Electronics Lab
Department of Physics and Astronomy
Brigham Young University
Provo, Utah 84602
(Version: 14.05)
Name
This manual was originally written in TEX by Harold Stokes for Physics 340, the electronics
lab for physics majors at BYU through Spring 2006. When the lab courses were rearranged
to put the electronics lab at the front of the sequence as Physics 140, the manual was
modified by Robert Davis and Grant Hart to better suit the new clientele. It has been
modified by Branton Campbell to adjust the labs and sequence. Many of the changes were
suggested by the student Laboratory Assistants as they have observed the difficulties some
of the exercises presented. Rob Davis and Branton Campbell, with students Kristian Sims
and Michael Rawlins developed and tested the Audio Amplifier exercise. Bryan Peterson
separated the AC Measurements and RC Circuits labs to allw more time to become acquainted with the oscilloscope. The introduction to the oscilloscope was rewritten, along
with some of the exercises, since we have changed from a Tektronix analog oscilloscope to
a LeCroy digitial oscilloscope for the exercises.
We appreciate the assistance of the many students that have contributed ideas and corrections to the manual.
Lab Kit Contents
51 resistors,
1
4
W, 5%
1 each: 120, 150, 180, 220, 270, 330, 390, 470, 560, 680, 820, 1.2k, 1.5k, 1.8k, 2.2k,
2.7k, 3.3k, 3.9k, 4.7k, 5.6k, 6.8k, 8.2k, 12k, 15k, 18k, 22k, 27k, 33k, 39k, 47k,
56k, 68k, 82k
2 each: 100, 100k, 1M, 10M
5 each: 1k, 10k
2
2
2
4
2
1
1
4
1
4
1
2
1
1
2
1
1
1
1
2
1
1
1
1
1
2
1
1
1
1
1
1
2
1
100 pF ceramic capacitors
1000 pF ceramic capacitors
0.01 µF ceramic capacitors
0.1 µF ceramic capacitors
1 µF electrolytic capacitors
10 µF electrolytic capacitor
100 µF electrolytic capacitor
1N4148, signal diodes
W04G or VM48, diode bridge
red LEDs
2N3904, npn small-signal transistor
2N3906, pnp small-signal transistors
IRF3704, power MOSFET
phone jack
pushbutton switches
#47 incandescent lamp
speaker
i
TL3472, dual op-amp
311, comparator
555, timers
7400, quad NAND gate
7402, quad NOR gate
7408, quad AND gate
7414, hex inverter
7432, quad OR gate
7474, D-type flip-flops
74193, up-down counter
74154, 4-bit decoder
DC10SRWA, 10 segment LED display
TDA7056B, audio amplifier
78L05, positive output 5 V regulator
wire kit
breadboards
9 VDC power supply
Resistor Color Code
The values of resistors are indicated by three bands of colors. Each color corresponds to a
number:
0
1
2
3
4
5
6
7
8
9
-1
-2
black
brown
red
orange
yellow
green
blue
violet
gray
white
gold (exponent only)
silver (exponent only)
The three bands represent three numbers: n1 , n2 , n3 . The value of the resistor is given by
R = n1 n2 × 10n3
(in ohms).
For example, if the three bands are yellow, violet, and red, the value of the resistor is
47 × 102 = 4700 Ω.
The fourth band on the resistor indicates tolerance: gold = ±5% and silver = ±10%. This
is rare in modern resistors, but if there is no fourth band the tolerance is ±20%.
If there is a fifth band on the resistor it generally indicates the quality of the resistor.
High precision resistors (2% or better tolerance) will have four bands to indicate the value
– 3 digits plus a multiplier. The tolerance band is then the fifth band with possible values:
brown = ±1%, red = ±2%, green = ±0.5%, blue = ±0.25%, and violet = ±0.1%.
ii
Resistor Values in Lab Kit
(Quantity in parentheses)
brown
red
orange
yellow
green
blue
brown-black
100 (2)
1k (5)
10k (5)
100k (2)
1M (2)
10M (2)
brown-red
120 (1)
1.2k (1)
12k (1)
brown-green
150 (1)
1.5k (1)
15k (1)
brown-grey
180 (1)
1.8k (1)
18k (1)
red-red
220 (1)
2.2k (1)
22k (1)
red-violet
270 (1)
2.7k (1)
27k (1)
orange-orange
330 (1)
3.3k (1)
33k (1)
orange-white
390 (1)
3.9k (1)
39k (1)
yellow-violet
470 (1)
4.7k (1)
47k (1)
green-blue
560 (1)
5.6k (1)
56k (1)
blue-grey
680 (1)
6.8k (1)
68k (1)
grey-red
820 (1)
8.2k (1)
82k (1)
iii
45
50
55
60
50
55
60
30
30
45
25
25
40
20
20
40
15
15
35
10
10
35
5
5
-+
+-
1
A BCDE
A BCDE
FGH I J
FGH I J
1
-+
+-
Breadboard Layout
Above is shown the layout of a breadboard similar to the one you will be using during this
course. There may be some minor differences between the diagram above and the actual
breadboard you will be using.
The dots represent holes into which you can insert the ends of wires and the leads of
resistors, capacitors, ICs, and other electronic components. The lines on the figure show
how the holes are internally connected. Voltages from the power supply and ground are
usually applied to the bus lines along the top and bottom edges of the board so that every
component on the breadboard can have easy access to them.
Your breadboard may not have numbers and letters to organize the rows and columns. For
most of the exercises you can ignore the numbers and letters.
iv
3-Terminal Packages (printed side up)
DRAIN
SOURCE
DRAIN
GATE
IRF
3704
COLLECTOR
BASE
EMITTER
2N
390x
IN
GND
OUT
78L05
IRF3704
N-Channel MOSFET
2N3904, NPN transistor
2N3906, PNP transistor
78L05
5-Volt Regulator
IC (integrated circuit) Packages
TL3472
op-amp
W04G
Diode bridge
AC
1
+
~
+
04
W
2
V+ 8
−
G
−
V+ 8
1
7
2
+
7
2
6
3
BAL
6
3
5
4 V-
BAL 5
4
−
+
~
-
1 GND
+
3
555
Timer
LM311
Comparator
AC
7400
NAND gates
4 V-
7408
AND gates
7402
NOR gates
1
8 2
3
7 4
5
6 6
7
5 8
GROUND
TRIGGER
OUTPUT
RESET
CONTROL VOLTAGE
THRESHOLD
DISCHARGE
V+
7414
Inverters
7432
OR gates
1
VCC 14
1
VCC 14
1
VCC 14
1
VCC 14
1
VCC 14
2
13
2
13
2
13
2
13
2
13
3
12
3
12
3
12
3
12
3
12
4
11
4
11
4
11
4
11
4
11
5
10
5
10
5
10
5
10
5
10
6
9
6
9
6
9
6
9
6
9
7 GND
8
7 GND
8
7 GND
8
7 GND
8
7 GND
8
74154
Decoder
7474
Type-D flip-flops
1 1R
VCC 14
2 1D
2R 13
1 Y0
VCC 24
DC10EWA
LED bar
74193
Counter
2 Y1
A0
23
3 Y2
A1
22
1 D1
1
20
VCC 16
2
19
1 n.c.
2 VP
4 Y3
A2
21
2 Q1
D0 15
3
18
5 Y4
A3
20
3 Q0
MR 14
4
17
6 Y5
E2
19
4 CPD TCD 13
5
16
E1
18
3 VIN
4 SGND
3 1CLK 2D 12
5 VCTRL
4 1S 2CLK 11
5 CPU TCU 12
6
15
8 Y7
Y15 17
6 Q2
7
14
9 Y8
Y14 16
7 Q3
10 Y9
Y13 15
8 GND D3
7 Y6
5 1Q
6 1Q
TDA7056B
Audio-amp
6 OUT+
2S 10
PL
11
7 PGND
2Q 9
D2 10
8
13
9
12
10
11
8 OUT-
7 GND 2Q 8
11 Y10 Y12 14
12 GND Y11 13
v
9
9 n.c.
Lab 1
Voltage and Current
Notes for Lab 1:
Voltage and Current
Electric Charge: Electric charge is a fundamental property of matter. The electric force
between charged objects varies like the inverse square of the distance between them, and
the more electric charge an object has, the more electric force it experiences in the vicinity
of other charged objects. Electricity and gravity are similar in this respect. But unlike
gravitational mass, electric charge comes in two varieties, which American physicist Benjamin Franklin discovered to be opposites that can cancel one other out; he coined the
terms positive and negative to describe them. Opposite charges attract one another, and
like charges repel one another. Though we can’t reverse the direction of gravity to make
objects fall up rather than down, it is possible to reverse the direction of an electric force
by changing the sign of one of the charges. Thus, electric forces can either push or pull,
while gravity can only pull.
It is often helpful to think of electric charge as a fluid like water. A cup of water can
be subdivided over and over again until arriving at a smallest indivisible building block,
the H2 O molecule. Similarly, electric charge has a smallest indivisible building block: the
electron charge. The standard international (SI) unit of electric charge was named in honor
French engineer Charles Coulomb, who first quantified the strength of the electric force.
A Coulomb (C) of electric charge is equivalent to the charges of 6.241 × 1018 electrons, so
that one electron charge (e) is equivalent to 1.602 × 10−19 C. Electric charges you may have
experience with include the charge on a balloon that you rubbed on your hair or sweater
(10−6 C), the charge transferred in a bolt of lightening (15 to 300 C), or the charge in a
AA battery (7.5 kC).
Electric Current: The flow of electric charge is analogous to the flow of a fluid, and it
is useful to think of the electric charge flowing through a wire in terms of water flowing
through a pipe. When electric charge flows through a circuit, we refer to the rate of flow as
the electric current. Just as the flow of water can be measured in liters per second, electric
current can be measured in Coulombs per second (C/s). In honor of French physicist,
Andre Ampere, who made major contributions to early study of electricity, the SI unit of
electric current is the ampere which is usually abbreviated as amp or just "A". One amp is
equivalent to one Coulomb/second (C/s).
Electric currents common to everyday experience might include those flowing through an
LED light bulb (20 mA), an incandescent bulb (1.0 A), a laptop computer (3.0 A), an
electric range (20-30 A), an electric car (200-500 A), or a bolt of lightening (35 kA). As you
know, a sufficiently large electric current can do damage to an object that it flows through.
For steady currents in humans, 1 mA generates a mild tingle, 10 mA hurts, 10-100 mA
results in loss of muscle control and potentially fatal heart fibrilations, 100-2000 A can stop
the heart and damage nerve tissues, and larger currents can further cause severe burns
(http://en.wikipedia.org/wiki/Electric_shock).
Conductors, insulators, and resistance: It is possible for electric charge to flow through
otherwise empty space, an example being the solar wind, a very hot but neutral collection of
fast-moving electrons and protons. But at ordinary temperatures, opposite electric charges
3
4
Notes for Lab 1, Voltage and Current
strongly prefer to come together to form electrically neutral atoms, which then further
coalesce into various forms of matter, such as gas, liquid and solid. It should probably
be surprising that electric charge can actually flow inside of a liquid or solid material. A
material that permits such an internal flow of electric charge is called a conductor, while
a material that does not is called an insulator. By surrounding conductors with insulating
materials such as air, glass, wood, plastic, etc., one can define very specific pathways for
the flow of electricity in a circuit.
In the context of electronics, the most common conductors that you will encounter are
metallic wires. But in reality, all materials (even those that we call insulators) conduct
electricity to some degree. The words metal and insulator are technical terms that identify
the physical mechanism of current flow in a material. For an object made of any given
material, the resistance of the object quantifies just how hard or easy it is for electric
current to flow through it. In recognition of the pioneering work of German physicist Georg
Ohm, the SI unit of electric resistance is the ohm, abbreviated as Ω. For a one-centimeter
length of rod or wire with a 1 mm diameter, the resistance of some common materials are
0.002 Ω for copper, 0.9 Ω for graphite, 76 × 106 Ω for pure silicon at room temperature,
and 1 × 1020 Ω for wood or glass. With such high values of resistance, it’s no surprise
that wood and glass are called insulators. Passive two-lead circuit elements designed to
have intermediate values of resistance in the range from about 0.1 Ω to 107 Ω are known
as resistors, and are typically made of carbon or of thin-film ceramic materials. A resistor
impedes or resists the flow of current – hence the name.
When a battery pushes electric current through a conducting wire, electric charge is transferred from one battery terminal to the other, while the conductor itself remains electrically
neutral. Franklin correctly hypothesized that this is possible because the conductor itself is
comprised of equal amounts of positive and negative charge, some of which is free to move
inside the conductor. He found that a that an object can become positively charged either
by adding positive charge or by taking away negative charge. From the outside, we can’t
tell the difference.
The electrically-charged mobile species within a conducting material are called charge carriers. The atomic nucleus of each atom is positively charged, whereas the electrons that
orbit the nucleus are negatively charged. In most metals, one or more of the outermost electrons of each atom become delocalized so as to travel freely throughout the entire conductor
rather than being associated with a specific atom. In contrast, the positively-charge atomic
nuclei are not mobile. Thus, in a metal, the charge carriers are negative! In insulators, the
charge carriers can be either positive or negative, and are not delocalized, but rather hop
from site to site when the temperature is high enough to occasionally kick them over local
energy barriers. Based on this conduction mechanism, semiconductors and ionic conductors
are actually classified as insulators, though they don’t insulate well in practice.
Consider a battery with its positive terminal on the left and its negative terminal on the
right, connected to each other by a conducting wire. Electric current then flows through
the wire from the positive terminal to the negative terminal (from left to right). If we could
see the microscopic activity inside the wire, we might see positive charge carriers flowing in
the direction of the current (left to right), or alternatively, negative charge carriers flowing
opposite the direction of current (right to left), or maybe even both. From outside the wire,
Notes for Lab 1, Voltage and Current
5
however, these scenarios look the same and have the same outcome. If the wire material is
a metal, we often conveniently pretend that the charge carriers are positive, flowing from
left to right in the direction of the electric current, though we know that this isn’t true.
The carriers in a metal are negatively-charged electrons, and must therefore flow opposite
the current (right to left). The decision to designate the electron charge as negative rather
than positive was arbitrary and historical, and certainly makes the study of electricity more
confusing. But as long as we all agree on the convention, it doesn’t matter. So don’t rock
the boat! Electrons are negative; and they flow left, we say that the current flows right.
Electric Potential or Voltage: Analogous to pressure in a water-flow circuit, voltage is
what causes electric charge to flow in an electric circuit. Just as increasing the pressure
at one end of a hose increases the flow of water, increasing the voltage across a conductor
increases the flow of charge. And reversing the direction of the voltage reverses the direction
of flow. In recognition of Italian chemist/physicist Alessandro Volta’s invention of the
electrochemical battery, the SI unit of voltage is the Volt (V).
Electric potential is a more formal name for voltage. Electric potential (voltage) does not
have an absolute zero, and can only be defined as a difference. The zero-potential point
in any system is arbitrary. Whenever we say voltage or potential, we really mean voltage
difference or potential difference. If the electric potential of a power line is 1 megavolt (MV)
higher than that of the earth, we could say that the earth is at 0 V and the power line at
+1 MV, or we could say that the power line is at 0 V and the earth at −1 MV. Or we could
arbitrarily choose the zero-potential point to be somewhere in the middle.
Going back to the gravitional-mass analogy of electric charge, electric potential (voltage) is
highly analogous to gravitational potential, which is approximately proportional to elevation
near the surface of the earth. When hiking, for example, the energy required to climb from
one elevation to another is equal to ∆E = m∆Vg = mg∆h, where m is your mass, ∆Vg is
the change in gravitational potential, g ≈ 9.80 m/s2 is the gravitational acceleration, and
∆h is the change in elevation from base to peak. Similarly, the energy required to move an
electric charge Q from a low voltage to a high voltage is ∆E = Q∆V . A battery or other
power supply is like a pump that uses pressure to move a positive electric charge uphill on
the voltage landscape, and thereby increases the electric potential energy of the system.
When this positive charge is released, it will roll downhill again towards regions of lower
voltage, converting its potential energy into other forms of energy, such as kinetic or heat
energy, along the way.
Electrical Ground: For safety reasons, it is common to use a wire to connect some
reference point in a circuit to the earth, and refer to this reference point as electrical
ground, assigning it a zero voltage. Even for circuits that are not electrically connected to
the earth, we typically choose a reference point to call ground anyway. It doesn’t matter
which point we call ground, so long as we all agree on the choice. Having an agreed-upon
reference point allows us to discuss the voltage at any point in the circuit without using
the cumbersome language of differences. For example, when we say that the voltage is +10
V at some point in the circuit, we really mean that it is 10 V higher than the agreed-upon
ground point. In circuit diagrams, we use the
symbol to indicate ground.
AC vs DC circuits: Our course will focus primarily on circuits in which the currents and
6
Notes for Lab 1, Voltage and Current
voltages are constant in time, which are commonly known as direct current or "DC" circuits.
Circuits with sinusoidally-varying currents and voltages, on the other hand, are known as
alternating current or "AC" circuits. Most complicated electronic devices use DC circuits,
whereas most electric power is distributed via AC circuits.
Ohm’s Law states that the voltage V across a circuit element is proportional to the electric
current I flowing through that circuit element, where the constant of proportionality is the
electrical resistance R of the element, so that we have V = IR. Thus, at fixed voltage, a
larger resistance to flow results in a smaller current. Despite the fact that Ohm’s law is
a simple linear relationship, it tends to be the subject of considerable confusion and angst
amongst beginning electronics students. The challenge lies in the fact that Ohm’s law can
either be applied to an individual circuit element or to a network of circuit elements, and
that it can be tempting to try to relate the current through one portion of the network to
the voltage across another portion of the network. You’ll generally be safe if use Ohm’s law
to relate the voltage difference between two points in a circuit to the current flowing through
the network between those same two points, via the resistance of the network between those
same two points.
Electric power: A battery or generator must do work to push electric current through a
resistor. The work is expended in lifting electric charge uphill to the positive-voltage side
of the resistor, which energy is then dissipated as heat as the charge flows downhill inside
the resistor. The physical mechanism of energy dissipation is the collisions of conduction
electrons with the thermal vibrations of the atomic lattice (imagine driving bumper cars
through a forest of dancing trees). The rate of energy loss per unit time in the resistor (i.e.
dissipated power) is calculated as
P = IV = I 2 R =
V2
.
R
These three expressions are equivalent, and can all be used to calculate power. Which one
we use in a given situation is a matter of convenience, and depends on which quantities we
know at the time that we perform the calculation.
Kirchhoff’s Laws of Circuit Analysis: The goal of a circuit analysis is to determine
the voltage and current at every point in the circuit. Before we state the basic principles
used to perform such an analysis, we need to define some elementary circuit structures:
junctions, loops, and branches. A circuit junction is a point where three or more wires come
together, so that current can flow into or out of the junction along multiple paths. Current
doesn’t normally just flow out the end of a wire and into empty space. Rather, it circulates
around a closed conducting path called a loop. An electric circuit is always comprised of one
or more loops. Each conducting pathway that connects two adjacent junctions is called a
branch of the circuit. A single-loop circuit has one branch and no junctions. A double-loop
circuit can have either two branches and one four-way junction (if the two loops connect at
only one point), or three branches and two three-way junctions.
Kirchhoff’s Current Law states that the sum of the currents flowing into a junction equals
the sum of the currents flowing out of a junction. In other words, no electric charge is
created or destroyed at a junction – what goes in must come out.
Notes for Lab 1, Voltage and Current
7
Kirchhoff’s Voltage Law states that the sum of the voltage changes encountered around
any closed loop is zero. In a circuit, batteries and generators act like pumps that push the
flow of electric charge uphill on the voltage landscape, whereas the current in a resistor is
like a river running dowhill. When traversing a circuit loop in the direction of the current,
an electric charge experiences a positive (uphill) voltage change while passing through a
battery, and a negative (downhill) voltage change when passing through a resistor. On a
closed-loop hike, the sum of the uphill and downhill elevation changes add up to zero, so
that starting and ending elevations are the same. And in a closed-loop circuit, the sum of
the positive and negative voltage changes add up to zero, so that the starting and ending
voltages are the same. This is not a complicated principle.
Wires as Equipotential Objects: In simple eletronic circuits, the electrical resistance of
a highly-conducting metallic wire is often so small as to be neglected. Ohm’s law then tells
us that the flow of current through such a wire is also negligible. In our circuit analyses,
we will accept this simplification, and effectively assume that voltage is the same at every
point on a wire or a network of directly-connected wires, no matter their length, shape,
or configuration. When we want the voltage to be the same at two points in a circuit, we
connect them with a wire. In the gravitational-potential analogy, a metallic conductor is
like a lake; the elevation is constant across the lake because any attempt to change elevation
on one side would be immediately counteracted by an an internal flow from one side to the
other. Ditto for the voltage landscape of a metallic conductor; an internal flow of charge
immediately counteracts any attempt to create a voltage gradient.
Resistance of a Network: When a voltage is applied between two points of a network
of connected resistors, the current must split across the various branches of the network, so
that some resistors carry more current than others. In such cases, it normally takes more
than mental algebraic manipulation to determine the current through and voltage across
each individual resistor. The first step in the analysis of a complicated resistor network
is the reduction to a single effective or equivalent resistance Reff that relates the overall
voltage across the network to the total current through the network. One then attempts
to sequentially divide the network into smaller and smaller branches until the current and
voltage of each circuit element is known. When such a step-by-step reduction is not possible,
one can always turn to the more involved matrix method. But we will concern ourselves
primarily with simple networks that are easy to reduce by hand in just a few steps.
Resistors connected between points A and B, such that an electric charge flows first through
one and then through the next are said to be connected in series, and are equivalent to a
single resistor whose resistance is obtained by a simple sum of the consituent resistances.
We often call this series addition. A corollary of Ohm’s law is that resistors in series must
pass the same current, but may experience different voltages.
=
R1
R2
R = R1 + R2
Resistors connected between points A and B, such that an electric charge can travel from A
to B through any one of the resistors, are said to be connected in parallel, are equivalent
8
Notes for Lab 1, Voltage and Current
to a single resistor whose resistance is obtained by inverting the sum of the inverses of the
constituent resistors. We often call this parallel addition. A corollary of Ohm’s law is that
parallel resistors experience the same voltage difference, but may pass different currents.
R1
R2
=
1
1
1
=
+
R1
R2
R
To obtain the effective resistance of a multi-loop circuit between two points, A and B, first
perform series addition on any network branches that contain resistors in series. Then
combine parallel branches via parallel addition. Iterate this process until a single branch
with a single resistor is obtained. Take care not to attempt to treat a pathway that passes
through point A or point B as a branch.
Name
In-Class Exercises for Lab 1:
CID
Voltage and Current
1. The top circuit shows a single light bulb (shown as a resistor) connected to a battery.
Predict the brightness of both bulbs in the bottom circuit (labeled B and C) compared
to the original bulb (labeled A). Assume all the bulbs are identical.
(a) Both B and C are brighter than A.
(b) Both B and C are the same as A.
12 V
A
(c) Both B and C are dimmer than A.
(d) B is the same as A. C is dimmer.
(e) B is the same as A. C is brighter.
12 V
B
C
(f) C is the same as A. B is dimmer.
(g) C is the same as A. B is brighter.
2. The circuit below shows two identical light bulbs in series. How does their brightness
compare with A in exercise 1?
(a) Both B and C are brighter than A.
(b) Both B and C are the same as A.
B
(c) Both B and C are dimmer than A.
(d) B is the same as A. C is dimmer.
12 V
(e) B is the same as A. C is brighter.
C
(f) C is the same as A. B is dimmer.
(g) C is the same as A. B is brighter.
3. For the circuit diagram below, assume that A is 660 Ω, and that both B and C
are 120 Ω. Determine the equivalent resistance of the resistor network, the total
current delivered by the voltage source, and the voltage at the junction connecting
the three resistors (relative to ground, which is the low voltage side of the source).
A
12 V
B
C
Additional exercises on the next page
9
In-Class Exercises for Lab 1, Voltage and Current
11
4. Consider the circuit below, where all resistances are measured in Ω. Assume that
no external voltage is applied to point B, or in other words, that the resistor is not
connected to anything at point B. Find the voltages at the points A and B.
10 V
200
A
100
0V
B
100
Lab 1 Exercises:
Voltage and Current
For these exercises, you will need a 5-V DC power supply, a hand-held multimeter, a breadboard, and several 1/4-W 100-Ω resistors. You will also need Internet access to the Circuit Construction Kit (an interactive Java applet) available from the Physics Education
Technology (PhET) web site located at http://phet.colorado.edu/en/simulation/circuitconstruction-kit-dc. Before you begin, read the “Breadboard Layout” section in this lab
manual. Resistors are labeled with bands of color that are coded as described in the “Resistor Color Code” section in this lab manual.
Consider the following four circuits below. Each one has a DC voltage source (i.e., a battery
or a DC power supply), and several resistors.
#1:
V
#2:
R
V
#3:
R
V
#4:
R
R
V
R
R
1. In the circuit diagrams above, assume that the voltage source delivers V=5 V, that
ground (0 V) is defined to be the low-voltage side of the source, and that all resistors
have the same value of 10 Ω. Then use Ohm’s law to answer the following questions.
(a) On each of the circuit diagrams above, record the voltage expected at each point
indicated by a black dot, and also record the current expected in each circuit
element that connects two dots. Use correct units.
(b) Why is the voltage the same on both sides of the resistor in circuit #1, but not
in circuit #2?
(c) Observe that the voltage differences across the two resistors in circuit #3 are
the same as that of the resistor in circuit #2. How can the current through the
source in circuit #3 then be different from that of circuit #2?
(d) Why is the current through the source in circuit #4 half that in circuit #2?
Additional exercises on the next page
13
3
Exercises for Lab 1, Voltage and Current
15
2. Using the interactive PhET applet, build each of the four circuits described above
within the same workspace. Measure all of the same voltages and currents examined
in the previous exercise, and verify that they agree with the values calculated in the
previous exercise. Don’t forget to adjust the voltage of your source to be 5 V, and the
resistors to 10 Ω each. Use the voltmeter tool for the voltages, always connecting the
black lead of the voltmeter to ground. Use the non-contact ammeter for the current
measurements, and observe that the electron-flow animations visually reinforce the
measured results. You must also use the contact ammeter for at least one current
measurement, and show it to your instructor.
Just for fun, try replacing the resistors in circuits #3 and #4 with light bulbs, and
turn the source voltages up to 10 V in order to make the bulbs glow more brightly.
The current differences should now be readily apparent.
#1:
V
#2:
R
V
#3:
R
V
3
#4:
R
R
V
R
R
3. Explain to your TA in detail that in voltage-measurement mode, the multimeter is
like an infinite resistance (actually more like 10 MΩ), and therefore has no effect on
current in the circuit when attached in parallel to the circuit element to be probed.
But if attached in series to the element, the current would be shut down. Also explain
that in current-measurement mode, the multimeter is like a wire with zero resistance,
which would completely bypass a circuit element if attached to it in parallel; a current
meter must be connected in series with the circuit element to be probed, so that the
current remains unaffected.
4. Build each of the four circuits shown above one at a time on your breadboard. Use
your 5-V DC power supply output as the source, and use 100 Ω for each resistor.
For each circuit, use your hand-held digital multimeter to measure the same voltages
examined in the previous exercises, and record them on the circuit diagrams above.
5. Use your multimeter to measure the current in circuit #2, treating it just like the
contact ammeter from the PhET simulation. CAUTION: the ammeter must be placed
in series with the element for which the current is being measured. That means
that replace one of the wires with the ammeter. Attach the red multimeter lead
to Amp input rather than the Volt input of the meter. Because this measurement
can easily DAMAGE the multimeter if not performed correctly, show your setup to
your instructor BEFORE turning on the power supply! Aside from the fact that the
currents are 10 times lower, describe any differences that you see compared to the
previous exercises.
4
Lab 2
Input and Output
Impedance
Notes for Lab 2:
Input and Output Impedance
Let’s begin by defining some new terms:
Impedance: In electronics, a circuit element that resists or impedes the flow of current
is said to have impedance. In the context of DC voltages and currents, impedance and
resistance are the same thing. But when we start dealing with AC voltages and currents,
there will be a subtle difference.
Signal: A signal is any voltage that is being generated, transmitted, or used by an electric
circuit (or part of a circuit). A signal can be used to transmit information, power, or both.
Thévenin Equivalent Circuits: Any two-terminal network of linear resistances and voltage sources is equivalent to a single ideal voltage source VT in series with a single effective
resistance RT . This equivalent circuit is called the Thévenin equivalent circuit. As an example, the big, complicated circuit below on the left is equivalent, at the terminals, to the
simple circuit below on the right.
RT
terminal 1
VT
terminal 2
terminal 1
terminal 2
Sometimes all we are told about a network is its Thévenin equivalent circuit. And in many
cases, that’s all that we need to know. Notice that a network consisting only of resistors
will have a Thévenin equivalent circuit as well with VT = 0.
Input and output devices: An output device (also called a source) produces an electric
voltage or signal. Examples include electrical wall outlets, power supplies, batteries, mp3
players, and cell-phone chargers. An input device (also called a load) receives an electrical signal from an output device, either to consume power, accept information, make a
measurement, or for some other purpose. Examples of input devices include light bulbs,
speakers/headphones, motors, multimeters, and oscilloscopes.
Signal-modifier devices: A signal-modifier device can be viewed as a combination of
an input device and an output device. It receives a signal on one end, modifies it, and
19
20
Notes for Lab 2, Input and Output Impedance
produces a new signal at the other end. Amplifiers are great examples of signal-modifier
devices. Perhaps the simplest signal-modifier device is the voltage divider, which we’ll talk
about soon.
Input and output impedance: To a DC output device (source), an simple input device
(load) looks like a Thévenin equivalent resistance that we refer to as the input impedance. An
ideal input impedance would be infinite, so as not to burden the source by drawing current
or energy. To an input device (load), an output device looks like a Thévenin equivalent
pure-voltage source and resistance in series; we refer to resistance as the output impedance.
The ideal output impedance would be zero, so as not so dissipate energy or cause an internal
voltage drop that would reduce the voltage output. Of course, real input impedances are
not infinite, and real output impedances are never zero. This lab intends to teach you how
to match real impedance and output impedances. If mismatched, the two won’t work well
together.
Voltage Dividers
Recall that in Lab 1, we built a circuit powered at 5 V with two resistors in series and found
that the voltage between the two 100 Ω resistors was 2.5 V. We call this a voltage divider;
we effectively divided our input voltage by two and measured it. Voltage dividers can be
drawn using an alternate form of circuit diagrams like this:
Vin
Vout
R1
R2
The new symbol that you see at the bottom of the diagram represents ground, the point
defined as being at zero voltage. This circuit is very similar to circuit #4 in the Lab 1
exercises. Vin was just the voltage of the power source, and Vout was the voltage at the
point between the two resistors. Let us assume that all of the current which flows through
R1 also flows through R2 (That is, no current is drawn at Vout . If we design our circuits
carefully, this assumption is valid). From Ohm’s law, the current is given by
I=
Vin
,
R1 + R2
where R1 + R2 is the total resistance between Vin and ground. Vout is the voltage across
just R2 and is given by Vout = IR2 . Using the value of the current found above, we obtain
R2
Vout
=
.
Vin
R1 + R2
With an appropriate choice for resistor values, we can obtain a value for Vout anywhere in
the range between 0 and Vin .
Notes for Lab 2, Input and Output Impedance
21
Impedance Matching
The figure below represents an arbitrary output element connected to an arbitrary input
element, both of which are represented by their simple Thévenin equivalent circuits. V0
represents the ideal source voltage of the output device, while Vt represents the terminal
voltage that is passed between the output and input device. Here, Rout is the output
resistance of the output device, and Rin is the input resistance of the input device.
Rout
V0
Vt
Rin
Observe that this circuit is essentially a voltage divider, where voltage Vt is related to V0
according to
Vt
Rin
=
.
V0
Rout + Rin
This tells us what fraction of the ideal source voltage actually made it to the input device
(e.g., motor, speaker, oscilloscope).
If we don’t want to lose useful voltage inside the output device, then we need to have
Rin >> Rout , which sets the value of this fraction close to one. For example, a 100 W light
bulb designed for use on a 100 V source has a 1 Ω input resistance and draws 1 A of current.
If you try to illuminate such a bulb with a 100 V source that happens to sport a hefty 2 Ω
output resistance, the output voltage that actually reaches the light bulb will only be 33 V,
and the bulb probably won’t light up at all. This is one of the most important lessons that
you will learn in Physics 140! Please reread the preceding paragraphs until you understand
what went wrong. What would be a more appropriate output resistance for this scenario?
Let’s take a moment (or a single paragraph) to tangentially explore a completely different
application. If you have a fixed source voltage and a fixed output resistance, and your goal
is to transmit the maximum amount of power possible to the input device, the choice of
input resistance is very different from before. Recalling that the power is given by V 2 /R,
you can write an expression for the power delivered to Rin as a function of Rout and Rin .
Then, setting the derivative with respect to Rin equal to zero to find the maximum, you
find that Rin = Rout . If your output device is a power plant, this condition is the relevant
one. But if you are just trying to power a load in the lab without overburdening the source
(i.e. causing the output voltage to drop too low), then the previous paragraph is the one
to pay attention to.
22
Notes for Lab 2, Input and Output Impedance
Calculating Equivalent Circuits
When the internal details of a complex circuit are known, one can apply straightforward
rules to calculate the Thévenin-equivalent circuit. Here is a very simple example that
involves the output resistance of a circuit containing 1-Ω and 7-Ω resistors in series.
1Ω 7Ω
8Ω
1V
1V
Because the resistors in the circuit below are not in series, the problem is more challenging.
A short-cut method for calculating the output resistance is to simply replace all ideal voltage
sources with wires and then calculate the resistance across the two output terminals.
1Ω
1V
1Ω
5Ω
5Ω
This yields a 1 Ω and a 5 Ω resistor in parallel, so that the output resistance is
Rout =
1
1
+
1Ω 5Ω
−1
=
5
6
Ω.
Alternatively, we can separately calculate the closed-circuit terminal voltage VL across an
arbitrary load resistor RL , and the open-circuit terminal voltage VOC in the absence of a
load resistor.
1Ω
1V
5Ω
5Ω
Without the load resistor, the two internal resistors function as a voltage divider, so that
the open circuit (unloaded) voltage is
V0 =
5Ω
1V =
1Ω + 5Ω
5
6
V.
Notes for Lab 2, Input and Output Impedance
23
When a 5-Ω load resistor is attached, the two parallel 5-Ω resistors are equivalent to a single
5
2 -Ω resistor. The output of the equivalent voltage divider is then
VL =
5
2
Ω
1V =
1 Ω + 52 Ω
5
7
V.
The ratio of the open and closed-circuit voltages can then be used to determine the equivalent output resistance
Rout =
V0
− 1 RL =
VL
5
6
5
7
!
V
− 1 5Ω =
V
5
6
Ω,
which is the same result obtained above using the short-cut approach.
24
Notes for Lab 2, Input and Output Impedance
Measuring Input and Output Resistance
When the output resistance and voltage of an output device are not known, they can be
determined via the following two-step procedure:
1. Because the Thévenin equivalent voltage V0 is the same as the open-circuit voltage
in the absence of a load, measure V0 with a high-quality voltmeter or an oscilloscope.
Such a measurement (input) device has a very large input resistance, and draws such
a small current that it is usually like not having any load at all. Note: if the output
device you are measuring has a very large output resistance, i.e., the output resistance
may be comparable to the input resistance of the measuring device, this measurement
will be incorrect.
Rout
V
VOC
2. Then use the same measurement device to determine the closed-circuit voltage across
a known load resistor RL
Rout
V
RL
VOC
The voltage divider equation yields
RL
VL
=
V0
Rout + RL
and
Rout =
V0
− 1 RL .
VL
To measure the input resistance (or load resistance) of an unknown input element, only one
measurement is required. Connect the load RL to a characterized output device (i.e., one
where V0 and Rout are already known), and measure VL . Then solve the equation above for
RL .
Rout
VOC
VL
RL
Name
In-Class Exercises for Lab 2:
CID
Input and Output Impedance
1. For the input and output devices connected in the diagram below, we have V0 = 5 V
and Rin = 150 Ω. If want to make sure that most of the source voltage is delivered to
the load, what would be an appropriate value of Rout ?
Rout
Vt
V0
Rin
2. To characterize an input device with unknown input resistance RL , you use a calibrated output device with V0 = 5 V and Rout = 1 Ω. When the input and output
device are connected, you find that the voltage across the load is VL = 2.5 V. Calculate
the value of RL .
3. To characterize an output device with unknown Rout , you use a calibrated input device
with RL = 100 Ω. You first make an open-circuit measurement of the output voltage
to discover that V0 = 10 V. Then, with the input and output device connected, you
find that the voltage across the load is VL = 1 V. Calculate the value of Rout .
4. We want to get the most light possible out of a light bulb and four 1.5 V batteries that
each have 1 Ω of internal resistance. If the bulb has a resistance of 8 Ω, which battery
configuration, series or parallel, will result in the most current, and therefore, the
most light? Recall that light output is roughly proportional to the power dissipated.
(a) If the three batteries are connected in series, what output voltage will actually
reach the bulb, and what current will flow through the bulb? Hint: Draw the
Thévenin equivalent circuit.
(b) If the three batteries are connected in parallel, what output voltage will actually
reach the bulb, and what current will flow through the bulb? Hint: Draw the
Thévenin equivalent circuit.
25
Lab 2 Exercises:
Input and Output Impedance
1. Design and construct a voltage divider using Vin = 5 V from the power supply. The
output voltage should be 1.0 ± 0.1 V. Use only two resistors, one of which should be
at least 1 k.
2
2. Construct the following voltage divider. Use the 15-V output from the DC power
supply. Measure Vout with a multimeter in the absence of a load and compare it to
the value that you calculate using the voltage divider equation.
15 V
10 k
Vout
4.7 k
Calculate the Thévenin equivalent output voltage and resistance of this output device,
and use these values to calculate the value of Vout when a 33 kΩ load resistor is added
to the circuit. Then add the load resistor to your physical circuit and measure the
value of Vout to see how it compares to your calculation.
2
3. Use the voltmeter and the resistor board to measure the output resistance, Rout , of
one of the following voltage sources: (a) D-cell, (b) 9-volt battery, or (c) solar cell. If
you choose the solar cell, make sure it is under constant illumination. We will compile
these measurements on the board in the lab so we can compare the different sources.
3
4. Use the potentiometer (“pot”) in your lab kit. It has three wires connected to it.
Your TAs will instruct you on how to connect it. If connected incorrectly, it is very
easy to break. Determine with an ohmmeter which of the three wires is connected to
the “wiper”.
(a) Construct the following voltage divider using this pot.
15 V
10 k
Vout (wiper)
View the output on the voltmeter, which should vary continuously from 0 to 15 V
as you turn the knob.
(b) Now design a circuit using the 10-k pot so that the output varies between −1 V
and +1 V when you turn the knob of the pot from one extreme to the other. You
will need to use both the +15 V and −15 V power supplies, and you will need to
use some resistors besides the pot.
Additional exercises on the next page
27
3
Exercises for Lab 2, Input and Output Impedance
29
5. Optional: At your lab station, there should be a variable lab power supply and a set
of Christmas light bulbs. The power supply has two knobs, a current limit and a
voltage limit. If the power supply is not attached to anything, it will output the set
voltage and no current. If a load resistor is attached to the power supply, then it will
continue to output that same voltage as long as the current is less than the current
limit (constant voltage mode). If the load resistance is low enough that the current
would exceed the current limit, then the output voltage drops until the current is
equal to the current limit (constant current mode).
Set the variable lab power supply voltage to 8 V and the current limit to 0.2 A. Don’t
trust the meters! To set the maximum voltage, disconnect everything from the power
supply, connect a voltmeter, and adjust the voltage limit knob until the reading gives
8 V. To set the maximum current, turn the current limit knob down to 0, connect
an ammeter on the 10-A scale across the output of the power supply, and adjust
the current limit knob until the current reads 0.2 A. We use the 10-A input of the
ammeter for two reasons. First, it has a smaller input resistance than the milliamp
input. Second, some of the meters we use have a 200 mA fuse on the milliamp input
and setting the current to the specified value will result in blowing the fuse. The key
to these methods are the very high and low resistances of the voltmeters and ammeters
respectively. Make sure you understand what you just did and why it works.
Measure the current and voltage output of the power supply when 1, 2, 3, 4, and 8
bulbs in series are connected to it. Use one multimeter to measure the voltage and a
second multimeter to measure the current (remember that the ammeter must be in
series with the bulbs). Plot the measured power supply current versus voltage in the
space below. Explain to the TA what the plot means.
10
voltage (V)
8
6
4
2
0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
current (A)
How many bulbs should be connected in series to the power supply to draw the most
power from the supply? How much is the power in this case?
Lab 3
AC Measurements
Notes for Lab 3:
AC Measurements
Up to this point, we’ve only dealt with constant, steady voltage sources like batteries
or power supplies. However, you probably know already that it’s common for signals to
vary with time for various reasons. Examples of this include mains current, the electricity
delivered to your home and accessed through wall plugs; digital signals such as those used
in computers and other digital devices; and analog signals like audio or old video signals.
Generally, we refer to any of these non-constant voltage signals as “AC” signals. As you may
know, AC stands for alternating current. In the case of mains current, this is because the
voltage changes sinusoidally (like a sine wave). Since the voltage periodically changes sign,
the current sometimes flows from the “live” lead to ground (also known as the “neutral”
lead), and sometimes in the opposite direction—hence the “alternating current.” This allows
for the use of electrical transformers that can convert electricity to very high voltages (for
long-distance power transmission) or to lower voltages (this is how you can charge a phone,
which requires 5 V, with a charger plugged into a 120 VAC outlet). However, sometimes
there is no neutral lead, and both leads have voltages that oscillate 180◦ out of phase (that
is, they are opposites). In addition, other signals can be all positive—oscillating between
0 V and some peak voltage, like most digital signals. These are still referred to as AC
signals in the context of this lab, but that’s not universal.
Properties of Sinusoidal Signals
The essential characteristics of any sinusoidal signal are its frequency (or period), and
amplitude. When dealing with multiple sine waves with the same frequencies, we also are
interested in their relative phase—whether one wave precedes or lags behind another. You’ll
see this in the next lab. An AC signal can also have a nonzero DC offset, which means that
the signal is more positive than negative, or vise versa.
Frequency and Period: Frequency, f , which is also known as cyclic frequency, is the
number of times that a wave is repeated in a second. A wave that goes through a full positive
and negative cycle 1,000 times each second is said to have a frequency of 1 kHz. “Hz” is
the abbreviation of Hertz, named after Heinrich Rudolf Hertz, who built the first radio
transmitter, thus proving the existence of electromagnetic waves and confirming Maxwell’s
equations. Ironically, his opinion of the non-scientific applications of his discovery was, “It’s
of no use whatevsoever.”
While the frequency characterizes a wave’s cycles per second, the period, T , relates the
actual duration (in seconds) of each cycle. They are inversely related: f = T1 . In the
following lab, we will also discuss and use the angular frequency: ω = 2πf .
The AC electric power distributed in the United States has a cyclic frequency of 60 Hz.
Other parts of the world use 50 Hz. Typical electrical signals for consumer level audio have
a maximum frequency of about 22 kHz. Signals for sending and receiving radio waves can
have a wide range of frequencies. Familiar examples include the AM radio band (1610 kHz
33
34
Notes for Lab 3, AC Measurements
to 1700 kHz), the FM radio band (87.9 MHz to 107 MHz), cellular frequencies (various
bands between 700 MHz and 2690 MHz), and WiFi (2.4 GHz and 5 GHz).
Voltage: There are several non-equivalent but useful ways to indicate the voltage amplitude
of an AC signal, including the peak voltage (also called amplitude), Vpeak , which is the
maximum deviation from the center of the wave; the peak-to-peak voltage, Vpp = 2Vpeak ;
and the root-mean-squared (RMS) voltage, VRMS . These three quantities are illustrated in
the diagram below.
V
V
rms
Vpeak
Vpp
t
It’s worth nothing that the peak-to-peak voltage isn’t a physical voltage that could be
measured between V and ground; it is a voltage “difference” that only exists in time.
However, there are circuits that can take advantage of this and “double” the voltage. We
will build something close to this in Lab 5 (Diodes and Rectification).
The RMS voltage is the square-root of the average of the voltage squared, which is a measure
of the average deviation from the center of the wave; but it is not strictly the average voltage,
which would be zero for a wave centered at zero. Like all function averages, the RMS voltage
is computed as an integral:
s
Z
1 T
VRMS =
V (t)2 dt .
T 0
√
For a sinuoidal signal, this expression reduces to VRMS = Vpeak / 2 ≈ 0.7Vpeak . When we
say that the voltage of a common wall outlet is 120 VAC, we mean that VRMS = 120 AC
Volts, and we almost never mention that the peak voltage is about 170 V.
Note that Vpeak and Vpp are independent of the DC offset, or mean, of the signal. That
is, a signal oscillating between −5 V and +5 V has the same Vpeak and Vpp as a signal
that oscillates between 0 V and +10 V. This is not true of the RMS voltage, but we don’t
usually calculate that for a sine wave with a DC offset, so the previous derivation is OK for
sinusoidal signals with a mean of 0 V.
Power: In Lab 1 (Voltage and Current), we learned that the power dissipated by a resistor
is calculated as P = V 2 /R. But for an AC voltage, which varies with time, the power also
varies with time. In most cases, any reference to the power dissipated by a resistive load in
an AC circuit is actually a reference to the average power, which is calculated as
Pavg =
1
T
Z T
P (t)dt =
0
1 1
RT
Z T
0
V 2 dt =
2
2
VRMS
1 Vpeak
=
.
R
2 R
Notes for Lab 3, AC Measurements
35
The convenience of this relationship between Pavg and VRMS is one of the main reasons that
we use the RMS voltage when referring to an AC voltage becuase we can use the same
formula for calculating power dissipated as we do for DC voltages. However, this is only
valid when we can use Ohm’s law to change P = IV to P = V 2 /R for any instant in time.
However, as we’ll see in the next section, there is sometimes a phase difference between the
voltage and the current, which affects how much power is actually delivered. In general, the
average power consumed by a circuit with a phase delay like this can be calculated using a
more general formula:
Z
1 T
V2
Pavg =
P (t)dt = RMS cos φ .
T 0
R
Other Types of Signals
Aside from sinusoidal signals, there are other common periodic signals used in electronics.
Square waves oscillate between their high and low voltage extremes almost instantaneously
and are used in digital circuitry, as we’ll see in Lab 10 (Digital Gates). Triangle waves look
almost like sine waves, but the voltage rises and falls with a constant slope. They are used
in generating and measuring AC signals. Sawtooth waves have voltages that rise with a
constant slope and then drop very quickly to their minumum value and rise again. They
are mostly used in the implementation of old CRT televisions and oscilloscopes, but they
can also be used to measure signals or in the creation or amplification of other signals. All
of these waveforms are also used in electronic music, and they have very distinct sounds.
Square waves also have an additional important parameter: duty cycle. A square wave’s
duty cycle is the percentage of time that the voltage is at the maximum (or high) value. A
symmetric square wave has a duty cycle of 50%, while a square wave with a duty cycle of 0%
or 100% would just be a DC voltage. Duty cycle can be varied continuously, allowing digital
circuits to make signals with different average voltages. This is called Pulse-width modulation. It is used in power conversion (all modern device chargers work this way), changing
the average power going to lights or motors, and rough digital-to-analog conversion.
Another important property of electric signals that has been mentioned but not clearly
defined is DC offset, also known as DC bias. The DC offset of a signal is its mean voltage,
and can be thought of as the signal’s baseline, or voltage about which the signal oscillates.
For example, a signal that oscillates between −5 V and +5 V has zero DC offset, while a
signal that oscillates between 0 V and +10 V has a DC offset of +5 V. It’s the same AC
signal, but “shifted up.” These signals are often considered to be the superposition of an
AC signal and a DC voltage—hence the term. In some cases where the DC offset of a signal
is such that the voltage is always positive, it can be regarded as a DC signal that changes
in time. This is especially the case for digital signals, since they oscillate between 0 V and
some positive voltage, but their current never changes direction.
Equipment
In this lab, we’ll be using a few new pieces of equipment. A function generator (also called
a signal generator) is a bench-top AC voltage source that allows the output amplitude and
frequency to be adjusted. It usually also has the ability to generate different waveforms
(sine wave, square wave, triangle wave, etc.) and to add a DC offset. An oscilloscope is
36
Notes for Lab 3, AC Measurements
a powerful tool for measuring AC voltage signals over a very wide range of frequencies.
It allows you to actually “see” the voltage in the signal as a function of time. You will
learn more about the capabilities of an oscilloscope by reading the “Introduction to the
Oscilloscope” section in Appendix I of this manual.
Name
In-Class Exercises for Lab 3:
CID
AC Measurements
1. Given a sinusoidal 1 kHz signal that oscillates between −5 V and +15 V, what are
the peak-to-peak voltage, the RMS voltage, the amplitude, and the DC offset of the
signal?
Voltage (V)
2. Let a 5 V peak-to-peak signal with a frequency of 1 kHz and no DC offset be the
input to a voltage divider constructed from two 1 kΩ resistors. Sketch two periods
of the input and expected output signals on the same graph, being careful to get the
relative amplitudes right.
3
2
1
0
-1
-2
-3
0
0.5
1.0
Time (ms)
1.5
2.0
3. A signal generator is set up to generate a 1 V peak-to-peak sine wave. A 4 Ω speaker
is connected to the 50 Ω output of the signal generator. What peak-to-peak voltage
is actually applied to the speaker? How much power is dissipated by the speaker?
37
Lab 3 Exercises:
AC Measurements
1. Read the “Introduction to the Oscilloscope” section in Appendix I in this lab manual.
You should experiment with adjusting the knobs and choosing menu options so you
understand what each one does. You should have the output of the signal generator
connected to channel 1 to provide an input signal as you go through the exercises.
After you have gone through the “Introduction to the Oscilloscope” complete the
following exercises.
(a) Connect the high-level output (HI) of the signal generator to the channel 1 input
of the oscilloscope. Take care that grounds on both instruments are connected
together (if you’re using a BNC cable, this is done for you, but it’s something
that you should be aware of). Look at the three types of waveforms (square,
triangle, and sine).
(b) Demonstrate that you know how to set and interpret the horizontal (time) and
vertical (amplitude) scales as well as the vertical offset knob.
(c) Set the frequency of the signal generator to 1 kHz and the peak-to-peak amplitude
to 5 V. Measure the amplitude and period with the oscilloscope, which is far more
accurate than the dials on the signal generator. Quantify any frequency error
that you observe.
(d) Turn on the DC offset switch on the signal generator and check the effect of
the offset knob with oscilloscope channel 1. Select “DC” input coupling from
the Channel 1 menu, observe the behavior of varying the DC offset, then select
“AC” coupling and observe the signal. Explain what you see.
(e) Use this sine wave as input to a voltage divider constructed on your breadboard
from two 10 kΩ resistors. Simultaneously display the input and output of the
voltage divider on the oscilloscope using the channel 1 and channel 2 inputs. Is
the output as you would expect?
7
2. Use the function generator to obtain a 1 kHz sine wave with 5 V peak-to-peak amplitude. Connect a 470 Ω resistor between its output and ground. What has happened
to the output of the function generator? Why? Use this information to calculate the
output resistance of the function generator.
3
39
Lab 4
RC Circuits and Filters
Notes for Lab 4:
RC Circuits and Filters
Capacitors were briefly introduced in Lab 3 (AC Measurements). In this lab we will look
more closely at how a capacitor behaves in a circuit. Much of the time a capacitor will be
used in conjunction with a resistor, creating what is known as an “RC Circuit.”
Whenever a current, I, flows into one lead of a capacitor, the same amount of current flows
out of the other lead. Thus, for the purpose of understanding circuits, we can consider
the current to flow through the capacitor, into one lead and out of the other. The main
difference is how the voltage across the capacitor changes as the current flows through it.
For a resistor the voltage is always just V = IR. But for a capacitor the voltage will change
as charge is deposited in or removed from the capacitor by the flow of current. You should
recall the equation
dV
I=C
dt
from Lab 3.
Just as with resistors, we can replace combinations of capacitors with an equivalent capacitor. But the combination rules are different. Two capacitors in series are equivalent to a
single capacitor:
=
C1
C2
1
1
1
=
+
C
C1
C2
Similarly, two capacitors in parallel are also equivalent to a single capacitor:
C1
=
C = C1 + C2
C2
The circuit below contains both a resistor and a capacitor in series. We will treat the voltage
coming from the left as an input, and the voltage going out to the right as an output.
Vin
Vout
R
C
43
44
Notes for Lab 4, RC Circuits and Filters
If we assume that all of the current which flows through R must also flow through C, then
we can write
Vin − Vout
dVout
=C
.
R
dt
The expression on the left-hand side of the equation is the current through R (the voltage
across the resistor is the difference of the voltages at the two ends), and the expression on
the right-hand side is the current through C. This is a differential equation for Vout . Let
us look at solutions of this equation for various limiting cases.
First, suppose that Vin is equal to a constant and we wait a long enough time so that Vout
is also constant. Then dVout /dt = 0, and, from the above equation, Vout = Vin . No current
flows through R, so there is no voltage drop across R, and the capacitor is simply storing
a charge Q = CVout .
Now suppose that we suddenly change Vin . The voltage across C cannot suddenly change.
That would require an infinite amount of current through R which would require either an
infinite voltage at Vin or R = 0. Instead the current is limited by the finite voltage difference
across R, resulting is a finite rate of change in Vout (the voltage across C). We can see from
the above differential equation that as Vout approaches the new value of Vin , the difference
in voltage across R decreases, the current through R decreases, and the rate at which Vout
approaches its final value decreases. In fact, Vout approaches its final value exponentially.
If we solve the above differential equation for this situation, we obtain
Vout = Vin(final) + (Vin(initial) − Vin(final) )e−t/RC ,
where the values of Vin before and after the sudden change at t = 0 are given by Vin(initial)
and Vin(final) , respectively. The product RC is called the “time constant” of the exponential
decay. Below we sketch the waveforms of Vin and Vout .
Vin
Vout
t
t
If Vin is a square wave, then the capacitor simply rounds off each of the sharp corners, as
shown in the sketch below:
Vout
Vin
t
t
Notes for Lab 4, RC Circuits and Filters
45
The time constant, RC, can be determined from any of these exponential Vout decay curves.
Just remember that an exponential decays by a factor 1/e during one time constant RC,
as illustrated below:
−t/RC
e
V
V/e
RC
The curved lines are exponentials of the form e−t/RC . When the time changes by one time
constant (i.e., t = RC), the exponential drops to 1/e (0.368) of its original value, as shown
in the above figure.
Instead of an abrupt change of voltage, we will now consider a smooth sine-wave input
V = V0 sin(ωt) (recall that ω = 2πf where f is the frequency of the signal in Hz). Because
this is a more difficult case to analyze, we will start with only a capacitor C (no resistor).
The current through C is given by
I=C
dV
= ωCV0 cos(ωt) = I0 cos(ωt) .
dt
We note here that (1) the current is 90◦ out of phase with the voltage, and (2) the amplitude
of the current I0 = ωCV0 is proportional to the amplitude of the voltage V0 . This is Ohm’s
law for a capacitor:
V0 = I0 Z ,
where Z is called the impedance and is given by
Z=
1
.
ωC
The impedance of a capacitor is measured in units of Ohms (Ω) and depends on the frequency of the sinusoidal voltage across it: large for low frequencies and small for high
frequencies. Despite the fact that impedance has units of Ohms, like a resistor, a capacitor
cannot dissipate power. Power is computed as P = IV . But for a capacitor, the current
and voltage are 90◦ out of phase, so that the power averages to zero.
Now let us return to the circuit we have been considering:
Vin
Vout
R
C
46
Notes for Lab 4, RC Circuits and Filters
If we view the capacitor to be like a resistor, except that the impedance depends on the
frequency, then this circuit looks very much like a voltage divider. But now Vout depends
on the frequency. At low frequencies, where the impedance of C is much greater than R,
we have Vout ≈ Vin . At high frequencies, where the impedance of C is much less than R,
we have Vout Vin . This circuit allows signals with low frequencies to pass through, but
blocks signals with high frequencies. We call this a low-pass filter.
Of course, a capacitor does not behave exactly like a resistor. Its current is 90◦ out of phase
with its voltage. (A resistor’s current and voltage are in phase with each other.) Thus, the
above circuit only approximately behaves like a true voltage divider. The correct expression
for Vout in the case of the above circuit is given by
1
Vout = p
Vin .
(ωRC)2 + 1
Below, we display this expression graphically:
1
0.7
Vout/Vin
0
0.01
0.1
1
ωRC
10
100
Note that when the angular frequency ω is equal to 1/RC, we have Vout /Vin =
This is called the cutoff frequency:
ωcutoff =
p
1/2 ≈ 0.7.
1
.
RC
Now consider a similar circuit:
Vin
Vout
C
R
This circuit also looks much like a voltage divider. At low frequencies, where the impedance
of C is much greater than R, we have Vout Vin . At high frequencies, where the impedance
of C is much less than R, we have Vout ≈ Vin . This circuit allows signals with high frequencies
Notes for Lab 4, RC Circuits and Filters
47
to pass through, but blocks signals with low frequencies. We call this a high-pass filter. The
correct expression for Vout in the case of this circuit is given by
ωRC
Vout = p
Vin .
(ωRC)2 + 1
Below, we display this expression graphically:
1
0.7
Vout/Vin
0
0.01
0.1
1
ωRC
10
100
Observe that for the same R p
and C values, the low and high-pass filters have the same
cutoff frequencies (Vout /Vin = 1/2), but opposite trends.
Name
In-Class Exercises for Lab 4:
CID
RC Circuits and Filters
1. What resistance in series with a 100 µF capacitor will yield a 1 second time constant?
2. Suppose that we send a 10 V peak-to-peak 250 Hz square wave with zero offset into
the input of the circuit below. Calculate the RC time constant and use the expression
for exponential decay to determine the expected maximum peak-to-peak voltage at
the output. Sketch several periods of the input signal together with the expected
output signal on the same graph, taking care to get the relative amplitudes right.
Vin
Vout
1 kΩ
1 µF
3. You have an audio amplifier, and you wish to amplify signals with frequencies of
2000 Hz or greater. Unfortunately, there is a lot of unwanted 60-Hz noise on your
microphone input. Design a filter circuit that will attenuate the 60-Hz pickup without affecting the desired frequencies too much. Note that filter circuits usually use
capacitors of less than about 1 µF and resistors of at least a few kΩ.
(a) Draw your circuit and label the values of all components.
(b) Determine the ratio Vout /Vin for both 60 Hz and 2000 Hz for your design.
(c) How would you modify this circuit to achieve a low-pass filter with the same
cutoff frequency?
49
Lab 4 Exercises:
RC Circuits and Filters
1. Design and construct a low pass filter for which Vout /Vin = 0.7 at 100 ± 20 Hz.
(Since the output resistance of the signal generator is about 500 Ω, you should use
R ≥ 10 kΩ.) It is best to choose the capacitor first and then adjust the resistors to
give the desired frequency, since you have a lot more options available with resistors
than with capacitors. Again, display your input and output simultaneously on the
oscilloscope. Measure the frequency at which Vout /Vin = 0.7 by measuring the period
on the oscilloscope.
3
2. Design and construct a high pass filter for which Vout /Vin = 0.7 at 10 ± 2 kHz. Use a
0.001-µF capacitor.
2
3. The object of this exercise is to practice setting up the oscilloscope to observe a signal
and to measure some of the characteristics of that signal. The signal we are going to
use is a square wave that is modified by passing it through the following RC circuit:
Vin
Vout
10 k
0.01 µF
The combination of a resistor and a capacitor in this manner will modify the input
square wave because the voltage on the capacitor cannot change immediately. In fact,
the output voltage, Vout , will undergo an exponential decay from the previous value
to the new value according to the equation on page 44 of the lab notes.
(a) Feed a 500-Hz square wave from the signal generator into the circuit above (amplitude does not matter). Simultaneously view both the input and the output
on the oscilloscope.
(b) In the Trigger menu set the trigger type to Edge and the source to whichever
channel is connected to the output of the above circuit. Set the trigger mode to
Auto. Now adjust the trigger level control up and down. Pay particular attention
to what happens when the trigger level is set too high or too low. Now alternate
the trigger Slope between positive and negative and note how it changes the
oscilloscope display.
(c) Now set the trigger mode to Norm (normal). Adjust the trigger level control up
and down, noting what happens when the trigger level is set too high or too low.
Explain to your TA what is happening in parts (b) and (c).
Additional exercises on the next page
51
Exercises for Lab 4, RC Circuits and Filters
53
(d) Do the same thing as in parts (b) and (c) with the trigger source set to whichever
channel is connected to the input of the above circuit. Explain why things are
different.
(e) In order to obtain the most accurate measurements of time and voltage, you
should expand a single RC decay signal so that it fills the screen as much as
possible, both horizontally and vertically.
(f) Now on to business. Measure the RC time constant from the exponential decay
that you observe on the oscilloscope. Also use your multimeter to independently
measure R and C and calculate the time constant (the components must be
removed from the circuit before attempting to measure their values). Record the
calculated and measured time constants in the margin of your notebook.
3
4. Use the signal generator to obtain a 1-kHz square wave. Connect a 100 kΩ resistor
from the output of the signal generator to the input of the oscilloscope and view
the result. (Connect the ground of the signal generator directly to the ground of
the oscilloscope.) You should see exponential decay at the edges of the square wave,
similar to what you saw in the previous exercise. This is evidence of capacitance to
ground. Use the oscilloscope display to determine the amount of capacitance. Part
of that capacitance is due to the input capacitance of the oscilloscope (13 pF, printed
on the oscilloscope near the input connector), and part of it is due to capacitance
between the inner wire and the outer grounded shield of the coaxial cable. How much
capacitance is in the cable?
2
5. Optional: The circuit below is a notch filter. It greatly attenuates signals with frequencies near f = 1/(2πRC). Construct this circuit using R = 10 k and C = 0.1 µF.
You can obtain R/2 by putting two 10 k resistors in parallel and 2C by putting two 0.1µF capacitors in parallel. Test this circuit. Does the “notch” occur at the frequency
you calculate using the known values of R and C?
R
R
2C
Vin
Vout
C
C
R/2
Additional exercises on the next page
Exercises for Lab 4, RC Circuits and Filters
55
6. Optional: Capacitors always have some internal resistance between the two leads. If
you charge up the capacitor and then disconnect the power supply, the capacitor will
discharge itself through its own internal resistance R with a time constant RC. If
you measure the time constant of the discharge, you can use the known value of C to
measure R. Try this with a 100 µF capacitor. Note that this capacitor has polarity
and must be charged up with the positive voltage applied to the end marked with
a plus sign. Charge the capacitor to 15 V. Remove the power supply leads and let
the capacitor sit undisturbed for about a minute. Then measure its voltage with the
digital voltmeter. The voltage decays exponentially with time: V = V0 e−t/RC , where
V0 is the initial voltage (15 V). From the measured values of V and t, calculate R.
Be careful not to touch the leads of the capacitor while it is discharging. Your body
also has resistance and can cause an extra discharge of the capacitor, resulting in false
data.
Lab 5
Diodes and
Rectification
Notes for Lab 5:
Diodes and Rectification
A diode is a two-lead device which allows current to flow in only one direction. Below is its
symbol:
V1
V2
When V1 < V2 , the diode has infinite resistance and allows no current to flow. When
V1 > V2 , the diode has very small resistance and current flows freely in the direction of the
arrow in the symbol: from V1 to V2 .
The diode actually requires a small amount of voltage (usually about 0.6 V) to be turned
on. Thus, when current is flowing from V1 to V2 , there will be about 0.6 V across the diode.
This voltage drop increases slightly as the current through the diode is increased. We will
ignore this voltage drop in the following discussion, but we must keep in mind that it is
present and will affect the actual performance of the circuits.
A light-emitting diode (LED) gives off light when current is flowing through it. In the
following circuit, the diode will light up when Vin > 0. (You may see either of the two
symbols to indicate an LED.)
Vin
Vin
LED
Like other diodes, the LED requires a certain amount of voltage to turn it on. For the LED,
this minimum voltage is usually about 1.5 V. The LED lights up quite brightly if we put
about 20 mA through it. However, even 2 mA is sufficient to produce a visible amount of
light. The resistor in the above circuit controls how much current flows through the LED.
Quite often the resistor and the LED are contained in the same package and are referred
to as an RLED. The LEDs in your kit may have resistors built into their package. It is
hard to tell if your LED does have an integral resistor. It is usually best to use a 150-Ω (or
larger but not too large) resistor in series with the LED if you are going to connect it to
5 V unless the exercises instruct you otherwise.
The following circuit is called a half-wave rectifier, and is like a voltage divider. When
Vin > 0, the diode is turned on and behaves like a small resistor, resulting in Vout ≈ Vin .
When Vin < 0, the diode is turned off and has infinite resistance, resulting in Vout = 0. We
show the resulting waveform when the input is a sine wave.
Vin
Vout
59
60
Notes for Lab 5, Diodes and Rectification
Vin
Vout
t
t
The objective of this lab is to use diode rectification to convert an AC voltage source into
a DC power supply capable of delivering 100 mA of current. Whereas your signal generator
has too large of an output impedance (600 Ω) for such a large current, we will instead employ
a transformer that converts 110 Vrms ac wall power into a smaller (Vrms ≈ 15 V) AC signal.
A transformer consists of two coils of wire wound around opposite sides of a solid ring of
magnetic material (e.g., iron), and is represented by the symbol below.
Through a phenomenon called electromagnetic inductance, which we will not explain in
detail here, an AC voltage, Vp , in the primary (input) coil creates an AC voltage, Vs ,
in the secondary (output) coil. The number of coil windings on the primary (Np ) and
secondary (Ns ) sides then dictate the transformer voltage and current ratios according to
the transformer equation:
Vs
Ns
Ip
=
=
.
Vp
Np
Is
This equation assumes that the power delivered on the primary side is all transferred to the
secondary side, based on the realistic assumption that most transformers are nearly 100%
efficient. A simple rearrangement of the equation yields
Ps = Is V s = Ip V p = Pp .
There are actually three wires in a modern wall socket. The two “slots” are the “hot” and
“neutral” wires, while the round hole is the safety ground, which is connected to a rod
that has been driven down into the dirt below the building – hence the name ”ground”).
The neutral wire is also connected to ground at some nearby location, usually at the main
electrical service entrance. But unlike the safety ground, the neutral wire is intended to
carry the return current that flows out from the hot wire. One should never purposely
cause current to flow through the ground wire. A safe high-power circuit will be enclosed
in a grounded metal case, so that if the hot wire accidentally makes contact with the case,
the current can safely travel through the ground wire and thereby cause a fuse to blow or
a circuit breaker to trip, rather than flowing through you.
Any time power is brought from a wall socket into an important circuit, a fuse should be
inserted into the hot wire in series with the circuit in order to limit the amount of current
Notes for Lab 5, Diodes and Rectification
61
and power that can be dumped into it in the not entirely unlikely event of a short circuit.
The wall socket itself is often fused at 20 A. But this still allows over 2000 W of power
dissipation at 110 VAC, enough power to melt components and even start a fire. Thus, we
will use a much smaller fuse in order to further limit the maximum current allowed to pass
through our power supply. Remember that it is much easier to replace a burnt-out fuse
than to replace an entire circuit. We first determine the current requirements of the load,
which lies on the secondary side of the transformer, and then use the transformer equation
to compute the corresponding primary current that dictates the size of the fuse located on
the primary side.
Building a DC Power Supply
There are a variety of ways to rectify the AC output of a transformer. Shown below is a
simple half-wave diode rectifier.
Vout
110 VAC
Vout
RL
t
If the secondary coil of the transformer has a “center tap” (a connection to the center of
the secondary coil), we can construct a full-wave rectifier as shown below. Note that the
secondary coil has no direct connection to the primary coil. We can ground the secondary
coil anywhere we want. When we ground the center tap, we obtain two sine waves from the
two ends of the secondary coil, 180◦ out of phase with each other and each with half the
amplitude of the sine wave from the secondary coil in the previous circuit where one end of
the coil was grounded. When the voltage on the top of the secondary coil is positive, the
top diode conducts. When the voltage on the bottom of the secondary coil is positive, the
bottom diode conducts. This circuit is actually two half-wave rectifiers connected together.
Vout
110 VAC
Vout
RL
t
The most efficient way to rectify voltage from the secondary coil is shown below. Here,
neither end of the transformer is grounded. When the top of the secondary coil is positive
with respect to the bottom, diodes 1 and 3 are turned on. The load is connected to the top
of the coil, and ground is connected to the bottom. When the bottom of the coil is positive
with respect to the top, diodes 2 and 4 are turned on. The load is connected to the bottom
62
Notes for Lab 5, Diodes and Rectification
of the coil, and ground is connected to the top. This results in full-wave rectification with
twice the voltage obtained in the previous case. The four diodes in the above circuit are call
a “diode bridge”. Diode bridges for power supplies are available in a single package with
four leads.
V
110 VAC
out
4 1
3 2
Vout
RL
t
Now let us connect a capacitor between the output of the above circuit and ground, as
shown below.
110 VAC
Vout
4 1
3 2
Vout
C
RL
t
Each time the rectified voltage reaches its peak, the capacitor is charged up. In between
peaks, the capacitor discharges through the load. This results in a DC voltage with some
“ripple” on it. The time constant of the capacitor discharge is RL C, so the amplitude of
the ripple can be reduced by using a larger capacitor.
The last stage of the power supply is the voltage regulator. This is a miniature integrated
circuit (IC) which contains some zener diodes, resistors, and transistors. For example, the
78L05 is a 5-volt regulator. It has three leads: input, output, and ground. If we put any
voltage between 7 and 30 V on the input, we get 5 V from the output. In the power supply,
we connect the input of the regulator to the unregulated voltage with ripple on it, and we
connect the output to the load, as shown below. We now have a regulated power-supply.
110 VAC
Voltage
Regulator
4 1
3 2
IN
Vout
OUT
GND
RL
The 78L05 has self-protection built into it. If it gets too hot, it automatically turns itself
off. Similarly, if we try to draw too much current from the output, it also turns itself off.
The 78L05 will provide up to 100 mA.
Name
In-Class Exercises for Lab 5:
CID
Diodes and Rectification
1. For the diode circuit below, Vin = (2 V) sin(ωt). Sketch Vout as a function of t. You
may ignore the diode’s threshold voltage.
Vin
Vout
2. The circuit below shows a configuration of diodes called a diode bridge. Consider the
path of current through the bridge circuit when point A is at a higher voltage than
point B. Draw arrows in the direction that current flows through the diodes and label
them with a “1”. Now consider what happens when point B is at a higher voltage
than point A. Label these currents with a “2”. What is the sign of the voltage Vout
in both of these cases? Draw a qualitative sketch of the output voltage as a function
of t if the voltage on the output of the transformer is as shown (VAB = VB − VA ).
A
110 VAC
Vout
B
RL
VAB
t
Additional exercises on the next page
63
In-Class Exercises for Lab 5, Diodes and Rectification
65
3. Make an estimate (an upper bound, actually) for the amount of ripple on a 100 µF
capacitor in a power supply that has 10 mA of current drawn from it. Remember that
1
the time between peaks is 120
second.
Lab 5 Exercises:
Diodes and Rectification
1. At your lab station you have a transformer with three outputs. The two red terminals
are connected to each end of the secondary coil of the internal transformer. The one
black terminal is connected to the center tap of the internal transformer. Plug in your
lab transformer and turn it on. Use your oscilloscope to view the voltage amplitude
(not peak-to-peak voltage) of either one of the red outputs relative to the center tap
and record it here in your lab manual. Then measure the voltage amplitude of the
signal between the two red outputs and record it (this should be twice the value
obtained from the center-tapped configuration; be sure you understand why that is
the case).
1
2. (a) Using your transformer, construct the half-wave rectifier shown below. View
the voltage between the two red outputs of the transformer on channel 1 of the
oscilloscope, while simultaneously viewing the rectifier output relative to ground
on channel 2. Graph these signals in your lab manual, and indicate the amplitude
of each signal.
Vout
110 VAC
1 kΩ
(b) Using your transformer, construct the full-wave rectifier shown below. View
the input voltage (the voltage across the secondary of the transformer) and the
output voltage with the oscilloscope after reading the entire caution below. Graph
these signals in your lab manual and indicate the amplitude of each signal. How
do the results compare to those of the half-wave rectifier?
Vout
110 VAC
1 kΩ
3
CAUTION: You must be careful how you make this measurement. If you
simultaneously connect one channel of the scope to the output of the transformer
(a red clip lead on one of the red transformer terminal and a black clip lead on
the other of the red transformer terminal) and the second channel of the scope
to the output of the diode bridge (a red clip lead on Vout and a black clip lead on
ground), you will NOT get what you expect. The reason is that the two black
leads on the scope cables are connected together (and to ground) at the scope.
Caution is continued on the next page...
Additional exercises on the next page
67
Exercises for Lab 5, Diodes and Rectification
69
This means that you will have connected a wire between the center tap (since it
is grounded) and one end of the transformer windings, shorting out one half of
the transformer.
You can, however, use the advanced features of the scope to view both signals
simultaneously. Turn on channel 1 of the scope and attach the leads to the
output of the transformer. Adjust the trigger level to be at about 2 to 3 V (be
sure it is set to a positive value). Adjust the vertical and horizontal scales to get
a good picture of the transformer output – get at least one complete cycle on the
screen. Now press the Ref button to save that trace as a reference by pressing
the “Save” button (be sure the source is set to channel 1). Now remove the leads
from the transformer output and attach them to the output of the rectifier circuit
(Vout to ground). You will be able to simultaneously view the transformer signal
that was saved as a reference with the output of the rectifier.
3. (a) A full-wave rectifier can also be constructed using the W04G diode bridge (or
other equivalent component), which actually contains four diodes connected in
the bridge configuration (see p. 62 in the lab notes). Study the pin diagram of
your chip, and then construct the circuit shown below. Do not connect the
center-tap of the transformer output to anything!
Graph the transformer output voltage and Vout on the accompanying graph. Be
sure that you get the relative amplitudes correct.
110 VAC
W04G
~ +
-
~
Vout
10 k
How do these results compare to the half-wave and full-wave rectifiers in the
previous exercise?
NOTE: You will need to use the same technique as in the previous exercise to
simultaneously view the input and output voltages on the diode bridge circuit.
If you try to connect the scope to both the input and output simultaneously you
will effectively connect a wire across either diode 3 or diode 4 (see the diagram
on p. 62) depending on how you attach the scope leads.
You will probably want to turn off the reference waveform before proceeding to
the next exercise.
Additional exercises on the next page
Exercises for Lab 5, Diodes and Rectification
71
(b) Filter the output by connecting a 1 µF capacitor between Vout and ground. Be
aware that the 1 µF capacitor has polarity – connect the “minus” side as indicated by the symbols on the case to ground. Repeat for 10 µF and 100 µF and
explain what happens to the peak-to-peak magnitude of the “ripple” as the capacitance increases. You can magnify the ripple on the oscilloscope by switching
the coupling on the scope channel to “AC”.
3
4. Construct the circuit shown below. The 78L05 is a positive output 5-volt regulator.
The identification of the pins on the 78L05 are shown on the pinouts sheet in your
packet. The 0.01 µF capacitor on the output of the voltage regulator is required for
stability.
110 VAC
W04G
~ +
-
~
78L05
100 µF
IN
Vout
OUT
GND
0.01 µF
You now have a complete power supply. Measure Vout with the digital voltmeter.
How accurate is the 5-volt regulator? Connect a 1-k load resistor between Vout and
ground. Does this affect Vout ? Replace the 1-k resistor with 50 Ω. (You can obtain
50 Ω by putting two 100-Ω resistors in parallel.) Calculate the current through the
50-Ω resistor and explain what causes the output to droop under such a heavy load.
Simultaneously view Vout and the voltage on the 100 µF capacitor on channels 1 and
2 of the oscilloscope. Set the vertical position of both channels to be at the same
place when the inputs are grounded, and view the signals with “DC” input coupling,
making sure that both channels are at the same volts/div setting. What problem do
you see and how could you solve it?
Additional exercises on the next page
3
Exercises for Lab 5, Diodes and Rectification
73
5. Optional: Now construct the diode-bridge circuit shown below. Use the diodes in your
kit, not the packaged diode bridge. Pay attention to the directions of the diodes! Do
not connect the center-tap of the transformer output to anything! The 1-k resistors are
in the circuit to protect the diodes until you are sure they are in the circuit correctly
and operating properly. If you get one of the diodes in backwards it will get very hot
and you will likely let the smoke out. Once the circuit is working, remove the 1-k
resistors and replace them with wires.
1k
110 VAC
Vout
1k
10 k
You want to compare the transformer output with the bridge output. You should
look at the instructions for exercise 3 to properly view the two waveforms. Graph
these signals in your lab manual and indicate the amplitude of each signal. How do
the results compare to those of the half-wave and full-wave rectifiers?
6. Optional: Connect an LED between the output of the signal generator and ground.
Drive it with a square wave. The output resistance of the signal generator is large
enough so that the current through the LED will be properly limited. Adjust the
frequency of the signal generator so that the LED blinks once per second. Now
connect an additional LED so that one LED is on while the other is off, i.e., the
LEDs blink 180 degrees out of phase with each other.
Lab 6
Transistors and
MOSFETs
Notes for Lab 6:
Transistors and MOSFETs
A bipolar junction transistor (BJT) has three leads labeled as emitter, base, and collector.
There are two different types of BJTs, NPN and PNP, as shown below.
NPN
collector
PNP
emitter
base
base
emitter
collector
The base-emitter junction is like a diode with the orientation indicated by the arrow in
the symbol. The transistor is “turned on” by applying a forward-biased voltage across the
diode junction so that a small current, IB , flows between the base and the emitter in the
forward diode direction. This small current then causes a much larger current, IC , to flow
between the collector and the emitter (also in the forward diode direction). IC and IB are
roughly proportional so that IC = βIB , where the value of β is typically at least 100, though
this value can vary widely from one transistor to the next. Thus, the BJT is essentially a
current amplifier. We put a little current through the base, and get a lot of current through
the collector.
Two simple rules govern the behavior of a transistor. For NPN BJTs, these are:
1. VC ' VE + 0.2 V when the transistor is fully on (but not when turned off).
2. When the transistor is on, VB ≈ VE + 0.6 V. The 0.6 V here is the expected diode
voltage drop across the base-emitter junction. The transistors that we are using
experience IB ' 1 mA and IC ' 100 mA when fully on.
For PNP BJTs, the corresponding rules are:
1. VE ' VC + 0.2 V when the transistor is fully on (but not when turned off).
2. When the transistor is on, VE ≈ VB + 0.6 V. The 0.6 V here is the expected diode
voltage drop across the emitter-base junction. The transistors that we are using
experience IB ' 1 mA and IC ' 100 mA when fully on.
Note that IC , IB , and |VC − VE | must not exceed maximum (safe) values or smoke will
result. More advanced transistor analysis requires the use of transistor curves, but this is
beyond what we need here.
77
78
Notes for Lab 6, Transistors and MOSFETs
In the circuit below, an NPN transistor is being used as a switch.
+VCC
RL
Vin
The purpose of this switch is to turn on the current through the load RL . When the
transistor is turned on, it draws current from the power supply (+VCC ) through the load
and into the collector. This current causes a voltage drop across the load. When the current
reaches the value IC = VCC /RL , the voltage drop across the load is equal to VCC and the
voltage at the collector is near 0 V. The base current required to do this is IB = IC /β. If we
put more current into the base than this amount, IC will not increase any further because
the collector voltage cannot drop below the emitter voltage. We say that the transistor is
“saturated” and fully turned on. Thus, turning on the switch requires that IB > IC /β. For
example, if we have a load which requires 100 mA, we can turn on this current with only a
few milliamps into the base.
Here is the analogous switch made with a PNP transistor.
+VCC
Vin
RL
The following NPN transistor circuit is called an emitter follower.
+VCC
Vin
Vout
RL
The input voltage turns on the transistor. Current flows from the power supply (+VCC )
into the collector, out of the emitter, and through the load RL into ground. This current
Notes for Lab 6, Transistors and MOSFETs
79
causes a voltage drop across the load. As the current increases, the voltage at the emitter
increases until it becomes about 0.6 V below the voltage at the base. It cannot increase any
further because the base-emitter junction requires 0.6 V to be turned on. Therefore Vout
“follows” Vin . If Vin increases, so does Vout . If Vin decreases, so does Vout . Vout stays 0.6 V
below Vin . A current IC through the load only requires an amount of current IB = IC /β
into the base. Thus, the input resistance of the base is βRL , much larger than the resistance
of the load itself.
Note that the circuit above can only drive the load with positive voltage. If Vin goes negative,
the transistor is turned off. If we want to drive the load with a negative voltage, we can do
so by employing a PNP transistor in our emitter follower.
Vout
Vin
RL
-VCC
A metal-oxide-semiconductor field-effect transistor (MOSFET) also has three leads. They
are labeled source, gate, and drain, as shown below:
drain
gate
source
drain
gate
source
The symbol on the right is the “correct” symbol for the MOSFET we will be using since it
is an n-channel enhancement mode MOSFET (i.e., the conduction region is made of n-type
semiconductor and the conductivity is enhanced, or the resistivity is reduced, by applying a
voltage to the gate). The symbol on the left, either with or without the arrow in the center,
is a more generic symbol for a MOSFET transistor.
Current flows into the drain and out of the source in an n-channel device. This MOSFET is
turned on by applying a few volts to the gate (measured with respect to the source). Unlike
a BJT, no current flows through the gate of an ideal MOSFET because it is well insulated
from both the drain and source. In practice, there is a little bit of capacitance between the
gate and the other two leads, so that a tiny current flows each time the voltage on the gate
is changed. But we will ignore this effect for now.
80
Notes for Lab 6, Transistors and MOSFETs
In the figure below, we illustrate a switch circuit based on a MOSFET. Because no gate
current is required, such a switch is generally superior to the corresponding transistor circuit.
With this MOSFET, we can switch a large current on and off merely by changing the gate
voltage at zero gate current.
+VCC
RL
Vin
Name
In-Class Exercises for Lab 6:
CID
Transistors and MOSFETs
1. In the transistor switching circuit below, assume that R is 100 kΩ. How much current
goes through the resistor when the switch is turned on, and what will the light bulb
do? Reconsider these questions for R = 10 kΩ and R = 1 kΩ. Ignore the transistor’s
forward-voltage bias (i.e. assume a perfect transistor with β = 100).
+10 V
100 Ω
when lit
b
R
c
e
2. The following circuit is an emitter follower. Vin = (3 V) + (1 V) sin(ωt). Sketch Vout
as a function of t. What happens to Vout if a 500 Ω load is attached to the follower
output?
+10 V
Vin
Vout
1k
3. When an output device is used to directly drive a load resistor RL , the input resistance
of the load is simply equal to RL . However, when the load is incorporated into a FET
switch as shown below, the input resistance is increased. Estimate the effective input
resistance of this new configuration.
+VCC
RL
Vin
81
Lab 6 Exercises:
Transistors and MOSFETs
1. Briefly test the incandescent lamp (#47) in your lab kit by connecting the two leads
between 5 V and ground. It should light up brightly. Next, obtain a 1-Hz square
wave from the signal generator that switches between 0 V and 5 V as shown below.
This will require that you adjust the DC offset while observing the output with your
oscilloscope. Once this is working, connect the two leads of the lamp between the
signal generator output and ground. Does the lamp blink brightly? Why or why not?
5V
0V
Next, construct the following two transistor switches and connect the signal generator
output to the Vin of both circuits. The identifications of the pins on the 2N3904 and
2N3906 transistors are given in the “Pinouts” section in this lab manual. Because you
only have one incandescent lamp, you will only be able to test one circuit at a time.
5V
5V
#47 Lamp
Vin
Vin
2N3906
#47 Lamp
2N3904
In either circuit, the lamp should blink brightly. Explain why. Observe the signal
generator output with your oscilloscope while watching the lamp blink. For the NPN
circuit, does the lamp turn on when the Vin is high or low? Why? How is the PNP
circuit different? Why?
Additional exercises on the next page
83
4
Exercises for Lab 6, Transistors and MOSFETs
85
2. Construct the following transistor switch using a power MOSFET. Use the same input
from the signal generator as in exercise 1 above. The identification of the pins on the
IRF3704 MOSFET is given on the pinouts sheet in your packet.
5V
#47 Lamp
Vin
IRF3704
The lamp should blink brightly as before. Now disconnect the signal generator from
Vin . Connect the ground of the power supply to the ground of the circuit. Connect a
wire to Vin and hold the end of it between two fingers of one hand. With the other
hand, touch the end of the wire connected to the 5 V power supply. The lamp should
turn on. Touch the ground. The lamp should turn off. Explain why this happens.
You may also find it interesting to hold the wire connected to Vin with one hand and
then move your other hand around, but without touching anything, and observe what
happens.
Additional exercises on the next page
3
Exercises for Lab 6, Transistors and MOSFETs
87
3. This next exercise illustrates the uses and limitations of a voltage follower circuit.
Obtain a 1-kHz sine wave from the signal generator. Adjust its amplitude to be 5 V
peak to peak. Adjust its offset to be about 5 V above ground. Now connect a 100-Ω
resistor between the output of the signal generator and ground. View the result on
the oscilloscope. What happened to the signal? Why? Next disconnect the 100Ω resistor from the signal generator and use the sine wave as input for the emitter
follower shown below. The capacitor is necessary for stability. Note that there is no
connection between the capacitor and Vout .
+15 V
2N3904
Vin
100
0.1 µF
Vout
View Vin and Vout simultaneously on the oscilloscope. Adjust the oscilloscope channel 1
and channel 2 vertical offsets to zero with the inputs grounded so that you can measure
both the DC and AC components of these signals accurately. What does the follower
do to the signal? What happens when the DC offset switch on the signal generator
is turned off? Why? Qualitatively, how large is the AC output impedance of the
follower compared to 100 Ω. How is a follower useful?
4. Optional: Devise a method to measure the current gain β = IC /IB of the 2N3904
transistor in your lab kit. Be sure that the voltage between the collector and emitter
is at least 2 V when you make this measurement.
3
Lab 7
Operational Amplifiers
Notes for Lab 7:
Operational Amplifiers
An operational amplifier (also called an op-amp) has two inputs: the noninverting
input (Vin+ ) and the inverting input (Vin− ), as well as one output (Vout ). It has the
following symbol:
Vin-
−
Vin+
+
Vout
Like a transistor, an op-amp is an active device that requires power to operate. Two of
its pins, usually labeled V + and V − on a pin diagram, must be connected to positive and
negative power supplies, respectively. These positive and negative power supply voltages
are often called +VCC and −VCC , respectively, though the power-supply connections are
not usually shown explicitly on circuit diagrams.
An op-amp is essentially a differential amplifier, which means that its output voltage is
proportional to the difference of the two input voltages:
Vout = A(Vin+ − Vin− ) .
The proportionality constant A is called the open-loop gain and is usually greater than
105 at low frequencies. Thus, a small difference in input voltages is amplified to become a
large voltage at the output.
If we ground Vin+ , the amplifier output will be Vout = −A Vin− , so that the input and
output have opposite signs. We call this an inverting amplifier. If we instead ground
Vin− , the amplifier output will be Vout = A Vin+ which yields a non-inverting amplifier.
The following circuit includes negative feedback and is called an inverting amplifier.
R2
R1
Vin
−
+
Vout
Because the resistor R2 allows the voltage at the output to influence the voltage at the
input, it is called a feedback resistor.
For the inverting amplifier above, first observe that when Vin is 0 V, Vin+ , Vin− and Vout
will also be zero. Then consider that raising Vin to a small positive voltage will raise Vin−
to a small positive voltage, which causes Vout to swing to a large negative voltage, which
91
92
Notes for Lab 7, Operational Amplifiers
then pulls a current to the right through R2 and thereby pulls Vin− back down towards
zero. Similarly, lowering Vin− to a small negative voltage causes Vout to swing to a large
positive voltage, which will then pushes a current to the left through R2 and thereby pushes
Vin− back up towards zero. Because the reaction of the output always opposes the original
change at the input, we refer to the feedback as “negative”.
Negative feedback always opposes fluctuations at the input and thereby stabilizes the output
at an equilibrium value somewhere between +VCC and −VCC . Positive feedback (achieved
by attaching a feedback resistor from Vout to Vin+ ), on the other hand, reinforces input
fluctuations and thereby destabilizes the output, causing Vout to swing towards infinity
until it saturates at either +VCC and −VCC .
When a negative feedback resistor is present, the large open-loop gain, A, guarantees that
the op-amp will do everything it can to force the two inputs to be equal. And the large
input resistance of the op-amp guarantees that this happens without drawing a substantial
current through either of the inputs. Even if you don’t fully appreciate how remarkable this
op-amp magic is, remember the following two rules for working with a negative feedback
op-amp circuit.
1. Vin− = Vin+ (as long as the configuration allows the amplifier to achieve this).
2. Iin− = Iin+ = 0.
With these two rules in mind, We can now determine the stable value of Vout that results
from a given value of Vin . Because Vin+ is connected to ground, the first rule requires that
Vin− also be zero. The second rule requires that all of the current that passes through R1
must pass through R2 . The application of Ohm’s law to each resistor then yields
(Vin − Vin− )
(Vin− − Vout )
=
,
R1
R2
which can be reduced to
R2
Vout
=−
.
Vin
R1
The gain (amplification factor) is equal to the ratio of the two resistances, and can be set
by selecting appropriate values for R1 and R2 . The minus sign reminds us that the input
and output of an inverting amplifier have opposite polarities.
You should also notice that the input impedance of an inverting amplifier is fairly easy
to approximate. Since the op-amp will do everything it can to hold Vin− at 0 V, the input
impedance appears to be just R1 to ground. This is fairly accurate for low frequency signals.
As the frequency of the input signal increases, the input impedance will vary from this value
due to the non-ideal characteristics of a real op-amp.
By exchanging the Vin and ground connections we can make a noninverting amplifier.
Notes for Lab 7, Operational Amplifiers
R1
93
R2
−
+
Vin
Vout
We can clearly see that Vin = Vin+ . And the first op-amp rule further requires that Vin+ =
Vin− , so that Vin = Vin− . Because the second op-amp rule allows no current to flow into the
inverting input, the R1 and R2 resistors form a simple voltage divider where
Vin
R1
=
,
Vout
R1 + R2
which can be arranged as
Vout
R2
=
+1.
Vin
R1
The input impedance of a non-inverting amplifier is also simple to estimate. If you look
at the data sheet for your amplifier, it will usually list the differential input resistance or
the input resistance of the amplifier. For an op-amp that uses BJT input stages, this is
typically 100 to 200 MΩ. If you find a higher-quality op-amp that uses FET input stages,
this value will be more like 1013 Ω.
A special case of this circuit, shown below, is a voltage follower.
−
Vin
+
Vout
Here, R2 = 0 and R1 = ∞, so that Vout = Vin . The follower has a very large input resistance
and a very small output resistance. It easily outperforms a follower constructed from a single
transistor. Internally, the op-amp is actually a rather complicated circuit containing many
transistors.
Op-amps have limitations which are important to understand.
1. The output current of an op-amp is limited. If we try to draw too much current from
an op-amp, its output voltage will rise until its output current maxes out and then
go no further. This condition can, for example, be achieved by shorting the output to
ground, and does not harm the op-amp.
2. Regardless of the resistor values that you use, the gain of an amplifier circuit can’t be
greater than the open loop gain of the op-amp, which is very large, but not infinite.
94
Notes for Lab 7, Operational Amplifiers
3. The open loop gain of an op-amp decreases at high signal frequencies. The frequency
at which the open loop gain drops to 1 (i.e., no gain at all) is called the bandwidth of
the op-amp, and can vary from about 1 MHz up to 10 GHz depending on the quality
(and cost). This means that an amplifier made from an op-amp will only be good up
to a certain frequency, beyond which its output voltage will drop. Note that high-gain
op-amps tend to have small bandwidths.
You should refer to the data sheet for the TL3472 in Appendix II. This is the specific op-amp
that we will be using for the exercises in this lab. The data sheet lists the limitations on the
operating parameters of the op-amp as well as typical parameters and some performance
graphs. The data sheet also includes a representative schematic of the internal circuitry of
the op amp on p. 2. Some of the specifications are given in units of dB (decibels). If you
have not encountered these units before, you will find a short discussion in Lab 13(Audio
Amplifier). dB is a logarithmic unit that is commonly used in electronics.
We can modify the inverting amplifier to create a summing amplifier.
Vin1
Vin2
R
R1
R2
−
+
Vout
Since Vin+ is grounded, the op-amp adjusts its output so that Vin− = 0. The total current
into Vin− must be zero, so we have
Vin1 Vin2 Vout
+
+
=0.
R1
R2
R
Rearranging terms, we obtain
Vout = −
R
R
Vin1 −
Vin2 .
R1
R2
If R1 = R2 = R, this becomes
Vout = −(Vin1 + Vin2 ) .
The output is equal to the sum of the two inputs (with a minus sign). For other choices of
resistors, we can obtain a “weighted” sum of the two inputs.
With a little more rearranging we arrive at a difference amplifier.
Notes for Lab 7, Operational Amplifiers
R
Vin1
R
−
R
Vin2
95
Vout
+
R
The voltage at Vin+ is 12 Vin2 . The op-amp adjusts its output so that Vin− is also equal to
1
2 Vin2 . Since the total current into Vin− must be zero, we have
Vin1 − 12 Vin2 Vout − 12 Vin2
+
=0.
R
R
Rearranging terms, we obtain
Vout = Vin2 − Vin1 .
The output is equal to the difference of the two inputs. If we choose values of resistors
which are different from each other, we can also obtain various “weighted” differences from
this kind of circuit.
The following circuit is a constant current source.
+VCC
R
+VCC
R1
−
+
R2
RL
We are very acquainted with voltage sources. A power supply is usually a voltage source.
It puts out a certain voltage no matter what the load is (within limits). The current it puts
out depends on the load. The smaller the load resistance, the larger the current. A current
source is just the opposite. It puts out a certain current no matter what the load is (within
limits). The voltage it puts out depends on the load. The larger the load resistance, the
larger the voltage.
In the above circuit, the resistors, R1 and R2 , form a voltage divider and determine the
voltage at Vin+ . The op-amp adjusts its output so that Vin− = Vin+ . Thus, the emitter of
the transistor will be at Vin+ and its base 0.6 V lower than Vin+ . Note that in this case,
the negative feedback does not come directly from the output of the op-amp. The negative
96
Notes for Lab 7, Operational Amplifiers
feedback can come from any point in the circuit as long as the op-amp can somehow control
the voltage at that point. In this case, the op-amp controls the voltage on the emitter of
the transistor via the voltage on the base. The output of the op-amp will automatically
adjust its value so that the emitter is exactly at Vin+ .
Since the voltage across R is fixed, +VCC on one end and Vin+ on the other end, the current
through R is also fixed. More than 99% of this current comes out of the collector of the
transistor into the load RL . A small amount comes out of the base into the output of the
op-amp. IC (the collector current) is what is provided to RL . This current can be related
to I, the current through R:
I = IC + IB = IC + IC /β = IC (1 + 1/β)
I
β
IC =
=I
.
1 + 1/β
β+1
Recall that β ' 100, and doesn’t vary significantly for a given transistor. Thus the current
through the load is fixed. It depends only on the choice of the other resistors in the circuit.
This circuit is called a clamp.
+VCC
R1
R
Vin
Vout
−
+
R2
The resistors, R1 and R2 , form a voltage divider and determine the voltage at Vin+ . The
op-amp tries to adjusts its output so that Vin− = Vin+ . When Vin is less than Vin+ , The
output of the op-amp increases, trying to bring Vin− up to Vin+ . But the diode does not
conduct current in that direction, and no matter how high the output of the op-amp goes,
it cannot affect Vin− . In this case, the op-amp does not succeed in forcing Vin− = Vin+ . The
highest voltage that the output of the op-amp can go to is +VCC , and it just stays there,
frustrated in its attempt at negative feedback. Since no current flows through the diode,
Vout and Vin are simply connected by a resistor, R, and we have Vout = Vin .
When Vin is greater than Vin+ , the output of the op-amp decreases. Now current is flowing
through the diode in the correct direction, and the op-amp is able to pull Vin− down to the
value of Vin+ . Thus, we have Vout = Vin+ . The circuit “clamps” the voltage at Vin+ . Vin+
is called the clamp voltage and is determined by the two resistors, R1 and R2 . This circuit
allows signals which are less than the clamp voltage to pass through it. If a signal greater
than the clamp voltage tries to pass through it, the circuit holds the output at the clamp
voltage. The purpose of resistor R is to limit the current through the diode when the circuit
is clamping the voltage.
Notes for Lab 7, Operational Amplifiers
97
Replacing one of the resistors in the input/feedback circuit on an inverting amplifier results
in some very useful circuits. If you replace the feedback resistor with the capacitor you have
an integrator.
C
Vin
R
−
Vout
+
Since Vin+ is grounded, the op-amp adjusts its output so that Vin− = 0. The total current
into that point must be zero, so we have
Vin
dVout
+C
=0.
R
dt
Solving for Vout , we obtain
1
Vin dt .
RC
The output is proportional to the integral of the input. If the input is some constant
voltage, then the output is a “straight line,” i.e., the voltage ramps linearly, either up or
down, depending on the sign of the input. A square wave on the input produces ramps
alternating between upward and downward directions, i.e., a triangle wave. This circuit
could also be used to obtain the area of a voltage pulse.
Z
Vout = −
The integrator circuit shown above has a problem. If the input is not exactly balanced
between positive and negative voltages, its integral will tend to drift, and eventually, the
output will saturate at either +VCC or −VCC . Drift can also be caused by a small bias
current from the (non-ideal) op-amp inputs. This can be remedied by connecting a large
resistor, R0 , in parallel to C, as shown below.
R′
C
Vin
R
−
+
Vout
98
Notes for Lab 7, Operational Amplifiers
If, instead of replacing the feedback resistor, you replace the input resistor with a capacitor
the resulting circuit is a differentiator.
R
C
−
Vin
+
Vout
Since Vin+ is grounded, the op-amp adjusts its output so that Vin− = 0. The total current
into Vin− must be zero, so we have
C
dVin Vout
+
=0.
dt
R
Solving for Vout , we obtain
dVin
.
dt
The output is proportional to the derivative of the input.
Vout = −RC
Name
In-Class Exercises for Lab 7:
CID
Operational Amplifiers
1. Calculate the gain Vout /Vin for the circuit shown below.
10 kΩ
Vin
1 kΩ
−
Vout
+
2. Calculate the gain Vout /Vin for the circuit shown below.
1 kΩ
10 kΩ
−
+
Vin
Vout
3. Calculate the gain Vout /Vin for the circuit shown below.
−
Vin
+
Vout
Additional exercises on the next page
99
In-Class Exercises for Lab 7, Operational Amplifiers
101
4. Calculate the output of the circuit below when V1 = 5 V, V2 = 2 V, R1 = R2 = 1 kΩ
and R = 2 kΩ.
R
R1
V1
R2
V2
−
Vout
+
5. Assuming that the load resistance RL is appropriate to the application below, determine the current delivered to the load when VCC = 15 V, R1 = 10 kΩ, R2 = 5 kΩ and
R = 1 kΩ.
+VCC
R
+VCC
−
R1
+
RL
R2
6. Let the input voltage be a 10 V peak-to-peak sine wave. Sketch several periods of the
output voltage when VCC = 15 V, R1 = 10 kΩ, R2 = 5 kΩ and R = 1 kΩ.
+VCC
R1
R
Vin
Vout
−
+
R2
Lab 7 Exercises:
Operational Amplifiers
1. (a) Using the op-amp chip in your kit (either the LM741 or the TL3472), design and
construct an inverting amplifier with a gain of −10±1. The input resistor should
be at least 10 kΩ to avoid loading down the signal generator too much. Use a
2 V peak-to-peak sine wave with no DC bias and a frequency of about 1 kHz as
input. View both the input (channel 1) and output (channel 2) simultaneously
on the oscilloscope. Trigger the oscilloscope externally using the TTL output of
the signal generator. Set the V/div and the vertical offset of each channel, and
also Invert channel 2 (on the Channel 2 menu), so that the input and output
overlay each other on the oscilloscope.
CAUTION: Don’t forget to connect the V + and V − pins of the op-amp to the
+15-V and −15-V power supplies respectively. Have you ever wondered why
your computer isn’t working, only to discover that you forgot to plug it in? Also
take care not to insert the op amp chip upside down. If you connect +15 V and
−15 V to the wrong pins, you can destroy the chip.
(b) Test the voltage limit of your op-amp by increasing the input voltage to 5 V
peak-to-peak. Explain the difference between what you saw before and what you
see now.
(c) Adjust the input voltage to 1 V peak-to-peak, and test the current-driving ability
of your op-amp by adding a 100-Ω resistor between the output and ground.
If your output appears fuzzy, add a 0.1 µF capacitor between the output and
ground. Use the result to determine the maximum output current of your opamp.
(d) Now use another op-amp to construct a second inverting amplifier with the same
gain (-10) and send the output of the first amplifier into the input of the second
amplifier.
Note that the TL3472 contains two amplifiers in the same package - you do not
need a second chip if you are using this amplifier. Check the pinout drawings at
the front of the manual to see how to wire the second amplifier.
View the input of the first op-amp (channel 1) and the output of the second
op-amp (channel 2) simultaneously on the oscilloscope (you will want to turn
off the Invert on channel 2). Use a 0.1 V peak-to-peak sine wave as input. The
resulting gain should be +100.
4
2. Design and construct a noninverting amplifier with a gain of 5.0 ± 0.5. As in the
previous exercise, verify the amplification factor by overlaying the input and output
signals on the oscilloscope.
3
Additional exercises on the next page
103
Exercises for Lab 7, Operational Amplifiers
105
3. (a) Set your signal generator amplitude knob about 1/4 of full amplitude, set the
frequency to about 2000 Hz, and view the output on the oscilloscope. Attach a
speaker to the signal generator output and again measure the output signal and
record its amplitude. What happens to the amplitude of the signal when you
attach the speaker?
(b) Now build an op-amp voltage follower and insert it between the signal generator
and the speaker. Measure the voltage amplitude across the speaker. Since the opamp destabilizes when it reaches its current limit, it may help to connect a 1-µF
capacitor between the terminals of the speaker (in parallel with the speaker).
What do these measurements tell you about the relative output resistances of
the signal generator and the follower?
4. Optional: Construct a summing amplifier, as shown in the notes for this lab. Use R1 =
R2 = R = 10 k. The output voltage of this circuit should be Vout = −(Vin1 + Vin2 ).
Test this, using 5 V and −15 V as the two inputs. Also try adding together 5 V from
the power supply and a sine wave from the signal generator. How would you change
this circuit if you wanted the output to be Vout = −(Vin1 + 2Vin2 )? Make the change
and try it. Use the +5-V power supply for the input with gain = 1 and a sine wave
for the input with gain = 2.
5. Optional: Construct a clamp circuit, as shown in the notes for this lab. Choose R1
and R2 so that the clamp voltage is 1 V. For Vin , use a 1-kHz sine wave from the
signal generator. Turn the DC offset off and set the amplitude to its maximum value.
Demonstrate that you obtain the desired clamping effect.
6. Optional: Construct a current source, as shown in the notes for this lab. Choose R1
and R2 so that the noninverting input of the op amp is 10 V. The op-amp will adjust
its output so that its inverting input is also 10 V. Thus the resistor R will have 15 V at
one end and 10 V at the other end. Choose R so that the current through it is 1 mA.
Since more than 99% of this current comes out of the collector of the transistor, the
current through the load will also be 1 mA. Adjust the resistors, if necessary, so that
the current through the load is 1.0 ± 0.1 mA. Test this circuit for various values of RL .
Does this circuit put out 1 mA no matter how large RL is? What is the limit on RL ?
7. Optional: Design and construct a noninverting amplifier with variable gain. Use a 10k
pot to control the gain. When the knob on the pot is fully turned counterclockwise,
the gain should be 1. When the knob on the pot is fully turned clockwise, the gain
should be 10.
8. Optional: Construct an integrator circuit, as shown in the notes for this lab. Put
a 1-MΩ resistor in parallel with the capacitor. (This is the resistor R0 shown in the
figure on p. 97 of the notes for this lab.) If the input is a square wave, the output
should be a triangle wave. You should choose RC to be larger than the period of the
input signal.
Additional exercises on the next page
3
Exercises for Lab 7, Operational Amplifiers
107
9. Optional: Construct a differentiator circuit, as shown in the notes for this lab. If the
input is a triangle wave, the output should be a square wave. If the input is a square
wave, the output should be a series of positive and negative “delta functions” (spikes).
You should choose RC to be smaller than the period of the input signal.
Lab 8
Comparators
Notes for Lab 8:
Comparators
Consider the following circuit:
−
Vin
Vout
+
The first thing to notice about this circuit is that there is no feedback from the output to
the inputs. The output cannot influence the input. The output is simply given by
Vout = A(Vin+ − Vin− ) = AVin .
This circuit amplifies Vin with a gain equal to A, the open loop gain of the op-amp. The
value of A is typically greater than 105 . What happens when Vin = 1 V? We do not get
an output of 100 kV! The highest voltage anywhere in the circuit inside the op-amp is VCC .
Therefore its output cannot exceed VCC . When we try to require the op-amp to put more
than VCC on the output, it just puts out VCC . We say that the op-amp is “saturated.”
The output of the op-amp in the above circuit will be saturated at +VCC whenever Vin is
greater than VCC /A, which is a very small voltage, typically much less than a millivolt.
Similarly, its output will be saturated at −VCC whenever the Vin is less than −VCC /A, also
a very small voltage. Thus, except for a very narrow range of input voltages, Vout will be
either +15 or −15 V. We usually neglect that small range of voltages and write, as a good
approximation,
(
+VCC ,
Vin > 0;
Vout =
−VCC ,
Vin < 0.
The output tells us whether or not the input is greater than 0 V. It does not tell us how
much greater (or less). It just tells us yes or no. This kind of circuit is called a comparator.
It compares Vin with the voltage at Vin− . The voltage that Vin is compared to is called the
threshold voltage. In this case, the threshold voltage is 0 V.
In the circuit below, we connect the output of a voltage divider to Vin− .
+VCC
R1
−
+
Vin
R2
111
Vout
112
Notes for Lab 8, Comparators
Depending on the values of R1 and R2 , the threshold voltage at Vin− can be anything
between +VCC and ground. This circuit compares Vin with the voltage at Vin− . We obtain
(
Vout =
+VCC ,
−VCC ,
Vin > Vin− ;
Vin < Vin− .
We can reverse the polarity of the output by interchanging the two inputs:
+VCC
R1
−
Vin
Vout
+
R2
We are now comparing Vin with the threshold voltage at Vin+ , and we have
(
Vout =
+VCC ,
−VCC ,
Vin < Vin+ ;
Vin > Vin+ .
There is always some “noise” on a signal. Sometimes this noise can cause a very fast
comparator to switch back and forth several times as Vin moves across the threshold voltage.
We can solve this problem by adding a little positive feedback to the circuit.
+VCC
R1
−
Vin
Vout
+
R2
R
-VCC
Suppose that Vin is below the threshold voltage. The output is at +VCC . As Vin crosses
the threshold voltage, Vout switches to −VCC . Since Vout is connected to Vin+ through the
resistor R, this decrease in Vout causes a decrease in Vin+ . The threshold voltage jumps to
a lower value, and Vin is now suddenly above the threshold voltage. As long as the jump in
Notes for Lab 8, Comparators
113
the threshold voltage is greater than the amplitude of the noise on Vin , the output will only
switch once. Similarly, when Vin crosses the threshold voltage coming down, Vout switches
from −VCC to +VCC . This increase in Vout now causes the threshold voltage to jump to a
higher value, and Vin is now suddenly below the threshold voltage. The amount that the
threshold voltage jumps as the output switches depends on the size of R. Lower values of
R result in larger jumps. A comparator with positive feedback is called a Schmitt trigger.
Note that the feedback must be to Vin+ , not Vin− .
In some applications, the speed with which a comparator switches its output from +VCC
to −VCC and back again is important. ICs are available which are made especially for this
purpose and have very fast switching speeds. We will use the LM311 (sometimes referred
to as just the 311) comparator for some of our exercises.
You should note that the LM311 has a ground connection indicated on the pinouts, unlike
the typical op amp. This is because the LM311 has what is often referred to as an “open
collector” output. Below is the schematic for the output stage of the LM311.
(to the rest
of the LM311)
Output
(pin 7)
4Ω
GND
(pin 1)
This is just the schematic for the transistor switches we encountered in Lab 6 (Transistors
and MOSFETs). It is called “open collector” because the output is an NPN transistor with
no internal connection to the collector. This is advantageous because it allows the LM311
to switch up to 50 V and and 50 mA loads. It is a disadvantage because there is no voltage
source present on the output.
In order for the LM311 to work properly you will have to connect a resistor between some
positive voltage source (typically 5 V in our case) and the output (pin 7). This resistor
usually needs to be large with respect to the resistor between the emitter of the internal
transistor and ground – 4 Ω in the case of the LM311. In our exercises we usually use 100 Ω
to 5 V. Those values can be changed depending on the application.
Name
In-Class Exercises for Lab 8:
CID
Comparators
1. Sketch the output Vout for the op-amp circuit below when the input is a sine wave of
amplitude VCC . Note the limits on the voltage divider.
+VCC
1k
−
Vin
Vout
+
2k
-VCC
2. (a) When no feedback resistor is present in the circuit shown below, what nominal
reference voltage Vref is produced by the voltage divider on the left?
(b) When the feedback resistor is present, Vout alters the value of Vref . When the
Vin rises above Vref , the output swings to ground causing Vref to change to what
value?
(c) Because this op-amp has an open collector output, it requires a 100-Ω pull-up
resistor on the output. When Vin drops below Vref , the output swings up to 5 V
causing Vref to change to what value (ignore the small contribution of the pull-up
resistance)?
+15 V
+5 V
2k
Vin
Vref
1k
−
+
100
Vout
10 k
115
Lab 8 Exercises:
Comparators
1. Construct the following comparator circuit using an op-amp. For Vin , use a 5-kHz
square wave from the signal generator. Adjust the amplitude to the highest possible
setting.
Vin
−
Vout
+
View Vout on the oscilloscope. Determine the time required for the op-amp to switch
its output between −15 V and +15 V. Show your result to the lab instructor.
Now replace the op-amp with an LM311 comparator. This is an IC made especially
for this kind of circuit. Be careful: the pin assignments for the LM311 are not the
same as for your op-amp. Also note that there is a ground connection in addition
to the usual connections to +15 V and −15 V . Connect a 100-Ω resistor between the
output and +5 V from the power supply. View Vout on the oscilloscope. Determine
the time required for the LM311 comparator to switch its output between 0 and 5 V.
Save this circuit for the next exercise.
3
2. Use the LM311 comparator circuit from exercise 1. For Vin use a 100-Hz sine wave
from the signal generator. Adjust the amplitude to be 1 V peak to peak. View Vout
on the oscilloscope. Look carefully at the transitions between 0 and 5 V. What do you
see? Show it to the lab instructor. Now add some positive feedback, as shown below.
+5 V
Vin
−
+
1k
100
Vout
22 k
View Vout on the oscilloscope. How is Vout now different from before?
Additional exercises on the next page
117
4
Exercises for Lab 8, Comparators
119
3. Design and construct a comparator circuit, using the LM311, such that
(
Vout =
5 V,
0 V,
Vin > 2.0 ± 0.2 V;
otherwise.
For input use a 1-kHz sine wave from the signal generator. Adjust its amplitude
to be 6 V peak to peak. It is not necessary to include the 22k positive feedback
resistor in this circuit. The 100-Ω pull-up resistor is still necessary. View Vin and Vout
simultaneously on the oscilloscope. Trigger the oscilloscope externally, using the TTL
output of the signal generator. Turn on the DC offset on the signal generator and
observe what happens to the output as the offset is moved up and down.
4. Optional: Design and construct a comparator circuit, such that
(
Vout =
5 V,
0 V,
Vin < −5 V;
otherwise.
3
Lab 9
555 Timer
Notes for Lab 9:
555 Timer
The 555 timer is an IC with eight pins. Their functions are as follows. (The input/output
in parentheses indicates whether the pin is an input to the IC or an output from the IC.)
GROUND
V+
Connect to ground.
Connect to the power supply, VCC . It can be any voltage between 5 and 15 V.
(output) This is the output of the timer. It has only two states, called high and
low. The output is high when it is at VCC and low when it is at ground. Note that the
output of the 555 timer can source or sink 200 mA.
OUTPUT
TRIGGER
(input) When the voltage at the trigger input goes below VCC /3, the output goes
high. The output is not affected when the trigger input goes above VCC /3. It remains high.
THRESHOLD
(input) When the voltage at the threshold input goes above 2VCC /3, the
output goes low. The output is not affected when the threshold input goes below 2VCC /3.
It remains low.
(output) The discharge pin is grounded by the IC whenever the output is low.
It is an open circuit whenever the output is high.
DISCHARGE
RESET
(input) When the voltage at the reset input is zero, the output is low, no matter
what the voltages at the trigger and threshold inputs are. Normally we connect the reset
input to VCC to allow the IC to function.
CONTROL VOLTAGE
(input) The control voltage input can be used to change the voltages
at which the trigger and threshold inputs affect the output. The trigger input causes the
output to go low when its voltage goes below one-half the voltage at the control voltage
input. The threshold input causes the output to go high when its voltage goes above the
voltage at the control voltage input. When the control voltage input is not connected to
anything, its voltage is set internally by the IC to be 2VCC /3.
For uses of the 555 timer, study the data sheet in Appendix II.
123
124
Notes for Lab 9, 555 Timer
Below is an oscillator circuit using a 555 timer.
VCC
VCC
VCC
V+
R1
DISCHARGE
RESET
555
R2
TRIGGER
OUTPUT
Vout
THRESHOLD
C
GND
When the trigger input goes below 13 VCC , the output goes high. The capacitor charges up
through R1 + R2 , gradually raising the voltage on the threshold input. When the threshold
input goes above 32 VCC , the output goes low and the discharge pin is grounded. This
discharges the capacitor through R2 alone, gradually lowering the voltage on the trigger
input. When the trigger input goes below 13 VCC , the cycle repeats itself again. The amount
of time the output spends in the high state depends on R1 + R2 . The amount of time the
output spends in the low state depends on R2 alone. Since R1 cannot be zero (the IC would
not be able to ground the discharge pin if it were connected directly to VCC ), the output
will spend more time in the high state than in the low state, resulting in an asymmetrical
square wave. The period, T , of oscillation (sum of time in low state and time in high state)
is given by
T = (ln 2)(R1 + 2R2 )C ≈ (0.7)(R1 + 2R2 )C .
If we control the reset input, instead of just connecting it to VCC , we can turn this oscillator
on and off. The oscillator will be off (Vout = 0) when the reset input is at 0 V.
Notes for Lab 9, 555 Timer
125
Below is a pulse generator (also called a monostable multivibrator) using a 555 timer.
VCC
VCC
VCC
V+
Vin
TRIGGER
RESET
555
R
DISCHARGE
OUTPUT
Vout
THRESHOLD
C
GND
When Vin is high (above 13 VCC ), the output just stays low. The discharge pin also stays
at ground, keeping the capacitor discharged. When Vin goes low, the output goes high,
and the capacitor charges up through R, gradually raising the voltage on the threshold
input. When the threshold input goes above 23 VCC , the output goes low again, and the
capacitor is quickly discharged. Nothing else happens as long as Vin has gone high again in
the meantime. This circuit puts out a single pulse of length T , given by
T = (ln 3)RC ≈ (1.1)RC .
Name
In-Class Exercises for Lab 9:
CID
555 Timer
1. Design a circuit using a 555 timer that will oscillate at a frequency of 5 kHz. Give the
values of all resistors and capacitors.
2. Design a circuit using a 555 timer that will give a pulse 25 ms long when it is triggered.
Give the values of all resistors and capacitors.
127
Lab 9 Exercises:
555 Timer
You should connect a 1 µF capacitor between 5 V and ground on your breadboard and leave
it there for each of the exercises in this lab.
1. (a) Design and construct an oscillator with a frequency of 1.0 ± 0.1 kHz using a 555
timer. Use 5 V for VCC . Display the output signal on your oscilloscope to verify
the frequency.
(b) Connect the output of your oscillator through a 1-k current-limiting resistor to
a speaker with the second terminal on the speaker connected to ground. This
will produce a soft 1-kHz tone. Use the oscilloscope to measure the peak output
voltage to an accuracy of 0.05 V and divide this voltage by the load resistance
(i.e., 1 k + speaker resistance) to determine the amount of current drawn by the
speaker. Record the output voltage and current here in your lab notebook. Then
refer to the room-temperature curve in the figure in the bottom left-hand corner
of page 5 of the 555 data sheet in Appendix II of your lab manual and verify that
the current-voltage point that you measured lies on or close to this curve (i.e.,
the chip is performing as designed). Note that the vertical axis in the figure is
actually VCC − Vout .
Save this oscillator circuit for use in the exercises that follow, some of which will
require two 555 timers. You should have an extra 555 timer in your lab kit.
3
2. Use a 555 timer to construct a pulse generator that produces a 4 ± 1 s 5 V pulse.
Connect the trigger input of the 555 timer to a switch that connects to ground when
pressed and 5 V when released. Connect the output of the pulse generator to the
Reset input of the oscillator you built previously (first disconnect the Reset pin from
5 V). Each time you press the button, the 1-kHz tone from the speaker will sound for
4 s.
3
Additional exercises on the next page
129
Exercises for Lab 9, 555 Timer
131
3. Use a single 555 timer to construct a metronome. The output of your circuit should
have the following waveform.
The short pulses should be about 0.8 s apart. The width of each short pulse should
be as small as possible. If you connect this to a speaker, you will hear a click, click,
click. . .
Recommendations: Use a large capacitor in series with the speaker rather than a 1-k
resistor to eliminate DC bias current, and make R2 as small as possible (even zero)
in order to keep the pulses short.
4. Optional: Design and construct an “alarm” circuit that produces the following waveform. To accomplish this, disassemble the pulse generator from the previous exercise
and reuse its 555 timer to build a 2-Hz oscillator. Then connect its output to the
Reset input of your previously constructed 1-kHz oscillator.
≈0.5 s
≈1 ms
This figure is not drawn to scale. There should be about 250 cycles in each “burst”.
When the speaker is attached to the output of the 1-kHz oscillator (don’t forget the
1k series resistor), you should hear beep, beep, beep. . .
Next, add a 311 comparator to your circuit (like the one from exercise 2 of the comparator lab). Choose the resistor values so that the inverting input is fixed at 2.5 V,
while using a potentiometer to deliver a variable voltage between 0 and 5 V to the
non-inverting input. Once you connect the output from the comparator to the Reset
input of the 2-Hz oscillator, the alarm should beep continuously whenever the variable
comparator input is greater than 2.5 V.
Additional exercises on the next page
4
Exercises for Lab 9, 555 Timer
133
5. Optional: Design and construct a dimmer switch. It should include a 10k pot. When
the knob on the pot is turned fully counterclockwise, the output should look like the
following.
≈100 ms
≈8 ms
When the knob on the pot is turned fully clockwise, the output should look like the
following.
≈100 ms
≈80 ms
In other words, the pot controls the length of each pulse, which varies from 8 ms to
80 ms as the knob on the pot is turned. The pulses are about 100 ms apart. Connect
this output to an NPN transistor switch and use it to turn a #47 incandescent lamp
on and off. (You will not be able to visually detect the flicker, though. Why not?)
Turning the knob on the pot should cause the brightness of the lamp to change.
Hint: Build a 100-Hz oscillator with very short negative-going pulses, like the
metronome circuit. Use the output of this oscillator to trigger a pulse generator.
Use the pot in series with a resistor in this pulse generator to produce pulses with
widths that vary from 8 ms to 80 ms.
Lab 10
Digital Gates
Notes for Lab 10:
Digital Gates
The voltage in a digital signal has only two states. One state is 0 V, called the “low” state
or logical “0”. The other state is 5 V, called the “high” state or logical “1”. A digital gate
is a circuit with one or more digital inputs and a digital output. The output is some logical
function of the inputs. Below, we list the most common digital gates found on ICs.
Inverter. If the input is 0, the output is 1. If the input is 1, the output is 0. The symbol for
the inverter is shown below. The little circle on the output indicates the NOT operation.
Also shown below is the “truth table” for the inverter. In a truth table, the output is shown
for every possible combinations of inputs.
A Q
0 1
A
Q
1 0
AND gate. The output is 1 if all inputs are 1. The output
A
0
A
0
Q
B
1
1
is 0
B
0
1
0
1
if any of the inputs is 0.
Q
0
0
0
1
OR gate. The output is 1 if any of the inputs is 1. The output
A B
0 0
A
0 1
Q
B
1 0
1 1
is 0 if all inputs are 0.
Q
0
1
1
1
NAND gate. This is an AND gate with an inverter on the
the symbol indicating the NOT operation.
A
0
A
0
Q
B
1
1
Q
1
1
1
0
output. Note the small circle in
B
0
1
0
1
NOR gate. This is an OR gate with an inverter on the output.
A B
0 0
A
0 1
Q
B
1 0
1 1
137
Q
1
0
0
0
138
Notes for Lab 10, Digital Gates
XOR gate. This is an exclusive OR gate. The output is
both inputs are 1.
A
0
A
0
Q
B
1
1
1 if either input is 1, but not if
B
0
1
0
1
Q
0
1
1
0
Now let us consider some simple combinations of digital gates. First, suppose we tie together
the two inputs of a NAND. This is simply an inverter. Similarly, we can also make an inverter
from a NOR.
=
=
Next, suppose we invert each input of a NAND. This example is straightforward if we don’t
try to do too much in our head. First, we label the outputs of the inverters to be C and D.
Then we construct a truth table showing A, B, C, D, and Q. We write down all possible
combinations of A and B. Then we write down the results of the inverters at C and D.
Then we write down Q by performing a NAND function on C and D. The result is shown
below. We recognize that Q is simply A OR B. A NAND with both inputs inverted is
simply an OR. Similarly, a NOR with both inputs inverted is simply an AND.
A B C D Q
C
0 0 1 1 0
A
Q
0 1 1 0 1
D
1 0 0 1 1
B
1 1 0 0 1
We can do an even more complicated example (see below). As in the last example, we
label all intermediate steps, in this case, the outputs of the two inverters, C and D, and
the outputs of the two ANDs, E and F . We then work through the truth table, as shown
below. When we are done, we recognize that Q is simply A XOR B. We have built an
exclusive OR.
A
F
D
Q
C
B
A
0
0
1
1
B
0
1
0
1
C
1
1
0
0
D
1
0
1
0
E
0
1
0
0
F
0
0
1
0
Q
0
1
1
0
E
The two most common families of digital ICs are TTL (transistor-transistor logic) and
CMOS (complementary MOSFET). The TTL ICs are made of BJTs. Each input is connected to the base of a BJT in the IC. If an input is not connected to anything, it usually
“floats” up to 2 or 3 V and is interpreted by the IC to be a logical 1.
The CMOS ICs are made of MOSFETs. The inputs have very high input resistance and
must be protected against static charge. They require much less current than TTL.
Notes for Lab 10, Digital Gates
139
Digital ICs are labeled with an alphanumeric code. This code indicates the function of the
IC as well as its family. For example, the 7400 is standard TTL using BJT technology and
contains four NAND gates. The 74LS00 also contains four NANDs in the same configuration, but the LS indicates that the circuit is low-power Schottky TTL. This is a popular
type of TTL IC. It requires less power than the 7400. The 74HCT00 uses CMOS technology
and also contains four NANDs in the same configuration. All of these ICs have the same
pinouts and are usually interchangeable in a circuit.
The 74xx families of digital gates usually define a “0” to be an input voltage between 0 V
and 0.8 V. A “1” is an input voltage between 2 V and 5 V. For the outputs a “0” is 0 V to
0.5 V and a “1” is 2.7 V to 5 V.
There are several combinations of letters that may be present after the “74” in the number.
You have already been introduced to “LS” and “HCT.” You may also find “L” (low-power),
“H” (high-speed), “S” (Schottky), “F” (fast), “ALS” (Advanced low-power Schottky), and
“AS” (Advanced-Schottky). Each of these variants has particular characteristics with respect to power consumption and switching speeds but they are usually interchangeable for
general use.
Name
In-Class Exercises for Lab 10:
CID
Digital Gates
1. In the space below, design transistor circuits (with NPN transistors) that perform
each of the following three logic operations: NOT, AND, OR.
2. The truth table below includes each of the 16 possible two-input (A,B) binary logic
functions. In the space below each column (you may need to write sideways), assign
a name to the function in terms of NOT, AND and OR operations. Q0 is FALSE and
Q15 is TRUE.
A
0
0
1
1
B
0
1
0
1
Q0
0
0
0
0
Q1
1
0
0
0
Q2
0
1
0
0
Q3
1
1
0
0
Q4
0
0
1
0
Q5
1
0
1
0
Q6
0
1
1
0
Q7
1
1
1
0
Q8
0
0
0
1
Q9
1
0
0
1
Q10
0
1
0
1
Q11
1
1
0
1
Q12
0
0
1
1
Q13
1
0
1
1
Q14
0
1
1
1
Q15
1
1
1
1
3. Complete the truth table for the circuit shown below. The circuit is called a full adder
because it adds two binary digits A and B (from two different binary numbers) as
well as a carry bit Ci−1 from the addition of the previous two digits and produces
their sum and a carry bit (Ci ) to be applied to the next two digits.
intermediate
values
Ai
Bi
Ci-1
Si
Ci
141
Ai
0
0
1
1
0
0
1
1
Bi
0
1
0
1
0
1
0
1
Ci−1
0
0
0
0
1
1
1
1
Si Ci
Lab 10 Exercises:
Digital Gates
1. Find the 7414 IC in your lab kit. This IC contains six inverters. The connections from
these inverters to the pins is shown on the pinouts sheet in your manual. Connect
the input of one of the inverters in this IC to a push-button switch as shown below.
The switch should connect the inverter input to 5 V when pressed and to ground when
released. You should use an ohmmeter to determine which leads of the switch perform
these functions before you connect the switch to the circuit. (The black wire on the
switch does not mean ground.) Write on the circuit diagram the colors of the three
wires you determined. Ask a TA to check your result before you connect the switch to
the circuit. Connect the output of the inverter to an LED as shown below. The LED
should go on when the output of the inverter is logical 1. Don’t forget to connect the
appropriate pins of your IC to 5 V and ground to power it.
5V
1k
LED
When the switch is released, the LED should be on. When you press the switch, the
LED should go off.
1
2. Repeat exercise 1 for an AND gate. There are two push-button switches in your lab
kit. Connect each of them to an input of an AND gate. Now the LED should not go
on unless both switches are pressed.
1
3. Construct an XOR gate using the circuit shown in the notes for this lab. Connect a
push-button switch to each input, and connect an LED to the output, as in exercise 1.
2
Additional exercises on the next page
143
Exercises for Lab 10, Digital Gates
145
4. Design and construct a circuit with two inputs, A and B, and four outputs, Q0 , Q1 ,
Q2 , Q3 . The circuit should obey the following truth table. This circuit is called a
decoder and allows you to turn on a separate output based on what binary number is
encoded in the two digits A and B. Hint: The logical statement to produce line Q1
is [A AND (NOT B)]. One way of building this circuit involves 4 AND gates and
2 inverters, although there are other ways to do it as well.
B A Q0 Q1 Q2 Q3
0 0 1
0
0
0
0 1 0
1
0
0
1 0 0
0
1
0
1 1 0
0
0
1
Connect A and B to push button switches, and connect the four outputs
Q0 , Q1 , Q2 , Q3 to LEDs as in exercise 1 so that the truth table can be easily verified.
3
5. (a) We want to investigate the effect of propagation delay in a digital circuit. Use a
7414 IC and wire the output of each inverter on the IC to the input of the next
inverter until you have six inverters in series, as in Fig. (A) below. Put a 10-kHz
signal from the TTL output of the signal generator into the first of the inverters.
Display on the oscilloscope simultaneously the signal into the first inverter and
the signal from the last inverter. Trigger on the input signal and look very closely
at the first transition you see on the scope. Measure and record the propagation
delay (the delay between a change in the input and the response in the output)
per TTL inverter.
(A)
signal from
signal generator
signal out
to scope
(b) Now disconnect the signal generator from the first inverter and connect the output of the fifth inverter in the chain to the input of the first inverter, as in Fig. (B)
below. Display on the oscilloscope the signal from the sixth inverter and explain
what you observe. How is the period of oscillation related to the propagation
delay of a single inverter? Now feed the output of the sixth inverter rather than
the fifth inverter back into the first inverter. How is the output of the sixth
inverter different than before? Why?
signal out
to scope
(B)
3
Additional exercises on the next page
Exercises for Lab 10, Digital Gates
147
6. Optional: Use AND, OR and NOT gates to design and build a data-selector with two
binary data inputs, A and B, a logical selector input S, and a logical output Q. The
circuit should behave so that setting S = 0 causes Q = A, and setting S = 1 causes
Q = B. Hint: The circuit may have some similarities to the XOR gate in the previous
exercise.
Lab 11
Flip Flops
Notes for Lab 11:
Flip Flops
The symbol for a type-D flip-flop is shown below.
S
Q
D
>CLK
Q
R
The flip-flop has two outputs, Q and Q. The Q output is simply Q inverted. If Q = 1,
then Q = 0. If Q = 0, then Q = 1. The flip-flop is positive-edge triggered, i.e., when the
clock input makes a transition from 0 to 1. D is the data input. Whenever the flip-flop is
clocked, whatever is at D is transferred to the Q output. D only affects the output at the
instant when the clock goes from 0 to 1. The output then “stores” the value of D until the
next time the clock triggers the flip-flop.
S is the set (or preset) input. Whenever S = 0, we have Q = 1, no matter what D is or
what the CLK input is doing. R is the reset (or clear) input. Whenever R = 0, we have
Q = 0, no matter what D is or what the CLK input is doing. In order for the CLK input
to be enabled, both S and R must be 1. Often, we just connect the S and R inputs to 5 V.
The little circles on the S and R inputs indicate a logical NOT function. The circle indicates
that we set or reset the flip-flop with a 0. If the circle were not present in the symbol, we
would know that we set or reset the flip-flop with a 1. The symbol tells us how to set or
reset the flip-flop without having to read a description of the IC.
A flip-flop can be used to capture the state of a data line at a given instant. When the
flip-flop is clocked, the value at D is captured and sent to Q. It is held there, no matter
what D does next, as long as we don’t clock the flip-flop again. This function is called a
latch.
If we connect several flip-flops together, as shown below, we have a shift register.
data in
D
>CLK
Q
D
Q
D
>CLK
>CLK
Q
D
Q
data out
>CLK
clock
Note that the S and R inputs are not shown. We assume that they are connected to 5 V.
All clock inputs are tied together so that all of the flip-flops are clocked at the same time.
151
152
Notes for Lab 11, Flip Flops
When clocked, the first flip-flop sends the data input to its Q output. Note that the Q
output of the first flip-flop is connected to the D input of the second flip-flop. When the
second flip-flop is clocked, it captures the Q output of the first flip-flop before it has a
chance to send its D to its Q. There is a delay between the time of the clock transition and
the time that D is finally sent to Q. By the time that the first flip-flop has sent its D to its
Q, the second flip-flop has already captured its D. The same is true of each flip-flop in the
chain. Each flip-flop captures the output of the previous flip-flop, and then each flip-flop
sends that value to its own output.
The shift register can be used to store a binary number. For example, suppose we want to
store 1010. We send a 0 to the data input and then clock the flip-flops. The 0 is now at
the output of the first flip-flop. We then send a 1 to the data input and clock the flip-flops
again. The 0 is now at the output of the second flip-flop, and the 1 is at the output of
the first flip-flop. We repeat this procedure, sending another 0 and then another 1, each
followed by a clock, so that finally 1,0,1,0 are at the outputs of the first, second, third, and
fourth flip-flops, respectively. We can retrieve the number we have stored by clocking the
flip-flops some more times. Each time we clock the flip-flops, the next digit appears at the
data output.
The following is a divide-by-two counter.
D
>CLK
Q
Q
Suppose that initially Q = 1 and Q = 0. Then D = 0, and when the flip-flop is clocked, this
value will be sent to the output, and now Q = 0 and Q = D = 1. Note that the sequence
of events is important here. The value at D is captured by the flip-flop before it is sent to
the output. By the time Q changes, D has already been captured, and the new value of Q
will not be captured until the next time the flip-flop is clocked. The next clock sends the
new value of D to the output, and we have Q = 1 and Q = 0 again. If the input is a square
wave, the output will also be a square wave, but with half the frequency, as shown below.
Note that the output toggles only on the positive-going edge of the input.
input
output
We can connect N flip-flops together and obtain a divide-by-2N counter. With four flipflops, as shown below, we have a divide-by-16 counter.
Notes for Lab 11, Flip Flops
D
input
>CLK
Q
Q
Q
D
>CLK
Q
D
>CLK
Q
Q
153
D
>CLK
Q
output
Q
1 th
If the input is a square wave, the output will be a square wave, but with 16
the frequency.
Suppose we monitor the Q output of each flip-flop in the chain. We will call the output of
the first flip-flop Q0 , the output of the the second flip-flop Q1 , etc.
input
Q0
Q1
Q2
Q3
The signal that clocks a counter can either be generated by another digital circuit element,
or by a manual switch. Manual switches (e.g., buttons, dials, flip switches) tend to deliver
lousy clock signals due to a phenomenon called “bounce”. When the switch is thrown from
ground to 5 V, the moment of signal lead must first break contact with ground and then
make contact with 5 V. This transition is anything but clean, and often produces a confused
intermediate state that effectively oscillates back and forth between 0 and 5 V many times.
The same can also happen when the switch is thrown back to ground. When this happens,
a single flip of the switch can advance the clock through many (possibly hundreds) of cycles.
5V
>CLK
There are a variety of “debouncing” strategies available, though we will discuss only one:
the SR latch. This device can either be constructed from two NAND gates (the circuit
below on the left) or from the S and R inputs of a flip-flop (the circuit on the right), and
never requires you to leave a friend stranded in the forest.
154
Notes for Lab 11, Flip Flops
5V
1k
5V
1k
A
Q
S
5V
Q
5V
Q
>CLK
1k
1k
B
R
C
In the NAND-gate circuit on the left, intermediate points are labeled as A, B, and C for
convenience. With the switch in the position shown in the figure, A is connected to 5 V
through a 1-k resistor and is therefore 1, and B is grounded through the switch and is
therefore 0. If B = 0, then C = 1, no matter what Q is. Since both inputs of the upper
NAND are 1, we have Q = 0. Now suppose we throw the switch. As the switch is traveling,
neither A nor B are grounded by the switch and are therefore both 1. Since Q = 0, C is
still 1, even though B is now 1. And since both inputs to the upper NAND are still 1, Q is
still 0. In other words, nothing happens. When the switch first touches the upper contact,
A goes to 0. This causes Q to go to 1. Since both inputs of the lower NAND are 1, we have
C = 0. If the switch bounces, A will go back and forth between 0 and 1. But this has no
effect. Now that C = 0, one of the inputs of the upper NAND is 0, and Q will stay at 1,
no matter what A does. Q cannot change back to 0 until the switch is thrown again and
touches the lower contact. Using similar arguments, we can show that bounce on the lower
contact also has no effect on Q. This latch completely removes the bounce from the switch.
When we throw the switch, Q makes a single transition from 0 to 1 or from 1 to 0.
In the SR latch on the right, the clock is grounded. Pulling the switch upward causes
Q = 1 the instant that S first becomes grounded. The state won’t change again, regardless
of the subsequent behavior of S until the switch is pulled downward, which causes Q = 0
the instant that R first becomes grounded. In practice, either circuit does an adequate job
of switch debouncing.
Name
In-Class Exercises for Lab 11:
CID
Flip Flops
1. Use the figures in the lab notes to make a chart of the values of each of the
four flip-flop outputs during each of the 16 clock cycles of a divide-by-16 counter.
Start with all-zero outputs. Why do you think we call this circuit a “counter”?
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Q0 0
Q1 0
Q2 0
Q3 0
2. Design and draw a four-bit shift register that recycles its data rather than bringing it
in and out. Include an LED on each Q output and a push-button switch for injecting
the bit pattern into the register. Make sure that your switch doesn’t do battle with
any of the flip-flop outputs.
3. The following circuit is an RS switch debouncer. Start in the state shown, with the
switch connected to the top terminal, and consider what happens at points A, B, Q
and Q as the switch first touches the bottom contact and then briefly bounces (i.e.,
breaks and makes contact many times). Similarly, follow the switch back up to the
top terminal and let it bounce again. Fill in the table below.
5V
1k
A
A
Q
5V
1k
Switch
Switch
Switch
Switch
Switch
up
moved down
bounces
moved up
bounces
Q
B
Additional exercises on the next page
155
B
Q
Q
In-Class Exercises for Lab 11, Flip Flops
157
4. Use logic gates and four flip-flops to make a shift register that can be parallel-loaded.
The circuit will have the normal D and CLK inputs as well as a parallel-load input PL
and four inputs L0 -L3 , When PL is low, the shift register should function normally.
When PL is high, either the S or R inputs on each flip-flop should go low depending
on the value of their respective L input, “loading” that value to the flip-flop. The
best implementation of this uses one inverter, four OR gates and four NAND gates,
but there are many other working implementations as well.
Lab 11 Exercises:
Flip Flops
In this lab, whenever we tell you to connect an input to a switch, we mean to wire the
switch so that when it is released, the input is connected to ground, and when it is pressed,
the input is connected to 5 V.
1. Find the 7474 IC in your lab kit. This IC contains two type-D flip-flops. The connections from these flip-flops to the pins is shown on the pinouts sheet in your packet.
Connect 5 V and ground to this IC. Connect the S and R inputs of one of the flip-flops
to 5 V. Connect the CLK and D inputs to switches. Connect the Q and Q outputs to
LEDs. Press the switch connected to CLK. The LED connected to Q should go off,
and the LED connected to Q should go on. Now press the switch connected to CLK
again, this time while holding down the switch connected to D. The LED connected
to Q should go on, and the LED connected to Q should go off.
2
2. (a) Using four flip-flops (two 7474 ICs), make a 4-bit shift register as shown in the
notes for this lab. Use a 1-Hz TTL signal from the signal generator as the clock
signal, and use a push-button switch (5 V when pressed, ground otherwise) to
set the input values. Use LEDs to display the values of each of the Q outputs of
the shift register.
2
(b) Optional: Connect the Q output of the last flip-flop to the input of the first
flip-flop so that the shift-register can recycle its data. Find a way to use the
switch to continue injecting data values into the register without doing battle
with any of the flip-flop outputs (i.e., do NOT connect the switch directly to an
output). Demonstrate that your circuit works as expected. Hint: remember the
S and R pins.
3. Construct a divide-by-16 counter, using four flip-flops, as shown in the notes. There
are two 7474s in your lab kit, each of which contains two flip-flops. Connect a 16-kHz
square wave to the input. The signal emerging from the last flip-flop should be a
1-kHz square wave. Use the oscilloscope to simultaneously view both signals. Then
use an LED to display each of the Q outputs and verify the the counter is counting
from 0 to 15 in binary.
3
4. Replace the TTL input signal to your divide-by-16 counter with a push-button switch
(5 V when pressed, ground otherwise), and attempt to increment the counter manually.
Does the result appear random? If so, you have switch bounce problems. Use two
NAND gates to convert your switch into a bounceless switch. Verify that your counter
can now be incremented from 0 to 15 in a controlled way.
3
159
Lab 12
Counters and Displays
Notes for Lab 12:
Counters and Displays
The symbol below represents a divide-by-16 counter. The four logic outputs (Q0 , Q1 , Q2 , Q3 )
form a four-bit binary number that runs from 0 (binary 0000, hex 0) to 15 (binary 1111,
hex F).
Q0
Q1
Q2
>CLK Q3
R
The 4-bit output advances by 1 each time the clock (CLK) input is triggered. After reaching
the maximum value (binary 1111, hex F), the outputs all start over at zero on the next clock
cycle. Typically, the counter is triggered when the CLK input passes from LOW to HIGH.
Some ICs, however, are specifically engineered to trigger on a HIGH to LOW transition.
Counters can be “chained” in series to provide a larger counting range. This can be accomplished by connecting the Q3 output of one 4-bit-counter to the clock input of the next 4-bit
counter as shown below. More sophisticated ICs have a “carry” input and output that allow
one counter to advance the next counter each time it passes from its maximum value back
to zero. Chaining two 4-bit (divide-by-16) counters together yields an 8-bit (divide-by-256)
counter that starts at 0 (binary 00000000, hex 00) and ends at 255 (binary 11111111, hex
FF).
Q0 Q1 Q2 Q3
clock
Q4 Q5 Q6 Q7
Q0
Q1
Q2
Q0
Q1
Q2
>CLK Q3
>CLK Q3
R is the reset input of the counter. When R = 0, the counter functions normally. When
R = 1 (no circle on this input), the counter resets to Q0 = Q1 = Q2 = Q3 = 0 and stays
there until R = 0 again. While R is usually connected to ground, some applications require
it to be used for control purposes. For example, consider the following circuit.
Q0
Q1
clock
>CLK
R
163
Q0
Q1
164
Notes for Lab 12, Counters and Displays
The instant that the output advances to Q0 = Q1 = 1, the AND gate sends a 1 to the reset
input, which causes the counter to be immediately reset. Thus, the output never reaches 3
(binary 11), but instead counts as
. . . 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2 . . .
This is a divide-by-3 counter. Similarly, we could divide by any other integer by wiring an
appropriate logic combination to the reset input. Some 4-bit counter ICs come internally
wired for division. Divide-by-10 (“decade”) and divide-by-12 counters, for example, are
quite common.
Just as divide-by-16 counters can be chained together, two decade (divide-by-ten) counters can also be chained to build a divide-by-100 counter. Such a circuit requires special
consideration. Because the four-bit output of each decade counter has a maximum value
of 9 (binary 1001), the maximum value of a divide-by-100 counter is 99, which is then
represented as 10011001. But wait. 100110012 is 15310 and not 9910 . Because this interpretation of an 8-bit output is inconsistent with the binary number system, we give it a special
name, “binary-coded-decimal” (BCD). For BCD displays, each set of four bits represents
one decimal digit.
In many electronic devices, binary numbers are displayed as hexadecimal digits with a device
known as a “7-segment display.” In the figure below, the seven segments of the display are
LEDs. In practice, most devices (e.g., watches and calculators) now use liquid crystal
display (LCD) elements, which draw far less current than LEDs. The 4-bit binary input
D0 , D1 , D2 , D3 goes into a “decoder” chip, whose job is to determine the corresponding
hexadecimal digit and to selectively illuminate the display segments that best represent
it. The number “4”, for example, is displayed by illuminating segments B, C, F and G.
Study the figure and observe that each of the hexadecimal digits [0-F] can be displayed. We
typically use a lower-case “b” and “d” to distinguish them from “8” and “0”, respectively.
Decoder
D0
D0
D0
D0
A
B
C
D
E
F
G
Display
A
B
C
D
E
F
A
F
B
G
E
G
C
D
GND
The 7-segment display is only one of many devices for displaying a number. In the lab
exercises below, you will use a 10-element LED bar graph for which the nth of 10 LEDs
indicates the number n. Each type of display device needs a corresponding decoder that
knows how to convert binary numbers into the format required by the display.
Name
In-Class Exercises for Lab 12:
CID
Counters and Displays
1. Use a 4-bit counter and logic gates to design a divide-by-6 counter. Draw your circuit
in the space below.
2. Use a 4-bit counter and logic gates to design a divide-by-10 counter. Draw your circuit
in the space below.
3. Fill in the table below.
Digits
00010001
00000011
10010110
01011101
11111111
Binary-Decimal interpretation
17
BCD-Decimal interpretation
11
4. Fill in the table below.
Decimal
17
3
150
93
255
8-bit binary
00010001
165
8-bit BCD
00010111
Hex
11
Lab 12 Exercises:
Counters and Displays
Do not take your counter circuit apart after building it! It will be used as part of the future
audio-amplifier lab.
If you follow these instructions carefully, you will be able to build your audio amplifier
entirely on one breadboard. This will leave your second breadboard for the final exam in
case you don’t want to disassemble the amplifier after you complete lab 13.
1. (a) First, we need to set up the power supply rails. Orient your board vertical with
the text on the board right-side up and the row labeled “1” at the top. You
should attach 5 V to the two red (“+”) supply rails. Attach ground to the two
blue (“-”) supply rails. You should have a jumper connecting the two red rails
and another connecting the two blue rails. Don’t turn on the power supply yet.
(b) Locate the following parts in your kit and arrange them carefully onto your
breadboard.
First place the 74193 (4-bit binary up/down counter) with pin 1 in the upper
left-hand corner and located in row 28. Below the 74193 place the 74154 (4-bit
decoder) with pin 1 in the lower right-hand corner and located in row 50. This
should leave three blank rows between the 74193 and the 74154 (rows 36, 37, and
38). The placement of the 74154 is quite critical for later exercises. Below the
74154 place the 7474 (D-type flip flop) with pin 1 in the upper left-hand corner.
Above the 74193 place a 7408 (AND gate) with pin 1 in the upper left-hand
corner and the LED bar graph with the notched corner in the upper left-hand
corner.
(c) Connect the proper pins on each IC to power and to ground. Connect the
cathodes (i.e., negative sides) of the first four (closest to the 74193) bar graph
LEDs together and then connect them all to ground with a single wire – if the
notch is in the upper-left-hand corner, then the cathodes are on the right-hand
side of the IC. Don’t connect any power to the LED bar graph. Have your TA
check your board before turning the power on! Upon powering up, immediately
check each IC for signs of overheating. If you detect a problem, quickly power
down and reexamine your connections.
Additional exercises on the next page
167
1
Exercises for Lab 12, Counters and Displays
169
2. Connect the binary outputs of the counter to the first four LED inputs on the bar
graph. You probably want to put them in order according to Q0 , Q1 , Q2 , and Q3 . Use
a 1-Hz signal from the TTL output of your signal generator to trigger the Count-Up
(CPU) clock input of the counter. Each of the Master Reset (MR), Parallel Load
(PL), and Count-Down (CPD) pins must either be connected to 5 V or to ground.
Study pages 5 and 6 of the 74193 data sheet to learn how to configure these inputs
for count-up operation. Verify that the display is counting from 0 to 15 in binary.
3. (a) Remove the wires connecting the Q0 -Q3 outputs of the counter to the LED bar
graph, and use them instead to connect the Q0 -Q3 outputs of the counter to the
A0 -A3 inputs of the decoder.
We are going to use an unusual way to attach the LED bar to the 74154, but it
will save you a lot of wiring. Remove the bar graph from your board. Rotate
it so the notch is in the lower right corner and move it down to the right of the
74154. You want to place it so that pins 1-10 (the anodes) plug into the 5/,V
bus on the right (the red one). Pin 1 should line up with row 49 on the board
– the same row where pin 2 of the 74154 is located. In this position the LEDs
will attach to pins 2-11 (Y 1 through Y 11) on the 74154. You will notice that
there is not a hole in the breadboard for pin 6 on the bar graph. We cure this by
carefully bending pin 6 so that it lays next to, and in contact with, pin 5. Try
not to bend the pin where it exits from the plastic case. If you leave a little space
you will get a gentler curve in the wire and avoid breaking it off.
This orientation is chosen because the decoder outputs are active low rather than
active high, and this will allow the current to flow from the 5 V rail to the 74154
outputs to illuminate the LEDs on those outputs that are active.
(b) Study the 74154 data sheet in order to correctly wire the enable pins (E1 and
E2) of the decoder to either 5 V or ground. Use a 1-Hz TTL signal to clock the
Count-Up input of the counter. The 10 bar graph LEDs should now repeatedly
light up in sequence from bottom to top. Why are the LEDs all off for a time
before the cycle repeats itself?
(c) Use AND gates (7408) in conjunction with the MR input of the counter IC to
convert your divide-by-16 counter into a divide-by-11 counter. Again using the
1-Hz TTL signal to clock the counter, verify that the counter resets to 0 instead
of advancing beyond 10.
Note: Due to the vagaries of gate delays you have to order the AND gates
properly to get this to work. The most rapidly varying output, Q0 in this case,
must be connected to one of the inputs on the AND gate whose output is actually
connected to the MR input of the counter IC.
Additional exercises on the next page
2
Exercises for Lab 12, Counters and Displays
171
(d) There’s a more elegant way to prevent the counter from resetting when it advances beyond 10. First, remove the AND gate and associated wires from the
breadboard. Replace this IC with a 7414 (hex inverter). Then connect the Y11
output of the decoder to the MR input of the counter through an inverter. The
inverter is necessary because the decoder outputs are active low, whereas the
MR input is active high. Verify that the counter operates as before.
Note: If you want to save space on your board you can replace the one gate
of the 7414 that we are using with a 2N3904 transistor. This transistor will fit
nicely in the three rows between the 74193 and the 74154. You will need to
consider carefully how a single NPN transistor can be used to invert a signal. It
will require a 1 kΩ resistor in series with the base of the transistor to limit the
base current and a 1 kΩ pull-up resistor between the collector of the transistor
and 5 V.
3
4. (a) Instead of using a TTL signal, wire a button switch to the Count-Up input of
the counter so that the switch output is LOW when pressed and HIGH when
released. Locate the switch at the bottom of the board below the 7474 flip-flops.
Press the button a couple of times and watch the output of the LED bar jump
randomly due to switch bounce problems. Then build an SR latch to debounce
the switch and verify that releasing the button advances the counter by 1 in a
controlled way.
(b) Add a second debounced switch to clock the Count-Down input. You should
now be able to increment the counter with one button and decrement it with the
other.
Additional exercises on the next page
2
Exercises for Lab 12, Counters and Displays
173
5. Instead of allowing the counter to cycle around when it reaches an endpoint (from 10
to 0 or from 0 back to 15), we now want it to simply stop at the endpoint. Imagine
being magically confined to your bedroom (by an angry parent, roommate or spouse)
so that every time you set foot outside, you immediately find yourself just inside the
door again. That’s the idea.
On the low end, detach the Y11 decoder output from the inverter (or the base of the
2N3904 transistor) and replace it with the Y15 output.
On the high end, we will use the parallel-load (P L) feature of the counter to reload
the value 10 whenever it attempts to advance to 11. Connect the counter’s parallel
(D0 -D3 ) inputs to 5 V and ground in such a way as to represent the decimal value 10
in binary. Then connect the Y11 decoder output directly to the P L input.
The completed circuit will appear as shown below. Verify that the counter is now
confined to the range from 0 to 10 and unable to cycle around at the endpoints.
Explain the operation of the new features to your TA.
Do not disassemble this circuit. It will serve as the digital volume control
for your audio amplifier in Lab 13.
74193
D0
D1
D2
D3
Q0
Q1
Q2
Q3
switch
CPU
switch
CPD
PL MR
74154
A0
A1
A2
A3
E1
E2
Y0
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Y9
Y10
Y11
Y12
Y13
Y14
Y15
DC10EWA
VCC
2
Lab 13
Audio Amplifier
Notes for Lab 13:
Audio Amplifier
In this lab, you will take a relatively weak signal from a high-impedance output device
(portable music player) and use it to drive a low-impedance input device (loudspeaker). In
the past, you have connected small speaker drivers to a variety of output devices including a
signal generator, a transistor voltage follower and an op-amp voltage follower. In each case,
the output device was less than equal to the task, either because of high output impedance
or intrinsic current limitations.
Most good loudspeaker drivers have very low input impedances (4 Ω or 8 Ω) and therefore
require a source that has an exceptionally low output impedance as well as the ability to
deliver a lot of current (and power). The output impedance of a line-level audio source
from a DVD player or computer is often as large as 100 Ω (very bad). Even the output
impedance of a source designed for headphones will be about 5 Ω, which is still comparable
to the impedance of a loudspeaker. In order to match such a weak source to a heavy
loudspeaker load, we will need an impressive power amplifier between them. For this lab,
you will use an audio amplifier with an input impedance of 20 kΩ and an output impedance
less than 0.5 Ω.
You may opt to try stereo sound at the end of the lab, but will first be required to work
with a single channel (i.e., mono). Some care must be taken in summing the two channels
of a stereo signal into a mono signal. They cannot simply be connected together, which
would cause one channel to fight against the other. In general, never directly connect two
device outputs together which can cause a large current to flow out of one input and into
the other. This not only causes substantial signal loss, but can also permanently damage
(i.e., melt) your output device.
One way to sum two signals is to use an op-amp summing amplifier. A simpler approach,
however, is also effective and less prone to instability. Simply connect the stereo channels
together through separate but equal resistors, and then pass the mono signal arising at the
junction between these resistors on to the audio amplifier. Just make sure that resistor used
on each output is far smaller than the 20 kΩ input impedance of the audio amplifier.
The “dB” unit
You have probably noticed that some of the operating characteristics of amplifiers are
expressed in dB. The decibel (dB) is a logarithmic unit that expresses the relative magnitude
(i.e., ratio) of two quantities. Because the response of the human ear to sound waves is
approximately logarithmic rather than linear, the decibel is a natural and intuitive unit
for describing sound wave amplitude (A) and intensity or power (I ∝ A2 ). It is also used
in specifying electronic signals because it is easier to handle a large dynamic range with a
logarithmic unit.
177
178
Notes for Lab 13, Audio Amplifier
A decibel is 1/10 of a bel (note the deci- prefix), where a bel is the base-10 logarithm of
the ratio of two signal intensities. If the gain of signal-transmission device is described as
an output-over-input ratio of voltage amplitudes, the decibel gain can be computed as:
Iout
Iin
!
A2out
= 10 log10
A2in
Aout
.
= 20 log10
Ain
Gdb = 10 log10
When the denominator is a standard reference magnitude rather than an input, absolute
measurements can be also be described in decibel units. In the case of sound intensity,
the reference magnitude is the quietest sound that a typical human can hear. Because
log10 (1) = 0, the decibel measure of the reference signal itself is defined to be 0 dB.
Each successive 10-fold increase in the intensity of a signal increases its logarithmic value
by 10 decibels. Thus, a 100-fold amplitude increase (a 10000-fold intensity increase) is also
described as a 40 decibel increase. Though the fractional amplitude gain, Aout /Ain , and the
fractional intensity gain, Iout /Iin , are not equivalent, we see that they both yield the same
decibel gain.
A Digital-to-Analog Converter for Volume Control
The audio amplifier IC in your kit is unusual in that it has a volume-control pin that takes an
analog input voltage. We plan to use this feature to adjust the volume of music delivered to a
loudspeaker. The control voltage, however, comes from the digital 4-bit counter constructed
in the previous lab. This will require us to construct a specialized circuit that accepts a
digitally-encoded value (usually binary) as input and returns an analog voltage. We call
such a device a digital-to-analog (D/A) converter, or simply a DAC. DACs can be found in
many devices, particularly in any digital device that outputs analog audio, such as an mp3
player.
DACs are available as single ICs that have N digital inputs and one analog output. The
number of input bits determines how finely the output voltage can be specified. The DAC
can specify 2N distinct voltages. These voltage levels are generally evenly distributed between the minimum and maximum output voltages.
Notes for Lab 13, Audio Amplifier
179
The technique for building a DAC is reasonably simple. One of the simplest DAC circuits
is the binary-weighted summing amplifier.
R0
D0
R0/2
D1
R0/4
Rf
D2
R0/8
D3
−
Vout
+
Each of the four logical inputs, D0 , D1 , D2 , D3 , have two possible values: 0 or 5 V. We
analyze this circuit by summing the current entering the point at Vin− and setting that sum
to zero:
D0
D1
D2
D3
Vout
+ 1
+ 1
+ 1
+
=0.
R0
Rf
2 R0
4 R0
8 R0
Solving for Vout , we obtain
Vout = −
Rf
(D0 + 2D1 + 4D2 + 8D3 ) .
R0
The minus sign reminds that this configuration is actually an inverting amplifier. Assuming
that R0 = 5k and Rf = 1k, we calculate the analog output voltage below for every possible
combination of digital inputs.
D3
0V
0V
0V
0V
0V
0V
0V
0V
D2
0V
0V
0V
0V
5V
5V
5V
5V
D1
0V
0V
5V
5V
0V
0V
5V
5V
D0
0V
5V
0V
5V
0V
5V
0V
5V
Vo ut
−0 V
−1 V
−2 V
−3 V
−4 V
−5 V
−6 V
−7 V
D3
5V
5V
5V
5V
5V
5V
5V
5V
D2
0V
0V
0V
0V
5V
5V
5V
5V
D1
0V
0V
5V
5V
0V
0V
5V
5V
D0
0V
5V
0V
5V
0V
5V
0V
5V
Vo ut
−8 V
−9 V
−10 V
−11 V
−12 V
−13 V
−14 V
−15 V
Note that the value of Vout is numerically equal to the binary number represented by the
four digits D3 D2 D1 D0 (except for the minus sign). For example, the binary number 1101
is equal to 13. Thus, when D0 , D2 , D3 are logical 1 (5 V) and D1 is logical 0, the output
is −13 V. The circuit in the example above is a 4-bit DAC, and therefore has an output
resolution of 1 part in 16. Similarly, an N -bit DAC would have an output resolution of 1
part in 2N .
180
Notes for Lab 13, Audio Amplifier
By changing the values of the resistors above, we can easily change the output voltage range.
For example, if we increase R0 to = 75 kΩ, then Vout would only get as large as −1 V. In
general, we see that the output voltage is proportional to rather than equal to the binary
number represented by the digital input.
Consider the non-inverting summing amplifier circuit below, which also serves as a DAC.
This circuit has the advantage that it can be powered with a positive voltage, and still
produces a positive output voltage. Because the negative feedback loop has zero resistance,
we could just as well call this type of DAC a non-inverting summing follower.
R0
D0
R0/2
−
D1
Vout
+
R0/4
D2
R0/8
D3
No current flows into the non-inverting input. The network of digital inputs and input
resistors can be treated as a multi-resistor voltage divider that produces the desired output
voltage. If you apply Kirchoff’s current law to the node attached to the non-inverting input
of the op-amp, and rearrange the equations sufficiently, you arrive at the equation for the
output voltage
(D0 + 2D1 + 4D2 + 8D3 )
Vout =
.
15
The result is, once again, a sequence of 16 uniform voltage steps from 0 to 5 V.
The analog output of the DAC above swings between Vmin = 0 V and Vmax = 5 V. Your
audio amplifier, however, has an analog volume-control input that requires a smaller voltage
0
0 ) than what the DAC provides. The three-resistor network below is a
range (Vmin
to Vmax
relatively straightforward means of shrinking the output range on both top and bottom.
VCC
R0
D0
R0/2
D1
R0/4
D2
R3
−
+
Vout
R1
V′out
R2
R0/8
D3
To determine the resistor values required, use Ohm’s law to determine the current though
each resistor, and recall that the total current flowing into a circuit node must equal the
Notes for Lab 13, Audio Amplifier
181
total current flowing out of the node (Kirchhoff’s current law). With a DAC voltage of Vmax
0
coming in from the left and a modified voltage of Vmax
going out on the right, we have
0 )
0 )
0 )
(VCC − Vmax
(0 − Vmax
(Vmax − Vmax
+
+
=0.
R1
R2
R3
0
Similarly, with Vmin coming in from the left and Vmin
going out on the right, we have
0 )
0 )
0 )
(VCC − Vmin
(0 − Vmin
(Vmin − Vmin
+
+
=0.
R1
R2
R3
The first thing to notice about these equations is that there are three unknowns (R1 , R2 ,
and R3 ) but only two equations. We will only be able to find a solution involving ratios of
the resistances because the system is underdetermined.
The solution of these equations is tedious but can be done by hand with some effort. First,
solve both equations for 1/R2 and set them equal. With some rearranging you arrive at
0
0 )
R1
(Vmax
− Vmin
= VCC
.
0
0 V
R3
(Vmax Vmin
− Vmax
min )
Next, you can select either of the original equations that have been solved for 1/R2 and
0
multiply that equation by R1 . If we choose the second equation involving Vmin and Vmin
we
get
R1
VCC
Vmin
R1
= 0 +
−1
−1.
0
R2
Vmin
Vmin
R3
Use the first expression to compute R1 /R3 , and then substitute the result into the second
expression to obtain R1 /R2 .
Though the overall scale of these resistor values is not fixed, their relative values are. The
scale will be a compromise between three competing needs. First, the Thévenin-equivalent
impedance of the modified output Rout = R1 ||R2 ||R3 , which will be somewhat smaller
than the smallest of the three resistors, must be much smaller than the impedance of the
amplifier’s volume-control input. Secondly, each resistor must be large enough to prevent
excessive V 2 /R power consumption. Thirdly, you have to choose available values for the
three resistors, understanding that you may have to use series or parallel resistors to arrive
at the desired values.
Name
In-Class Exercises for Lab 13:
CID
Audio Amplifier
1. An amplifier receives a sinusoidal input signal whose rms amplitude is 0.1 V and produces a sinusoidal output signal whose rms amplitude is 10 V. What is the fractional
gain? What is the decibel gain?
2. (a) Choose resistor values for a 4-bit inverting summing amplifier that will yield an
output range 0 to −5 V as the digital inputs vary from 0000 to 1111. Assume
that digital High and Low are 5 V and ground, respectively.
(b) What voltage range is realized if the digital inputs can only go up to 1010?
Additional exercises on the next page
183
In-Class Exercises for Lab 13, Audio Amplifier
185
3. (a) Suppose that we have a 4-bit DAC based on the non-inverting summing follower
in the notes. What voltage would you expect to see at the output when the
digital inputs are 1010?
(b) Assume that digital High and Low are 5 V and ground, respectively. Draw the
corresponding four-resistor voltage divider between 5 V and ground, and demonstrate that it produces the expected voltage.
4. Given a DAC output that swings between Vmin = 0 V and Vmax = VCC , we want to
0
0
< Vmax instead. Study
modify that output to swing between Vmin
> Vmin and Vmax
the datasheet for the TDA7056B audio amplifier, and figure out what range of voltages
should be applied to the volume-control (VC) pin in order to utilize the entire linear0
output range. Choose a minimum voltage Vmin
that is about 0.05 V below the cutoff
voltage so that the volume can be turned completely off, and choose a maximum
0 ,
voltage Vmax that is about 0.05 V beyond the top of the linear range. At VC = Vmax
what does the figure predict for the fractional amplitude gain? Record your result
here in the lab manual.
0
0
5. Let Vmin and Vmax be the range from 0 V to 3.3 V (digital 1010). Let Vmin
and Vmax
be the modified values obtained from exercise 4. Using the equations in the notes,
determine the seven numerical resistor values that should be used for the DAC and
the output modifier. Are any of the values arbitrary (i.e., not fixed by the problem)?
If so, what practical constraints might we apply?
Lab 13 Exercises:
Audio Amplifier
The counter and display breadboard from lab 12 should still be assembled for use in this
lab.
1. Power up your counter circuit from the previous lab and make sure that it still works
(be careful not to get VCC and ground mixed up).
Position the breadboard vertically with the text right side up again. The 74154 and
the LED bar graph should be in the lower half of the breadboard.
Find the regulated 9-VDC “wall-wart” power supply in your kit, and use a multimeter
to determine which of two leads (striped or plain) is 9 V and which is ground. Use
tape to label the two leads (manufacturers are not consistent in choosing which lead
to stripe).
2
2. (a) Using the circuit diagram in the notes, build a non-inverting summing amplifier
(DAC) just above the 74193 and use it to convert the 4-bit digital output of the
counter into an analog output that increments from 0 to 5 V in 24 = 16 uniform
steps. Place the TL3472 op-amp just above the 74193 on the board. If you are
VERY careful in how you lay out your resistors for the summing amplifier you
will only need two rows between the 74193 and the TL3472. Do NOT place the
amplifier in any of rows 1-10.
Based on your rather limited collection of resistors, we recommend the following
resistor combinations: 4 kΩ = 3.9 kΩ + 100 Ω, 8 kΩ = 6.8 kΩ + 1.2 kΩ, 16 kΩ =
15 kΩ+1 kΩ, and 32 kΩ = 22 kΩ+10 kΩ. All four resistor paths should connect to
the same row on the breadboard. Use wires to connect each resistor combination
(D0 , D1 , D2 , and D3 of the DAC) to the corresponding counter output (Q0 , Q1 ,
Q2 , and Q3 ). Power the TL3472 op amp with VCC+ = 9 V and VCC− = 0 V.
At this point you will just connect the 9 V output from the “wall-wart” power
supply directly to the appropriate pin on the TL3472.
(b) In order to allow the counter to cycle through the natural range from 0 to 15,
temporarily move either of the E1 and E2 enable pins of the decoder from ground
to 5 V (i.e., disable the decoder), and temporarily disconnect the CPU clock input
of the counter from its debounced switch, clocking the counter with the signal
generator (TTL output) instead.
(c) Using a 1-kHz clock signal, view the analog output of the DAC with the oscilloscope. You should see a voltage staircase that runs from 0 to 5 V. If the voltage
increments are not uniform, you may need to further fine-tune your resistor combinations. If the staircase is fuzzy, add a 0.1 µF capacitor between VCC (9 V) and
ground. Show the output to your TA.
Additional exercises on the next page
187
4
Exercises for Lab 13, Audio Amplifier
189
3. (a) Reverse the changes made in step (b) of the previous exercise, which will restore
the counter to its normal mode of operation. Verify that the button switches now
allow you to manually step the analog output between 0 and (10/15) ∗ 5 = 3.3 V.
(b) This analog output will eventually be used to control an audio amplifier
(TDA7056B) that accepts a volume-control input voltage. Refer to Fig. 3 on
page 6 of the audio amplifier datasheet in order to identify the range of control
voltages that produce a linear response. Record these voltages here in your lab
manual. Choose a minimum voltage Vmin that is about 0.05 V below the cutoff
voltage so that the volume can be turned completely off, and choose a maximum
voltage Vmax that is about 0.05 V beyond the top of the linear range. At Vmax ,
convert the decibel gain into a fractional amplitude gain and record the result
here.
(c) Using a network of three resistors as illustrated in the notes, convert the summing
amplifier output (0 to 3.3 V) into a voltage that varies from Vmin to Vmax . As
you determine the resistance values (R1 , R2 , R3 ), try to keep all of the values
in the range between 100 Ω and 5 kΩ. Verify that the modified output now has
approximately the desired range.
4. (a) Caution: Read the following instructions before you begin. You don’t want to
destroy your most expensive IC.
Power down the breadboard and insert your audio amplifier IC (TDA7056B).
Ground the power ground (PGND) and signal ground (SGND) leads close together with short wires.
Connect power (VP) to 9 V (i.e., connect a jumper from the 9 V supply attached
to the TL3472 to the correct pin on the TDA7056B). For improved stability,
connect a small 0.1 µF capacitor and a large capacitor (use 100 µF rather than
220 µF) between VP and ground as shown in Fig. 13 of the amplifier datasheet
(be sure to read the footnote).
Connect the modified DAC output from the previous exercise to the volume control (VCTRL) pin. Also connect a 1 µF capacitor between VCTRL and ground
(as shown in Fig. 14). Note that Fig. 14 illustrates a far simpler method of
volume control.
Unbias (i.e., remove any DC voltage components) the signal input (VIN) by
attaching a capacitor (use 0.1 µF rather than 0.47 µF) in series and a resistor
(use 10 kΩ rather than 5 kΩ) to ground as shown in Fig. 13. In subsequent
exercises, we will always send the input signal into this anti-bias capacitor rather
than directly to the VIN pin.
The signal generator has a HI-voltage output (0 to 10 V) and a LO-voltage output
(0 to 100 mV). Connect the LO output of the signal-generator to the unbiased
audio input of the amplifier – don’t bypass the capacitor. Be very careful NOT
to accidentally send a high voltage (> 1 V) into the audio input, which could
damage the amplifier.
Show your TA before powering up again.
Additional exercises on the next page
4
Exercises for Lab 13, Audio Amplifier
191
(b) Simultaneously view the audio input (VIN) and output (OUT+) signals relative
to ground on channels 1 and 2 of the oscilloscope. Adjust the 1-kHz input signal
to be 100 mV peak-to-peak. Use the counter to step the amplifier gain up and
down between its limits and observe the gain variation with the oscilloscope. Do
NOT attach the ground lead of the oscilloscope to OUT−. This lead really is
ground; and forcing OUT− to ground triggers an MCL sensor (see page 2 of the
datasheet) that temporarily disables the IC. Just be content to observe only half
the output signal.
(c) Use the counter to set the amplifier gain (VCTRL) to its maximum value (Vmax ).
Measure the fractional amplitude gain (Vout /Vin ) and then double it to account
for the fact that you are only looking at half the signal. Compute the corresponding decibel gain and compare your result to what you should expect based
on Fig. 3 of the amplifier datasheet.
4
5. Construct a 5 V power source on the breadboard using the 5 V regulator chip (78L05)
in your kit. Connect IN to 9 V, GND to ground, and OUT to an unused row of the
breadboard. Measure the 5 V output of the regulator with a multimeter to ensure
that it is working properly. If so, completely disconnect the breadboard from its large
external 5 V power supply, and connect the 5 V rails to the OUT pin of the regulator
instead.
Hint: if you place the 78L05 along the left side of the board and consider the pin
connections you should be able to build the power supply using a single row of the
board and the OUT pin will be in the 5 V rail of the breadboard. You may want to test
the 78L05 using 3 rows of the board and then transfer it to a single-row configuration
after you are sure it is working correctly to prevent applying the wrong voltage to
your digital ICs.
Now you are powering the entire audio amplifier circuit with the wall-wart. The
regulator is necessary because the digital ICs would be damaged if you attempted to
power them with 9 V. Verify that your circuit still functions as before. Then disconnect
the signal generator from the circuit.
Additional exercises on the next page
3
Exercises for Lab 13, Audio Amplifier
193
6. (a) Insert a 1/8-inch stereo-input jack close to your audio amplifier. Ground its
ground lead (improve stability by connecting it directly to the SGND pin of the
amplifier), and use separate 1-kΩ resistors to connect each of its channels to the
unbiased audio input of the amplifier. Use a 1/8-inch stereo cable to connect a
portable music player to your amplifier. Set the player volume to a reasonable
listening level and simultaneously view the amplifier input (VIN) and output
(OUT+) on the oscilloscope. Estimate the average peak-to-peak input voltage.
(b) Turn the the audio amplifier gain all the way down and disconnect the oscilloscope from the circuit. Then connect an 8-Ω loudspeaker across the OUT+
and OUT− pins of the amplifier. Slowly turn the gain back up again. If you
have good taste in music, you should now find yourself in a transcendental state
of aural bliss. Feel free to try this at home, but make sure that you have 8-Ω
loudspeakers. 4-Ω speakers are also common and could draw enough current to
damage the amplifier IC.
7. Optional: Borrow a second audio amplifier and set it up on a second breadboard.
You can connect the second breadboard to the first by using the tabs and slots on the
sides to hold them together. If you are very careful, you may be able to fit the second
amplifier on the first breadboard.
As before, power down first and make sure that everything is connected correctly.
Connect the volume control pins of both amplifiers to the same analog control voltage.
Connect each amplifier to a different loudspeaker. Connect one channel of the stereo
jack to the audio-input of one amplifier and the other channel to the audio-input of
the other amplifier. Use a separate anti-bias circuit for each audio input, but remove
the 1-kΩ resistors that were originally used to sum the two channels. Power up again.
How does stereo compare to mono?
3