Introduction - Facstaff Bucknell

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ELEC 350L
Electronics I Laboratory
Fall 2002
Lab 5: Relaxation Oscillator (Astable Multivibrator)
Introduction
Many electronic devices and systems require a source of pulses, often called a “clock signal,” for
proper operation. One example is the clock circuit found in all computers that controls the
timing of the digital signals within the CPU. Another is the signal generator on your lab bench.
The waveform most often required in a large number of applications is the square wave (and
sometimes the nonsymmetrical square wave, also called a rectangular wave). There are many
ways to generate square waves, but one of the easiest and least expensive approaches is to use an
op-amp. In this lab experiment you will build a square wave generator circuit known as a
relaxation oscillator (also called an astable multivibrator).
Theory of Operation
A common square wave generator circuit based on an op-amp is shown in Figure 1. One of the
first things you might notice about this circuit is that it has no input port other than the
connections to the power supply. Rather than serving as processors of signals (like amplifiers
and filters, which have inputs and outputs), oscillators instead are sources of signals (which have
only outputs). Note that there are resistors (R1 and R3) connected from the output port of the opamp to both its inverting and noninverting terminals. Basically, this circuit is a Schmitt trigger
that creates its own input voltage (vc in this case). There is no independent source that drives the
circuit. The design is unusual in that it incorporates both positive and negative feedback. The
positive feedback dominates, primarily because the capacitor voltage cannot change
instantaneously, whereas the voltage at the junction of the voltage divider formed by R1 and R2
can. Thus, the assumption that a virtual short exists between the two inputs is not valid. We will
see that the circuit behaves cyclically, switching regularly between two stable states.
Because of the positive feedback, the output of the op-amp will be near either the positive power
supply voltage VPOS (+ 15 V in Figure 1) or the negative voltage VNEG (-15 V). The reference
voltage at the noninverting input will be VPOS or VNEG, where

R2
.
R1  R2
For example, if R1 and R2 are equal in value, the reference voltage will be 0.5VPOS. Thus, the
state of the output (VPOS or VNEG) will depend upon the value of the capacitor voltage (which sets
v+) and the voltage at the junction of the voltage divider (v-). The capacitor charges through
resistor R3 with a time constant of R3C (no current flows into the inverting input of the op-amp).
The capacitor voltage vc would eventually reach VPOS or VNEG if the circuit were not to react to
the change in vc before then.
1
R3
+
C
+15 V
vc
_
_
vo
+
-15 V
R1
R2
Figure 1. Relaxation oscillator.
Consider what happens when vo is saturated at VPOS. The voltage at the noninverting input (v+) is
VPOS. The output voltage charges the capacitor through resistor R3. The capacitor voltage
becomes more positive, rising toward the value VPOS. However, the capacitor voltage never gets
that high, because the output voltage changes abruptly to VNEG when vc reaches VPOS. In
response, the capacitor voltage begins to drop and then becomes negative as the capacitor
charges toward the new negative output voltage VNEG. But once vc reaches the value VNEG, the
output voltage changes state again and switches back to VPOS. The capacitor voltage rises again,
becoming less negative and then more positive, and the cycle repeats. The output voltage cycles
between VPOS and VNEG as long as power is applied to the circuit. The capacitor alternately
charges, discharges, and charges again with the time constant R3C.
Figure 2 shows a plot of the output voltage vo and the capacitor voltage vc vs. time. The figure
illustrates why the circuit is sometimes called a relaxation oscillator; the capacitor voltage tries to
“relax” to a steady-state value after every transition in the output voltage. The circuit is also
called an astable multivibrator because the output continuously alternates between two states,
where each state is initially stable but then becomes unstable when the capacitor voltage
approaches the reference voltage v+. The term multivibrator is just another name for an
oscillator, specifically one that generates pulses.
2
V
POS
V
POS
0
t
V
NEG
V
NEG
T/2
T
Figure 2. Output voltage (solid line) and capacitor voltage (dashed line) produced by the
relaxation oscillator. The capacitor voltage does not actually go above VPOS or below VNEG.
The dashed lines are extended simply to illustrate the exponential behavior of the capacitor
voltage.
The period T of the square wave is controlled by the values of R3, C, and . A formula for T can
be derived from the equation that gives the voltage across the capacitor. When the output
voltage is at the value VPOS, the time-varying capacitor voltage is given by
v c (t )  VPOS  VNEG  VPOS  e  t / R3C ,
where t is measured from the instant the output voltage made its most recent transition from VNEG
to VPOS. Note that at t = 0 the capacitor voltage is VNEG, the value of vc that causes the output to
switch from VNEG to VPOS. As t   , vc would reach the value VPOS; however, it never gets there
before the output changes state again. During the other half-cycles, when the output voltage is
VNEG, the capacitor voltage is
v c (t )  VNEG  ( VPOS  VNEG ) e  t / R3C ,
where again t measures the time that has passed since the most recent output transition. If we
make the assumption that VNEG = -VPOS (that is, that the power supply voltages are equal in
magnitude and opposite in sign), then the equations for vc become


when vo  VPOS


when vo  VNEG .
v c (t )  VPOS 1    1 e  t / R3C
and
vc (t )  VNEG 1    1e  t / R3C
3
Note from Figure 2 that the output voltage spends equal amounts of time in the positive state and
in the negative state. We can therefore use either equation for the capacitor voltage to find the
period T of the square wave. We’ll use the first one (valid when the output is positive).
The output voltage remains positive until the capacitor voltage rises to VPOS. At that instant in
time the output switches to VNEG, and the capacitor voltage begins to drop again. Thus,
vc (0.5T )  VPOS . (The switch in output voltage occurs after half a period.) Substituting this
“end condition” (as opposed to “initial condition”) into the equation for the time-varying
capacitor voltage yields


v c (0.5T )  VPOS  VPOS 1    1 e 0.5T / R3C .
In the right-hand equality the VPOS factors cancel out, leaving
  1    1 e 0.5T / R C .
3
A bit of algebra yields
1 
 e  0.5T / R3C .
1 
Taking the natural logarithm of both sides leads to the formula for the period of the square wave,
1  
T  2 R3C ln 
 ,
1  
which can also be written
1  
T  2 R3C ln 
 .
1  
We would have obtained the same result if we had started with the equation for the capacitor
voltage for the case when the output voltage is negative.
Quite often the resistors R1 and R2 are chosen to be equal, which sets  = 0.5 and results in a
simplified equation for the period given by
T  2.2 R3C for   0.5 ,
because
 1  0.5 
2 ln 
  2.2 .
 1  0.5 
4
Experimental Procedure
Assemble a relaxation oscillator like the one shown in Figure 1. Use power supply voltages of
±15 V, and make the resistors in the positive feedback network (R1 and R2) equal in value.
Choose reasonable values for R1 and R2 based upon your experience with op-amps and the
requirement to limit the output current of the op-amp. Add the circuit shown in Figure 4 to the
output of the op-amp. The devices D1 and D2 are called light-emitting diodes (LEDs). Like
regular diodes, LEDs conduct in one direction only. Since the LEDs shown in Figure 4 are
oriented in opposite directions, one LED should light when the output of the op-amp is positive,
and the other should light when the output is negative. The LEDs used in this lab glow well
when they draw around 20 mA. Assuming that each LED has a 1-V drop across it when it
conducts and that the op-amp’s output voltage at saturation is around 14 V, find an appropriate
value for the LED current-limiting resistor R4. Select reasonable values for C (0.1 F minimum)
and R3 to produce a square wave with a period of approximately 0.5 sec. This period is long
enough to allow you to see the two LEDs blinking alternately when you apply power to the
circuit. Verify that your circuit is working properly by counting the number of blinks that occur
in several (15 or more) seconds and then using this result to estimate the frequency of the square
wave. The longer the time during which you count pulses, the more accurate the calculated
frequency will be.
R4
to vo
D1
D2
Figure 4. Light-emitting diodes used to indicate output voltage polarity.
Increase the frequency of the oscillator by replacing the capacitor with one having a value of
0.001 F. Calculate the new frequency and then verify your calculation by displaying the output
waveform on the oscilloscope. Measure the length of one cycle (one period) of the square wave,
and make sure it corresponds to the frequency you measured. Display the capacitor voltage and
the output voltage together using the same scale, making sure that both waveforms have the same
zero-volt reference. Sketch or print out the display. Is the capacitor voltage at the proper levels
when the output voltage changes states?
Decrease the time/division setting of the oscilloscope so that the slope of the output voltage
transition from positive to negative (or negative to positive) is exaggerated. Calculate the time
rate of change of the voltage during the transition. How does this result relate to the slew rate
specification of the op-amp?
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