2 class days
Lab 10: Feedback
Lab Assignment
1. Two Comparators
Source: Eyler lab 8 (HH 10.1)
Start with Horowitz and Hill, lab 10-1, but do the section on “defective
comparators” only using an LM311 (at the bottom of pg. 234). This “defective”
approach is good enough for many purposes.
Now build a Schmitt trigger using an LM311 comparator. Before arriving in the
lab, design your own circuit to provide roughly 1V of hysteresis, with threshold
levels centered at approximately 3V. Note that there is a worked example for a
similar problem in the Student Manual on pp. 228-229. The output should swing
between ground and the positive supply voltage.
Schmitt triggers are very widely used to receive digital data over long
transmission lines, since they provide good immunity against noise transients.
They are highly recommended for general laboratory use, and standard serial
computer interfaces usually use Schmitt triggers at the inputs.
Answer the questions for the HH 10.1 Schmitt Trigger section. Compare your
measured hysteresis to your design goal. Estimate thresholds for on and off, and
compare with measurements.
2. RC Oscillator
HH 10.2. Answer all questions.
3. IC Oscillator
HH 10.3. Answer all questions.
4. Alarms with IC
Source: Hands-On chapter 9
You can configure the 555 to sound an alarm when prompted by an external
signal. The alarm is simply a 555 oscillator with the output connected to a
speaker. To prevent the alarm from sounding continuously, ground is applied to
the RESET input pin, which overrides the TRIGGER and THRESHOLD pins and
forces the output near ground. When used in this way, the RESET line is said to
‘enable’ the 555, enabling oscillation when high while disabling oscillation when
low.
Build the circuit as shown in the figure below. Use a long wire, which simulates a
security loop. Cut or unplug the wire and hear the alarm!
555 timer configured as an alarm.
The alarm can also be configured to sound for a specified duration using two
555s by combining the timer from the figure below with the alarm circuit. Connect
the alarm’s RESET input to the timer’s output line. Trigger the timer and the
alarm will sound for the duration of the timer’s output signal.
555 timer IC used as a one-shot or timer.
You can also make a pulsing alarm by using two oscillators. The first should
oscillate with a period of a few seconds. The output of this oscillator is then
connected to the RESET line of the second oscillator. The second oscillator’s
output is connected to the speaker and can oscillate with any audio frequency of
your choice.
Prove to the instructor during class time that this circuit works.
5. Active Band-Pass Filter
Source: Hands-On chapter 9
Active filters have significant advantages over passive ones, including
* lower cost due to replacement of expensive inductors by capacitors phaseshifted via feedback;
* high input impedance and low output impedance;
* ease of tuning over a wide frequency range.
They also have significant disadvantages such as frequency response limited by
the bandwidth of the op amp and the need to provide power to the op amp. There
are a variety of configurations that can be used for low-pass, high-pass, and
bandpass applications; for a more extensive discussion, see e.g. Horowitz and
Hill, The Art of Electronics, or Simpson, Introductory Electronics for Scientists
and Engineers.
Wire up the circuit below, apply power, and drive it with a sine wave of about 1 V
peak-to-peak. Vary the driving frequency until you find the maximum output
amplitude. What are the center frequency f0 and upper and lower −3 dB points fH
and fL of the frequency response? If you adjust the output voltage to be 14.1 V
peak-to-peak at f = f0, you can easily find the −3 dB points by varying the
frequency until the output voltage is 10.0 V peak-to-peak. Measure and graph the
gain vs. frequency for a reasonable frequency range.
Active bandpass filter.
Compare f0 with the theoretical value
f0 = [ 1 / (2πC) ] * [ √(2 / (R1R2) ) ]
and compare the voltage gain at the center frequency with the theoretical value
A0 = -(1/2) * (R2 / R1)
The ‘quality factor’ for a bandpass filter is
Q = f0 / (fH - fL )
Thus, a narrow passband corresponds to high Q and a broad passband to low Q.
Compare your observed Q to the theoretical value for this filter,
Q = √ (R2 / 2R1)
Try to explain how the circuit works.