dB point has the same definition in both the low
and high pass filters but that for the low pass filter the response rolls
off at high frequencies (i.e., the output voltage decreases as frequency is
increased) at a rate of 20 dB per decade of frequency (6 dB per octave)
while for the high pass filter the response rolls off at low frequencies.