Instrumentation for Scientists
Linking the physical/chemical and electrical worlds
in the scientific laboratory
Physical and
chemical
domains
Scale
position
Number
Potential
difference,
V
Conductivity, ρ
Charge, q
Non-electrical
domains
Electrical domains
Current, I
Paul W. Zitzewitz
Department of Natural Sciences
University of Michigan-Dearborn
June, 2008
i
© Paul W. Zitzewitz, 2008
ii
Table of Contents
Note to students............................................................................................................................... v
1. Simple sensors ............................................................................................................................ 1
A. Electric circuit models ........................................................................................................... 1
B. Introduction to laboratory equipment..................................................................................... 2
C. Resistive sensors .................................................................................................................... 2
1. Laboratory explorations ...................................................................................................... 2
2. Textbook and classroom explanations ................................................................................ 3
3. Questions and problems ...................................................................................................... 9
D. Converting resistance to potential difference ...................................................................... 10
1. Laboratory explorations .................................................................................................... 10
2. Textbook and classroom explanations .............................................................................. 10
3. Applications ...................................................................................................................... 11
4. Questions and problems .................................................................................................... 12
E. Loading the voltage divider.................................................................................................. 13
1. Laboratory explorations .................................................................................................... 13
2. Textbook and classroom explanations .............................................................................. 13
4. Questions and problems .................................................................................................... 16
2. Complex sensors ....................................................................................................................... 19
A. Sensors that produce a voltage or a current ......................................................................... 19
1. Laboratory explorations .................................................................................................... 19
2. Textbook and classroom explanations .............................................................................. 19
3. Questions and problems .................................................................................................... 20
B. Introduction to additional laboratory equipment.................................................................. 20
C. Signal and noise: RC filters.................................................................................................. 22
1. Laboratory explorations .................................................................................................... 22
2. Textbook and classroom explanations .............................................................................. 23
3. Applications ...................................................................................................................... 29
4. Questions and problems .................................................................................................... 29
3. Increasing the size (amplitude) of signals................................................................................. 31
A. The operational amplifier..................................................................................................... 31
1. Laboratory explorations .................................................................................................... 31
2. Textbook and classroom explanations .............................................................................. 33
3. Questions and problems .................................................................................................... 37
B. Other uses of operational amplifiers. ................................................................................... 38
1. Laboratory explorations .................................................................................................... 38
2. Textbook and classroom explanations .............................................................................. 39
3. Applications ...................................................................................................................... 43
4. Further textbook and classroom explanations................................................................... 46
5. Questions and problems .................................................................................................... 49
C. Extending the use of the op-amp: diodes and transistors .................................................... 50
1. Laboratory explorations .................................................................................................... 50
2. Textbook and classroom explanations .............................................................................. 54
3. Applications ...................................................................................................................... 70
4. Questions and problems .................................................................................................... 70
iii
4. Entering the digital world ......................................................................................................... 73
A. From analog to digital .......................................................................................................... 73
1. Laboratory explorations .................................................................................................... 73
2. Textbook and classroom explanations .............................................................................. 75
3. Questions and problems .................................................................................................... 78
B. Digital logic.......................................................................................................................... 79
1. Laboratory explorations .................................................................................................... 79
2. Textbook and classroom explanations .............................................................................. 80
3. Applications ...................................................................................................................... 87
4. Questions and problems .................................................................................................... 87
C. Sequential logic.................................................................................................................... 87
1. Laboratory explorations .................................................................................................... 87
2. Textbook and classroom explanations .............................................................................. 90
3. Applications ...................................................................................................................... 91
4. Questions and problems .................................................................................................... 92
5. Conversions between the analog and digital worlds................................................................. 95
A. Fundamentals of analog-digital conversion ......................................................................... 95
1. Laboratory exploration...................................................................................................... 95
2. Textbook and classroom explanation................................................................................ 95
B. Digital to analog converters ................................................................................................. 96
1. Laboratory Exploration ..................................................................................................... 96
2. Textbook and classroom explanation................................................................................ 98
3. Questions and problems .................................................................................................. 101
6. Interfacing the computer with the laboratory.......................................................................... 103
A. The National Instruments USB-6009 digital acquisition system....................................... 103
1. Laboratory exploration.................................................................................................... 103
2. Textbook and classroom explanations ............................................................................ 106
B. Using the USB-6009 with LabView .................................................................................. 106
1. Laboratory exploration.................................................................................................... 106
2. Applications .................................................................................................................... 106
Appendices.................................................................................................................................. 107
Appendix A: Resistor color code............................................................................................ 107
Appendix B: RC Filters using complex numbers ................................................................... 109
Appendix C: A/D, D/A game ................................................................................................. 115
Appendix D: Data sheets ........................................................................................................ 119
iv
Note to students
This book is the product of teaching an electronics course, Instrumentation for Scientists, at the
University of Michigan-Dearborn many times since 1974. In the first few years the course
concentrated on dc and ac circuits and those involving diodes and transistors. Integrated circuits,
both analog and digital, were taught in the last few weeks. The transistors and integrated circuits
were plugged into sockets; all connections were made by soldering wires. The oscilloscopes
were assembled from Heathkits. Circuits built and tested mostly demonstrated how signals could
be filtered and amplified, how feedback worked, and how regulated dc power supplies could be
designed.
During the 1970s grant money enabled the purchase of Heathkit digital and analog circuit
designers that allowed students to plug wires in rather than soldering them. Integrated circuits
became a greater focus of the course and digital circuits were taught before analog. Counters
were built that could count pulses and display numbers up to 999. In 1982 digital-to-analog
(DAC) and analog-to-digital (ADC) converters were introduced and digital and analog light
measurements were made.
In 1987 additional grant money allowed the purchase of Commodore 64 computers. Students
could now learn how the computer could analyze digital signals that came from laboratory
measurements and how the computer could transmit digital signals to laboratory equipment. The
ADCs and DACs built with integrated circuits allowed analog signals to be used as inputs to the
computers and analog signals to be sent back to the equipment. Students wrote short BASIClanguage programs to receive and transmit the signals, although machine-language programs,
supplied by the instructor, were needed for high-speed transfers.
In 1995 the Commodore 64s were retired and replaced with IBM PCs. Commercial interface
cards had multiple ADCs and DACs as well as digital inputs and outputs. BASIC-langauge
programs could still be used, although commercial software handled most of the detailed
communication between the interface cards and the computers.
Sensors were investigated that could measure light intensity and temperature, and stepper motors
demonstrated how electronics could create controlled motion in the laboratory.
Between 1995 and 2004 Professor Vaman Naik improved the laboratories and introduced
LabVIEW to the course. This widely-used commercial program greatly simplifies
communication with interfaces and analysis of laboratory data.
In 2005 I revamped the course in several ways. First, I shifted the focus from a collection of
electronic circuits that were useful in the laboratory to sensors that convert physical quantities
(light, magnetic field, temperature, etc.) to electronic signals (voltages) and to devices that
convert voltages to physical quantities (motion, heat, etc.). Electronic circuits to be investigated
are selected on the basis of whether or not they are useful in these conversions.
Second, I introduced new flexible, low-cost, high quality data acquisition systems from National
Instruments. These communicate with the computer over a USB cable and work directly with
LabVIEW.
Third, I changed the way the course was taught from separate lectures and laboratories, where
the lectures explained what would be done in the next four-hour laboratory, to a combined
lecture/laboratory where the lectures explained what had just been done in the last laboratory
v
activity. That is, the course is now based on the Learning Cycle (Engage, Explore, Explain,
Expand, Examine).
I have found that no commercially available textbook fits either the content or the pedagogy of
this course. Further, most books written for electronics courses taught to physics students are at
least ten years out of date. For that reason, this book includes laboratory explorations, textbook
explanations, and questions and problems for students. It contains only slightly more material
than can be covered in a thirteen-week course that meets seven hours per week and allows for
student projects in the last two weeks.
This book (and the course) are clearly a work in progress. Student and instructor experiences will
most certainly lead to changes in the laboratory explanations, the textbook explanations, and the
questions and problems. Students are likely to find it useful to select a traditional, comprehensive
textbook as a reference. There is no better reference than The Art of Electronics by Paul
Horowitz and Winfield Hill (Cambridge University Press, 1989). Trade books written for the
hobbiest and often available at Radio Shack are also very useful. There are also excellent
resources on the Web for circuits, explanations, and datasheets. I sincerely invite suggestions for
improvements.
Paul W. Zitzewitz
Dearborn, MI
June, 2008.
vi
1. Simple sensors
A. Electric circuit models
How can one model voltage, current, and resistance: the fundamental quantities describing
electric circuits?
Building models is an important part of any science. Models for electricity abound. You’ve
probably heard of a flowing water model. You’ve also used mathematical models (V = IR and
Kirchhoff’s laws) in physics classes. Later in this course you’ll be introduced to several other
mathematically-based models.
For now, however, you will investigate a physical model involving air. You’ll find some straws
and coffee stirrers on the table. Blow through one and use your hand to feel the air flow coming
through it. Create a quantitative measure of the amount of air flow that uses a sheet of paper.
Make a dark mark in the middle of the paper. Using tape, hang the paper from the edge of the
table. Now blow directly at the dark mark. Note that the deflection of the paper is a good
measure of the amount of air flow. What do you have to do to change the air flow?
The following table lists many of the analogies between electric circuits and this model
Electric Circuit
Battery
Strength of battery push (measured in volts)
Blowing-through-straw Model
Lungs and mouth
Strength of blowing through straw (air pressure
in lungs/mouth)
Straw
Air in lungs, mouth, straw and hitting paper
Air current: the rate of flow of air
Resistor
Electric charges in entire circuit
Electric current: the rate of flow of electric
charges
Electric current meter (ammeter)
measures value of electric current
Value of electric current measured in amps
(milliamps)
Air current meter (hanging paper)
measures value of air flow
Value of air flow measured by how far end of
paper moves from its original position
Work with your group to develop demonstrations of the following aspects of electricity
1. For a constant resistance, current is proportional to battery voltage.
2. A resistor consists of a material (often a wire) characterized by its diameter and length. The
resistance is proportional to the length of the wire.
3. The resistance is inversely proportional to the cross-sectional area (πr2).
4. When resistors are placed in series, for a constant battery voltage the current drops.
5. When resistors are placed in parallel, for a constant battery voltage the current increases.
Your lab group will be asked to show the class one of these demonstrations.
What are some limitations of this model?
1
B. Introduction to laboratory equipment
What are the design workstation (ProtoBoard) and digital multimeter (DMM), tools you will be
using throughout the course?
•
Use the DMM to measure the voltages that the power supplies on the ProtoBoard produce.
Note that connections to the workstation are made by pushing the stripped ends of #22 solid
insulated wires into the small holes in connection blocks on the workstation. DMM probes
can then touch the other ends of the stripped wires.
•
Obtain a resistor from the stocks. Pick one with a resistance of several thousand ohms (kΩ
or just k). Use the DMM to measure its resistance. Determine whether or not the resistance is
within the stated tolerance of the resistor. To interpret the color code used for resistors see
Appendix A.
•
Using the workstation, hook up a simple circuit with the resistor and a power source.
Measure the voltage drop across the resistor and the current through it. Check to see that the
resistance calculated using R = V/I agrees with the measured resistance within the
uncertainties of your measurements.
C. Resistive sensors
What are the properties of electronic devices that can sense physical conditions such as light,
temperature, force, position, angle?
Photoresistor
(light)
Thermistor
(temperature)
Strain gage
(force)
Linear potentiometer
(position)
Potentiometer
(angle)
1. Laboratory explorations
You will be given a selection of sensors that are in the resistive or conductive domain. That is,
their resistance depends on a specific physical condition, light, temperature, force, position, or
angle. You are to design experiments to measure the resistance of each under three or four
different conditions. You should make these conditions reproducible so that you can test
different sensors under the same condition at a later time. For example, when testing a
temperature sensor you could use the ice point (0oC), room temperature, and hot water at or near
the boiling point (100oC). When testing a light sensor use a point light source (LED or diode
laser) and two polarizing sheets whose angles can be measured to 5o or better. Light,
temperature, and force sensors are two-terminal devices, but potentiometers are not. When using
the rotary and linear potentiometers measure the resistance between the one end and the movable
contact. The force sensor will show a small, but detectable change in resistance. Wait to make
quantitative measurements until later.
Chapter 1. Simple sensors
2
•
Record the resistance of the device under each physical condition.
•
Determine how the resistance changes in response to the input condition.
•
Draw a graph of the relationship of either resistance or its inverse, conductance, as a function
of the physical condition.
•
How linear is the relationship? How can you tell?
2. Textbook and classroom explanations
Basic principles of direct-current circuits
You have studied the principles of electric circuits in previous physics courses, so we need only
make a brief review here. Circuits are made up of materials the permit the motion of electric
charges. A circuit is a complete loop that allows the charges to cycle around for as long as the
circuit is complete.
Current
In most metals the mobile charges are electrons, which have a charge of –1.602×10–19 coulombs.
The rate of flow of charges is called current, represented by the symbol I. The SI unit of current
is the ampere (A), which is equal to one coulomb (C) per second (s).
While an ampere is a useful unit to use in measuring current in household circuits, in electronic
instrumentation, currents are much smaller and the following units are normally used:
Unit
mA (milliamp)
µA (microamp)
nA (nanoamp)
pA (picoamp)
Value
10–3 A
10–6A
10–9 A
10–12 A
Electrons are the mobile charge carriers in most metals, but in semiconductors, as we will learn,
either electrons or “holes” (the absence of an electron) carry the current, depending on the type
of material. For historical reasons, physicists and engineers work with conventional current.
They assume that positive charges flow in the opposite direction from the flow of electrons.
Potential
Charges flow in the direction that minimizes their energy. Therefore they move from a region of
high potential energy to a region of lower energy. Rather than considering potential energy, we
normally consider electric potential, or energy per charge. The unit of electric potential is the
volt (V), which is equal to one joule (J) per coulomb (C). So, charges flow from high to low
potential.
Energy must be added to charges in order to increase their potential. Therefore a circuit must
have a source of electrical energy if there is to be a current. Because energy can be neither
created nor destroyed, the energy source is really a device that converts another form of energy
to electrical energy. The most common source is the battery, which converts chemical to
electrical energy. While we will usually analyze circuits as if they had a battery, in practice we
will use a “power supply,” a device that is able to raise the potential of charges to a selected
value and supply as much current as needed.
Chapter 1. Simple sensors
3
Words about language
Most people use “voltage” in place of potential difference. They will say “A battery has a
voltage of 1.5 volts.” We will follow this usage here. Power, which is measured in watts, if often
treated the same way. Wattage and power are often used interchangeably. The same is often done
for current and amperage. One might even talk about ohmage instead of resistance and jouleage
instead of energy, faradage for capacitance and henryage for inductance. Please resist the
temptation! Well use voltage, but go no further!
Does current flow? Current is the flow of charge, so a flow can’t really flow. There is a current
through a wire, and there is a potential difference (ok, voltage) across a resistor.
Ground
A charge moves in response to a difference in potential, so the voltage at one location in a circuit
must be used as a reference point, and all other voltages are expressed relative to that point. The
reference point is then assumed to be at ground, that is, zero volts. Strictly speaking, ground is
the potential of Earth. In practice, the ground potential is that of a cold water pipe or a metal rod
driven several feet into the ground. The rounded or grounding pin of a three-prong electrical
outlet is connected to ground. In a properly designed and constructed the metal case will be
grounded.
Power
As you can see by inspecting units, the rate at which a battery or power supply produces
electrical energy is the product of the voltage it maintains and the current drawn from it. That is,
P(J/s) = V(J/C) × I(C/s).
Resistance
A device that causes the potential of a charge moving through it to drop is called a resistor and
has the property of resistance. Resistance, R is defined as the potential drop, V, divided by the
current, I. Resistance is measured in ohms (Ω). Thus, since R = V/I, an ohm equals a volt
divided by an amp.
An ohm is a very small amount of resistance. In most circuits resistances have magnitudes of
kilohm, (103 Ω) kΩ or simply k, and megohm (106 Ω), MΩ or simply M. Note that
1 kΩ = 1 V/1 mA, and 1 MΩ = 1 V/1 µA. Thus for voltages commonly used in the laboratory,
the current through resistors in the kilohm range will be milliamps, while those through megohm
resistors will be microamps.
When there is a current through a resistance the potential drops, and so the energy of the
electrical charges drops. That means that energy must be converted to another form. Indeed, it
becomes thermal energy—the resistor’s temperature increases. The resistor, in turn, transfers this
energy to its surroundings, it heats the air around it and anything it touches.
The rate at which a resistor converts electrical to thermal energy, that is the power it “dissipates”
(or converts to thermal energy) is, by the definition of electrical power,
P = IV.
This equation can be rewritten if either the voltage and resistance are known, or if the current and
resistance are known as
P = V2/R or P = I2R.
Chapter 1. Simple sensors
4
Commercially available resistors are most often made of a thin carbon film deposited on a rod
made of an insulator. Resistances from 0.5 Ω to 10 MΩ are available in sizes designed to
dissipate power from 1/8 W to 1 W. In the laboratory we will generally be using ¼ W resistors.
The power rating can be determined from the size of the resistor. Resistors are sold with specific
tolerances, which are typically between 10% and 1%. For example, the 5% tolerance resistors in
the laboratory should be within 5% of the advertised resistance.
Resistors are labeled using a color code, a series of colored bands painted on the body of the
resistor. The color code is described in Appendix A.
In addition to fixed resistors, we will be using “potentiometers” or devices designed to allow the
resistance to vary depending on the position of a knob or slider. We will also use devices
designed so the resistance depends on temperature (thermistor) or light intensity falling on them
(photoresistor).
Potentiometers
Potentiometers are most often strips of resistive material (for example, wire, graphite, a
conductive polymer) with a contact that can be moved along the material from one end to the
other. The resistance of the device depends on the resistivity of the material, ρ, and the
dimensions (length, l, and cross-sectional area A) of the device: R = ρl/A. If the strip has a width
w and a thickness t, then the resistance is R = ρl/wt. How does this equation correspond to what
you experienced when you changed the location of the contact? How could you create a
potentiometer with a larger or smaller resistance if you had to make it the same physical size?
Strain Gage
If a piece of resistive material is stretched, its length will increase and its cross-sectional area
will be reduced. Thus its resistance will increase. A device that is designed to show a change in
resistance resulting from a small change in its length is called a strain gage. If the gage is
attached to a metal strip that is bent as the result of forces on it, for example, the gage can
measure the change in length of that strip.
A common configuration of a strain gage is shown below. It is made of a thin foil, about 0.025
mm thick, and from 5 to 100 mm long. The foil is bonded to a thin plastic film that in turn is
glued to the metal strip.
Strain gage
When stretched horizontally its resistance increases. The amount of increase is given by the
Gage Factor,
GF =
Chapter 1. Simple sensors
5
∆R / R
.
∆L / L
A typical strain gage has a gage factor of about 2%. That means that the change in resistance is
extremely small. Moreover, the resistance depends on temperature, and if too much current is put
through it, it will become warm. Nevertheless, strain gages, when used with modern amplifiers,
can provide linear, reproducible, and stable measures of the deformation of the material on which
they are mounted.
Thermistors
Photoresistors and thermistors are devices constructed from semiconductors. Resistivity
(measured in ohm-meters) and conductivity are reciprocal properties: σ = 1/ρ. The conductivity
of such a device depends on the number of free charge carriers that exist in the material. A
thermistor uses a semiconductor selected to have a large change in the number of free carriers
over a particular temperature range. The dependence of conductivity on temperature is nonlinear.
The industry-standard equation that provides an excellent fit to the resistance is the so-called
Steinhart-Hart equation
1 = A + B ln( R ) + C ln( R )3 .
T
The thermistor is a negative-coefficient device—that is,
its resistance decreases with increasing temperature, as
shown on the graph to the right. The resistance is
obviously nonlinear, but, over a small enough
temperature range, a linear approximation is often used.
Thermistors are inexpensive and can be made very
small (1-2 mm in diameter) so that they not only
respond quickly to changes but can be used to measure
the temperature of small devices. Resolutions of 5×10–4
o
C are easily obtained. A serious problem is self
heating. Current through the device causes it to dissipate power. When it warms, its resistance
drops, which increases the heating. The temperature can increase rapidly, destroying the device.
Photoresistors
A typical cadmium sulfide (CdS) photoresistor is shown at the right:
The device is connected to a source of potential difference and the current
(limited by a resistor placed in series) through it is measured. In the dark there is
a very small current, called the dark current. When light falls on the device the
conductivity increases, resistance falls, and the current increases.
CdS is a semiconductor. That is, at room temperature in the dark essentially all the electrons are
involved in covalent bonding; they are in the “valence band.” The minimum energy needed to
create a free electron is 2.41 eV, the energy of a 515-nm photon. If light or this or a smaller
wavelength is incident on the material, free electrons are created that are removed from the
material by the applied potential difference (and replaced by that potential source).
Chapter 1. Simple sensors
6
Each photon can generate as many as
900 free electrons, making CdS very
sensitive to light. Sensitivity also
depends on construction of the cell. It
should have minimal separation of
electrodes (l) and maximum length (d),
which can be obtained by creating
many zig-zags.
The conductivity, σ, of the device
depends on the intensity of the light, E,
according to the following equation:
σ = kEγ. The characteristic γ is the
slope of the log-log plot of the
conductivity as a function of light
intensity, or illumination. If the resistance rather than the conductance is plotted, then γ is the
negative of the slope. For the graph shown above γ = 1.15.
Note that many details have been omitted in this simple explanation. See any solid-state physics
textbook or http://www.aicl.com.tw/index.htm.
Kirchhoff’s Laws
Two fundamental conservation laws of physics are given new names when applied to electronic
circuits. They are called Kirchhoff’s laws after the German physicist of the 19th century, Gustav
Robert Kirchhoff, who published the laws in 1845.
If the potential of a charge is increased by an amount V in a battery or power supply, then the
potential must drop an equal amount in the remainder of the circuit. If it did not, then the energy
of the charge would keep rising (or falling) as the charged continued to loop around the circuit.
Conservation of energy would be violated. To conserve the energy of a charge in a closed circuit
loop, the net change in potential difference around the loop must be zero. That is ∑V = 0. This
equation is called Kirchhoff’s loop law.
Electric charge cannot be created or destroyed. As a result, at any junction in a circuit, the
number of charges entering the junction (or node) must equal the number leaving. That is,
∑I = 0. This equation is called Kirchhoff’s current law.
Let’s illustrate the use of Kirchhoff’s laws by deriving the equations for resistances in series and
parallel.
Consider three resistors connected in series with a voltage source:
Chapter 1. Simple sensors
7
+
I
+
R1
Vs

+

R2

+
R3

Resistors in series
According to Kirchhoff’s loop law, the changes in potential going around the loop sum to zero.
The current through the three resistors is the same and is labeled by I. When you move through a
resistor in the direction of the current the potential falls by an amount ∆V = IR. Thus the first end
of the resistor encountered is more positive than the second end. When moving through a voltage
source from the negative to the positive terminal the potential increases. Therefore Kirchhoff’s
loop law gives
∑V = +V
s
− IR1 − IR2 − IR3 = 0
or Vs = I(R1 + R2 + R3).
Now if the three resistors were replaced by a single, equivalent resistor, Vs = IReq. Therefore, you
can see that
Req = R1 + R2 + R3.
In the general case,
Req = ∑Ri.
Consider now three resistors connected in parallel:
I
I1
+
Vs
+ I2
+ I3
+
R1
R2
R3




Resistors in parallel
According to Kirchhoff’s node or current law, ∑I = 0. Currents into a junction are considered
positive, those leaving the junction are negative. Therefore, at the upper junction
I – I1 – I2 – I3 = 0.
Chapter 1. Simple sensors
8
That is, I = I1 +I2 + I3. Now the potential drops across the three resistors are the same and are
equal to the potential increase across the voltage source. That is,
Vs = I1R1, Vs = I2R2, and Vs = I3R3.
Therefore
I1 = Vs/R1, I2 = Vs/R2, and I3 = Vs/R3.
Thus
I = Vs(1/R1 + 1/R2 + 1/R3).
Now, if the three were replaced by a single equivalent resistor, the current through that resistor
would be I = Vs/Req.
Therefore
1/Req = 1/R1 + 1/R2 + 1/R3.
Or, in general
1/Req = ∑(1/Ri).
NOTE: ∑(1/Ri) is NOT the same as 1/∑Ri!
3. Questions and problems
1. What are some major limitations to the sensors you explored? For example,
a) What are the minimum and maximum values of force/light
intensity/temperature/position/angle to which they will respond?
b) How fast do they respond? How could you tell?
c) If you attached the sensor to an object, would the sensor change the property of the object
being measured? How could you tell?
To answer some of these questions you will need to do some research either in the library or
on web.
2. Resistors stocked in the laboratory come in a small number of different values. In general,
the values are 1.0×10n, 1.5×10n, 2.2×10n, 2.7×10n, 3.3×10n, 4.7×10n, 6.8×10n, and 9.1×10n
where n varies from 1 to 6. The minimum value is 10Ω and the maximum 3.3 M. You have
a 100k resistor, but need a total resistance of 150k. What combination of resistors in series
would create this resistance?
3. What is the resistance of 15-k and 10-k resistors in parallel?
4. What is the resistance of 3.3-k, 4.7-k, and 2.2-k resistors in parallel?
Chapter 1. Simple sensors
9
D. Converting resistance to potential difference
How can you convert a a change in resistance to a change in voltage? How can the electrical
domain of a sensor be changed from resistance or conductivity to voltage?
1. Laboratory explorations
We’ll use the fact that when there is a current through a resistor a potential difference exists
across it. Because a battery (or “power supply”) is a source of constant potential difference, there
must be a resistor in series with the sensor. A circuit containing a voltage source and two
resistors in series can be called a “voltage divider.”
Vsource
R1
Vout
R2
The voltage divider
A potentiometer effectively contains two resistors, one on each side of the moveable contact.
Connect the potentiometer across a voltage source, Vsource. Warning: the power dissipation of
potentiometers is limited. If you connect too large a voltage across them they will overheat and
be ruined! The 1k pots on the ProtoBoard will be destroyed if you put 15 volts across them.
•
Measure the output voltage, Vout, for three or four positions of the potentiometer. Compare
the quotient Vout/ Vsource with the position or angle of the potentiometer knob.
Now make voltage dividers using the resistive temperature or light sensors as one of the two
resistors in a divider. How do you choose the value of the second resistor?
Here is one approach. Use the data you collected in part C to find the resistance of sensor
roughly in the middle of the range of conditions you explored. Make the other resistor in the
divider approximately equal to the resistance of the sensor.
•
Measure the output voltage for the three or four conditions you previously selected.
Change the resistor to one with a value approximately ten times as large or one-tenth as large.
•
Measure the output voltage for the three or four conditions.
•
Summarize your results by describing how the choice of the second resistor affects the use of
the sensor.
2. Textbook and classroom explanations
We can use Kirchhoff’s laws to find the output voltage of the voltage divider. The voltage
divider is a series circuit. According to Kirchhoff’s current law, because there are no junctions,
the current is the same everywhere in the circuit. Call that current I. The current is given by the
battery voltage, Vsource, divided by the total resistance, R1 + R2. That is,
Chapter 1. Simple sensors
10
I=
Vsource
.
R1 + R2
The “output” voltage, Vout, is the potential drop across the sensor with resistance R2. That is,
Vout = IR2. Putting the two equations together gives
Vout
R2
=
.
Vsource R1 + R2
The ratio of the output to input voltage is equal to the ratio of the sensor’s resistance to the sum
of the resistance of the sensor and the added resistor.
3. Applications
Selecting the second resistance value.
Should the sensor or fixed resistor be R2? In either case the output voltage depends on the value
of R2 in a non-linear way. Let the sensor be R2 and write the equation above in terms of R2/R1.
You obtain
Vout
R2 / R1
=
.
Vsource 1 + R2 / R1
If R2/R1 >> 1 then the output voltage is essentially equal to the source voltage.
If R2/R1 << 1 then Vout/Vsource ≈ R2/R1.
Now let the sensor be R1 and write the equation above in terms of R1/R2. You obtain
Vout
1
=
.
Vsource R1 / R2 + 1
For R1/R2 >> 1 Vout/Vsource ≈ 1/(R1/R2)
while for .R1/R2 << 1, output voltage is essentially equal to the source voltage.
So, if you want an output voltage that is proportional to the resistance of the sensor, place the
sensor in the R2 location and make the fixed resistance much larger than the sensor resistance.
Note, however, that the output voltage will be small. If you want an output voltage proportional
to the conductance of the sensor, place the sensor in the R1 location and again make the fixed
resistance much larger than the sensor resistance.
Measuring very small voltage changes
There are two force sensors glued to the steel strip. They can be connected as a voltage divider
across a power supply. One sensor is R1, the other R2. When the strip is bent one resistance
increases, the other decreases. You would find that it is very difficult to get precise
measurements of the output voltage of the divider because of the limited precision of the digital
voltmeter. An excellent method of increasing the measurement precision of a varying voltage is
to subtract a fixed voltage so that variation is around zero. A Wheatstone bridge is a circuit that
accomplishes this subtraction. It consists of two voltage dividers. The output voltage, Vout, is just
the difference between the output voltages of the two voltage dividers. That is,
Chapter 1. Simple sensors
11
Vout = Vsource (
R2
R4
−
).
R1 + R2 R3 + R4
The sensors form one divider while the other divider has two resistors selected so that when the
force is zero, the output voltage is also zero.
Vsource
R1
R3
Vout
R2
R4
Wheatstone bridge
Construct a Wheatstone bridge for the force sensors using a 1-k pot as the second voltage
divider. Repeat the force measurements that you made above and compare the precision of your
results.
4. Questions and problems
1. A voltage divider consists of R1 = 1.0k and R2 = 3.3 k. The source voltage is 5.0 V. What is
the output voltage?
2. Find a pair of resistors that would produce an output voltage of 5.0 V from a source voltage
of 15 V.
3. You are working with Vsource = 15 V. A thermistor has a room-temperature resistance of 30k.
What value of R1 would produce an output voltage of 5 V. Replace R1 with a value available
in the laboratory and calculate the new output voltage.
4. The same thermistor is used with Vsource = 9 V and R1 = 27k. Find the output voltage at room
temperature and when the thermistor’s resistance has risen by 10% to 33k.
5. How does the choice of R1 affect the amount the output voltage changes, ∆Vout, when the
sensor resistance changes by an amount ∆R2? Consider the three cases R1 ≅ R2, R1 ∼0.1 R2
and R1 ∼10 R2. Let ∆R2/R2 be 10% and find ∆Vout/Vout. Describe the advantages of using
each of the three values of R1 in a voltage divider containing a sensor.
6. Four 120.4-Ω strain gages, electrically connected in a Wheatstone bridge, are glued to a
flexible metal strip, two on one face and two on the opposite. R1 and R3 are stretched when
the strip is bent, R2 and R4 are compressed. A 3.0 V potential is placed across the bridge. The
gage factor is 2.10%. Suppose the strip is bent so that ∆L/L = 0.0031. What is the change in
resistance of each gage and the output voltage of the bridge?
Chapter 1. Simple sensors
12
E. Loading the voltage divider.
What happens to the output of the voltage divider when current is drawn from it?
1. Laboratory explorations
The DMM used as a voltmeter acts like a 10M resistor. Thus there is very little current through
it. Other devices, however, have lower resistances and therefore draw more current.
•
Choose one of the voltage dividers including a sensor that you have investigated. Add a “load
resistor” as shown in the diagram below.
•
Measure the output voltage as you lower the value of the load resistor in steps from 10M to
1M, 100k, 10k, and 1k. Report your results.
Vsource
R1
R2
V out
Rload
Loaded voltage divider
2. Textbook and classroom explanations
The most straightforward method of analyzing the loaded voltage divider is to recognize that R2
RR
and Rload are in parallel, to calculate the equivalent resistance of the two using Req = 2 load ,
R2 + Rload
and then to recalculate the output voltage of the voltage divider.
A more general method is to model the voltage divider by creating a Thévenin equivalent circuit.
Such a circuit is a combination of an ideal voltage source and a single series resistance. The first
statement of a method of modeling a complex circuit in this way was developed by the German
physiologist and physicist, Hermann von Helmholtz (1821-1894) in 1853 (Őber einige Gesetze
der Verteilung elektrischen Strım in kırperlichen Leitern mit Anwendung auf die tierischelektrischen Versuche Concerning some laws about the distribution of electrical currents in solid
conductors with applications to animal-electricity experiments, Annalen der Physik und Chemie
89 211-233, 1853). The idea was rediscovered and named after Léon Charles Thévenin (18571926), a French telegraph engineer who published the results in 1883 (Sur un nouveau théorème
d’électricitié dynamique On a new theorem of current electricity, Comptes Rendus de l’Académie
des Sciences 97 159-151, 1883)
The steps in creating a Thévenin equivalent are
•
Isolate the circuit element (in this case the voltage divider)
•
Calculate the output voltage with no load and call it VTH. In this case
VTH = Vout.
•
Replace the battery with a wire and calculate the equivalent resistance of the circuit. This
resistance is called RTH. In this case after the battery is replaced R1 and R2 are in parallel, so
Chapter 1. Simple sensors
13
RTH =
•
R1R2
.
R1 + R2
The Thévenin equivalent is then the voltage source VTH in series with the resistance RTH.
RTH
V out
VTH
Thévenin equivalent circuit
To find the effect of the load resistor, you connect it across the terminals of the equivalent
circuit. Note that it now acts like a new voltage divider with the two resistors RTH and Rload and
you can easily find the output voltage:
Vout = VTH
Vsource
RLoad
.
RTH + RLoad
RTH
RLoad
V out
Loaded voltage divider using
the Thévenin equivalent
Consequences of the Thévenin equivalent:
1) To obtain the largest output voltage make sure the resistance of the load is much larger than
the equivalent resistance: RLoad >> RTH.
A. Use high input-resistance (RLoad) meters, instruments, or circuits.
B. Design sources with low Thévenin equivalent resistances (RTH)
2) Because a load that draws current (low RLoad) will reduce the output voltage, measure the
output voltage under the conditions it will see in service.
A. Test batteries by drawing current, not using a high-resistance voltmeter.
B. The equivalent resistance of batteries increases with age and when temperatures drop. Car
batteries have problems starting cars in winter.
3) The power delivered to a load is given by
2

 VTH R Load 
VTH
VTH
RL

 =
P = IV = 
.
2
R
+
R
R
+
R
(
R
+
R
)
Load  TH
Load 
 TH
TH
Load
Chapter 1. Simple sensors
14
a. When RLoad is very small in comparison to RTH the power delivered is small and
proportional to RLoad.
Reason: current is large (VTH/RTH) and independent of RLoad. Voltage is very small
(VTHRLoad/RTH) but proportional to the load resistance.
b. When RLoad is very large in comparison to RHT the power is small and proportional to RL–1.
Reason: current is very small (VTH/RLoad) and inversely proportional to RLoad. Voltage is the
source voltage (VTH) and independent of the load resistance.
c. There is a maximum power at some value of RLoad that can be found by differentiating the
power equation with respect to RLoad and setting the derivative equal to zero. At this value
VTH2
power delivered to the load is given by P =
. Half the power is dissipated in the load,
4 R Load
half in the equivalent resistor.
d. Application: selecting speakers to match amplifiers. Speaker impedance should be equal to
the amplifier output impedance. Connecting two speakers in series increases the effective
impedance; connecting two in parallel decreases it. Both connections make power delivery
less efficient.
Here’s a second example of using the Thévenin equivalent. You have already used the
Wheatstone bridge. What happens if you use a low-resistance device to measure the output
voltage? Then the circuit looks like this:
RL
R1
Vsource
R3
Vout
R4
R2
Loaded Wheatstone bridge
Without the load resistor, RL, the output voltage, Vout, is
Vout = Vsource (
R2
R4
−
).
R1 + R2 R3 + R4
You can use Kirchhoff’s laws to calculate the voltage with the load resistor, but it is easier to
replace each voltage divider with its Thévenin equivalent circuit:
RL
RTH1
RTH2
VTH1
Vout
VTH2
Wheatstone bridge Thévenin equivalent
Chapter 1. Simple sensors
15
First, use the results of the voltage divider example to find the equivalent voltages and
resistances.
VTH1 = Vsource(R2/(R1 + R2)),
VTH2 = Vsource(R4/(R3 + R4)) and
RTH1 = R1R2/(R1 + R2),
RTH1 = R3R4/(R3 + R4).
It’s now convenient to rearrange the voltage sources and resistors so you see that the circuit
consists of two voltage sources in series (with opposite signs) and two resistors in series.
RTH1
RTH2
VTH1
Vout
VTH2
RL
Simplified Thévenin equivalent
Now add the load resistor and use Kirchhoff’s loop law to find the current through this series
circuit:
– VTH2 + VTH1 – I(RTH1 +RTH2 + RL) = 0.
That is, I = (VTH1 – VTH2)/(RTH1 +RTH2 + RL).
Now note that Vout = IRL.
Or, Vout = RL
VTH1 − VTH2
RL + RTH1 + RTH2
Notice that this relatively simple procedure doesn’t produce a simple equation, especially when
you substitute back the values of the equivalent voltages and resistances. Note that when the load
resistance is much larger than the four resistors in the bridge the earlier result is obtained.
Vout = VTH1 – VTH2,
4. Questions and problems
1. Return to problem 3 in the previous section. Calculate the Thévenin equivalent circuit.
2. What would be the output voltage if a 1k resistor were connected across the output
terminals? A 10k resistor?
3. Two different thermistors are being considered for a circuit. One has a room-temperature
resistance of 30k, the other 3k. What would be the advantages and disadvantages of using
each one in a portable instrument that is powered by batteries? Assume the device connected
to the voltage divider has a 1k resistance.
4. Prove the result given in the text that power delivered is maximum when P =
Chapter 1. Simple sensors
16
VTH2
4 R Load
5. Show that in a Wheatstone bridge when the load resistance is much larger than the four
resistors in the bridge the earlier result, Vout = VTH1 – VTH2, is obtained.
6. In grade school, students often wide wire around a nail to create an electromagnet. A
scientific supply house delivered kits containing rechargeable batteries to save money and
reduce waste. The teacher discovered, however, that the electromagnets now got hot enough
to burn fingers. The alkaline and rechargeable batteries supplied the same voltage. Why
would they act differently? A quick search of battery properties showed that the internal
resistance of rechargeable batteries is much smaller than that of alkaline batteries. Why
would this cause the overheated magnets?
Chapter 1. Simple sensors
17
Chapter 1. Simple sensors
18
2. Complex sensors
A. Sensors that produce a voltage or a current
How can more complex sensors be used to measure the same or different physical conditions?
1. Laboratory explorations
Obtain three new sensors, a phototransistor to measure light intensity, an integrated-circuit
temperature sensor, and a hall-effect magnetic field sensor. For each one in turn, connect it as
shown below, then measure the output under the three-four different physical conditions you
used in the previous unit.
V = +10V
Light
V=5V
V =+10 V
Red
R = 10k
Phototransistor
Black
UGN3503U
Vout
R = 10k
Black
LM335
Green
Phototransistor
Red V
out
Temperature sensor
(Viewed from
"branded" side)
Vout
Magnetic field sensor
For the magnetic field sensor you will need to create new conditions. Make sure you stay below
the limits of the device, ±900 gauss or ±0.09 T. Determine how each sensor’s output voltage
changes in response to the physical input. Draw graphs. Describe limitations of the sensor.
2. Textbook and classroom explanations
We’ll explore transistors in more depth later. For the purposes of this experiment you need to
know that the current through the phototransistor is proportional to the light intensity incident on
it. What happens when you add a resistor in series with the phototransistor? As the current
increases, the voltage across the resistor increases proportionally (to what maximum voltage?).
But, as the voltage across the resistor increases, the voltage across the transistor decreases. This
affects the proportionality constant between light intensity and current. We’ll learn later in the
course how to convert the current to a voltage in a way that avoids this problem.
The temperature sensor contains a number of transistors, diodes, and resistors on an integrated
circuit chip. The datasheet for the LM335 is in Appendix D. The LM335 has an output voltage
proportional to the absolute (Kelvin) temperature in the amount of 10 mV/K. The claimed
accuracy is 1 K after the green lead is used to calibrate the device at one temperature. The
resistor in series with the power supply limits the current through the device to safe levels. The
output voltage should not depend on the power supply voltage (or the current through the device,
as long it is between 0.4 and 5 mA).
Chapter 2. Complex sensors
19
The magnetic field sensor uses the Hall effect—the appearance of a potential difference across a
current-carrying conductor in a magnetic field. The device creates a constant current through a
conductor (or semiconductor). The tiny Hall-effect voltage is amplified in the device. Because
the voltage depends on the temperature of the conductor, additional circuitry is needed. Typical
sensitivities are between 1.25 and 2.50 mV/gauss. These devices have been used extensively in
industry in recent years, and so the price has fallen dramatically. They come in two flavors—
linear and digital. The latter produce zero volts if the field is below a certain value and 5 V if it is
above the trigger value. They are often used to measure the rotational speed of motors. See the
datasheet and application guide in Appendix D.
3. Questions and problems
You have explored two different light and two different temperature sensors.
1. From your work so far, describe the strengths and weaknesses of each.
2. After doing appropriate research, design experiments to measure properties such as
•
response speed
•
linearity
•
thermal capacity (heat required to raise its temperature) of the two thermal sensors
•
wavelength dependence of the two light detectors.
B. Introduction to additional laboratory equipment
The oscilloscope is the instrument of choice for measuring voltages that vary in time, such as
those produced by a fluorescent lamp. It can display complex waveforms, permitting you to
determine the amplitude, period, and shape visually.
At the heart of most oscilloscopes is a cathode-ray tube (CRT). A CRT is a vacuum tube in
which a beam of electrons is produced, controlled, and directed to the tube’s face on which a
phosphorescent coating is deposited. Electrons striking the coating produce light pulses.
Normally the electron beam is swept from left to right across the face while the signal to be
measured moves the beam vertically. Thus a graph of voltage versus time is displayed.
The oscilloscope has four major systems as shown schematically on the next page. The vertical
amplifier conditions and amplifies the signal so that it can deflect the beam over a suitable range
of positions. The amplifier’s gain and offset can be adjusted.
The horizontal sweep generator creates a voltage that increases in time to move the beam from
left to right. The rate at which the voltage increases determines how rapidly the beam moves and
thus the scale of the time axis of the display. That rate can be adjusted.
The trigger system detects a suitable point on the waveform of the signal and starts the horizontal
sweep. The level of the signal, whether its slope is positive or negative, and other details can be
selected. Proper adjustment of the sweep rate and trigger is necessary to create a graph that is
both stable and displays the desired details of the waveform.
Finally, voltage supplies are needed to create the electron beam, accelerate and focus it, and
provide the high voltages needed to deflect it vertically and horizontally.
Chapter 2. Complex sensors
20
dc
ac
Input
gnd
Vertical
Amplifier
Vertical deflection
Gain (V/cm)
Offset
Trigger circuit
Level
Slope (+/-)
Type
deflection (cm)
Sweep circuit
Horizontal deflection
time (ms)
Sweep rate
(cm/ms)
You learned how to use an oscilloscope in your introductory physics course.
•
Practice these skills by first measuring the voltages produced by the power supplies. Set the
sweep and triggering so that you obtain a constant trace.
•
Explore and then describe the difference between “dc” and “ac” inputs.
•
Use the oscilloscope to see the outputs of the function generator on the workstation. Explore
how the triggering controls work to create static displays. Sketch and describe what you see.
•
Use the oscilloscope to view the output of the light and/or magnetic field sensors. Develop
ways to vary the light intensity (hint—view the fluorescent lamps) and magnetic field. Sketch
and describe what you see.
•
Connect a small electret microphone as shown below. Decide whether to use an oscilloscope
or DMM (or both) to measure the output voltage as you sing low and high pitches into it.
V=5V
R = 2.2k
Microphone
Capsule
Red
vout
Microphone connections
Chapter 2. Complex sensors
21
C. Signal and noise: RC filters
How can you distinguish signal from noise? In fact, what is the difference between signal and
noise? Is constant dc signal and variations in it noise or the other way around? Is the flickering
light from the fluorescent lamps noise or signal? It obviously depends on the task you are trying
to accomplish.
1. Laboratory explorations
A voltage divider that consists not of two resistors but a capacitor and a resistor has an output
that depends on the frequency of the input voltage. You are to explore the operation of this new
kind of voltage divider, called an RC filter. The input voltage will be provided not by the
workstation but by a function generator.
Period (T = 1/f)
Amplitude
(peak-to-peak)
Introduction to the function generator and ac signals.
A function generator is a versatile instrument that can produce
signals of different frequency, amplitude, and shape. Connect
the output of the function generator to the oscilloscope using a
coaxial cable with BNC connectors on each end. Explore how
to vary the frequency, amplitude, and shape (sine, triangle, or
square waves). Your goal is to create a sine-wave signal with
an amplitude of 10 V peak to peak with frequencies between 50 Hz and 50 kHz (periods between
20 ms and 20 µs).
Measuring the output of an RC voltage divider.
Explore the way the two dividers below work using R = 15 k and C = 0.01 µF.
vin
C
R
vin
vout
R
C
vout
•
The input voltage should be 10 V peak to peak. You will use both channels of the
oscilloscope, using Channel 1 to view the input voltage, Channel 2 to view the output. The
‘scope should be triggered by Channel 1 only. Be sure to adjust the sensitivity of Channel 2
so that the trace is as large as possible.
•
Measure the output voltage at frequencies of 50, 100, 200, 500, 1000, etc. up to 50,000 Hz.
Make a table of the ratio vout / vin versus the frequency.
•
One of these dividers is called a “low-pass filter” and the other a “high-pass filter.” Which is
which?
•
Add a high-pass filter to the microphone circuit using C = 10 µF and R = 2.2 k. Describe the
results. Can you detect any change in the low-frequency sound response of the microphone?
Chapter 2. Complex sensors
22
2. Textbook and classroom explanations
As you have seen, when you put a sinewave signal into a circuit containing a resistor and
capacitor the output is a sinewave with exactly the same frequency, but a different amplitude
and, perhaps, a different phase. We will explore the sinewave in detail, not only because it is the
most common alternating voltage, but also because Fourier showed that almost any repetitive
signal can be created by a series of sine waves of different frequencies. So, our goal is to find for
such a circuit its frequency response, that is, a graph of the ratio of output to input amplitude and
phase as a function of frequency.
The sinewave
Let’s start with a mathematical representation of a sinewave. The instantaneous voltage, v(t) is
given by
v(t) = Vp sin(2π ft),
where Vp is the peak or maximum voltage, and f the frequency in hertz (cycles per second). The
period is the inverse of the frequency: T = 1/f. A more general form of the sine wave allows for
an arbitrary phase:
v(t ) = V p sin( 2πft + φ) .
Suppose you observe a wave such as the one below on the oscilloscope. What is the equation of
the voltage versus time?
The vertical scale is set at 2V/cm, the horizontal at 2 ms/cm. The maximum value of the trace is
2.5 cm, or 5 V and the minimum is –5V.
Given the information above the amplitude
Vp = 5V.
The period is measures 2.5 cm on the screen which is a time of 5 ms. Thus the frequency
f = 1/(5 ms) = 200 Hz.
Find the phase by noting that
v(0) = Vpsin(φ).
Because v(0) = Vp, sin(φ) = 1 and φ = π/2.
Chapter 2. Complex sensors
23
Therefore
v(t) = (5 V) sin(2π(200Hz)t + π/2).
If the generator producing this signal were connected to a resistor there would be a current
i(t) = v(t) / R = Ip sin(2πft + φ)
where Ip = Vp/R.
The current has the same phase as the voltage.
i(t)
v(t)
R
Alternating current through resistor
Power dissipation
What power is dissipated in the resistor? Power is the product of voltage and current, or
P(t) = VpIp sin2(2πft + φ).
The power is varying between 0 and VpIp, but always positive. What is its average value over one
period? If we assume that the phase angle is zero, then we need to calculate
P (t ) = V p I p
1T 2
∫ sin ( 2πft ) dt .
T0
Using standard techniques we find the value of the integral is ½.
Therefore
P = ½VpIp.
In similar ways one can find P = ½Vp2/R and P = ½Ip2R.
It is helpful to define a value ac current or voltage that can dissipate the same power as the dc
current. Inspecting the equations above suggests that for a sine wave
Veffective = Vp/√2.
The proper name for the effective voltage (or current) is the rms, or root-mean-square value,
coming from the sequence of operations of squaring, taking the mean value, and then taking the
square root. Thus, for a sine wave,
Vrms = Vp/√2 and Irms = Ip/√2.
In the United States wall sockets deliver Vrms = 120 V. The peak voltage is then 170 volts. The
peak-to-peak voltage is twice that, or 340 volts.
Chapter 2. Complex sensors
24
The rms values for waveforms other than sine waves can be found by following the same
method: integrating the square of the voltage (or current) over one cycle, then taking the square
root.
Capacitors and inductors
The relationship between charge on and voltage across a capacitor is
Q = C V.
Because current is the time rate of change of charge, then
i = C dv/dt.
Therefore, if we let
vC(t) = VC cos(2πft + φ)
then
i(t) = –2πfCVC sin(2πft + φ)
But sin(θ) = cos(θ–π/2). That is sin(θ) = cos(θ–90o). Therefore
i(t) = –2πfCVC cos(2πft + φ – π/2).
That is, the voltage across a capacitor is 90o behind the current through it.
If we use an analogy to the definition of resistance: R = V/I, the equivalent resistance of a
capacitor would be 1/2πfC . This quantity is called the capacitive reactance, XC. Its magnitude is
given by
XC =
1
.
2πfC
Reactance has units of ohms, but is obviously frequency dependent. At dc the reactance of a
capacitor is infinity (it blocks constant currents) while at very high frequencies it vanishes. Thus
a capacitor acts like a good conductor to high-frequency currents.
The relationship between current and voltage in an inductor is v = –(di/dt)L. This leads to an
inductive reactance
|XL| = 2πfL
o
and the voltage being 90 ahead of the current. An inductor acts like a short circuit (low
resistance) to dc and like an open circuit (high resistance) for high frequencies.
Chapter 2. Complex sensors
25
Resistors, capacitors, and inductors in circuits
When you use two of the three devices, resistor, capacitor, and inductor, in a circuit (or all three),
the different phase relationships between current and voltage in the devices means that the
algebra and trigonometry becomes complicated. A method that uses only trigonometry will be
discussed here. Another method is to represent the voltage as a complex number. This method is
developed in Appendix B.
RC Circuits: Now let’s consider an RC circuit such as this:
C
vin
R
vout
RC circuit: high-pass filter
Kirchhoff’s node law says that the current is the same through both the resistor and the capacitor.
The voltage across the resistor is given by the equation vR = iR, or
vR(t) = –2πfRCVC sin(2πft + φ).
Note that the amplitude of the voltage across the resistor is
VR = 2πfRCVC.
Kirchhoff’s loop law demands that vin = vC + vR or
Vin cos(2πft) = VC cos(2πft + φ) –2πfRCVC sin(2πft + φ).
Now in order to solve this equation for the two unknowns, VC and φ, we need to use some
trigonometry to relate cos(2πft) to cos(2πft + φ) and sin(2πft + φ).
Use the relationships
cos(A+B) = cosA cosB – sinA sinB
sin(A+B) = sinA cosB + cosA sinB
with A = 2πft and B = φ to obtain
Vin cos2πft = VC {cos2πft cosφ – sin2πft sinφ –2πfRC sin2πft cosφ –2πfRC cos2πft sinφ}.
For this equation to be true, the coefficients of cos2πft and sin2πft must be the same on both
sides of the equation.
That is,
Vin = VC {cosφ – 2πfRC sinφ} and
0 = VC { –sinφ –2πfRC cosφ}.
Chapter 2. Complex sensors
26
The second equation can be satisfied if sinφ = –2πfRC cosφ, or tanφ = –2πfRC.
Substitute for sinφ in the first equation to obtain
Vin = VC cosφ (1 + (2πfRC)2).
Finally, we know what tanφ is, and we have to find out what cosφ is. Start with the definition
tanφ = sinφ/cosφ. Then use sin2φ = 1 – cos2φ, to replace sinφ with cosφ and follow through with
the algebra to obtain the relationship cos φ = 1 / 1 + tan 2 φ . Substituting for tan φ gives
cos φ = 1 / 1 + (2πfRC ) 2 .
We use this relationship and the equation above for Vin to find VC and VR in terms of V0.
VC = Vin
VR = Vin
`1
1 + (2πfRC ) 2
2πfRC
1 + (2πfRC ) 2
.
Do the amplitudes obey Kirchhoff’s loop law? That is, does VC + VR = Vin? From an inspection
of the equations above you see that the answer is no!
Does that mean Kirchhoff’s loop law is violated? Not if the two voltages are not in phase. How
are phases of the two voltages related? Recall that sinθ = cos(θ – π/2) and that
cosθ = –cos(θ – π). Therefore –sinθ = cos(θ + π/2) and
vC(t) = VC cos (2πft + φ)
vR(t) = VR cos (2πft + φ + π/2).
So, the voltage across the resistor leads the voltage across the capacitor by π/2 (90o).
In the filter we are considering vout is the voltage across the resistor, vR. Therefore
vout = vin
2πfRC
1 + (2πfRC ) 2
Let’s examine this equation. At high frequencies, where
2πfRC >> 1, vout = vin.
That is, there is no reduction, or attenuation, of the signal.
At low frequencies, where
2πfRC << 1, vout = 2πfRC vin.
That is, the output voltage is very small and increases linearly with frequency. When the
frequency doubles, the output voltage doubles.
The frequency where 2πfRC = 1, that is, f = 1/(2πRC), is a good way to characterize the filter. At
this frequency the output voltage is 1/ 2 times the input voltage. Engineers call this frequency
Chapter 2. Complex sensors
27
the “–3dB breakpoint” of the filter. Below this frequency the response is almost linear in
frequency, above it, it is almost constant. Thus
f 3dB =
1
.
2πRC
The phase angle between the output and input voltages is given by the tangent of the ratio of the
imaginary to the real parts tan φ =
1
.
2πRC
At high frequencies the phase angle approaches zero.
At low frequencies the phase angle approaches π/2 or +90o.
At the –3dB frequency tan φ = 1, or φ = 45o.
These results are summarized in these graphs:
In the low-pass filter the roles of the resistor and capacitor are interchanged, so the analysis is
very similar.
vin
R
vout
C
RC circuit: low-pass filter
The output voltage is
vout = vin
Chapter 2. Complex sensors
28
1
1 + (2πfRC ) 2
and the phase angle is given by
tan φ = −
1
2πRC
At very low frequencies the output voltage is now approximately equal to the input voltage and
the two are in phase. At very high frequencies the output drops proportional to 1/f (it is cut in
half for every doubling of the frequency) and the phase approaches –π/2 or –90o. The –3dB
frequency is the same,
f 3dB =
1
.
2πRC
The resultant graphs are shown below:
3. Applications
Does the measured phase relationship between input and output voltages agree with the
theoretical results? Your first task will be to measure it for either the low-pass or high-pass filter.
Use the same method you used to measure the amplitude. Remember that if the phase difference
is 0o, the peaks will be at the same time; when it is 90o the peak of one will occur when the other
signal is zero.
4. Questions and problems
1. A sinewave signal has a frequency of 1 kHz and an amplitude Vp = 10.0 V. It is connected to
a 10-kΩ resistor. Find the peak and rms currents through the resistor. What power is
dissipated in the resistor?
2. Consider a square wave signal with a period of 10 ms that alternates between –Vp and +Vp.
What is the average voltage? The rms voltage? What power would this signal dissipate in the
10-kΩ resistor?
3. What is f3dB for the low-pass and high-pass filters whose graphs are shown above?
4. If the capacitor in the low-pass and high-pass filters is 0.01 µF, what was the resistor?
Chapter 2. Complex sensors
29
5. Design an RC high-pass filter so that at f = 120 Hz Vout = (1/20) Vin. Make the resistance
about 10k.
6. Plot Vout/Vin for the low-pass and high-pass data you obtained in the laboratory versus the log
of the frequency. Also plot the log of Vout/Vin versus the log of the frequency.
7. According to theory, the –3dB points of the high-pass and low-pass filters are the same. Was
that true of your filters? Show your data and calculations.
8. Return to the question about Kirchhoff’s loop law. Show that when the phase it taken into
account at any instant in time Kirchhoff’s law is indeed satisfied.
Chapter 2. Complex sensors
30
3. Increasing the size (amplitude) of signals
A. The operational amplifier
1. Laboratory explorations
Introduction
The operational amplifier, or “op amp” for short, has been the workhorse of analog electronics
for over 30 years. It has two inputs and one output with the voltages related by the equation
vout = Av(v+ – v–) where Av , the “open loop gain” is a large number, typically 105. Rather than
using this large gain directly, we use negative feedback to reduce the gain but make the gain
depend on a pair of resistors, not on the amplifier itself. The reason for this decision and the
positive aspects of feedback will be explored in the next few units of the course.
We will be using one of the best known op amps, the 741. It is in an 8-pin device dual-inline
package (DIP). A dot on one corner of the package identifies pin 1. The internal connection to
each pin is shown below together with a schematic diagram of the input and output of the op
amp.
8
V+ vout
7 6
5
v–
vout
v+
1
2 3
4
v– v+ V–
Op amp connections
Op amp schematic
The op amp requires external power. Set the +5-15 V supply to +15 V and the –5-15 V supply to
–15 V. Connect the +15V to pin 7, the –15V to pin 4.
First measurements with an operational amplifier
The circuit below shows one way of connecting an op amp as an amplifier. Both vin and vout are
measured with respect to ground.
R2
vin
R1
vout
Inverting amplifier
Chapter 4 Entering the digital world
31
Choose one pair of resistor values with R2 > R1 and R1 at least 1k. First make the input voltage a
variable dc value that can be either positive or negative. A good way to do this shown below.
Use the 10-k potentiometer.
+15 V
Connect to vin
–15 V
Adjustable dc source
•
Determine the maximum output voltage of the amplifier (both positive and negative).
•
Measure the gain of the amplifier, where the gain is defined as
A = ∆Vout/∆Vin.
•
Make sure that the output voltage changes as you change the input voltage and isn’t “pinned”
at either the maximum positive or negative value.
•
Remove the 10-k pot and vin to ground to make the input voltage is zero. Measure the output
voltage. Any deviation from zero is called the voltage offset. Is there a measurable offset?
•
Find the gain of the amplifier for low-frequency (10-100Hz) ac. Connect the output of the
function generator to the input connection of the op amp. Is the dc gain, found above, and the
ac gain the same? If different, report the differences.
•
Repeat for at least two different pairs of resistor values to explore how the gain depends on
the resistors.
A second way of connecting an op amp as an amplifier is shown below.
R2
vout
vin
R1
Non-inverting amplifier
•
Repeat your gain measurements with the same set of resistors.
•
What differences are there between the two ways of connecting the op amp?
•
Why is one connection called “inverting” and the other “non-inverting”?
Chapter 4 Entering the digital world
32
2. Textbook and classroom explanations
History of the op amp
The term operational amplifier refers to their origin as the building blocks of analog computers
in the 1940s. These devices used voltages as analogs for numerical quantities. The computer then
performed mathematical operations such as addition, multiplication, differentiation, and
integration on the voltages. As an example of a problem such a computer could solve, you could
assign voltages to represent the parameters of an artillery shell (mass, air resistance, initial speed
and angle, initial position, and wind speed) and the computer could calculate its height and
horizontal position as a function of time. These computers were roomsized, used tremendous amounts of electrical power, and were very
slow.
In 1952 George A. Philbrick Researches, Inc. introduced the “small”
K2-W commercial op amp. It cost $24 ($160 in 2006 dollars) and
required sources of +300 V and –300 V. Ten years later they
introduced the first version that replaced the vacuum tubes with
transistors. In 1965 Robert Widlar, working at Fairchild
Semiconductors, invented the first integrated-circuit op amp, the
µA709 and it sold in 1966 for $22.50 ($120 now). Voltages sources of
+ and –15 volts made this device much easier to use. In 1968 Widlar,
now at National Semiconductors, developed an improved version, the
741. Today the 741, and many improved versions, are widely available
for about $0.40.
An ideal amplifier
What would be the properties of an ideal amplifier? It would have a gain that would easily
chosen, independent of frequency over a wide frequency range, and a gain that didn’t change if
the amplifier aged or if its power sources changed. It would have a very high input impedance so
that it wouldn’t load down its source, and it would have a very low output impedance so it
wouldn’t be loaded down by a circuit it was driving. It would be able to accept both input signals
reference to ground, and differential signals (like from a Wheatstone bridge).
How could the desired characteristics of gain be achieved? The answer came from a much earlier
invention, negative feedback. In the 1920s Bell Telephone worked hard on reducing distortion in
telephone amplifiers. One of the problems was that the vacuum-tube amplifiers of that time aged
rapidly, and it would be much better if the gain depended on simple components, like resistors,
rather than vacuum tubes. The solution came to Western Electric engineer Harold S. Black while
he was commuting to work on a ferry boat in August, 1927. He wrote the equations on the only
piece of paper available, the New York Times he was reading. His idea was to feed part of the
output of the amplifier back to the input, but in a way to reduce the gain. He showed that the
final gain would then depend more on the resistors used to feed back the signal than on the
amplifier. The patent office thought it was a stupid thing to do. Black said “Our patent
application was treated in the same manner as one for a perpetual-motion machine.” (IEEE
Spectrum, December, 1977) We will explore these ideas in more detail later in the course. The
important point to note here is that the gain of the ideal amplifier should be much higher than the
gain you want to achieve with your circuit.
Here, then is a schematic of an idea op amp. There is a very high resistance, Rin between the two
Chapter 4 Entering the digital world
33
input terminals. The op amp produces a voltage proportional to the difference in the two input
voltages, vo = A(v+ – v–). This voltage is connected to the output through a small resistor, Rout, the
output resistance.
v–
Rout
Rin
vo
vout
v+
The ideal op amp
Note that in the circuit above, and all circuits to follow, power must be supplied to the op amp. In
most cases both +15 V and –15 V are required.
Let’s apply the characteristics of the ideal amplifier and see how closely they can be achieved.
Infinite gain. Gain is the ratio of the output voltage, v0, to the input (v+ – v–). That is,
A = v0/(v+ – v–).
If A = ∞, then no matter what the output voltage is, the input voltage is zero! In practice, A > 105,
so if the output voltage is around 10 V, then the input voltage is well under a millivolt.
Infinite input impedance. The input impedance, represented by a resistor, Rin, between the two
inputs, determines the current into (or out of) the amplifier:
i = (v+ – v–)/Rin.
While infinite resistance cannot be obtained, resistances of 1012 Ω are possible, so input currents
can be infinitesimal. Not shown on the diagram are resistances between each of the inputs and
ground, which can lead to more input current than Rin. Even so, the input current can usually be
neglected.
Zero output impedance. If Rout is zero, then vout = vo for any current. In practice circuit design
limits the current an op amp can produce to about 25 mA.
The first two lead to the following “Golden Rules” that permit easy analysis of op amp circuits:
1. The output attempts to do whatever is necessary to make the voltage difference
between the two inputs zero: v+ – v– = 0
2. The inputs draw no current.
Rule 1 only works if there is way for the output to feed back a signal to the input. The op amp
can’t do it itself! Let’s apply these rules to some common op amp configurations
Chapter 4 Entering the digital world
34
The voltage follower
vout
vin
Voltage follower
According to the first golden rule, the two inputs of the op amp must be at the same voltage.
Because the negative (or inverting) input is at the same voltage as the output, then the positive
(or non-inverting) input must also be at the output voltage. That is,
vout = vin.
What is the use of a voltage follower? According to the second golden rule, op amp inputs draw
no current. Thus the input impedance of the follower is infinity. If you connected the follower to
a voltage divider and connected a load to the op amp, the voltage produce by the divider would
not be changed by the load. The device acts as a buffer between the signal source and load.
Inverting Amplifier
You have already explored the inverting amplifier.
I2

+
R2
I1

+
vin
R1
A
vout
Inverting amplifier
Let’s analyze this circuit. First redraw the two resistors as a voltage divider:
vin
I1
v-
R1
A
I2
R2
vout
Inverting op amp resistors
Chapter 4 Entering the digital world
35
According to the first golden rule, there is no voltage difference between the two inputs. For that
reason the potential at point A (v–) is zero. It is not physically connected to ground, so it is said to
be at a virtual ground. Therefore the current I1. the voltage across the resistor divided by its
resistance, is I1 = (vin – 0)/R1. In a similar manner I2 = (0 – vout)/R2.
But, according to the second golden rule, the op amp does not draw any current. That means that
I1 = I2. Therefore vin/R1 = –vout/R2, or
vout
R
=− 2
vin
R1
How does an op amp make vin and vout behave like this? Suppose R1is 1k and R2 is 2k and assume
the amplifier gain is ∞.
vin = 1 V
vin = 1 V
R1 = 1k
v- = 0.67 V
R1 = 1k
v- = 0 V
A
A
R2 = 2k
vout = 0 V
R2 = 2k
vout = -2 V
How an op amp responds to input voltages
Suppose you suddenly make vin = 1 V. The output, vout, is still 0V. Then what happens? Because
the currents through the two resistors is the same, point A is at 0.67 V. So the inputs, v+ and v–,
are not at the same voltage— v– is much more positive than v+, so vout goes negative. When vout
reaches –2V the v– is now 0 V, the two input voltages are the same, and the op amp is “happy.”
If the output were to go more negative, the inputs would again be unbalanced, but in the opposite
direction, which would bring the output back to –2V. If the amplifier gain were not infinite, then
the two input voltages would differ by a fraction of a millivolt, but the operation would not be
changed by much.
The Non-inverting amplifier
Consider the circuit below
vout
vin
I2
A
+

+
R2
I1
R1

Non-inverting amplifier
Chapter 4 Entering the digital world
36
The voltage divider formed by the two resistors is slightly different:
vout
I2
v-
R2
A
I1
R1
Non-inverting op amp resistors
The consequence of the first golden rule is that the voltage at point A, v–, is equal to the input
voltage, vin. The current through resistor R1 is given by I1 = (vin – 0)/R1. The current through the
resistor R2 is given by I2 = (vout – vin)/R2. But, the second golden rule means that I1 = I2.
Therefore vin/R1 = (vout – vin)/R2. This can be rearranged as vout = vin(1 + R2/R1). So, the gain of a
non-inverting amplifier is
vout R2
=
+1.
vin
R1
3. Questions and problems
1. Compare the measured and theoretical gains of the inverting amplifier at both dc and ac for
the pairs of resistors you chose. Are you measurements in agreement? If not, propose reasons
why they are not in agreement.
2. Repeat for the non-inverting amplifier.
3. Design an amplifier with a gain of –150.
4. The text above described how an ideal op amp works to keep the input voltage difference
zero. Repeat the analysis for a real op amp with a gain of 105. That is, vout = 105(v+ – v–).
5. Propose two ways to create an amplifier with a gain of +100. Note that you may use any
number of op amps in your circuits.
6. Use the two golden rules to find the gain of the circuit below.
100 kΩ
vin
10 kΩ
vout
9.1 kΩ
Chapter 4 Entering the digital world
37
B. Other uses of operational amplifiers.
What are other uses of operational amplifiers? What limitations do op amps have?
1. Laboratory explorations
1. How does the gain of an amplifier depend on frequency?
•
Assemble an inverting amplifier with a gain –10 and measure its gain from 100 Hz to 1MHz.
Make measurements at decade steps (100 Hz, 1 kHz, 10 kHz, etc.).
•
Assemble a gain –1000 amplifier and repeat the measurements.
•
Make plots of the log of the gain versus the log of the frequency, putting both plots on the
same graph. Report how the two graphs are similar and how they differ.
2. What load does op amp place on a signal source? That is, what is the input impedance?
Introduction
From your experience with voltage dividers, you know that if the two resistors are equal then the
output voltage is half the input voltage. To the signal source, the input of an amplifier “looks
like” (has a Thévenin equivalent) a resistor with a value of Rin. The function generator “looks
like” (has a Thévenin equivalent) an ideal voltage source in series with a Rout = 50-Ω. Connect a
resistance-substitution box (Rsub box) in series between the function generator and the amplifier as
shown below so that you can “dial in” an added resistance.
Increase the resistance of the sub box until the output of the amplifier drops to ½ its value with
no dialed-in resistance. Then Rin = Rsub box + Rout
Function
generator
Rsub box
Rout
Amplifier
vs
Rin
vout
Measuring input resistance
Measurements
•
Measure the input resistance of an inverting amplifier with two different values of R1.
Does the input resistance depend on gain? On frequency?
•
Repeat for a non-inverting amplifier. You may have to make an estimate or a lower limit
(as in, the input impedance is greater than …).
Chapter 4 Entering the digital world
38
2. Textbook and classroom explanations
Input impedance of op amp circuits
Return to the figure of the inverting amplifier. Because point A is at virtual ground, resistor R1 is
the input resistance. For that reason, an inverting amplifier is a poor choice if high input
impedance is important.
The input impedance of a non-inverting amplifier depends on the input impedance of the op amp
itself. So, for an ideal op amp the impedance is infinity.
Real operational amplifiers
Finite gain: The gain of a real op amp isn’t infinity. At dc and low frequencies it is typically 105
or higher. The gain is called the “open loop gain” because it is the amplifier’s gain before the
feedback loop is closed. Gain is usually specified in decibels (dB), which is a logarithmic scale.
For example, 100 dB is a gain of 105, 80 dB is a gain of 104, etc. Thus a decrease of 20 dB means
that the gain has decreased by a factor of 101, a decrease of 20 dB is a factor of 102, etc.
Open-loop Gain
If the amplifier is to be stable, that is, not prone to the type of oscillation that you often hear from
a public-address system, its gain must be limited at high frequencies. The high-frequency
behavior is the same as that of a low-pass filter. That is, if the frequency is doubled, the gain
drops by a factor of 2 (6 dB). If the frequency is increased by a factor of 10, the gain drops by 20
dB. The gain is 1 (unity gain) at a frequency called the gain-bandwidth product. A typical value
is 1.5 MHz. See the graph below
106
120
105
100
104
80
103
60
102
40
10
20
1
1
10
102
103
104
105
106
dB
0
107
Frequency (Hz)
741 frequency response
How does finite gain, especially at high frequencies, affect the gain of a circuit with feedback?
To find out we have to relax golden rule 1. Because A = vout/(v+ – v–), the two inputs cannot be
assumed to be at the same voltage. But golden rule 2 still holds; the op amp draws no current.
Chapter 4 Entering the digital world
39
I2

+
R2
I1

+
vin
v-
R1
vout
v+
Inverting amplifier
So, again writing I1 + I2 = 0 gives
(vin – v–)/R1 + (vout – v–)/R2 = 0.
But, with v+ = 0, A = –(vout/v–), or v– = –vout/A.
Therefore vin/R1 + vout/AR1 + vout/R2 + vout/AR2 = 0.
This simplifies to
vout
R
1
=− 2
.
vin
R1 1 + (1 + R2 / R1 ) / A
The table below shows how the kind of decrease in A typical of an op amp affects the closedloop gain for two different values of R2/R1
Open-loop gain
A
105
104
103
102
101
100
R2/R1 = 10
Ideal
Real
–10
–9.999
–10
–9.99
–10
–9.89
–10
–9.01
–10
–4.76
–10
–0.83
R2/R1 = 1000
Ideal
Real
–1000
–990
–1000
–909
–1000
–500
–1000
–90.8
–1000
–9.89
–1000
–0.998
Open-loop Gain
These results are shown graphically below:
106
120
105
100
104
80
103
60
102
40
10
20
1
1
10
102
103
104
Frequency (Hz)
Chapter 4 Entering the digital world
40
105
106
0
107
dB
Amplifier input errors
Offset voltage.
Input voltage offset: If you connect the two inputs together you’ll probably find the output
voltage not zero. This is called the input offset voltage. The offset voltage is the result of pairs of
resistors in the op amp that do not have precisely equal resistances. The input stages of an
inexpensive op amp can’t be perfectly balanced. This defect isn’t a problem with low-gain
inverting or non-inverting amplifiers, but it is in other applications that we’ll soon consider.
Op amps specify the offset voltage as the output voltage when the two inputs are at the same
potential. For the 741, VOS is about 2 mV. The closed-loop gain increases that voltage to
Vout = GdcVOS, where Gdc is the dc gain of the feedback circuit. So if the gain is 1000, then the
output voltage would be 2 V. The offset voltage can be measured using the circuit below.
200k
50Ω
Vout
VOS = Vout/4000
Offset voltage measurement
The voltage offset of most single op amps can be trimmed using a single potentiometer. For
example, the offset on a 741 can be made zero by connecting a 10-k pot between pins 1 and 5,
with the center (adjustable) terminal of the pot connected to –15 V. The OP177, which will be
used later requires a 20-k pot between pins 1 and 5 with the center terminal connected to +15 V.
Adjust the pot for zero output voltage. Unfortunately, the adjustment can change in time or if the
op amp’s temperature changes. For example, if the op amp is used to drive a high-current load,
then it can heat up, changing the offset voltage.
OP177
741
5
5
1
1
+15 V
Voltage offset adjustments
If the offset is a problem then use a precision op amp that was made with laser trimming
techniques to balance the two resistances individually for each device produced.
Chapter 4 Entering the digital world
41
Input bias current
Input bias current, IB, is the current into or out of both inputs. It produces a voltage drop across
the feedback resistors even when the input voltage is zero. For an inverting amplifier
Vout = (–R2/R1)IB(R1||R2), where R1||R2 is the parallel combination of the two resistors. Reducing
both the gain and the parallel combination of the two resistors decreases this problem but at the
cost of circuit input impedance. Placing a resistor between V+ and ground with a value equal to
the parallel combination causes an equal drop in the voltage at this input, reducing errors due to
input bias current.
Input offset current
Input offset current, IOS, is the imbalance in input bias current between the two inputs. It is
usually a factor of 2 to 20 times smaller than the input bias current. The solution to problems
caused by input offset current is to choose an op amp with the smallest possible current. Both
kinds of input currents increase rapidly with temperature. Running circuits at or below room
temperature is vital.
Input impedance of op amp
If the input impedance of the op amp is not infinity, then it forms a voltage divider with the input
resistor, R1, reducing the gain. But, feedback multiplies the op amp’s input impedance by the
closed-loop gain of the circuit, and low-cost op amps with astronomical values of input
impedances are widely available.
Selected Operational Amplifier Characteristics
Type
Supply
voltage
Gain
(dB)
fT
(MHz)
1.2
4
Voltage
offset
typ.
(mV)
2
0.8
Voltage
offset
max
(mV)
6
2
Current
bias
max
(nA)
500
0.2
Current
offset
max
(nA)
200
0.1
LM741
LF411
±18
±18
86
88
CA3140A
±18
86
3.7
2
5
0.04
0.02
LM308
OP07A
OP177G
AD509K
1436
±18
±22
±22
±20
±40
88
110
140
80
97
0.3
0.6
0.6
20
1
2
0.01
0.02
4
5
7.5
0.025
0.06
8
10
7
2
0.3
25000
30
1
2
1.2
30
Comments
Classic, low cost
Classic, low cost high input
impedance
Old very high input
impedance
Original low bias
Original precision
High precision
High frequency
High voltage
Input impedance of circuit
The input impedance of a 741 is about 2M. The effects of feedback increase this to such a high
value that it is almost never a problem. If necessary, use a high-input impedance op amp like the
LF411 or CA3140A. Of greater concern are the input bias currents and input offset currents.
Output impedance of circuit and current limits
Feedback reduces the output impedance to almost negligible values, but more important is the
maximum output current, which is typically 20 mA. When larger currents are needed either
individual transistors or a power op amp is called for.
Chapter 4 Entering the digital world
42
3. Applications
Current-to-voltage converters
As you have seen, a resistor essentially converts current to voltage: V= IR, but if a resistor and a
sensor that produces a current are connected in series across a voltage source, the voltage across
the resistor cannot be larger than the voltage source. Even before that occurs, the voltage across
the sensor can drop to such a small value that it no longer functions correctly.
The phototransistor is one such sensor. It develops a current that is proportional to the intensity
of electromagnetic radiation falling on it. In an earlier exploration you used a resistor to convert
the current to a voltage. But, in that device the voltage across the phototransistor varied as the
light intensity changed. The op amp circuit shown below keeps the voltage across the transistor
at a constant 15 V.
V = +15V
Light
I
R = 470k
Phototransistor
A
Vout
Photometer:
phototransistor and current-to-voltage
converter
Use the two golden rules developed to calculate the relationship between phototransistor current
and output voltage. Look up the characteristics of the phototransistor in Appendix D to obtain an
equation relating output voltage to input light intensity.
If this circuit is to be a useful photometer, its properties must be checked. Is its output voltage
zero when there is no light falling on the detector? Is the output voltage proportional to the light
intensity? That is, is it a linear device?
In order to test the linearity we must have a method that varies the light intensity in a known
matter. One method is to use the intensity of plane-polarized light passing through a polarizer.
Basic optics theory shows that for an ideal polarizer Iout = Iin cos2θ, where θ is the angle between
the plane of light polarization and the axis of the polarizer. There are two methods of obtaining
polarized light. One is to use a diode laser, which emits almost 100%-polarized light. The second
is to pass unpolarized light through a polarizer.
Develop an experimental apparatus and test the linearity of the photometer. You should carefully
plan your experiment so that you cover the range of cos2θ from 0 to 1 in three or four steps of
equal size in cos2θ. Also note that real polarizers do pass some light when θ = 90o and do not
pass all light when θ = 0o. Furthermore, some polarizers, specifically Polaroid ™ sheets, do not
polarize infrared radiation. How will your experiment address these limitations?
Chapter 4 Entering the digital world
43
Mathematical operations
As mentioned above, operational amplifiers were originally developed for use in analog
computers in which a voltage was an “analog” of a number, and op amps performed arithmetic
operations, such addition, subtraction, multiplication, and division, as well as more sophisticated
operations such as taking the logarithm and exponential, differentiating, and integrating. We will
explore three such operations.
Adding two voltages—the summing amplifier
Op amps can add two voltages at the same time that it amplifies them. One application is to shift
the central voltage of a sinewave. For example, you could either remove a non-zero central
voltage to center the result around zero, or introduce an offset to, for example, make the
sinewave totally positive. The circuit of a summing amplifier is shown below:
va
vb
R2
R1a
R1b
A
vout
Summing amplifier
As will be shown in the next section, vout = – (R2/R1a)va – (R2/R1b)vb.
An application that is related to the sensors you have previously explored is to convert the output
of the LM335 temperature sensor from kelvins to degrees Celsius. To make the device useful in
room temperatures when the output of an op-amp is limited to about +14V, you should design
the converter so that 1o C = 0.01 V. Your task is to choose the value of the second voltage and
the three resistors. If you choose to use a voltage divider to supply the second voltage remember
that the input impedance of the summing amplifier will load the divider and change its voltage.
Design the circuit so that loading is avoided (Hints: recall the value of a voltage follower or
avoid the problem all together by taking the second voltage from a power supply on you
Protoboard. Check your device with the three or four temperatures you previously used.
Subtracting two voltages—the difference amplifier
You have already encountered two sensors for which the output was a small variation of a
relatively large voltage. Two strain gages connected as a voltage divider created a voltage that
was approximately half the source voltage, with small variations due to the bending of the strip.
The magnetic field sensor produces approximately 2.5 V for no magnetic field. You explored the
use of a Wheatstone bridge in which the sensor(s) comprised half the bridge and a voltage
divider the other half. The output of the bridge was then a small signal varying about 0 V. Your
goal is to amplify that signal. Because one of the two leads is not at ground, op amp circuits
studied so far cannot be used. The solution is to use a difference amplifier for which the input is
the difference in voltage between the two outputs of the bridge.
You could use a 10-k pot as the voltage divider half of the Wheatstone bridge. But then it could
be very difficult to balance the bridge—to make the output zero when the strip is not bent or the
Chapter 4 Entering the digital world
44
magnetic field is equal to zero. A solution is to use a “trim pot” to make the small adjustments.
+15 V
R1
Rtrim
Vout
R1
-15 V
Voltage divider with
a trim pot
Suppose that you use a 1-k potentiometer as the trip pot, Rtrim. Further suppose you wanted a 1-V
adjustment in the output voltage, centered about 0 V. Then the current through the trim pot
would be 1 mA. The voltage across R1 would be 14.5 V, so a 14.5-k resistor would be called for.
A 15-k resistor is the closest standard value, which would slightly reduce the voltage across the
trip pot. The same size resistor would be used for R2. If you want Vout to be variable about a
voltage other than zero, then choose the voltages at the ends of the divider and the values of the
resistors accordingly. Construct a Wheatstone bridge with either the strain gages or the magnetic
field sensor as one half and the voltage divider with trim pot as the other.
Instead of using a 741 as a difference amplifier, you will amplify the output of the bridge using
an AD620 instrumentation amplifier. As can be seen from the datasheet in Appendix D and as
will be discussed later, this device has very high input impedance, low offset voltage and bias
currents, and, most importantly, needs no external feedback resistors! Gain is adjusted by a
single resistor connected between pins 1 and 8. The value is calculated from the formula
49.4kΩ
RG
G = 1+
+VS
VinRG
2
7
1
8
AD620
3
4
6
5
Vin+
-VS
Instrumentation amplifier
Chapter 4 Entering the digital world
45
Vout
Explore the output of your device with gains of approximately 100 and 1000. Find the sensitivity
of your device. That is, how small a force on the steel strip or how small a magnetic field can
you measure reliably? Note that pin 5 must be connected to ground for the amplifier to work.
Integrating a voltage
C
R
Vin
Vout
Integrator
You have already used one method of measuring magnetic field. A second method uses
Faraday’s law, which states that the EMF created across the terminals of a coil of wire is given
by E = – dΦ , where Φ is the magnetic flux through the coil and E the EMF generated.
dt
r r
Therefore Φ – Φ0 = –∫E dt. If the field is uniform across the coil, then Φ = nB ⋅ A = nBA cos θ ,
where B is the magnetic field, A the area of the coil, n the number of turns, and θ the angle
between the field and the normal to the coil. If the coil is flipped through 180o, then the initial
and final fluxes will be equal and opposite so that Φ – Φ0 = 2nB⊥A, where B⊥ is the component
of the magnetic field perpendicular to the coil.
You will design an experiment using a coil and the integrator that can measure the magnitude
and direction of Earth’s magnetic field by flipping the coil through 180o. Note that all three
components must be measured, then converted into a vector with magnitude and direction.
Consider the electronics side of the experiment. If the flux doesn’t change, then the output
voltage must remain zero. To test this condition first construct the integrator using a low-leakage
capacitor with C = 0.01 µF and a resistor of your choosing. If you find that the output voltage
drifts, it is most likely due to input bias current and input offset voltage in the op-amp. Replace
the 741 op amp with the OP177G that has a much lower IB and Vos than the 741. You may then
need to add the 20-k pot to adjust the voltage offset of the OP177 to zero.
Report results of tests you made of the output voltage of the integrator for zero input voltages
and the magnetic flux experiments you conducted.
4. Further textbook and classroom explanations
The current-to-voltage converter
How do the two golden rules permit us to analyze the operation of a current-to-voltage
converter? Consider the photometer circuit shown above. The second golden rule means that the
current through the phototransistor, I, is equal to the current through the resistor. The first golden
rule says that the summing point, A, is at virtual ground. That means that the current through the
resistor is given by I = –vout/R. Thus vout = –IR. Also because of the first golden rule the voltage
across the phototransistor is 15V, no matter what the current through it is.
Chapter 4 Entering the digital world
46
The summing amplifier
A version of the summing amplifier useful for analysis is shown below:
I2
I1a
va
vb
R2
R1a
R1b
A
vout
I1b
Summing amplifier
The first golden rule states that point A is a virtual ground. Therefore the three currents can be
easily found: I1a = va/R1a, I1b = va/R1b, and I2 = –vout/R2. The second golden rule says that
I1a + I1b = I2. Therefore va/R1a, + va/R1b = –vout/R2. Or,
vout = – (R2/R1a)va – (R2/R1b)vb.
Thus the summing amplifier can add two voltages, and the voltages can be separately scaled or
weighted.
The difference amplifier
How could you take the difference of two voltages? One way would be to use a unity-gain
inverting amplifier to produce the negative of one voltage. A better method is to use the circuit
below:
Ia
R2
Ia
va
vb
R1
vout
Ib
Ib
R1
R2
Difference amplifier
Let’s use the golden rules to analyze its operation. The first golden rule says that the voltages at
the op amp inputs are the same: v– = v+. The second says because the op amp draws no current
the currents into and out of each of the inputs are the same. They’re called Ia and Ib. Use the
equality of the currents to find the voltage drops across the resistors in the two voltage dividers.
Thus (va – v–)/R1 = (v– – vout)/R2 and (vb – v+)/R1 = v+/R2. Solve the second equation for v+ to
obtain v+ = vb R2/(R1 + R2). Now gather terms in the first and substitute v+ for v– to obtain
va/R1 + vout/R2 = vb R2/(R1 + R2)(1/R1 +1/R2) = vb/R1. The final result is
Chapter 4 Entering the digital world
47
vout = (R2/R1)(vb – va).
This result depends on the two resistors labeled R1 and the two labeled R2 being precisely
matched. Further, the input impedance of the two inputs is low.
The classic improvement was the three-op amp circuit shown below:
vin-
RG
R1
R2
vout
vin+
Three-op amp difference amplifier
The four unmarked resistors are all equal, so the third op amp has a gain of –1. The two added op
amps are connected in the non-inverting amplifier configuration, so the input impedance is very
high. The entire gain is set by the three resistors R1, R2, and RG.
G = 1+
R1 + R2
RG
If precision op amps are used this is an excellent difference amplifier.
Today you can purchase the three-op amp circuit as a single device called the instrumentation
amplifier. The AD620 that you used is a low-cost version, yet has extremely low offset voltages,
input currents, and a gain-bandwidth product over 1MHz. In this device R1 and R2 have been
laser trimmed to 24.7 kΩ each.
The integrator
Recall from the section on ac circuits that the definition of capacitance, C = q/V means that
i = C(dV/dt). Therefore V = V0 + (1 / C ) ∫ idt where V0 is the voltage at the start of the integration
period. When a capacitor is used as the feedback element, as in the circuit below, the voltage
across it is the negative of the output voltage –Vout and the current is the input current through R,
that is, Vin/R. As long as the capacitor has been discharged at the start (V0 = 0),
Vout = −(1 / RC ) ∫ Vin dt .
Chapter 4 Entering the digital world
48
C
I
Vin
R
A
Vout
The op amp integrator
There are several conditions on the op amp in order for the circuit to work well. First, there must
be minimal voltage offset. That is, when Vin = 0, Vout must be zero. Second, input bias currents
must be very small. For that reason an FET-input device such as the LF411 or OP177G is almost
a necessity. Third, the capacitor must be high quality, with no “leakage” (that is, charges on it
cannot leak away, reducing the voltage across it). Finally, it is useful to put a momentary-contact
switch across the capacitor so that it can be discharged immediately before the integrator is to be
used.
5. Questions and problems
1. The original K2-W op amp had a dc gain of 1.5×104 and a unity gain frequency of 105 Hz.
You choose the resistors to make a gain –1000 amplifier. What would be the highest
frequency for which gain the gain is greater than –100?
2. Derive the gain equation for the three-op amp difference amplifier: G = 1 + (R1+R2)/RG. Hint:
consider the three resistors that are in series. The gain of the final amplifier is 1, so the
voltage at the top of R1 is vout, the voltage at the bottom of R2 is zero. Thanks to the first
golden rule the voltage across Rg is the input voltage. The second golden rule says that the
current through the three resistors is the same.
3. Design a summing amplifier that can add 1.5 V to a sinewave to shift center of wave above
ground. Assume that you have available vb = –15 V.
4. You want to make the dc offset of the summing amplifier of the previous problem adjustable.
So, you use a 10-k pot between 0 and –15 V as a voltage divider and connect it to the same
resistor R1b that you used in the previous problem. Before connecting op amp you adjusted
output of pot to –6.5 V. But, when you connected it to the resistor you found that the voltage
had changed. Explain why. Hint: consider the input impedance of your amplifier.
5. You are using an op amp integrator to convert a square wave into a triangle wave. The square
wave has an amplitude of 10 V and a period of 20 ms. During the 10 ms that the square wave
is +10 V you want the output to rise from zero to 1 V. During the 10 ms that the square wave
is –10 V you want the output to fall 1 V. What
R
resistor and capacitor should you use?
6. Find how Vout depends on Vin in the circuit at the
right. Hint: start with the golden rules.
7. Design a Fahrenheit thermometer using the LM335.
That is, convert the output of the LM335 from 10
Chapter 4 Entering the digital world
49
C
Vin
Vout
mV/K to 10 mV/oF.
C. Extending the use of the op-amp: diodes and transistors
The output current of an LM741 is limited to 20 mA. How can you connect an operational
amplifier to a high-current device such as one that can produce light, sound, magnetic fields, or
heat?
1. Laboratory explorations
Rather than using special-purpose high-current op-amps, we will use the basic components from
which integrated circuits are built: diodes and transistors.
The LED
How are the current through and voltage across a light-emitting diode, or LED related?
The schematic diagram below represents a LED.
Schematic view of LED
The current must be limited to about 40 mA (0.04 A) to avoid burning out an LED, so you must
not connect an LED directly across the 0-15V power supply on the workstation. Instead a
current-limiting series resistor must be used. First, estimate the resistance needed to keep the
current from a 15-V supply to less than 0.04 A. Then create first a table and then a graph of the
current as a function of the voltage across the LED. On your table describe the LED brightness.
Test both polarities of voltage.
From your data answer the question: Is the diode ohmic? That is, does it obey Ohm’s law
(current through is proportional to voltage across, independent of the direction of the current)?
The transistor
The transistor is a three-terminal device with the terminals named (for historic reasons) the base
(B), emitter (E), and collector (C).
Chapter 4 Entering the digital world
50
E
C
B
B
C
E
Bottom view of 2N4401
transistor
Schematic view of npn
transistor
We will first explore the current-voltage relationship of the base-emitter junction. The arrow
suggests that this circuit contains a diode (with the base being more positive than the emitter to
create current in the direction of the arrow), so use the same method you used to measure the
properties of the LED, but keep the current below 1 mA. Rather than measuring the current
directly, put a 10k resistor in series with the base and measure the voltage drop across the
resistor.
Plot the potential difference between the base and emitter (base-emitter voltage), or VBE, as a
function of base current, IB, from zero to 1 mA.
As you will learn shortly, current in the base-emitter junction controls the current in the
collector-emitter circuit. Explore how this works with the following circuit.
+15 V
270 Ω
LED
+5 V
10k
1-k pot
2N4401
LED Driver
Adjust the 1-k pot to change the current through the base-emitter junction, and thus the current
through the LED (most easily determined by measuring the voltage drop across the 270-Ω
resistor). For each value of the collector (LED) current (try 10 to 50 mA in steps of 10 mA)
measure the base current (by measuring the voltage drop across the 10-k resistor) and VCE, the
voltage between the emitter and collector terminals of the transistor. Plot the collector current
versus VCE and note on each point on your graph the value of the base current.
You now see how the base current controls the collector voltage and current. We’ll use a
different transistor amplifier circuit, one in which the LED and its resistor is connected to the
emitter rather than the collector. The base will be driven by an op amp, as shown below.
Chapter 4 Entering the digital world
51
As is always the case, resistors R1 and R2 determine the op amp gain. Thus they set the amplitude
of the input signal that varies the brightness of the LED from zero to the maximum possible from
this circuit.
R2
+15 V
R1
2N4401
vin
270 Ω
LED
Op amp with transistor LED driver
For now choose resistors so that the op amp gain is 10 and put a very low-frequency (about 1
Hz) signal from the signal generator into vin. Adjust the generator output amplitude to vary the
LED brightness. Now turn the amplitude down to zero and change the dc offset of the signal
generator until the LED is at moderate brightness. Increase the amplitude and note that you can
make the brightness vary smoothly, without being either full-on or off during much of the cycle.
When you later use this driver you’ll use a summing amplifier to provide this dc offset.
Constant-current source
A battery or a buffered voltage divider can supply a constant-voltage source. Why would you
want a constant-current source? Here’s one example: the magnetic field of a solenoid is
proportional to the current through it. If it is connected to a source of constant voltage, as the
solenoid warms up due to the power dissipated in it, its resistance changes, and therefore the
current through it will change. For a second example, consider the thermistor we explored briefly
at the beginning of the course. The temperature can be found from the resistance. That in turn
can be measured by the voltage across the thermistor, but only if the current through it is
constant.
Consider the circuit below:
+15 V
R
+15 V
10k
2N3906
10k
Thermistor
Constant-current source
To see why the current through the thermistor is independent of its resistance, analyze the op
Chapter 4 Entering the digital world
52
amp circuit using the two golden rules.
First calculate the voltage at the non-inverting input. What is the voltage at the inverting input?
What is the voltage across the resistor R? What value of R is needed for a current of 0.1 mA?
The transistor used is pnp rather than npn. The current through the collector-emitter circuit is in
the direction of the arrow; down in this case. The base is about 0.7 V more negative (less
positive) than the emitter. What is the voltage at the base of the transistor and the output of the
op amp?
The current through the thermistor is the collector current. Because of the base current, the
collector current is not equal to the emitter current, but, for low-power transistors like the
2N3906, the ratio of emitter-to-base current is at least 50:1, so to a good approximation the
collector and emitter currents are equal. If there is a current of 0.1 mA through R, there is almost
exactly 0.1 mA through the thermistor.
Note that the current through the thermistor depends on the voltage set by the voltage divider
connected to the non-inverting input of the op amp, the resistance R, and the ability of the op
amp to keep the two input voltages the same. The role of the transistor is to provide a source of
current beyond what the op amp can supply.
The major limitation of this circuit is the need for the collector voltage to be less positive than
the emitter voltage. Therefore the voltage across the thermistor is limited. If it gets too high
because the thermistor resistance is too large, then the transistor will not be able to conduct and
perform its task. Either connect the grounded end of the thermistor to a negative voltage or
reduce the current through it.
Construct the circuit and find the thermsistor’s resistance at a minimum of three temperatures.
This circuit will be used later as a digital thermometer.
Producing high-current, positive and negative voltages with an op amp
The single transistor driver has an important limitation: there is no way to have an output that
can produce either positive or negative currents. A combination of two transistors, one npn, the
other pnp called a push-pull pair, avoids this problem. It will be used later to drive a variety of
devices. For now, we’ll just explore how it works. Choose a resistor of about 10Ω for Rload.
+15 V
R2
TIP29
R1
1k
vout
vin
Rload
TIP30
B
C
E
–15 V
Op amp with push-pull driver
TIP 29, TIP 30 pins
Use a function generator to supply a 1-kHz sine-wave signal to the input. View the output on the
oscilloscope as you change the input amplitude. Why is the output not a faithful reproduction of
the input? (Hint: what is the minimum base-emitter voltage needed for a transistor to have
Chapter 4 Entering the digital world
53
significant collector-emitter current?)
Put the two transistors within the feedback loop by connecting the right-hand end of R2 to vout
rather than to the output pin of the op amp. Is the output now a faithful reproduction of the input?
Explain why this simple change in connections makes such a large difference.
2. Textbook and classroom explanations
In order to understand diodes and transistors, we will first examine how metals conduct
electricity. We will then explore conduction in semiconductors.
Metals
What are some characteristics of a metal? Hard, but malleable, not brittle. Opaque to light and
shiny (reflects light). Good conductor of electricity and heat.
How are these characteristics explained? A quick glance at the periodic table shows that many of
the most common metals are in columns IA and IB. Most metals have one or two electrons in
either an s, p, or d orbital; these are valence electrons. Their nuclei are arranged in a regular array
or lattice. The valence electrons are mobile and not associated with a particular atom. As a result
one can envision a metal as an array of positively charged ions in an electron gas. The metals are
held together by the attractive force between the ions and the electron gas. The gas is dense,
with, in the case of copper, 8.5×1028 m–3. For comparison, the STP density of air is 3.2×1025 m–3.
Quantum mechanics is needed to explain the properties of the electron gas. Electrons have a halfintegral spin, and so are called fermions. The probability of finding an electron with energy E
can be shown to be
f FD ( E ) =
1
e( E − E F ) / kT + 1
where T is the temperature in kelvins, k is the Boltzmann constant and EF is called the Fermi
energy or Fermi level. At T = 0 the probability fFD(E) = 1 for E < EF and fFD(E) = 0 for E > EF.
Thus all states with energies less than the Fermi energy are filled and all with higher energies are
empty. At temperatures above absolute zero the population of states with energies between
(EF – kT) and EF are empty and those between EF and (EF + kT) are filled. The Fermi energy is
large. For copper EF = 7.1 eV, corresponding to a velocity of 1.6×106 m/s or an equivalent
temperature of 8.1×104 K.
So a metal contains a dense gas of very fast-moving electrons. The electrons go in straight-line
paths until they scatter off an ion. A variety of experimental data show that the mean-free-path
between collisions is 39 nm, or about 150 times the distance between the ions. How can they go
so far without scattering? It’s because of the wave nature of matter. If the lattice were perfect, the
distance would approach infinity. The motion is disturbed by imperfections in the lattice,
including the thermal vibrations of the ions, which increase with increasing temperature.
How does a metal conduct electricity? When a potential difference is placed across a conductor,
there is an electric field in it. That field results in a force on the electrons. During the time
between collisions the electron is displaced a small distance in comparison to the mean-free-path.
The result is a slow drift velocity proportional to the electric field, of order of less than one
millimeter per second.
When an electron moves under the influence of an electric field its energy increases. How does
Chapter 4 Entering the digital world
54
this increase occur in light of the quantized energy levels in atoms? Copper has 29 electrons. The
n = 1, n = 2, and n = 3 levels are filled, and there is a single electron in the 4s state so we can
almost think of it as a single-electron atom.
When two copper atoms are far apart, the energies of the 4s
electrons is the same. But, as they are brought closer together the
two 4s energy levels split, one going up, the other down. The
splitting can be understood in terms of the wave nature of
electrons. As the wave functions overlap they form two mixed
states, one symmetric, with the probability of finding the electron
being larger between the two atoms. The other is antisymmetric,
with the probabilities largest between the atoms. The symmetric
state has a lower energy. The energy separation depends on the
amount of overlap of the charge, and thus on the spacing between
the atoms. Where the energy is lowest the two atoms would find
themselves at an equilibrium separation.
As more atoms are added, there are more ways of combining the
wavefunctions and more energy levels. All energies are between
the low and high energies of the two atoms. Six atoms would lead
to six states; a mole of atoms would lead to of order 1023 states.
The energy levels are so close together that they are called an
energy band.
ψ1
ψ2
ψ1 + ψ 2
ψ1 – ψ 2
Two atoms
Six atoms
Large number of atoms
Chapter 4 Entering the digital world
55
Each energy level can be occupied by two electrons, one with each
orientation of its spin. The energy bands of the lower levels, n = 1,
2, or 3, are all filled. The energy band of the 4s electrons in copper
is half full.
The energy of the highestenergy electron in this band
is the Fermi energy. At
absolute zero the boundary
between occupied and
vacant levels is sharp. The
function fFD(E) goes from 1
to 0 at a single energy, EF.
E
Empty levels
4p
N
4s
10N
3d
6N
3p
2N
3s
6N
2p
2N
2s
2N
1s
EF
fFD(E)
Filled levels
0
1
E
At temperatures above
Empty levels
absolute zero the function
EF
drops smoothly to zero over
Filled levels
fFD(E)
a width ∆E = 2kT (at room
temperature 2kT is about 0.05 eV while EF is about 8 eV). This
means that a few states below EF are empty, a few states above are
filled.
When an electric field is added there is an empty level only a tiny amount higher into which the
electron can go. Thus metals are good conductors because there are so many empty states close
to the filled energy states.
The fact that electrons in metals can readily absorb tiny amounts of energy explains why they are
opaque to light. The electric fields in a light wave accelerate the electrons, increasing their
energy, absorbing the energy of the light. If the accelerated electrons are on the surface, they can
reradiate the light, resulting in reflection.
Semiconductors and insulators
Most of the electrical insulators you use today are plastics. Because there is such a wide variety
of insulating materials, we’ll describe only one, the diamond. Diamond is a form of carbon,
which has six electrons. In an isolated atom two are in the 1s state, two in the 2s, and two in the
2p states. There are four vacant 2p states. When two or more carbon atoms are brought together
the energies of both the 2s and the 2p states split.
Chapter 4 Entering the digital world
56
6N states
Lattice spacing in Si, Ge
Energy
Lattice spacing in diamond
Conduction band 4N states
2p
2s
2N states
4N states
Valence band
Atomic spacing
When the atoms are a certain distance apart the lowest 2p state of an atom attains the same
energy as the highest 2s state. The “s” and “p” characteristics of the four electrons merge,
forming a hybrid state called sp3. The four electrons try to get as far apart as possible, which
results in a tetrahedral shape. Because it is these electrons that form the bonds between atoms,
the diamond crystal has a tetrahedral structure.
Tetrahedral structure of
diamond, silicon, and germanium
So diamond has 4N valence electrons binding atoms together. Their energies are in the valence
band. There are also 4N excited sp3 states that are not occupied. There is a gap of 5.45 eV
between the highest state in the valence band and the lowest state in the conduction band. If an
electron is given that much energy it can be detached from an atom and be free to move through
the crystal to conduct electricity.
As the temperature is increased the transition in the function fFD does soften, but never enough to
be able to free enough electrons to allow diamond to conduct electricity. It is an insulator.
E
Empty levels
EF
fFD(E)
Band gap
Filled levels
0
1
Energy level structure of insulator
Chapter 4 Entering the digital world
57
The element below carbon is silicon. The valence electrons in silicon are the 3s and 3p, which
again hybridize. But in this case the equilibrium separation between atoms is larger than that of
diamond. As a result the band gap is only 1.17 eV. In germanium, the element below silicon, the
gap is 0.744 eV.
Silicon and germanium are called semiconductors. Why? Because the gap is much smaller, at
room temperature there is a very small, but significant number of electrons in the conduction
band. How many are there? The function fFD(E) for Eg>>kT can be well approximated by
− E / kT
f FD ( E ) = e g . For an insulator this fraction is 10–44. Thus in a sample of about 1020 electrons,
there are none in the conduction band. For a semiconductor the probability that there is an
electron in the bottom of the conduction band is about 10–9, so there may be 1011 electrons.
Compare this to a conductor where there would be one per atom, or 1020.
If there are a few filled states in the conduction band, there must be a few empty states in the
valence band. That is, there must be just as many bonds missing an electron as there are free
electrons, 1011. This equality means that the Fermi energy is in the center of the band gap.
What happens when an electric field is applied to a semiconductor? The free electrons in the
conduction band respond as they do in a conductor, but the electrons in the valence band are also
affected. The only ones that can move are those very close to a vacant state. So, as a valance
electron moves one way, the vacancy moves in the opposite direction. The vacancy is called a
“hole” and therefore as electrons move in one direction, holes move in the opposite direction.
Although there are as many free electrons as holes, the free electrons can move much more
easily, and contribute about four times as much to the current as do the holes.
Conduction band
Free electrons
Valence band
Band gap
Free holes
Electric field
Conduction in a semiconductor
A semiconductor that conducts current in this way is called an intrinsic semiconductor. Because
so few electrons can contribute to the conduction process, the semiconductor must be extremely
pure in order that the impurities don’t affect the conduction process. It is extremely difficult, but
now possible to remove impurities from Si and Ge at the level of a part in 109, or 1 ppb. On the
other hand, if impurities with a known effect on the properties of a semiconductor are introduced
at the level of one part in 106 or lower, they can easily overwhelm the intrinsic conduction
process.
Chapter 4 Entering the digital world
58
Doping semiconductors
Suppose atoms of an element like P, As, or Sb is added to silicon or germanium. These elements
are in group V of the periodic table; they have five valence electrons. Four of these electrons
would be involved in bonding to the neighboring host atoms, but the fifth electron is not needed,
and so would be weakly bound to the impurity atom. The energy needed to remove this electron
is small, so on an energy diagram they would be have a discrete energy level just below the
conduction band.
Conduction band
Donor states
Eg ∼ 1 eV
Valence band
Donor doping
Because the energy separation between the discrete states and the conduction band is close to
thermal energy at room temperature, it is easy for these electrons to be excited into the
conduction band and be free to move. They are called donor states because the impurity atoms
donate their electrons to the conduction band. The impurities are called donors, and the process
of adding them to a pure semiconductor is called doping. The resulting semiconductor is called
an extrinsic n-type semiconductor because negative electrons carry the current.
Atoms of an element such as B, Al, Ga, or In can also be added. They are in group III and have
only three valence electrons. Those three electrons bond to the host atoms, but one of the
neighboring host atoms has an unbound electron. An electron can be easily captured to complete
the bonding. That is, thermal energy can excite it from the valence band into the discrete state.
The state is called an acceptor state because it accepts electrons from the valance band. This
process leaves a hole in the valence band. That hole can move easily, resulting in electric
conduction. Because the conductivity is due to positively-charged holes, it is called an extrinsic
p-type semiconductor.
Conduction band
Eg ∼ 1 eV
Acceptor states
Valence band
Acceptor doping
It should be noted that n-type and p-type semiconductors are electrically neutral. The n and p
refer only to the charge carriers, not the overall charge. Finally, just as in an intrinsic
semiconductor electrons are the majority charge carriers and holes the minority, in doped
semiconductors holes carry some current in n-type and electrons some current in p-type.
Chapter 4 Entering the digital world
59
At absolute zero the donor and acceptor states are empty, and so the Fermi level lies between the
valence band and the acceptor level for a p-type and between the donor level and the conduction
band for an n-type semiconductor. But, as the temperature rises, thermal energy causes these
states to be populated, and the Fermi level moves toward the center of the band gap. At high
enough temperatures an extrinsic semiconductor behaves like an intrinsic one.
The pn junction
In practice, a pn junction is typically formed by starting with an n-type semiconductor and
doping it from the surface with enough group III atoms to create a p-type region. The process is
shown in detail at http://jas.eng.buffalo.edu/education/fab/pn/diodeframe.html.
In principle, one can conceive of starting with separate pieces of n-type and p-type
semiconductors and bringing them together so that they touch.
n-type
n-type
p-type
p-type
free electrons
free electrons
free holes
free holes
Creating a semiconductor junction
Once they are in contact the free electrons in the conduction band in the n-type material are
attracted to the positively-charged holes on the p-type side. They fill the holes. But remember
that each side was electrically neutral. The free electrons came from the donor states; the free
holes from the acceptor states. The positively-charged donor states and the negatively-charged
acceptor states near the junction oppose further electron migration. They create a kind of battery,
with the n-type side more positive and the p-type side more negative. The region devoid of
electrons and holes is called the depletion zone.
n-type
n-type
p-type
p-type
free electrons
free electrons
free holes
free holes
Depletion zone
Formation of the depletion zone
A more accurate picture of the junction is obtained by considering the Fermi energy. The energy
must be the same on both sides of the junction, as shown below.
Chapter 4 Entering the digital world
60
p-type
n-type
eV0
EF
Fermi level in a pn-junction
As a result the valence and conduction bands are “bent.” This picture helps you see how
electrons and holes can move. You can think of the electrons as marbles rolling on the bottom of
the conduction band. They would need additional energy to go “up hill” from the n-side to the pside. Holes can be considered to be air bubbles in a liquid-filled valence band. They rise to the
highest level. Again, energy would be needed to pull them down so they could enter the n-side
of the junction.
The semiconductor diode
The two pieces of doped silicon joined at the junction are called a diode. How does a diode act in
a circuit?
p-type
n-
p-type
n-
eV0 +
eV0 –
EF
EF
–
+
+
VS
–
VS
Forward- and reverse-biasing a diode
Suppose that you connect the diode to a voltage source, Vs, with a series resistor to limit the
current. If you connect the negative terminal of the voltage source to the n-side of the diode,
electrons will flow from the source into the diode, pushing the free electrons toward the junction.
If they have enough energy they will climb the potential hill and reach the p-side. That is, there
will be a current through the junction. The electrons leave the p-side with less energy than they
entered in the amount eV0, that is, their potential is lowered by an amount V0. Holes will flow in
the opposite direction and will lose the same amount of energy. The depletion zone is narrowed.
The diode is said to be forward biased.
If the positive terminal of the source is connected to the n-side of the diode, free electrons will be
attracted to the voltage source and away from the junction. This action increases the width of the
depletion zone. Similarly, holes will be attracted toward the negative terminal of the source and
away from the junction. There will be no current through the diode. The diode is said to be
Chapter 4 Entering the digital world
61
reverse biased.
How does the current depend on the potential difference placed across the diode? It can be
shown that the current is well described by the equation
I (V ) = I R (eαeV / kT − 1) .
In this equation e is the electronic charge, k is Boltzmann’s constant, T is the absolute
temperature, α is a constant that depends on how the diode is made, and IR is the (small) current
when the diode is strongly reverse-biased. Note that when the voltage is more than a few tenths
of a volt negative (reverse bias) the current is –IR. When the voltage is positive the current rises
rapidly.
A good model for the current through a silicon diode is shown on the graph below. The current is
very small until the diode is forward-biased by 0.7 V: Then the current increases very rapidly.
Diodes made of different materials begin to conduct at different voltages and have differing
current-voltage slopes. But in essence, all are similar in the sense that they act as open circuits
when reverse-biased and very low resistance closed circuits when forward-biased. Further, they
all have a voltage drop across them when forward biased. This is called the forward voltage
drop. For silicon diodes it is about 0.6 V. It is smaller for Ge diodes, much larger for LEDs.
Diode characteristic IV-curve
Light-emitting diodes and light-sensitive diodes
Light-emitting diodes, or LEDs, and light-sensitive diodes, or photodiodes, are complementary
examples of the same physical process, the transfer of energy between an electron in the diode
and a light photon. In a forward-biased LED a conduction electron fills a hole. It loses energy
and a photon of the same energy is emitted. The voltage source and resistor supply electrons to
the n-side of the diode and add holes to the p-side. The materials from which the diode is made
are carefully selected so that the bandgap has the correct energy for the wavelength of photon to
be emitted. LEDs are available that emit photons from blue through near infrared energies. The
forward voltage drop depends on the material. The power of the light emitted is roughly
proportional to the current. The diode is packaged in a transparent case, often with a lens built in
to create either a narrow beam of light or a diffused source.
Chapter 4 Entering the digital world
62
p-type
n-type
eV0 – eVS
EF
–
+
VS
Light-emitting diode
LEDs are constructed from semiconductors consisting of two, three, or four elements. Small
variations in the proportions cause a change in color produced. For example, GaAs0.60P0.40 is a
red LED, GaAs0.35P0.65 is orange, GaAs0.15P0.85 is yellow, and GaAs0.00P1.00 is green.
In general GaAsP (gallium arsenide, phosphide) is used for red through green. AlGaInP
(aluminum gallium indium phosphide) produces the same colors, but with higher brightness.
InGaN (indium gallium nitride) is used for green through blue LEDs. AlGaAs (aluminum
gallium arsenide) produces red through infrared light. Violet and UV LEDs are also available.
In a photodiode the opposite process occurs. A photon is absorbed in the p-layer, freeing a bound
electron, which simultaneously creates a hole and a conduction electron. A photodiode can be
run at zero bias where it generates a potential difference and, as a result, a small current through
a resistor. More frequently the diode is reverse biased, and current through the resistor is
monitored. The current is proportional to the rate at which light energy is absorbed by the
junction.
The actual construction, shown below, looks somewhat different from the energy diagram shown
above. Again, selection of the materials from which the diode is made determines the minimum
photon energy that can be absorbed. For example, Si diodes are sensitive from 190 to 1100 nm
(peak sensitivity 780 nm), Ge diodes in the infrared from 800 to 1800 nm, and InGaAs diodes
even deeper infrared, 800-2600 nm. Actually all diodes, including LEDs are photosensitive, but
they are not manufactured to maximize conversion of light to current.
Transparent
cover
Depletion zone
Positive
connection
Negative
connection
Light
n-layer
p-layer
Photodiode construction
Chapter 4 Entering the digital world
63
In summary, a diode is a two terminal device. As you have seen it is a one-way conductor. That
is, it acts like a short circuit when one end is positive and like an open circuit when the other end
is positive. For that reason it is often used to “rectify” ac, converting it to dc (at least to voltages
that are either positive or negative, but not both). Diodes can emit light; other diodes can detect
light. But one thing that a diode cannot do is to increase the current through it (or voltage across
it). For that a third terminal is needed.
The transistor
In 1945, William Shockley, John Bardeen, and Walter Brattain, began working at Bell
Telephone Laboratories to develop a device constructed from solids (crystals). In December,
1947, Bardeen and Brattain constructed the first three-terminal point-contact amplifier.
Shockley, the team leader, had not been directly involved. He was both elated and furious, and in
a four-week period developed the theory for a version that is the one used today. Creating a
practical device took two more years!
J.R. Pierce at Bell Labs decided to call the device the transistor. It is both a shortened version of
trans-resistor and is similar to the word thermistor that was already in use.
A junction transistor, the only kind in use today can be considered to be two diodes placed backto-back. You could put either the n or the p sides together. In one case you would have a pnp
transistor, in the other a npn. We’ll discuss npn transistors.
As shown below, the junction between what is called the base and the emitter is forward biased,
while the base-collector junction is reverse biased.
n
+
Collector
+
p
Base
n
Emitter
Transistor biasing
The second diagram shows the energy levels for the biased transistor. The forward bias of the
emitter-base junction pushes electrons into the emitter. They are given enough energy to
overcome the barrier and go into the base. Therefore there is a small emitter-base current.
Chapter 4 Entering the digital world
64
n-type
p-type
n-
eV0 – eVBE
eV0 + eVBC
emitter
base
collector
Energy levels in transistor
But, the base is made very thin. A majority of the electrons don’t leave at the base but continue
into the collector, which they can because of the bias that is applied. Therefore the emittercollector current is much larger than the emitter-base current. We define two ratios: α = IC/IE
and β = IC/IB. = α/(1 – α). The value of α is very close to 1 (like 0.99) and β is large, typically
around 100. That is, a very tiny base current produces a large collector current.
How does a transistor amplify? Small changes in the forward bias voltage of the emitter-base
junction cause large changes in the emitter-base current because the IV curve of the diode is
almost vertical. Then, these changes in the emitter-base current cause changes in the emittercollector current that are larger by a factor of β. In other words, the base acts like a valve to
control the current from the emitter to the collector.
Transistor amplifiers
How then can the transistor be connected into a circuit to be an amplifier? We’ll consider only
one configuration, the so-called common-emitter amplifier. That is shown below.
RC
IC
+
RB
VCE
IB
VBB
_
+
VCC
VBE
Common-emitter amplifier circuit
Suppose you vary the base voltage (VBB). The base-emitter voltage is virtually constant, so
almost all the change in the base voltage creates a change in the base current as given by the
equation
IB = (VBB – VBE)/RB.
The transistor then changes the collector current. How does that change a voltage? Consider the
collector circuit in the diagram. Kirchhoff’s laws give
Chapter 4 Entering the digital world
65
VCC – ICRC – VCE = 0, or
VCE = VCC – ICRC.
Therefore as IC changes, so does the voltage across RC and VCE.
Let’s quantify this analysis. The graph below has a family of curves that show how the collector
current and collector-emitter voltage depend on base current. Note that at a fixed base current as
VCE rises, the collector current first rises rapidly, then much more slowly. Now consider holding
VCE fixed and increasing the base current. The collector current rises. Note that if you increase IB
by 10 µA, IC increases by roughly 1 mA. That is, this transistor has a current amplification factor
of about 100.
80
70
9
60
50
IC (mA)
5
IB (µA)
40
30
20
10
5
VCE (V)
9
Transistor IV characteristic curves
Now consider the collector circuit that obeys the equation VCE = VCC – ICRC.
When IC = 0, VCE = VCC. In the case shown above, VCC = 9V.
When VCE = 0 IC = VCC/RC. Therefore RC = VCC/IC, or 1kΩ. The diagonal line represents how VCE
and IC are related. The value of IB determines the value of IC, and VCC and RC determine the value
of VCE.
So, we could change an input voltage in such a way that the base current varies between 0 and
80 µA and the result would be a variation in the voltage across the transistor from about 1 to 9
volts. Note that this is an inverting amp: as VB and IB increase, VCE decreases.
The problem is that most signals are bipolar—they oscillate around V = 0. We solve this problem
to creating a fixed base bias current and have the signal vary the base current around the fixed
value. The fixed value is called the Q or quiescent point. It is best to choose the Q point in the
middle of the diagonal line. In this case IBQ = 40 µA would be a good choice. It would result in
VCEQ = 4.5 V and ICQ = 4.5 mA.
A practical common-emitter design doesn’t use two separate batteries. Rather, it uses a voltage
divider to bias the base. Further, you want a design that doesn’t depend on the properties of the
transistor itself (just as you want an op amp design that doesn’t depend on the gain of the op amp
itself). You want the gain to depend on resistors, not on the transistor. A practical circuit is
shown below.
Chapter 4 Entering the digital world
66
VCC
RC
RB2
C1
C2
Vout
Vin
RB1
RE
Practical common-emitter amplifier
How does it work? Because the base-emitter voltage is about 0.6 V, VE = VB – 0.6V. The
changes in the base and emitter voltages will be the same: vE = vB.
The variation in the emitter current is given by iE = vE/RE, but vE = vB, so iE = vB/RE.
Because the value of β of the transistor is very high, iB << iC, so Kirchhoff’s node law for
currents, iE = iB + iC becomes iE = iC and so iC = vB/RE.
The changes in VCE are equal (and opposite in sign) to the changes in the voltage across the
collector resistor. That is, the changes in the output voltage are given by –vC. But vC = iCRC. But,
as was shown above, iC = vB/RE. Therefore vC = vB RC/RE. Therefore the voltage gain of the
amplifier,
AV = –vC/vE
AV = RC/RE.
Here is a technique for choosing the value of the components:
1) Choose the gain. Make it no larger than 10. As you know, the gain is equal to RC/RE.
2) Choose the quiescent point. A good value for ICQ is a few milliamps. A good value for VCEQ is
roughly half the power supply voltage VCC.
3) The Q-point then determines the sum of the two resistances: RC + RE = (VCC – VCEQ)/ICQ. The
gain gives you the ratio of the resistors, so you can determine the two values. Adjust everything
so you can use available resistors.
4) Select the base resistors. First, from calculations above, find VE. Add 0.6 V to it to get VB.
Then use the voltage divider equation to find RB2/RB1. That is, since VB = VCC/(1+RB2/RB1),
RB2/RB1 = (VCC/VB – 1). The parallel combination of the two should be at least ten times RE,
which means, typically, that each resistor is in the 10-50k range.
5) Choose the capacitors. Each is part of a high-pass filter. The –3-dB point should be below the
lowest frequency you want to amplify. The high-pass filter resistor for C1 is the parallel
combination of βRE, RB1, and RB2. The resistor for C2 is the input resistance of the next device.
Typically C1 and C2 are 1-10 µF.
Chapter 4 Entering the digital world
67
Practical example:
1) Make the gain about 5.
2) Choose ICQ = 2.0 mA and VCC = 9.0 V. Then VCEQ = 4.5 V.
3) RC + RE = 4.5V/2.0 mA = 2.25k. If RE = 390Ω, then RC = 1.8kΩ would be acceptable.
4) VE = (2.0mA)(390Ω) = 0.8 V, so VB = 0.8 + 0.6 = 1.4V and RB2/RB1 = (9/1.4 – 1) = 5.4.
12k and 68k would be reasonable choices.
5) The parallel combination of three resistances is 8k. If C1 = 2µF the –3dB frequency is 10 Hz.
With RC = 1.8k, one would need C2 = 10 µF to achieve the same –3dB frequency.
Input and output impedances
How are the input voltage and current related? For the transistor alone, Rin = vB/iB. Because the
base-emitter voltage is essentially constant, the variations in the base and emitter voltages are the
same, so Rin = vE/iB. But vE = iERE, so Rin = RE(iE/iB). As long as β is large, Rin = βRE.
In the common-emitter amplifier the input impedance is the parallel combination of Rin = βRE
and the two biasing resistors, R1 and R2. The reason for making those two resistors large is to
roughly balance the input impedance among its three components, βRE, R1, and R2.
The output impedance is just the collector resistance, RC. Because this resistance determines the
gain and is usually fairly large, the common emitter amplifier has a high output impedance.
Power amplifiers
The high output resistance of the common-emitter amplifier means if you draw significant
current from it the output voltage falls. One way to provide the current needed for a high current
load is to use an emitter follower amplifier. This is like the common emitter, but with RC = 0. It
has a voltage gain of 1. We will use it to drive an LED, which requires a current larger than an
ordinary op amp can supply.
+15V
Vin
VB
VE
RE
Vd
LED-driver
Here are the steps to be followed in designing an emitter follower amplifier used with an LED:
Chapter 4 Entering the digital world
68
Step
1) Choose the value of RE.
a) The op amp will limit the maximum value of Vin.
This voltage will be VB (max)
b) Find the maximum value of VE = VB – 0.7 V
c) Decide on the maximum current through the diode.
d) From measurements of the IV curves of the diode,
find the forward voltage drop across the diode, Vd, at
this current
d) Select RE = (VE (max) – Vd)/I
2) Recheck the voltage at the base of the transistor, VB.
a) Calculate the voltage at the emitter,
VE = Vd + I RE .
b) Find the voltage at the base, VB = VE + 0.7 V
c) If this is too large, reduce the size of RE
3) Find the input voltage needed to turn off the diode.
a) Select a very small current, one that results in a very
dim output.
b) Find the voltage across the diode at that current.
c) Find the emitter voltage VE = Vd + I RE
d) Calculate the base voltage VB = VE + VBE
5) Summarize
Find the desired range of input voltages and the center
value, the “quiescent” voltage, around which the input
should swing.
Example
VB (max) = 13 V
VE (max) = 12.3 V
I = 40 mA
Vd = 2.5 V
RE = (12.3 – 2.5)/0.04 = 245Ω. Use
an available value like 220Ω.
VE = 2.5 + (0.04 A)(220Ω) = 11.3 V.
VB = 11.3 + 0.7 = 12.0 V
No change needed.
Imin = 2 mA.
Estimate Vd = 2 V
VE = 2 V + (2 mA)(220Ω) = 2.4 V
VB =2.4 + 0.7 = 3.1 V
I = 2 to 40 mA requires Vin between
3.1 and 12.0 V. Center value is
7.55 V.
The problem with this circuit is that the collector current when the varying signal is zero is quite
large, especially if very large currents are needed. For example, for a 10-W amplifier driving an
8-Ω speaker, this current would be tens of amps. Even with no signal the transistors and resistors
would be dissipating large amounts of power and getting very hot.
The solution to use a “push-pull” amplifier using matched npn and pnp transistors. When the
input is positive the npn transistor conducts and when the input is negative the pnp transistor
does the duty. When the input is zero neither transistor conducts. There is less wasted power!
Chapter 4 Entering the digital world
69
+15V
–15V
Push-pull speaker amplifier
The problem with this circuit is the voltage drops between the bases and emitters. As a result, the
output voltage is always about 0.6V smaller than the input voltage. When the input drops to
+0.6V the output is zero, and it stays zero until the input is below –0.6V. Small input signals,
those with peak voltages less than 0.6 V, produce no output signal at all. We’ll investigate two
solutions to this problem, one in a problem, one later.
3. Applications
You will design and test a light-wave communication system that can communicate across the
laboratory. The transmitter will use a microphone that is amplified by an op amp and uses a
transistor to drive an LED. The receiver will use a phototransistor that is amplified using an op
amp and a complementary-symmetry circuit to drive a loudspeaker.
Hints for transmitter design: You’ll need to make the dc voltage output of the microphone zero
with no sound. This will require a high-pass filter. The “R” is the input resistance of the
amplifier. Choose the capacitor so that the circuit is sensitive to frequencies as low as 100 Hz.
The op amp gain should be large enough that the microphone will modulate the LED output from
no light to full intensity. You’ll need to use a summing amplifier so that you can add a variable
dc offset voltage to adjust the LED output to just-barely-visible when you are not speaking into
the microphone. What wavelength LED should you use? Match it to the sensitivity of the
phototransistor in the receiver.
Hints for receiver design: You will have to choose the gain of the circuit to produce a reasonable
level of sound at the speaker. The gain will depend on the amount of light detected by the
phototransistor.
Test the transmitter/receiver pair and develop methods of reducing interference from the room
lights.
4. Questions and problems
1. Design a common-emitter amplifier with a gain of 10 that has VCC = 24 V. Select values of
C1` and C2 so that the –3 dB frequency of the amplifier is 10 Hz.
2. Calculate the input and output resistances of the amplifier of Problem 1.
3. Design an emitter follower that will drive an LED with currents up to 30 mA. The op amp
Chapter 4 Entering the digital world
70
can produce output voltages between –12.5 V and +12.5 V. The collector of the transistor is
connected to +15 V, the cathode of the diode to –15 V. What input voltages would be needed
to change the current through the diode from 2 mA to 30 mA?
4. Consider the circuit below. The diodes are both Si diodes that have a forward voltage drop of
0.6 V. The two resistors are each 10k. If Vin = 0 V, what is V1? V2? Now suppose that Vin is
connected to a signal generator that produces an ac voltage with a peak-to-peak amplitude of
0.5 V. What would V1 and V2 be?
+15 V
R1
V1
Vin
V2
R2
−15 V
Circuit for problem 4
5. The circuit in Problem 4 is used with a push-pull amplifier, with V1 and V2 connected to the
bases of the two transistors, and Vin is the input to the circuit as shown below. Explain how
this addition solves the problem of no output signal for small input signals.
+15 V
R1
Vin
Vout
R2
−15 V
Circuit for problem 5
Chapter 4 Entering the digital world
71
Chapter 4 Entering the digital world
72
4. Entering the digital world
A. From analog to digital
Analog signals vary continuously; digital are either off or on. How can you convert an analog to
a digital signal? For example, you might want an electronic thermometer to turn a heater on
when the temperature falls below a set point, a force-sensitive resistor held against an artery to
create a signal when the vessel pulses, a light detector to register the blockage of a light beam.
Any of these events might be used to produce an “on” or an “off” signal or to create a pulse that
turns “on” and then back “off” after a specified time.
1. Laboratory explorations
The comparator
At the heart of a device to convert analog to digital is a comparator; a circuit that compares a
signal to a fixed voltage value. A 741 op amp without the feedback resistor can be used as a
comparator because the output voltage swings from one extreme to the other when the difference
in input voltage changes from a few millivolts positive to negative. Rather than using the 741, an
op amp specially designed to be used as a comparator, the 311, will be employed.
Complete the circuit below to test the 311. The circuit has a reference voltage that is produced by
the voltage divider consisting of R1and R2. This voltage is compared with the voltage, from the
10-k potentiometer. Note that the pin assignments on the 311 are different from those on the 741
and a single +5 V supply is used instead of the +/– 15 V supplies.
+5 V
10-k
pot
3
1k
8
Vin
R2 = 10k
2
Vref
7
311
1
Vout
4
R1 = 10k
311 comparator test circuit
Connect the output to one of the logic indicator lamps on the ProtoBoard. Check to see that the
indicator goes on or off when the voltage from the potentiometer, Vin, equals Vref.
A major difficulty in using a comparator such as this is that if the input voltage (now controlled
by the potentiometer) oscillates near the value of the reference voltage (most likely due to noise),
the output will also rapidly oscillate. The solution is to provide some “hysteresis,” or memory
(cf. “history”). The idea is if the comparator is off and the input voltage drops below a threshold
and turns on the comparator, the voltage has to go up to a higher voltage to turn the output off
again. That is, the hystersis circuit lowers the “turn on” voltage and raises the “turn off” voltage.
Chapter 4 Entering the digital world
73
Adding a resistor that provides positive feedback produces hysteresis as will be explained later.
The result is the Schmitt trigger circuit shown below.
+5 V
10-k
pot
R2 = 10k
1k
8
3
7
311
2
Vout
1
4
R3 = 27k
R1 = 10k
Schmitt trigger test circuit
Determine the hysteresis voltage of this circuit. How could you change R3 to increase the
hysteresis? Test your prediction by making R3 smaller and measuring the new hysteresis.
Now replace the 10-k pot with an input signal from a sensor. You may use the LM335 to make a
thermostat, a phototransistor (with an op amp I-V converter) to detect light/dark changes, or an
integrated circuit magnetic field sensor to see if a magnet is close to the sensor. Replace the
voltage divider formed by R1 and R2 with the 10-k pot so that you can adjust the reference
voltage. Report your investigation and results.
From level change to a pulse: the monostable multivibrator
The comparator and Schmitt trigger create a digital, or on/off signal from a varying analog
signal. Frequently digital circuits demand a pulse, a signal that turns on and then back off a
specified time later. You will first explore a circuit that creates a single pulse of fixed time width
when the input is lowered, and then connect this circuit to the Schmitt trigger.
The technical term for the pulse generator is a monostable multivibrator. Monostable refers the
fact that the output has one stable state (off) and the on state is only temporary. A convenient
device to use in this application is the 555 multivibrator.
+5V
1k
10k
R
8
4
7
555
6
3
Vout
2
0.01µF
1
C
5
0.01µF
555 monostable multivibrator
Chapter 4 Entering the digital world
74
The pulse width is given by T = 1.1 RC. Choose a combination to create a pulse about 10 ms
long. The 555 requires that the voltage on pin 2 drop to 1/3 of the voltage on pin 4 in order to
start the pulse, but it must return to 2/3 of 5 V before the pulse ends. Explore how the RC circuit
attached to the pushbutton switch accomplishes this deed.
Once the pulse generator is working remove the pushbutton and 1-k resistor and connect the lefthand end of the capacitor to the output of the 311 Schmitt trigger. Report on the results.
Note that the pulse is generated when the Schmitt trigger turns off. How can the entire circuit be
modified so that the pulse is generated when the trigger turns on?
2. Textbook and classroom explanations
A comparator should compare an input voltage, Vin, with a reference voltage, VR, and produce
one output voltage (1 V) when the input is greater than the reference and a second voltage (0 V)
when the input is below the reference, as shown in the figure below.
Noise-free comparator operation
The high gain of an ordinary op amp like the 741 can produce output voltages of approximately
+12.5 and –12.5, depending on whether it is true or false that V+ is larger than V–. Yet, a 741 is
hardly ever used as a comparator. Why?
First, the speed with which a 741 can drive its output voltage from one extreme to the other is
limited by the “slew rate.” This term comes from the language of servomechanisms, where the
slew rate is the maximum speed at which a device can move when the feedback system is not
attempting to move the servo to a particular position. For the 741 the slew rate is about 0.5 V/µs.
Thus it requires about 50 µs for the output to go from one extreme to the other. The slew rate is
slow because the internal circuitry limits high frequency gain in order to avoid oscillations.
Second, most digital systems work with a different pair of voltages that signify “true” or “false.”
The digital systems that we will explore use +5V for “true” and 0V for “false.”
The LM311 is an op amp specifically designed for use as a comparator. Its output can be driven
from “off” to “on” in about 0.2 µs. Further, the output is extremely flexible. One can arrange the
device to produce almost any pair of high and low voltages. We will ground pin 1 and connect
pin 7 to +5V through a 1-k resistor. Then when the comparator is “off” there will be a virtual
short circuit between pins 7 and 1, so the output voltage will be 0V. When the comparator is
“on” there will be an open circuit between pins 7 and 1 so the output voltage will be +5V. If pin
Chapter 4 Entering the digital world
75
1 were to be connected to –15 V, and the 1-k resistor to +15V, then the output would change
between –15V and +15V.
What happens if the signal has noise on it? The figure below shows a fast, low amplitude sine
wave “noise” superimposed on the slower “signal.”
Comparator with signal and noise
Notice that the output is turned on and off by the noise. The solution to this problem is to add
hysteresis to the comparator. “Hysteresis” comes from the word “history.” It means that the
reference voltage depends on whether the output is on or off. If it is off, then the reference level
is raised a certain amount. That is, the signal must be somewhat larger to turn it on. Once it is on,
then the level the signal must drop to a lower level to turn it off. The hysteresis band—the
difference between the off and on levels, is adjust to be slightly larger than the amplitude of the
noise.
Comparator with hysteresis
Hysteresis is added to a household thermostat so that the furnace isn’t rapidly turned on and off
when the temperature is at the set point. For example, the temperature might have to fall to 68o
before the furnace turns on and above 72o before it turns off.
Hysteresis is added to a comparator by using positive feedback. Resistor R3 modifies the voltage
divider consisting of resistors R1 and R2. Because the end of R3 connected to the comparator
output is either at 0 V or 5 V, the circuit can be modeled as shown in the diagrams below
Chapter 4 Entering the digital world
76
+5 V
+5 V
R2
R2
R3
Vref
R1
R3
Vref
R1
Schmitt trigger analysis
The left-hand situation applies when the output, pin 7, is at 0V. Then analysis of the voltage
divider shows that Vref = (5V)(R1||R3)/(R1||R3 + R2), where R1||R3 means the parallel combination
of R1 and R3. When the output is at +5V, then the output voltage is given by
Vref = (5V)(R1)/(R2||R3 + R1). Straightforward algebra shows that the difference in the voltages is
R1R2
equal to ∆Vref = (5V )
. This can also be written as
R1R2 + R1R3 + R2 R3
∆Vref = (5V)RTH/(RTH + R3), where RTH is the Thevénin equivalent resistance of the original
voltage divider, the parallel combination of R1 and R2.
The multivibrator
The 555 (pronounced “five-fifty-five”) is one of the most versatile integrated circuits ever made.
An equivalent circuit of the 555 is shown below to the left.
V+
+5V
8
R
R
Threshold
Control
6
8
5
R
4
7
Trigger
2
R
6
555
3
Vout
R
Flip Flop
Discharge
2
S
Reset
Q
V+
7
1
Reset
5
4
Output
C
0.01µF
3
V- 1
555 equivalent circuit
555 one-shot multivibrator
The 555 contains a three-resistor voltage divider, establishing reference voltages of 2/3 V+ and
1/3 V+. The left-hand comparator compares the threshold input to 2/3 V+. The right-hand
comparator compares the trigger input to 1/3 V+.
Chapter 4 Entering the digital world
77
These two comparators go to a Flip Flop. The Trigger comparator turns the Flip Flop on, the
threshold turns it off. When the circuit is first powered up the capacitor C is discharged. The
unmarked resistor connected to pin 2 keeps the Trigger input high.
Trigger
voltage
Capacitor
voltage
Output
voltage
t = 1.1 RC
One-shot multivibrator operation
When the switch is briefly closed, pin 2, the Trigger input, is pulled to 0V. The Flip Flop is
turned on. As a result the Q output is set at 0V. This turns off the transistor connected to it,
allowing the voltage on the Discharge (pin 7) to rise. Meanwhile, the Output (pin 3) is raised to 5
V.
The current through R charges C. When the voltage across C reaches 2/3 V+ (3.3 V) the
threshold is reached. The Flip Flop is turned off. This raises Q to 5V. The transistor turns on,
making a low-resistance connection between Discharge (pin 7) and ground, shorting the
capacitor and discharging it. At the same time the Output (pin 3) is turned off. The time required
for the capacitor to reach the threshold voltage t = 1.1 RC.
It is important that the voltage at the Trigger input return to 5V before the output goes back to
0V, so in the circuit used in the lab the 10-k resistor recharges the capacitor connected to the
switch, raising the Trigger input back to 5 V.
The 555 has many other uses. The principal use is as an “astable” multivibrator. In this mode its
output oscillates, producing an asymmetric square wave (on and off times are not equal). There
are entire books written of applications for the 555.
3. Questions and problems
1. Design comparator circuit (without hysteresis) that could be connected to an LM335 sensor
and would turn off when the temperature rises to 23oC .
2. Derive the equation for ∆Vref given above.
3. Add hysteresis to the circuit of problem 1 that requires the temperature to be above or below
the set point by 0.5K for the comparator output to change.
4. Explain how the 1-k and 10-k resistors and the 0.01-µF capacitor work. Determine how long
pin 2 has the required lower voltage (1/3 V+ or 1.67 V).
5. You need to obtain a 10 µs-wide pulse using a 555 monostable multivibrator. If C = 1000 pF,
what is R?
Chapter 4 Entering the digital world
78
B. Digital logic
How do you make decisions based on two or more digital signals? The familiar logical
operations of NOT, AND, and OR can be implemented with electronic circuitry
1. Laboratory explorations
Digital logic
Digital circuitry requires definition of the two states 1/0, TRUE/FALSE, HIGH/LOW or
ON/OFF. We will use the convention that a high voltage (the power supply voltage) represents
ON, TRUE, or 1, a low voltage (ground) represents OFF, FALSE, or 0. A number of different
definitions of high and low voltages exist that depend on the electronic circuits used. Two
families of devices are in common use today: CMOS (complementary metal-oxide
semiconductor) and TTL (transistor-transistor logic). We will concentrate on the latter.
TTL devices are powered by 5-V power supplies. As a result, they can accept signal voltages
between 0 and 5.0 V. As shown in the explanations that follow, when a voltage is applied to a
device, it will interpret a voltage between 0 and 0.7 V as “low” and between 2.0 and 5.0 V as
“high.” When a device produces a signal, a “low” signal is between 0 and 0.4 V and “high” is
between 2.4 and 5.0 V. But it isn’t quite a simple as that, as we shall see.
Your ProtoBoards support digital circuitry in several ways
1. Sources of digital signals
Eight tiny “DIP” switches that put out either 0 or 5 V
Two “SPDT” (Single Pole Double Throw) slide switches that connect a
center contact to either of two end contacts
Two “debounced” pushbuttons (to be explained later)
A TTL output of the function generator.
2. Indicators of digital signals
Eight “logic indicators,” LEDs that glow when their input voltage is greater than 1.4 V.
3. Power supply
5.0 V at up to 1.0 A.
Exploring the ProtoBoard
Connect a wire from one of the DIP switches to an indicator and determine how to set the switch
to produce a 0 (light off) and how to set it to produce a 1 (light on). Then connect the output of
one of the push buttons to an indicator to see how it works. Finally connect the TTL output of the
function generator to an indicator and set the frequency to about 1 Hz. Verify that the generator
can produce digital signals.
Exploring digital gates
A gate is an electronic circuit that implements one of the logical operations. The gate circuits are
an integrated circuit that sold as a DIP (dual-inline) package similar to the 741 op amp. To save
space, each package contains several gates. For example, devices that implement the OR
Chapter 4 Entering the digital world
79
operation usually have four gates and thus are called “quad or gates.” Each package must be
connected to 5.0 V and ground in order to operate.
The Inverter or NOT
The 7404 is a hex inverter. The connections to the pins are shown at the right. Connect pin 14
(Vcc) to 5.0 V and pin 7 to ground. Verify that one of the six 14 13
12
11
10
9
8
gates inverts the signal: a high input produces a low output
Vcc
and visa versa.
Summarize your results in a truth table like the one shown
below:
Input
1
0
Output
The NAND gate
“NAND” means “NOT AND.” That is, the logical
operation “AND” is applied to two (or more) inputs and the
result is inverted. The reason for this seemingly bizarre
operation will be made clear later. The 7400 is a two-input
quad NAND gate. Connect this device to 5 V and ground
and test one of the gates, summarizing your results in a
truth table like the one shown below:
Inputs
0
0
1
0
0
1
1
1
GND
1
2
3
4
5
6
7
14
13
12
11
10
9
8
Vcc
GND
1
2
3
4
5
6
7
Output
The NOR gate
“NOR” means “NOT OR.” As above, the logical operation “OR” is applied to two (or more)
inputs and the result is inverted. The 7402 is a two-input
14
13
12
11
10
9
quad NOR gate. Connect this device to 5 V and ground and
Vcc
test one of the gates, summarizing your results in a truth
table like the one shown below:
8
Inputs
0
0
1
0
0
1
1
1
7
Output
GND
1
2
3
4
5
6
2. Textbook and classroom explanations
As you have seen, digital electronics is based on the notion that a signal can be in one of two
states. There are many ways of naming the states, 1/0, TRUE/FALSE, HIGH/LOW or ON/OFF.
For compactness we will use 1/0 to designate the state of a signal.
Chapter 4 Entering the digital world
80
Boolean algebra
George Boole (1815-1864) was an English mathematician who, in 1847 published “The
Mathematical Analysis of Logic” in which he argued that the discipline of logic belonged more
in mathematics than in philosophy. Over the next seven years he refined his “Analysis,”
developing a language system for encoding logical arguments. His system, today usually called
Boolean Algebra, involves two objects, yes/no or true/false, and three operations, AND, OR, and
NOT.
Boole’s system was introduced to the United States in 1866 by the logician/philosopher C.S.
Peirce, but not widely used until Claude Shannon (1916-2001) in his 1937 Ph.D. thesis at MIT
made the connection between Boolean algebra and electric circuits. It was first applied to
telephone circuits. In 1948 he published “A Mathematical Theory of Communication,” the work
that is the foundation of the discipline of information theory. The development of digital
computers is based on Shannon’s work.
A variable in Boolean algebra is represented by a letter that can have one of two values, 0 and 1.
NOT
Logically, one uses the operation NOT like this: “If a statement is true, then the opposite of the
statement is false.” If we use the letter A to represent the statement and A to represent its
opposite, then one would say “If A is true, then A is false.” Or, “If A is 1, then A is 0.”
The Boolean operation NOT, or negation, is represented by a bar over a letter A . That is, if
A = 1, A = 0.
The most useful way of displaying the results of Boolean algebra is to use a truth table. A truth
table has two or more columns. The left-hand column(s) display the inputs, the right-hand
column the result. Here is the truth table for NOT
A
1
0
A
0
1
An electronic device that implements the NOT application is the inverter. It has one input and an
output. If the input is 0, the output is 1; if the input is 1, the output is 0. The symbol is shown
below:
Gates
Digital circuitry is built from gates, which can be thought of as a switch that can be opened (to
reject a signal) or closed (to transmit a signal). Thus a gate also has two states: 1/0, closed/open,
on/off. Again, for compactness, we will use 1/0 to designate the state of a gate. Gates can be
used to implement the Boolean operations AND or OR.
AND
Logically, one usually combines “and” with “both” in the following way: “If BOTH A AND B
are true, then the result is true.” Thus the result is true (1) in only one case, when A is true (1)
and B is true (1). If either is false, the result is false.
Chapter 4 Entering the digital world
81
In Boolean algebra the operation AND is represented by multiplication, the “•” sign. Thus one
would write:
A•B = Q
The truth table for AND is the following:
A
0
1
0
1
B
0
0
1
1
A••B
0
0
0
1
One can purchase AND gates in a number of configurations. The most common gate has two
inputs and one output. Four are included on a single integrated circuit, the so-called quad AND
gate. You can also purchase gates with three or four inputs. The symbol for a two-input gate is
shown below:
A
A•B
B
OR
Logically, one usually combines “or” with “either” in the following way: “If EITHER A OR B
are true, then the result is true.” Thus the result is false (0) in only one case, when A is false (0)
and B is false (0). If either (or both) is true, the result is true.
In Boolean algebra the operation OR is represented by addition, the “+” sign. Thus one would
write:
A+B = Q
The truth table for OR is the following:
A
0
1
0
1
B
0
0
1
1
A+B
0
1
1
1
One can also purchase OR gates in a number of configurations. The most common gate has two
inputs and one output, usually grouped so that four are on a single integrated circuit, resulting in
a quad OR gate. You can also purchase gates with three or four inputs. The symbol for a twoinput OR gate is:
A
A+B
B
NAND and NOR
While OR and AND are the fundamental Boolean operations, for electronic purposes the most
useful operations are combinations of these two called NAND (for NOT AND) and NOR (for
NOT OR). That is, a NAND is an AND gate followed by an inverter, and similarly for the NOR
gate. Their symbols are shown below. Note that a circle at the output denotes inversion
Chapter 4 Entering the digital world
82
A
A•B
B
A
A+B
B
How are they more useful than AND and OR gates? The first way they can be used is to create
an inverter. If you connect the two inputs of a NAND gate together, then B = A. From the truth
table you can easily see that A•A = A or A • A = A . Thus the output is the opposite of the input.
You should verify that the same can be done with a NOR gate.
Further uses are based on a set of two equations called de Morgan’s theorems. In symbolic form
the two theorems are
A+ B = A• B
A• B = A+ B
The relationships would be said “NOT A OR NOT B equals A NAND B,” and “NOT A AND
NOT B equals A NOR B.”
These theorems can demonstrate how to construct AND, OR, and NOR gates from several
NAND gates. For example, to create NOR, first use a NAND gate to invert A to form A and use
a second NAND gate to negate B. Then put the two inputs, A and B into a NAND gate. The
result is A • B = A + B = A + B . A final inverter changes the OR into a NOR. That is,
A
A
A•B = A+B = A+B
A+B
B
B
In a similar way NOR gates can be used to create all other operations.
A table of theorems of Boolean algebra is shown below:
A+ A = A
A• A = A
A+ A• B = A
A +1 = 1
A •1 = A
A+ A• B = A+ B
A+0 = A
A•0 = 0
A+ A• B = A+ B
A+ A =1
A• A = 0
A+ A• B = A+ B
It might seem that creating OR from four NANDs is an incredible waste of space, wire, and time.
But, if the gates are produced using microelectronic techniques on integrated circuits there are
great advantages of having only one type of gate on a chip and very small costs of creating
additional interconnections between the gates.
Analyzing logic problems
Consider the following problem:
Light an LED if either the left or the right car door is open and the driver is seated.
What are the steps in solving the problem?
Chapter 4 Entering the digital world
83
1) First, you have to decide, for each input and output, the meaning of “1” and “0” (or “true”
and “false”). Either an open or a closed door can be “true”; the driver’s seat vacant or
occupied can be “true.” You can light an LED with the output either high (“true”) or low
(“false”) depending on how you connect it.
2) Note how the choice of true/false determines the logic. “Either the left or right door is open”
is logically equivalent to the negative of “Both the left and right doors are closed.” This is
DeMorgan’s theorem in language form.
3) Write down a truth table for the problem.
Choices: Door closed is “1”, Seat occupied is “1”, light on is “1”
Left
0
1
0
1
0
1
0
1
Right
0
0
1
1
0
0
1
1
Seat
0
0
0
0
1
1
1
1
Light
0
0
0
0
1
1
1
0
Concentrate on the rows in the table where the smallest number of rows have a unique output. In
this case five are 0, three are 1, so concentrate on the three that are 1.
4) Plan to use two-input NOR or NAND gates. While three- and four-input gates are available,
they are rarely used. Because you can create all possible logical operations from NAND or
NOR alone they are much more versatile than AND and OR gates. Write down the truth
tables for the two operations for reference:
A
0
1
0
1
B
0
0
1
1
A⋅B
1
1
1
0
A+B
1
0
0
0
5) If there are more than two inputs in the problem look for obvious groupings of inputs. In this
problem the two doors should be grouped together first. Examine the truth table. In this case
you can see that the light is 1 if BOTH doors are NOT 1. Comparing that statement with the
truth tables for NAND and NOR, you can see that NAND has the desired result: a 1 when the
light is 1 and a 0 when the light is 0.
Left
0
1
0
1
0
1
0
1
Chapter 4 Entering the digital world
Right L ⋅ R
0
1
0
1
1
1
1
0
0
1
0
1
1
1
1
0
84
Seat
Light
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
0
6) Combine the result of the step above with the remaining input(s). In this case you can see that
if you combine Right NAND Left with Seat using an AND you have the correct logic. Write
down the result: Light = ( L ⋅ R )•S.
7) Convert to electronic gates. Because you’re not supposed to use AND you replace it with a
NAND and either (1) change the definition of the output, making 1 represent light off, or (2)
you follow the NAND with an inverter (made from another NAND). The result is the
following:
Left
Right
Light
Seat
8) Check your results. You might also reconsider your input choices and see if different choices
could lead to a simpler solution (one that involves only two gates instead of three).
TTL Gates
As digital logic devices have been developed a number of product lines, or groups of compatible
devices, have been made available. We are using the TTL (transistor-transistor logic), first
developed in the early 1970s. Devices are identified by four- or five-digit numbers beginning
with 74. For example, the 7408 is a package containing four two-input AND gates. There are
compatible sub-families identified by letters between the “74” and the digits identifying the kind
of device. The most common are L (low power, but slow), LS (low-power Schottky-clamped),
and ALS (Advanced Low-power Schottky). By 1980 regular TTL was almost obsolete, replaced
by LS.
A simplified schematic of a TTL NAND gate is shown below:
5V
4k
1.6k
130
Q3
Q1
A
Q2
Output
B
Q4
1k
TTL NAND gate schematic
The first transistor, Q1, has multiple emitters. There can be a base-emitter current through either
(or both). The diodes on the inputs are designed to protect the circuit from damage if the inputs
are connected to negative voltages. To see how the gate works, consider the two situations
below:
Chapter 4 Entering the digital world
85
5V
5V
4k
1.6k
VB high
Q1
A
Q2
VB low
ON
4k
130
Q1
A
OFF
VB low
1k
130
VB low
Q3
ON
OFF
VB high
Output
B
1.6k
B
Q4 High (3.6 V)
OFF
1k
Q2
ON
VB high
Q3
OFF
Output
Q4 Low (0.2 V)
ON
TTL NAND gate with input low, output high TTL NAND gate with input high, output low
In the left-hand situation one of the two inputs is connected to ground. With a current path from
5V through the 4-k resistor and the base-emitter circuit of Q1 there is base current and Q1 is
turned on. As a result, the collector-emitter voltage is small (about 0.2 V), and the base voltage
for Q2 is below 0.6 V. As a result, Q2 is off, meaning there is no collector current, so the emitter
voltage is at ground. There can be base current for Q3 through the 1.6-k resistor. This turns on
Q3, meaning that there can be a current through the 130-Ω resistor, Q3, and the diode to the
output. Because the emitter voltage of Q2 is at ground, transistor Q4 is off, so there is no current
through it to ground. The output voltage depends on the current drawn from the gate. If the
current is small the voltage is about 3.6 V.
Now look at the right-hand schematic. When the input voltage is high, then there is no base
current through Q1, so the transistor is off. There is enough current through the 4-k resistor and
Q1 (backward from the base to the collector). This turns on Q2. As a result its collector and
emitter are at almost the same voltage, turning off Q3 and turning on Q4. Now current can come
in from the output and through Q4 to ground.
Note that TTL operates in an almost counter-intuitive manner. The output can supply little
current when high, but can sink large currents when low. The inputs require no current to be
high, but require currents to ground if they are to be pulled low. One output can sink enough
current to pull about 10 gates low. That is, the “fan out” capability is 10.
Also note deMorgan’s theorem in action. If either input to the gate is low, then the output is not
low. If, in a perverse way, you chose low to be true, then this gate is would be a NOR gate.
Indicator lamps
An LED is a perfect indicator for the state of a TTL gate. If you want the LED to be on when the
output is high you run into the problem of the limited ability of a TTL gate to provide current.
An LED connected between the output and ground will emit a dim glow when the output is high.
But, the TTL gate can sink a relatively large current when the output is low. Connecting an LED
in series with a 220-Ω resistor between 5V and the output will emit a bright light when the
output is low. The LEDs on the ProtoBoard have built-in current amplifiers so that TTL
limitations are no problem.
Chapter 4 Entering the digital world
86
3. Applications
Explore how at TTL gate responds to an input that is not connected to either ground or 5 volts.
Does it consider the input to be high (1) or low (0)?
The name “gate” is not accidental because either the NAND or the NOR can be used to “gate” a
signal, that is to allow it to pass or to block it, depending on the value of a control signal. To
explore this use, use the TTL output of the function generator as a signal. Design and test a
circuit that will allow the signal to pass (it can pass inverted or not) when the control is 0 and
another when the control is 1.
Implement a circuit that will perform the following task: You are designing a traffic light for a
small North-South street that intersects a major East-West road. The East-West light should be
green under all conditions except when there is a car going either north or south and there are no
cars going either east or west. Design the logic circuit for the traffic light. Notice that you can
define either a lane with a car or without a car as a “1” input and that you can define the EastWest light green or red as the “1” output.
4. Questions and problems
1. You have seen one way that a NAND can perform the NOT operation. How else could be in
connected as an inverter? Should the second input be held at the 0 or 1 level? Find all the
ways a NOR gate could perform the NOT operation.
2. Using the results of problem 1, construct AND and OR gates from NOR gates.
3. Show how to construct a 3-input NOR gate from 2-input NORs.
C. Sequential logic
Combinatorial logic has allowed you to make decisions based on two or more inputs. These
inputs are likely to be static, or at least slowly changing. How can you deal with changing signals
and make decisions based on which signal changed first and how many changes there were? In
other word, how can you deal with pulses, not levels?
1. Laboratory explorations
Mechanical switches have a problem. They are constructed of some springy material. When they
are turned on this springy material is pressed against a metal contact. Because of the flexibility,
the material vibrates, repeatedly closing and re-opening the switch. If that switch were used, for
example, so that the first closing started a stopwatch and the second closing stopped it, the results
would be unpredictable. What is needed is a device that is turned on, or “set” by the first closing
of a switch and remains on until a second switch turns it off, or “resets” it. A device that has two
stable states is called a flipflop.
Construct the following device from a quad NAND gate:
Chapter 4 Entering the digital world
87
S
Q
Q
R
RS Flipflop
As you know, a bar over a symbol means that it is negated. That is, if Q is 0, Q is 1. S performs
a “set” or makes Q true if its value is 0; R performs the “reset,” making Q false when its value
goes to 0. “RS” means “Reset-Set.”
Experimentally determine the truth table for the RS flipflop. For the input states 0 0 and 1 1
check to see if the previous state of the outputs plays a role in determining the result. What is
strange about the outputs when the inputs are both zero?
Inputs
R
S
0
0
1
0
0
1
1
1
Outputs
Q
Q
We will next use the RS flipflop to construct a “debounced” pushbutton. Use
a momentary-contact, single-pole, double-throw switch that has three wires
soldered to it. “NC” means normally closed. That is, the switch is normally
in this position. “NO” means normally open. This contact will be connected
to the moveable contact when the switch is activated. Determine which wire
is connected to which contact.
NC
NO
Add the switch to the RS flipflop as shown below.
Q
S
NO
NC
Q
R
Debounced pushbutton
Determine the outputs of the circuit when the switch is pressed and when it is not pressed. Leave
it on your ProtoBoard for future use.
The D Flipflop
When a number of digital circuits are used together the system should be “clocked” so that all
outputs are synchronized so that they change at the same time. A typical synchronizer is the data
or D flipflop. We’ll use the 7474 dual D-flipflop device. The connections to the pins are shown
below.
Chapter 4 Entering the digital world
88
14
13
12
11
10
9
8
Vcc
2R
2D
2CP
2S
2Q
2Q
7474
1R
1D
1CP
1S
1Q
1Q
Gnd
1
2
3
4
5
6
7
7474 dual D flipflop
Each flipflop has six connections, which are shown schematically below:
S
D
Q
CP
Q
R
D-flipflop connections
The set ( S ) and reset ( R ) inputs and two outputs are similar to those on a RS flipflop. “D” is the
data input and “CP” is the input for the clock pulse. Explore how the data and clock inputs work.
In particular, determine how the output (Q) depends on the value of the data input (D) and the
clock pulse. In particular, does the input change the output
(a) when the clock is high,
(b), when the clock is low,
(c) when the clock makes a low-to-high transition, or
(d) when the clock makes a high-to-low transition.
You may be able to determine the answer to this question by using the TTL function generator
set at the lowest frequency as the input to the clock pulse and connecting indicator lamps to CP
and Q as you use a switch to change D.
Dividing by two
Because the value of Q is always equal to that of D (at least at the time determined by the clock
pulse), that means that D is always the opposite of Q (again, at the time determined by the clock
pulse). As a result, the device can be used as a true flip flop, or binary counter. Wire the
following circuit.
Output
D
Q
Q
CP
Input
Divide-by-two
Monitor input and output with the indicator LEDs, reporting the results in a table like that below
and describe why the circuit is called a “divide-by-two.”
Chapter 4 Entering the digital world
89
Input
0
1
0
1
0
1
Output
The divide-by-two can be used to control a gate so that the first input pulse turns the gate on and
the second pulse turns it off. Construct such a device using the debounced pushbutton as input to
the divide-by-two and the same kind of gate you used to control pulses from the function
generator. Report the circuit you devised.
2. Textbook and classroom explanations
A flip flop is the basic device on which digital memory and counters are based. A variety of very
useful instruments can be constructed of them, instruments that can count pulses, determine the
time between pulses or the frequency of a series of pulses. Microcomputers are based on these
devices. Yet it is precisely the computer that has, in the opinion of the author, rendered an
extensive study of the flip flop and its uses obsolete. Most of the tasks that discrete flip flops or
small numbers of them put together once did are now done much more easily by a computer. For
that reason, we have confined our study to only two kinds, the RS, or set/reset, and the D, or data
flip flop.
RS flip flop
As you have seen, the basic RS flip flop can be “set” to one or “reset” to zero by grounding the
appropriate input. When both inputs are high the output is often said to be “indefinite.” As you
found, under these conditions both outputs are the same. Because one output is defined to be Q
and the other Q , this input condition produces a logical impossibility.
The RS flip flop can be improved by adding two additional NAND gates as shown below:
S
Q
C
Q
R
Clocked RS flip flop
When you work through the truth table you will find that when C = 1, the device works like the
normal RS flip flop. But, when C = 0, the outputs are not affected by the inputs. That is, when
C = 0 the device is disabled.
A flip flop that can be disabled is called a latch. The value of the flip flop when C was last high
is retained, or held, when C is made low. The circuit has another use. C is said to be the clock
input and the flip flop is said to be under clock control. The device counts when the clock is high
and is frozen when the clock is low. Controlling a number of flip flops in this way allows their
Chapter 4 Entering the digital world
90
outputs to be synchronized so they can all be read at the same time.
Data or D flip flop
A problem with the clocked RS flip flop is that the output can change while the clock level is
high. That’s why it is also called a transparent latch, because the output “sees through” to the
input while the clock level is high. A modification of the clocked RS flip flop results in what is
called a data flip flop, shown schematically below. There are three sets of flip flops. The lefthand two make sure that the two inputs to the right-hand flip flop are opposite of each other. But,
the values of these two inputs are not delivered until the clock pulse goes from low to high.
For that reason, this is called an “edge triggered” flip flop. The small wedge on the diagram of
the D-flip flop means “edge-triggered.” Both rising and falling-edge triggering is possible. A
small circle in front of the wedge denotes a falling-edge triggered device. Data flip flops also
have set and reset inputs. They work independently of the clock pulse, and so can override the
clocked input and, for example, force the output of the flip flop to be zero.
Q
CP
Q
Data
Edge-triggered data flipflop
Two-state counting system
The ability to count from 0 to 1 is obviously extremely limiting. In the same way that the
decimal system permits counting beyond 10, by assigning a value to the digit depending on its
position, the two-state or binary system can be used to count well beyond 2. For example, in the
table below, the number 58026 is decoded as being equal to 5×104 + 8×103 + 2×101 + 6×100 or
50000+8000+20+6. Note that the zero in the 102 position is a place holder.
104
5
103
8
102
0
101
2
100
6
Here is a similar example in the binary system, where we decode the number 10111:
24
1
23
0
22
1
21
1
20
1
You can see from the table that number is 24 + 22 + 21 + 20. In the decimal system this number is
16+4+2+1 or 23.
3. Applications
Extend the divide-by-two to a “count-to-16” circuit by making the output of the divide-by-two
Chapter 4 Entering the digital world
91
the input of a second divide-by-two, and so on until you have four circuits (with one input and
four outputs). Extend the table above to include all four outputs and record the results until the
outputs return to the initial state (all zero).
Pulse
number
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
Outputs
2 3 4
Leave the circuit on your ProtoBoard for future use.
4. Questions and problems
1. Develop an RS flipflop using NOR gates rather than NAND gates.
2. Your “count-to-16” counter results allow you to represent the numbers 0-15 as a set of four
binary digits (0 or 1). Write a table that allows you to convert between the decimal and fourbit binary representations of the numbers 0 through 15.
3. Make a small modification to your counter to make it count down from 15 to 0.
4. It takes a small, but measurable time, typically 4 ns, for the input to a flipflop to be reflected
at the output. Why would a counter like the one we made be called “synchronous”? Consider
what would happen if you had a 32-bit counter operating at high clock speeds (100 MHz).
What problems could be caused?
Chapter 4 Entering the digital world
92
Chapter 4 Entering the digital world
93
Chapter 4 Entering the digital world
94
5. Conversions between the analog and digital worlds
A. Fundamentals of analog-digital conversion
What is needed to create an accurate representation of an analog signal?
1. Laboratory exploration
You and your lab partner will play the A/D D/A game (Appendix C) to gain some experience in
what is needed to be able to convert a signal from analog to digital and back again.
You and your partner will each draw an analog signal on the graph. You will then digitize it and
transmit the digital values to your partner who will use them to try to reconstruct your analog
signal. You will both compare the transmitted signals with your originals. The game sheets are in
the appendix.
2. Textbook and classroom explanation
As you have seen, the two key parameters that describe analog-digital conversions are the
number of bits and the sampling rate. Eight digital bits allow a choice of 28 or 256 values.
Therefore a 0-5V analog signal would have steps of (5/256) V or about 20 mV. Ten bits reduces
the step size to 5 mV, twelve bits to about 1 mV. Is higher resolution always better? Not
necessarily, because the higher resolution is lost if the analog circuitry adds a random, or noise
voltage that is equal in size to the resolution set by the converter.
How fast do you have to sample the analog signal? The textbook answer to the minimum
required sampling frequency is usually called the Nyquist criterion. The criterion states that the
sampling frequency should be twice the highest frequency component in the signal. The Nyquist
criterion is based on a rather amazing fact about periodic, real-valued functions. Such a function
can be expressed as a series of trigonometric functions:
v(t ) =
∞
A0 ∞
+ ∑ An sin( 2πnft ) + ∑ Bm cos(2πmft )
2 n=1
m =1
Here m and n are integers and f = 1/T, the period of the function.
For a pure sine wave v(t) = A sin(2πft), A1 = A, and for n > 1, An = 0. All the Bm coefficients are
also zero. If a function has a non-zero average value (like a dc component on an ac signal), A0/2
is that component.
Such a description of a function is called a Fourier series. The expansion of functions as sums of
trigonometric functions has a long history. Charles Bossut (1730-1814) used a finite number of
terms so write a number of functions. Daniel Bernoulli (1700-1782) extended Bossut’s results to
an infinite number of terms. Leonard Euler, in 1744 expanded one function as an infinite number
of sine terms. In 1777 he was able to find the coefficients for an expansion in cosine terms.
Joseph Fourier (1768-1830) published expressions for both the sine and cosine terms in 1807 in
his The Mathematical Theory of Heat. It took many people several more years to put the series
on firm mathematical foundations.
We will explore only a few Fourier series descriptions of common waveforms. For a square
wave of amplitude A and period T that has a phase such that v(t=0) = +A is given by Bm = 4A/πm
if m is odd and Bm = 0 if m is even. All An are zero. The first three terms are
Chapter 5 Conversions between the analog and digital worlds 95

2πt  1 
3 ⋅ 2πt  1 
5 ⋅ 2πt
 4 A 
v(t ) = 
) +   sin(
) +   sin(
) + K
 sin(
T
T
T
 π 
3
5

How many terms are needed to achieve a reasonably good reproduction of a square wave? On
the next page is a figure that shows a progression of sums. The first graph is only the first term,
the second the first two, the third the first three, and so on to the first five terms. Notice that
even with frequencies as high as nine times the fundamental, the “square” wave has wiggles that
have an amplitude of about 10% of the amplitude of the square wave.
The triangle wave series also has only odd sine terms, but in this case the coefficients are given
( m −1) / 2

 8 A  (−1)
 . That is, they alternate in sign. A good reference is at
by Bm =  2 
2
m
 π 

http://mathworld.wolfram.com/topics/FourierSeries.html.
When trying to reproduce sound waves, it is not necessary to include frequencies beyond those
detectable by the human ear (15 kHz). That means a minimum sampling rate is 30 kHz. In
practice 44 kHz is used for CD players.
B. Digital to analog converters
1. Laboratory Exploration
The digital-to-analog converter (DAC) we will be using, either the DAC0808 or the MC1408L8,
uses an 8-bit digital number, n, and a reference voltage, Vin, that is converted into an input
current by a resistor to create an output current equal to (n/256)Iin, where n varies between 0 and
255.
The DAC, as shown below, requires a reference current supply consisting of a 12-V potential
source and a 6.8-k resistor. The eight digital inputs will initially be connected to the eight DIP
switches. You will need to determine whether a disconnected input is treated as a 0 (low) or 1
(high).
The analog output current is converted to a voltage by means of an op amp with a 2-k feedback
resistor. Note that the current is negative, so the voltage will be positive.
+12 V
+5 V
6.8k
Vref+
14
VrefComp 15
16
0.01 µF
3
Vee 5
27
2k
Vcc
13
4
MC1408L8
DAC0808
6
26
7
25
8
24
9
23
10 11 12
22
21
2
Iout+
Iout-
20
-12 V
Digital inputs
Digital-to-analog converter
Chapter 5 Conversions between the analog and digital worlds 96
Vout
Synthesizing a square wave from Fourier components
Chapter 5 Conversions between the analog and digital worlds 97
Manually step through the digital input range. You do not have to go through all 256 possible
inputs. Using the DMM, determine the range of output voltage and the resolution of the
converter.
Connect the four most significant digits to the output of the divide-by-16 counter and view the
output voltage on an oscilloscope. Describe what you see
2. Textbook and classroom explanation
Digital-to-analog converters
Consult the datasheet for your DAC in the appendix. The datasheet’s explanation for the
operation of the MC1408 reads, in part, “The MC1408 consists of a reference current amplifier,
an R-2R ladder, and 8 high-speed current switches. The R-2R ladder divides the reference
amplifier current into binarily-related components, which are fed to the remainder current which
is equal to the least significant bit. This current is shunted to ground, and the maximum output
current is 255/256 of the reference amplifier current, or 1.992 mA for a 2.0-mA reference
amplifier current.” The simplified diagram is shown below:
MC1408L8 simplified schematic diagram
What is an “R-2R ladder” and how does it work? Let’s look a simplified version that has only
three bits and works with positive current, rather than the negative in the MC1408.
2R
I/2
R
I/2
Vref
I/4
2R
I/4
R
2R
I/8
MSB
2R
I/8
LSB
Vout
Four-bit R-2R ladder
Chapter 5 Conversions between the analog and digital worlds 98
The first thing to note is that the potential drop across the op amp inputs is zero, so the bottom of
the resistors are at ground potential whether the switch is to the right, sending the current through
the op amp, or the left, sending the current to ground. Consider the resistors starting at the righthand end.
The two 2R resistors, having equal resistance and potential drop across them have equal currents,
labeled I/8. Together their effective resistance is equal to R, which means that the sum of the
current through them equals the current through the R resistor.
The two parallel 2R resistors are in series with the R resistor, making an equivalent resistance 2R.
That’s why the I/4 currents are equal. This argument is repeated as many times as there are bits
in the ladder.
In this example, the current into the op amp would be I/2 + I/4 + I/8 = 7I/8. If any of the switches
were in the other direction, then the current through the corresponding resistor would go to
ground, not through the op amp.
The DAC0808 and MC1408L2 are called multiplying DACs because the output current is the
product of the reference current and the digital switching circuitry. The input current can be
either positive or negative, and the output current can be taken from either the positive or
negative terminals. The DAC0808 has an accuracy of ±1 LSB (least significant bit) and takes
100ns for the output to achieve its final value. The MC14008L8 has an accuracy of ±½LSB and
a settling time of 70 ns.
More advanced DACs
The two DACs that we have used have eight digital bits, which means that the least significant
bit has a value of 1/28 or 1/256 of the reference voltage. One commercial application of DACs is
in analog computer monitors where they convert the digital signals that represent the intensity of
each of the three primary colors (red, green, and blue) to an analog voltage that controls the
brightness of the color on the screen. Three 8-bit DACs can provide 24-bit color images. Other
applications are in conversion of digital signals into sound (CDs, cell phones, mp3 players, etc).
Ten, twelve, fourteen, and even sixteen bit DACs are now readily available. Most have serial
digital inputs, which reduces the number of pins needed on the device. For a 16-bit DAC serial
data inputs require sixteen clock cycles to send a byte, so the clock rates must be very high. For
example, the AD5060 from Analog Devices can accept clock rates as high as 30 MHz. It updates
the digital bytes at a rate of 250k samples per second. Its analog output amplifier has a settling
time of 4µs and an accuracy of ±1 LSB. It must be noted that if the reference voltage is 2.5 V,
then the least significant bit is about 36 µV. Building an analog system that has noise (random
voltage variations) as small as 36µV is a difficult task.
Because the AD5060 is a serial device, it has only eight pins and dissipates 6mW of power. A
faster DAC from the same manufacturer is the AD5062 that updates at 1.3M samples per second,
but has the same settling time and accuracy, and consumes only 3.5 mW, which is important in
densely-packed instruments. Equally important is the ability of the DAC to go into a low-power
(micro watt) standby state when not being used.
The DACs in the data acquisition system you will soon be using are very slow. While having 12bit resolution, and a 0-5 V output voltage that can drive 5 mA, they can sample at a rate of only
150 Hz.
Chapter 5 Conversions between the analog and digital worlds 99
The Sampling Theorem
The Nyquist criterion, is that if the maximum frequency of components in an analog signal is f,
then you need to have a rate of equally-spaced samples at a rate of 2f to reconstruct the signal.
The attribution of this criterion to Harry Nyquist uncertain. A 1928 paper is usually cited, but
how the results of that paper can be applied to the sampling rate problem is unclear. If all the
people who independently “discovered” the result were to be named it should be called the
Whittaker-Kotel'nikov-Raabe-Shannon-Someya (WiKRaSS) sampling theorem. For that reason,
it is probably best to just call it the “sampling theorem.”
How do you reconstruct an analog signal? As you found in the game, you have to interpolate
between the digitized values. A linear interpolation is best, but sharp corners contain highfrequency components, and so are smoothed by the reduced high-frequency response of the
analog output amplifier.
Analog-to-digital conversion (ADC)
Your laboratory work did not include experience with a stand-alone ADC. The data acquisition
system that you will use contains a successive approximation ADC.
Vin
Comp.
Vref
DAC
Vout
High/Low
Conversion
complete
Control logic
Clock
Start conversion
Digital output
Successive approximation ADC
The successive approximation ADC contains a DAC, a comparator, and control logic. Basically,
the DAC starts with a selected digital value and generates an appropriate voltage. That voltage is
compared with the voltage to be digitized. If the voltage is too high, the digital value is reduced;
if it is too low, it is increased. When the two voltages agree to within the accuracy of the DAC,
the conversion is complete.
Let’s try an example to see how it works. Assume that the input voltage is 3.370V and the 8-bit
converter’s range is 0 to 5V. Then the voltage represented by each of the 8 bits is given by
Bit
1
2
3
4
5
6
7
8
Voltage
5V/2 = 2.50000V
5V/4 = 1.25000V
5V/8 = 0.62500V
5V/16= 0.31250V
5V/32= 0.15625V
5V/64= 0.07813V
5V/128=0.03906V
5V/256=0.01953V
Chapter 5 Conversions between the analog and digital worlds 100
When the “Start conversion” signal is sent to the ADC digital bit 1 is turned on. The comparator
reports that 2.5 V is lower than the input voltage (3.370 V), so bit 1 remains on and bit 2 is
tentatively turned on and 2.5V+1.25V is compared with 3.370V. The process continues:
Step
1
2
3
4
5
6
7
8
Bit
1
2
3
4
5
6
7
8
Voltage
2.5
2.5+1.25 =3.75
2.5 + 0.625 = 3.125
2.5+0.625+0.3125 = 3.4375
2.5+0.625+0.15625 = 3.2813
2.5+0.625+0.15625 + 0.07813 = 3.3594
2.5+0.625+0.15625 + 0.07813 + 0.03906 = 3.3984
2.5+0.625+0.15625 + 0.07813 + 0.01953 = 3.3789
Comparison
< 3.370
> 3.370
< 3.370
> 3.370
< 3.370
< 3.370
> 3.370
> 3.370
Decision
on
off
on
off
on
on
off
off
Digital output
1xxxxxxx
10xxxxxx
101xxxxx
1010xxxx
10101xxx
101011xx
1010110x
10101100
When the eighth bit has been determined the digital output is set to 10101100, the “conversion
complete” signal is sent, and the converter is ready for the next voltage. Assuming that no time is
needed for starting or reporting the result, eight clock cycles are needed to convert a voltage. The
ADCs in the digital acquisition system you will be using can convert 48k 13- or 14-bit samples
per second, and require about 21 µs to complete a conversion.
Note that the resolution is the size of the least significant bit, or about 0.020V. If the input
voltage had been 3.379V, the digital output would have been one bit higher. If the voltage had
been 3.340V the result would have been one bit lower.
3. Questions and problems
1. A 12-bit DAC is attached to an op amp that converts the output current to a voltage that
varies between –12.5V and +12.5V. What change in voltage results from a change of 1 least
significant bit in the digital input?
2. You are designing a digital thermometer using a thermistor. You’ll be using a 12-bit ADC
that has a 0-10V input range. Your thermistor has the following resistances
T (oC)
R (kΩ)
0
10
20
30
40
94.99
59.75
37.30
24.27
16.15
∆ R /∆ T
(kΩ/
o
C)
4.70
2.75
1.65
1.02
0.64
To keep the maximum voltage below 10V you use a 0.1 mA constant current source. What
temperature resolution does your digital thermometer have at each of these five temperatures,
assuming that all of the uncertainty in the temperature comes from the resolution of the
ADC?
3. An alternative to the successive approximation ADC is the counting ADC. As you learned, in
the successive approximation ADC the digital value started in the middle of the range, and
then went up or down depending on the comparator. In the counting ADC the digital value
starts and zero and then counts up, one bit at a time, until the analog output exceeds the input
Chapter 5 Conversions between the analog and digital worlds 101
voltage. The counting ADC has the advantage of simpler digital circuitry. What are its
disadvantages? Quantify them for two possible input voltages, 0.65V and 4.76V, assuming
that the 12-bit ADC has a 0-5V input range.
Chapter 5 Conversions between the analog and digital worlds 102
6. Interfacing the computer with the laboratory
A. The National Instruments USB-6009 digital acquisition system
What can the NI USB-6009 do? It has eight analog inputs and two analog outputs, 12 combined
digital input/output lines, and a counter/timer input/output.
1. Laboratory exploration
Examine the DAQ. Note that there are separate analog and digital sides. Read the instruction
manual to find the eight analog inputs, two analog outputs, 12 digital terminals that can be either
input or output, and the counter/timer connector. Find the ground(s), the 5-V output, and the 2.5
V output. Note how wires are to be connected to the terminals.
We will first use the simple software provided with the device to perform tests on all functions of
the DAQ. Before connecting the device to the external world we need to learn its voltage and
current limitations. Read the instruction manual, paying careful attention to the maximum
voltage permitted for analog and digital inputs and the maximum current that the outputs can
supply.
The function generator on the ProtoBoard can safely test the analog input. Either a DMM or
oscilloscope can safely test the outputs (either analog or digital). But digital outputs can be
displayed more readily with an LED. How can one be connected to the digital output?
Let’s explore the system. Plug the DAQ into a USB port on the computer, turn on the computer,
and click on the “Measurement and Automation” icon. When the program starts you will be
presented with the following screen:
Click on “Devices and Interfaces” and you’ll see this screen:
Chapter 6 Interfacing the computer with the laboratory 103
The USB-6009 device should be listed. If it is not, check that the USB cable is connected
properly. Click on USB-6009 to bring up this screen.
The next step is to run the test panels to see how they work. Click on Test Panels at the top of
the screen. You will get a window with four tabs, initially open to the Analog Input test panel.
Here are what you see when clicking each of the four tabs.
Chapter 6 Interfacing the computer with the laboratory 104
Chapter 6 Interfacing the computer with the laboratory 105
First test analog input. Use the function generator to generate a varying analog signal. Select one
analog input channel. Be sure to connect ground on the DAQ to ground on the ProtoBoard. Make
sure that the same analog channel is selected by the software. Try a low-frequency (50-200 Hz)
sine wave with a medium amplitude. Start recording. Try changing the amplitude and frequency.
See how triangle and square waves are recorded.
Continue testing the DAQ, going through all other functions. The Voltage Output can be checked
with the DMM as all you can do is to choose a voltage and then tell the DAQ to put it on the
output line.
Try configuring some digital lines to input, others to output. Finally check the counter.
2. Textbook and classroom explanations
We will explore the instruction manual that summarizes the capabilities and limitations of the
DAQ.
B. Using the USB-6009 with LabView
1. Laboratory exploration
You have seen that the test panels are very limited. Work through the exercises in the first,
second, and fourth chapters of the book LabVIEW Express. In the fourth chapter you will finally
use LabVIEW to interface with the DAQ. Then work through the six-hour tutorial on LabView
that is on your computer.
2. Applications
Use the DAQ with LabVIEW to gather data from a variety of the sensors we have studied: the
LM335 temperature sensor, the thermistor connected to a constant current source, the
phototransistor connected to a current-to-voltage converter, the magnetic field sensor, the strain
gage force sensor, and the microphone with amplifier.
You will have to decide how much the signal from the sensor has to be changed by electronic
hardware (usually an op amp) before it is connected to the DAQ. Connect both the momentary
SPDT switch and the debounced pushbutton (you can use the ones on the ProtoBoard) to the
counter input to explore switch bounce. Use the analog voltage output to drive a speaker and an
LED through an appropriate power amplifier.
Explore the hardware/software tradeoff. That is, when should electronic hardware perform
functions such as amplification, filtering, level shifting, and conversion to a digital signal with a
comparator, and when should these functions be done in software? Do the same function two
different ways to determine the advantages and disadvantages of hardware and software
implementations of a function.
Chapter 6 Interfacing the computer with the laboratory 106
Appendices
Appendix A: Resistor color code
First band × 10
Second band × 1
Multiplier 10n
Tolerance
Examples:
Brown Black Brown
(1×10 + 0)×100 = 100
Yellow Violet Red
(4×10 + 7)×102 = 4700 or 4.7k
In general look first at the third band.
Brown 100s of ohms
Red kΩ
Orange 10kΩ
Yellow 100kΩ
Green MΩ
Appendices
107
Colors
Values
Tolerance
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
White
0
1
2
3
4
5
6
7
8
9
Gold
Silver
5%
10%
Appendices
108
Appendix B: RC Filters using complex numbers
An alternative approach to the trigonometric representations used to analyze ac circuits is used in
most electrical engineering texts. This approach uses complex number to represent electrical
quantities. It is also often called the phasor approach. First we must develop the rules needed to
go back and forth between the actual currents and voltages and their complex representations.
Representations:
Voltage and current are represented by the complex quantities V and I. For example, the voltage
v(t) = Vp cos(ωt+φ). On the Gaussian plane, where the real part is on the horizontal axis and the
imaginary on the vertical, the voltage v(t) is a vector of length Vp. At time t=0 the vector at an
angle φ above the horizontal. It rotates counter-clockwise, revolving once in the time T = 2π/ω.
This voltage is represented by the complex number V = Vpejφ. The complex representation for V
has the same length, Vp, but doesn’t rotate. Rather it remains at the angle φ.
Note that we use j to represent − 1 rather than i to avoid confusion with current. For a more
compact notation we use, ω = 2πf. The Euler equation, ejφ = cos φ + j sin φ connects the
exponential and graphical representations.
What is the complex representation of i(t) = Ip sin(ωt)?
Remember that sinθ = cos(θ – π/2). Therefore sin(ωt) = cos(ωt – π/2). So i(t) = Ip cos(ωt – π/2).
The complex representation will then be I = Ip e–jπ/2.
But, from the Euler equation e–jπ/2 = cos(–π/2) + jsin(–π/2) = –j.
So, I = –jIp. It is a vector pointing downwards along the –j axis.
Actual voltages:
To find the actual (physical) voltage, which changes in time, multiply the complex quantity by
ejωt and then take the real part. Following our previous example, multiplying Vpejφ by ejωt gives
Vpej(ωt+φ) = Vp(cos(ωt+φ) + jsin(ωt+φ)).
The real part is Vpcos(ωt+φ).
Note that the complex representations are not time dependent while the actual voltages and
currents are.
For example, a current with the complex representation I = –3j corresponds to a real current
given by i(t) = Re(–3j ejω) = Re(–3jcos(ωt)+j(–3j)sin(ωt)) = +3sin(ωt).
Resistors:
Voltage and current are in phase. That is, just in the case of dc circuits, R = v/i, so R = V/I.
Capacitors:
From the relationship i = C dv/dt, and with v(t) = Vp cos(ωt), we have i(t) = –VpCω sin(ωt). The
complex representation of v(t) is just V = Vp. The complex representation of current is I = jVpωC.
By analogy with resistance, XC = V/I.
Appendices
109
Thus XC = 1/jωC = –j/ωC.
The negative sign in front of the j means that the voltage lags the current by π/2 or 90o.
Inductors:
Again, we will not go through the derivation, but
XL = jωL.
The positive sign in front of the j means that the voltage leads the current π/2 or 90o
Impedance:
Real world devices usually have some resistance, some inductive reactance, and some capacitive
reactance. The sum of the three is called impedance, Z, and is defined as
Z = R + XC + XL.
An ideal resistor has no reactance, so for it Z = R.
For an ideal capacitor Z = XC = –j/ωC.
For an ideal inductor Z = XL = jωL.
Magnitude and phase:
So far we’ve dealt with currents and voltages that have complex representations that are either
real or multiplied by j or –j. That is, they have phase angles of either 0, +π/2 or –π/2. A voltage
at a more general phase angle would be written in complex representation as Vejφ. But such a
quantity can also be written as a sum of a real and imaginary part: Vcosφ + jVsinφ. What would
such a voltage look like on an oscilloscope? It would have a magnitude and a phase angle. The
phase angle can be found from tan φ = sinφ/cosφ, that is, the imaginary part divided by the real
part. The magnitude is (V cos φ) 2 + (V sin φ) 2 , that is, the square root of the sum of the squares
of the real and imaginary parts.
RC Filters
The high-pass filter
How can we find the output voltage? All we need to do is to write the equation for a voltage
divider in terms of the impedances of the two components:
Vin R
or
R + XC
Vin R
=
R − j / ωC
Vout =
Vout
What is the magnitude and what is the phase of the output voltage?
We need to separate this fraction into a real and an imaginary part, that is, to write it as a + jb.
To do this we use the identity (a + jb)(a – jb) = a2 + b2. Thus we multiply the numerator and the
denominator by R+j/ωC. and obtain
Appendices
110
Vout = Vin R
R + ( j / ωC )
.
R 2 + ( j / ωC ) 2
The magnitude of the output voltage is the square root of the sum of the squares of the real part
(R/(R2 + (j/ωC)2)) and the imaginary part, ((1/ωC)/(R2 + (j/ωC)2)). That is
Vout = Vin
R
R 2 + (1 / ωC ) 2
= Vin
2πfRC
1 + (2πfRC ) 2
.
Let’s examine this equation. At high frequencies, where 2πfRC >> 1, Vout = Vin. That is, there is
no reduction, or attenuation, of the signal.
At low frequencies, where 2πfRC << 1, Vout = 2πfRC Vin. That is, the output voltage is very
small and increases linearly with frequency. When the frequency doubles, the output voltage
doubles.
The frequency where 2πfRC = 1, that is, f = 1/(2πRC), is a good way to characterize the filter. At
this frequency the output voltage is 1/ 2 times the input voltage. Engineers call this frequency
the “–3dB breakpoint” of the filter. Below this frequency the response is almost linear in
frequency, above it, it is almost constant. Thus f3dB = 1/(2πRC).
The phase angle between the output and input voltages is given by the tangent of the ratio of the
imaginary to the real parts
tan φ = 1 (ωRC ) = 1 (2πfRC ) .
At high frequencies the phase angle approaches zero. At low frequencies it approaches π/2 or
+90o. At the –3dB frequency tan φ = 1, or φ = 45o.
The low-pass filter
In this case the output voltage is taken across the capacitor so
Vout =
Vin X C
. Or
R + XC
Vout =
Vin (− j / ωC )
.
R − j / ωC
“Rationalizing the denominator” gives Vout = Vin (− j / ωC )
R + ( j / ωC )
.
R 2 + ( j / ωC ) 2
The magnitude of the output voltage is then
Vout = Vin
(1 / ωC )
R 2 + (1 / ωC ) 2
= Vin
1
1 + (2πfRC ) 2
and the phase angle is given by
tan φ = − 1 (ωRC ) = − 1 (2πfRC ) .
At very low frequencies the output voltage is now approximately the input voltage and the two
are in phase. At very high frequencies the output is reduced as 1/f (being cut in half for every
Appendices
111
doubling of the frequency) and the phase approaches –π/2 or –90o. The –3dB frequency is the
same, f3dB = 1/(2πRC). The resultant graphs are shown in the main portion of the textbook.
Appendices
112
Appendices
113
Appendices
114
Appendix C: A/D, D/A game
Converting an analog signal to a digital signal
Draw a waveform on the graph below. Make it vary between 0 and 7 Volts and have some slowchanging regions and some fast-changing regions.
Convert signal into a digital format. The simplest way to do this is to record the voltage every 2
ms, rounding down to the nearest volt. Then convert the number into a binary number.
Time Voltage 22
(ms) (V)
0
2
4
6
8
10
12
21
20
Transmit these data to your partner. Do it by saying the binary bits in order. For example, if your
seven numbers were 2, 4, 6, 2, 1, 2, 5, 7 you would say “010, 100, 110, 020, 001, 020, 101, 111.”
Appendices
115
Converting a digital signal to an analog signal
Write down the signal transmitted by your partner. Then convert the binary numbers to decimal
and record them.
Time 22
(ms)
0
2
4
6
8
10
12
21
20
Voltage
(V)
Mark the points on the graph below and then connect them with a smooth curve.
Compare the received signal with the original.
Appendices
116
Improving the accuracy of analog to digital conversion
As you have probably guessed, the two methods to increase accuracy are:
1. Sample more often.
2. Divide the voltage into a smaller chunks by using more bits!
Repeat the conversion of an analog waveform to digital by drawing a new waveform on the
graph below. The signal can now go between 0 and 15 volts. Again make it vary rapidly in some
regions and slowly in others.
Convert the voltages to binary bits in the same way that you did before. Note, however, that this
time you will be recording data every ms. Transmit the binary bits to your partner.
Time Voltage 23
(ms) (V)
0
1
2
3
7
5
6
7
8
9
10
11
12
Appendices
22
21
20
117
Improved digital to analog conversion
Repeat the conversion of a digital waveform to analog by recording the received binary bits,
converting to a decimal value, and then plotting the points and joining them with a smooth curve.
Time 23 22 21 20 Voltage
(ms)
(V)
0
1
2
3
7
5
6
7
8
9
10
11
12
Compare the reproduced waveform with the original. Grade the result for accuracy.
Appendices
118
Appendix D: Data sheets
Page Device
Description
1
LM335
Integrated circuit temperature sensor
5
UGN3503
Hall-effect magnetic field sensor
9
OP802
Phototransistor
11
LM741
General purpose operational amplifier
15
LF411
High input impedance operational amplifier
19
OP177
Precision, low offset voltage, low input bias operational amplifier
25
AD624
Instrumentation amplifier
33
2N4401
General purpose npn transistor
35
TIP 29/30
Power transistors, npn and pnp
37
LM311
Comparator
43
LM555
Multivibrator
51
DAC0808
8-bit digital-to-analog converter
57
USB6009
National Instruments Data acquisition system
Appendices
119
Appendices
120
LM135/LM235/LM335, LM135A/LM235A/LM335A
Precision Temperature Sensors
General Description
The LM135 series are precision, easily-calibrated, integrated
circuit temperature sensors. Operating as a 2-terminal zener,
the LM135 has a breakdown voltage directly proportional to
absolute temperature at +10 mV/˚K. With less than 1Ω dynamic impedance the device operates over a current range
of 400 µA to 5 mA with virtually no change in performance.
When calibrated at 25˚C the LM135 has typically less than
1˚C error over a 100˚C temperature range. Unlike other sensors the LM135 has a linear output.
Applications for the LM135 include almost any type of temperature sensing over a −55˚C to +150˚C temperature
range. The low impedance and linear output make interfacing to readout or control circuitry especially easy.
The LM135 operates over a −55˚C to +150˚C temperature
range while the LM235 operates over a −40˚C to +125˚C
temperature range. The LM335 operates from −40˚C to
+100˚C. The LM135/LM235/LM335 are available packaged
in hermetic TO-46 transistor packages while the LM335 is
also available in plastic TO-92 packages.
Features
n
n
n
n
n
n
n
n
Directly calibrated in ˚Kelvin
1˚C initial accuracy available
Operates from 400 µA to 5 mA
Less than 1Ω dynamic impedance
Easily calibrated
Wide operating temperature range
200˚C overrange
Low cost
Schematic Diagram
DS005698-1
© 1999 National Semiconductor Corporation
DS005698
www.national.com
Appendix D
1
LM135/LM235/LM335, LM135A/LM235A/LM335A Precision Temperature Sensors
May 1999
Connection Diagrams
TO-92
Plastic Package
SO-8
Surface Mount Package
TO-46
Metal Can Package*
DS005698-8
Bottom View
Order Number LM335Z
or LM335AZ
See NS Package
Number Z03A
www.national.com
DS005698-26
DS005698-25
Order Number LM335M
See NS Package
Number M08A
2
Appendix D
2
*Case is connected to negative pin
Bottom View
Order Number LM135H,
LM135H-MIL, LM235H,
LM335H, LM135AH,
LM235AH or LM335AH
See NS Package
Number H03H
Specified Operating Temp. Range
Absolute Maximum Ratings (Note 4)
Continuous
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales Office/
Distributors for availability and specifications.
Reverse Current
Forward Current
Storage Temperature
TO-46 Package
TO-92 Package
SO-8 Package
Intermittent
(Note 2)
150˚C to 200˚C
125˚C to 150˚C
100˚C to 125˚C
LM135, LM135A −55˚C to +150˚C
LM235, LM235A −40˚C to +125˚C
LM335, LM335A −40˚C to +100˚C
Lead Temp. (Soldering, 10 seconds)
TO-92 Package:
TO-46 Package:
SO-8 Package:
Vapor Phase (60 seconds):
Infrared (15 seconds):
15 mA
10 mA
−60˚C to +180˚C
−60˚C to +150˚C
−65˚C to +150˚C
260˚C
300˚C
300˚C
215˚C
220˚C
Temperature Accuracy (Note 1)
LM135/LM235, LM135A/LM235A
Parameter
Operating Output Voltage
Uncalibrated Temperature Error
Uncalibrated Temperature Error
Temperature Error with 25˚C
Conditions
LM135A/LM235A
TC = 25˚C, IR = 1 mA
TC = 25˚C, IR = 1 mA
TMIN ≤ TC ≤ TMAX, IR = 1 mA
TMIN ≤ TC ≤ TMAX, IR = 1 mA
LM135/LM235
Units
Min
Typ
Max
Min
Typ
Max
2.97
2.98
2.99
2.95
2.98
3.01
V
0.5
1
1
3
˚C
1.3
2.7
2
5
˚C
0.3
1
0.5
1.5
˚C
Calibration
Calibrated Error at Extended
TC = TMAX (Intermittent)
2
2
˚C
Temperatures
Non-Linearity
IR = 1 mA
0.3
0.5
0.3
1
˚C
Temperature Accuracy (Note 1)
LM335, LM335A
Parameter
Operating Output Voltage
Uncalibrated Temperature Error
Uncalibrated Temperature Error
Temperature Error with 25˚C
Conditions
LM335A
TC = 25˚C, IR = 1 mA
TC = 25˚C, IR = 1 mA
TMIN ≤ TC ≤ TMAX, IR = 1 mA
TMIN ≤ TC ≤ TMAX, IR = 1 mA
LM335
Units
Min
Typ
Max
Min
Typ
Max
2.95
2.98
3.01
2.92
2.98
3.04
V
1
3
2
6
˚C
2
5
4
9
˚C
0.5
1
1
2
˚C
Calibration
Calibrated Error at Extended
TC = TMAX (Intermittent)
2
2
˚C
Temperatures
Non-Linearity
IR = 1 mA
0.3
1.5
0.3
1.5
˚C
Electrical Characteristics (Note 1)
Parameter
Conditions
LM135/LM235
LM335
LM135A/LM235A
LM335A
Min
Operating Output Voltage
400 µA≤IR≤5 mA
Change with Current
At Constant Temperature
IR = 1 mA
Dynamic Impedance
Typ
Max
2.5
10
Min
Units
Typ
Max
3
14
mV
0.5
0.6
Ω
+10
+10
mV/˚C
Still Air
80
80
sec
100 ft/Min Air
10
10
sec
Stirred Oil
TC = 125˚C
1
1
sec
0.2
0.2
˚C/khr
Output Voltage Temperature
Coefficient
Time Constant
Time Stability
3
Appendix D
3
www.national.com
Electrical Characteristics (Note 1)
(Continued)
Note 1: Accuracy measurements are made in a well-stirred oil bath. For other conditions, self heating must be considered.
Note 2: Continuous operation at these temperatures for 10,000 hours for H package and 5,000 hours for Z package may decrease life expectancy of the device.
Note 3:
Thermal Resistance
TO-92
θJA (junction to ambient)
202˚C/W
400˚C/W
TO-46
θJC (junction to case)
170˚C/W
N/A
SO-8
165˚C/W
N/A
Note 4: Refer to RETS135H for military specifications.
Typical Performance Characteristics
Reverse Voltage Change
Calibrated Error
Reverse Characteristics
DS005698-27
Response Time
DS005698-28
Dynamic Impedance
DS005698-30
Thermal Resistance
Junction to Air
Noise Voltage
DS005698-31
DS005698-34
www.national.com
4
Appendix D
4
DS005698-32
Thermal Response in Still Air
Thermal Time Constant
DS005698-33
DS005698-29
DS005698-35
Typical Performance Characteristics
(Continued)
Thermal Response in Stirred Oil Bath
Forward Characteristics
DS005698-36
DS005698-37
To insure good sensing accuracy several precautions must
be taken. Like any temperature sensing device, self heating
can reduce accuracy. The LM135 should be operated at the
lowest current suitable for the application. Sufficient current,
of course, must be available to drive both the sensor and the
calibration pot at the maximum operating temperature as
well as any external loads.
If the sensor is used in an ambient where the thermal resistance is constant, self heating errors can be calibrated out.
This is possible if the device is run with a temperature stable
current. Heating will then be proportional to zener voltage
and therefore temperature. This makes the self heating error
proportional to absolute temperature the same as scale factor errors.
Application Hints
CALIBRATING THE LM135
Included on the LM135 chip is an easy method of calibrating
the device for higher accuracies. A pot connected across the
LM135 with the arm tied to the adjustment terminal allows a
1-point calibration of the sensor that corrects for inaccuracy
over the full temperature range.
This single point calibration works because the output of the
LM135 is proportional to absolute temperature with the extrapolated output of sensor going to 0V output at 0˚K
(−273.15˚C). Errors in output voltage versus temperature are
only slope (or scale factor) errors so a slope calibration at
one temperature corrects at all temperatures.
The output of the device (calibrated or uncalibrated) can be
expressed as:
WATERPROOFING SENSORS
Meltable inner core heat shrinkable tubing such as manufactured by Raychem can be used to make low-cost waterproof
sensors. The LM335 is inserted into the tubing about 1⁄2"
from the end and the tubing heated above the melting point
of the core. The unfilled 1⁄2" end melts and provides a seal
over the device.
where T is the unknown temperature and To is a reference
temperature, both expressed in degrees Kelvin. By calibrating the output to read correctly at one temperature the output
at all temperatures is correct. Nominally the output is calibrated at 10 mV/˚K.
Typical Applications
Basic Temperature Sensor
Calibrated Sensor
Wide Operating Supply
DS005698-2
DS005698-9
*Calibrate for 2.982V at 25˚C
DS005698-10
5
Appendix D
5
www.national.com
Typical Applications
(Continued)
Minimum Temperature Sensing
Average Temperature Sensing
Remote Temperature Sensing
DS005698-19
DS005698-4
Wire length for 1˚C error due to wire drop
IR = 1
mA
IR = 0.5 mA*
AWG
FEET
FEET
14
4000
8000
16
2500
5000
18
1600
3200
20
1000
2000
22
625
1250
24
400
800
DS005698-18
*For IR = 0.5 mA, the trim pot must be deleted.
Isolated Temperature Sensor
DS005698-20
www.national.com
6
Appendix D
6
Data Sheet
27501B*
3503
RATIOMETRIC, LINEAR
HALL-EFFECT SENSORS
The UGN3503LT, UGN3503U, and UGN3503UA Hall-effect
sensors accurately track extremely small changes in magnetic flux
density—changes generally too small to operate Hall-effect switches.
X
As motion detectors, gear tooth sensors, and proximity detectors,
they are magnetically driven mirrors of mechanical events. As sensitive
monitors of electromagnets, they can effectively measure a system's
performance with negligible system loading while providing isolation
from contaminated and electrically noisy environments.
Each Hall-effect integrated circuit includes a Hall sensing element,
linear amplifier, and emitter-follower output stage. Problems associated
with handling tiny analog signals are minimized by having the Hall cell
and amplifier on a single chip.
2
3
OUTPUT
1
GROUND
CC
SUPPLY
V
Dwg. PH-006
Pinning is shown viewed from branded side.
Three package styles provide a magnetically optimized package for
most applications. Package suffix ‘LT’ is a miniature SOT-89/TO243AA transistor package for surface-mount applications; suffix ‘U’ is a
miniature three-lead plastic SIP, while ‘UA’ is a three-lead ultra-miniSIP. All devices are rated for continuous operation over the temperature
range of -20°C to +85°C.
FEATURES
■
■
■
■
■
Extremely Sensitive
Flat Response to 23 kHz
Low-Noise Output
4.5 V to 6 V Operation
Magnetically Optimized Package
ABSOLUTE MAXIMUM RATINGS
Supply Voltage, VCC ............................. 8 V
Magnetic Flux Density, B ......... Unlimited
Operating Temperature Range,
TA ............................ -20°C to +85°C
Storage Temperature Range,
TS ........................... -65°C to +150°C
Always order by complete part number, e.g., UGN3503UA .
Appendix D
7
3503
RATIOMETRIC,
LINEAR
HALL-EFFECT SENSORS
FUNCTIONAL BLOCK DIAGRAM
VCC
1
REG.
X
3
OUTPUT
2
GROUND
Dwg. FH-007
ELECTRICAL CHARACTERISTICS at TA = +25°C, VCC = 5 V
Limits
Characteristic
Symbol
Test Conditions
Min.
Typ.
Max.
Units
Operating Voltage
VCC
4.5
—
6.0
V
Supply Current
ICC
—
9.0
13
mA
Quiescent Output Voltage
VOUT
B=0G
2.25
2.50
2.75
V
Sensitivity
∆VOUT
B = 0 G to ±900 G
0.75
1.30
1.75
mV/G
—
23
—
kHz
—
90
—
µV
—
50
220
Ω
Bandwidth (-3 dB)
BW
Broadband Output Noise
Vout
Output Resistance
ROUT
BW = 10 Hz to 10 kHz
All output-voltage measurements are made with a voltmeter having an input impedance of at least 10 kΩ.
Magnetic flux density is measured at most sensitive area of device located 0.016" (0.41 mm) below the branded face of the ‘U’
package; 0.020" (0.51 mm) below the branded face of the ‘UA’ package; and 0.030" (0.76 mm) below the branded face of the ‘LT’
package.
Appendix D
115 Northeast Cutoff,
8 Box 15036
Worcester, Massachusetts 01615-0036 (508) 853-5000
Copyright © 1985, 1999, Allegro MicroSystems, Inc.
3503
RATIOMETRIC,
LINEAR
HALL-EFFECT SENSORS
OUTPUT VOLTAGE AS A
FUNCTION OF TEMPERATURE
OUTPUT NOISE AS A
FUNCTION OF FREQUENCY
4.0
10
OUTPUT NOISE IN µ Vrms √HZ
VCC = 5V
3.5
OUTPUT IN VOLTS
B= +500G
3.0
B = 0G
2.5
2.0
8.0
VCC = +5V
TA = +25°C
6.0
4.0
2.0
B= -500G
0
1.5
40
20
0
+25
+85
10
+125
AMBIENT TEMPERATURE IN C
SUPPLY CURRENT AS A
FUNCTION OF SUPPLY VOLTAGE
1K
10K
Dwg. A-12,505
DEVICE SENSITIVITY AS A
FUNCTION OF SUPPLY VOLTAGE
12
2.5
TYPICAL SENSITIVITY IN V/G
B = 0G
11
TA = +25°C
SUPPLY CURRENT IN mA
100
FREQUENCY IN HZ
Dwg. A-12,573
10
9.0
TA = +25°C
2.0
1.5
1.0
0.5
8.0
7.0
0
4.5
5.0
5.5
SUPPLY VOLTAGE IN VOLTS
6.0
4.5
5.0
5.5
SUPPLY VOLTAGE IN VOLTS
6.0
Dwg. A-12,506
Dwg. A-12,507
OUTPUT NULL VOLTAGE AS A
FUNCTION OF SUPPLY VOLTAGE
LINEARITY AND SYMMETRY AS A
FUNCTION OF SUPPLY VOLTAGE
100
LINEARITY AND SYMMETRY IN PERCENT
5.0
OUTPUT NULL VOLTAGE IN VOLTS
B = 0G
TA = +25°C
4.0
3.0
2.0
1.0
0
OUTPUT SYMMETRY
99
98
W
ITY BELO
LINEAR
G
B = -900
NULL
VE NU
ITY ABO
LINEAR
G
B = +900
LL
97
TA = +25°C
96
95
4.5
5.0
5.5
SUPPLY VOLTAGE IN VOLTS
6.0
4.5
Dwg. A-12,508
Appendix D
9
5.0
5.5
SUPPLY VOLTAGE IN VOLTS
6.0
Dwg.A-12,509
3503
RATIOMETRIC,
LINEAR
HALL-EFFECT SENSORS
OPERATION
NOTCH SENSOR
The output null voltage (B = 0 G) is nominally one-half the supply
voltage. A south magnetic pole, presented to the branded face of the Halleffect sensor will drive the output higher than the null voltage level. A north
magnetic pole will drive the output below the null level.
In operation, instantaneous and proportional output-voltage levels are
dependent on magnetic flux density at the most sensitive area of the device.
Greatest sensitivity is obtained with a supply voltage of 6 V, but at the cost of
increased supply current and a slight loss of output symmetry. The sensor's
output is usually capacitively coupled to an amplifier that boosts the output
above the millivolt level.
In two applications shown, a permanent bias magnet is attached with
epoxy glue to the back of the epoxy package. The presence of ferrous material
at the face of the package acts as a flux concentrator.
Dwg. A-12,574
GEAR TOOTH SENSOR
The south pole of a magnet is attached to the back of the package if the
Hall-effect IC is to sense the presence of ferrous material. The north pole of a
magnet is attached to the back surface if the integrated circuit is to sense the
absence of ferrous matrial.
Calibrated linear Hall devices, which can be used to determine the actual
flux density presented to the sensor in a particular application, are available.
SENSOR LOCATIONS
N
SUFFIX “U”
ACTIVE AREA DEPTH
0.015"
0.38 mm
0.093"
2.37 mm
NOM
3 2
0.072"
1.83 mm
1
Dwg. A-12,512
A
CURRENT MONITOR
BRANDED
SURFACE
1
2
3
Dwg. MH-002-5C
SUFFIX “LT”
ACTIVE AREA DEPTH
0.030"
0.76 mm
NOM
SUFFIX “UA”
ACTIVE AREA DEPTH
0.018"
0.46 mm
NOM
0.091"
2.32 mm
0.084"
2.13 mm
0.042"
1.08 mm
0.056"
1.43 mm
A
A
1
2
BRANDED
SURFACE
3
1
2
3
Dwg. A-12,513
Dwg. MH-008-9
Appendix D
115 Northeast Cutoff,
10 Box 15036
Worcester, Massachusetts 01615-0036 (508) 853-5000
Dwg. MH-011-3C
Appendix D
11
Appendix D
12
LM741
Operational Amplifier
General Description
The LM741 series are general purpose operational amplifiers which feature improved performance over industry standards like the LM709. They are direct, plug-in replacements
for the 709C, LM201, MC1439 and 748 in most applications.
The amplifiers offer many features which make their application nearly foolproof: overload protection on the input and
output, no latch-up when the common mode range is exceeded, as well as freedom from oscillations.
The LM741C is identical to the LM741/LM741A except that
the LM741C has their performance guaranteed over a 0˚C to
+70˚C temperature range, instead of −55˚C to +125˚C.
Features
Connection Diagrams
Metal Can Package
Dual-In-Line or S.O. Package
00934103
00934102
Note 1: LM741H is available per JM38510/10101
Order Number LM741H, LM741H/883 (Note 1),
LM741AH/883 or LM741CH
See NS Package Number H08C
Order Number LM741J, LM741J/883, LM741CN
See NS Package Number J08A, M08A or N08E
Ceramic Flatpak
00934106
Order Number LM741W/883
See NS Package Number W10A
Typical Application
Offset Nulling Circuit
00934107
© 2004 National Semiconductor Corporation
DS009341
Appendix D
13
www.national.com
LM741 Operational Amplifier
August 2000
LM741
Absolute Maximum Ratings (Note 2)
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales Office/
Distributors for availability and specifications.
(Note 7)
LM741A
LM741
± 22V
± 22V
± 18V
500 mW
500 mW
500 mW
± 30V
± 15V
± 30V
± 15V
± 30V
± 15V
Output Short Circuit Duration
Continuous
Continuous
Continuous
Operating Temperature Range
−55˚C to +125˚C
−55˚C to +125˚C
0˚C to +70˚C
Storage Temperature Range
−65˚C to +150˚C
−65˚C to +150˚C
−65˚C to +150˚C
150˚C
150˚C
100˚C
N-Package (10 seconds)
260˚C
260˚C
260˚C
J- or H-Package (10 seconds)
300˚C
300˚C
300˚C
Vapor Phase (60 seconds)
215˚C
215˚C
215˚C
Infrared (15 seconds)
215˚C
215˚C
215˚C
Supply Voltage
Power Dissipation (Note 3)
Differential Input Voltage
Input Voltage (Note 4)
Junction Temperature
LM741C
Soldering Information
M-Package
See AN-450 “Surface Mounting Methods and Their Effect on Product Reliability” for other methods of
soldering
surface mount devices.
ESD Tolerance (Note 8)
400V
400V
400V
Electrical Characteristics (Note 5)
Parameter
Conditions
LM741A
Min
Input Offset Voltage
LM741
Min
LM741C
Typ
Max
1.0
5.0
Min
Units
Typ
Max
Typ
Max
0.8
3.0
2.0
6.0
mV
4.0
mV
TA = 25˚C
RS ≤ 10 kΩ
RS ≤ 50Ω
mV
TAMIN ≤ TA ≤ TAMAX
RS ≤ 50Ω
RS ≤ 10 kΩ
6.0
Average Input Offset
7.5
15
mV
µV/˚C
Voltage Drift
Input Offset Voltage
TA = 25˚C, VS = ± 20V
± 10
± 15
± 15
mV
Adjustment Range
Input Offset Current
TA = 25˚C
3.0
TAMIN ≤ TA ≤ TAMAX
Average Input Offset
30
20
200
70
85
500
20
200
nA
300
nA
0.5
nA/˚C
Current Drift
Input Bias Current
TA = 25˚C
30
TAMIN ≤ TA ≤ TAMAX
Input Resistance
80
80
0.210
TA = 25˚C, VS = ± 20V
1.0
TAMIN ≤ TA ≤ TAMAX,
0.5
6.0
500
80
1.5
0.3
2.0
500
0.8
0.3
2.0
nA
µA
MΩ
MΩ
VS = ± 20V
Input Voltage Range
± 12
TA = 25˚C
TAMIN ≤ TA ≤ TAMAX
www.national.com
± 12
Appendix D
2
14
± 13
± 13
V
V
Parameter
(Continued)
Conditions
LM741A
Min
Large Signal Voltage Gain
Typ
LM741
Max
Min
Typ
50
200
LM741C
Max
Min
Typ
20
200
Units
Max
TA = 25˚C, RL ≥ 2 kΩ
VS = ± 20V, VO = ± 15V
50
V/mV
VS = ± 15V, VO = ± 10V
V/mV
TAMIN ≤ TA ≤ TAMAX,
RL ≥ 2 kΩ,
VS = ± 20V, VO = ± 15V
32
V/mV
VS = ± 15V, VO = ± 10V
VS = ± 5V, VO = ± 2V
Output Voltage Swing
25
15
V/mV
10
V/mV
± 16
± 15
V
VS = ± 20V
RL ≥ 10 kΩ
RL ≥ 2 kΩ
V
VS = ± 15V
RL ≥ 10 kΩ
± 12
± 10
RL ≥ 2 kΩ
Output Short Circuit
TA = 25˚C
10
Current
TAMIN ≤ TA ≤ TAMAX
10
Common-Mode
TAMIN ≤ TA ≤ TAMAX
Rejection Ratio
25
35
Supply Voltage Rejection
TAMIN ≤ TA ≤ TAMAX,
Ratio
VS = ± 20V to VS = ± 5V
RS ≤ 50Ω
25
± 14
± 13
V
25
mA
95
86
96
90
70
90
dB
77
96
77
96
dB
µs
TA = 25˚C, Unity Gain
0.25
0.8
0.3
0.3
Overshoot
6.0
20
5
5
Bandwidth (Note 6)
TA = 25˚C
Slew Rate
TA = 25˚C, Unity Gain
Supply Current
TA = 25˚C
Power Consumption
TA = 25˚C
VS = ± 20V
0.437
1.5
0.3
0.7
80
%
MHz
0.5
0.5
V/µs
1.7
2.8
1.7
2.8
50
85
50
85
150
VS = ± 15V
LM741
dB
dB
Rise Time
LM741A
V
mA
70
80
RS ≤ 10 kΩ
Transient Response
± 12
± 10
40
RS ≤ 10 kΩ, VCM = ± 12V
RS ≤ 50Ω, VCM = ± 12V
± 14
± 13
mA
mW
mW
VS = ± 20V
TA = TAMIN
165
mW
TA = TAMAX
135
mW
VS = ± 15V
TA = TAMIN
60
100
mW
TA = TAMAX
45
75
mW
Note 2: “Absolute Maximum Ratings” indicate limits beyond which damage to the device may occur. Operating Ratings indicate conditions for which the device is
functional, but do not guarantee specific performance limits.
Appendix
D
3
15
www.national.com
LM741
Electrical Characteristics (Note 5)
LM741
Electrical Characteristics (Note 5)
(Continued)
Note 3: For operation at elevated temperatures, these devices must be derated based on thermal resistance, and Tj max. (listed under “Absolute Maximum
Ratings”). Tj = TA + (θjA PD).
Thermal Resistance
θjA (Junction to Ambient)
θjC (Junction to Case)
Cerdip (J)
DIP (N)
HO8 (H)
SO-8 (M)
100˚C/W
100˚C/W
170˚C/W
195˚C/W
N/A
N/A
25˚C/W
N/A
Note 4: For supply voltages less than ± 15V, the absolute maximum input voltage is equal to the supply voltage.
Note 5: Unless otherwise specified, these specifications apply for VS = ± 15V, −55˚C ≤ TA ≤ +125˚C (LM741/LM741A). For the LM741C/LM741E, these
specifications are limited to 0˚C ≤ TA ≤ +70˚C.
Note 6: Calculated value from: BW (MHz) = 0.35/Rise Time(µs).
Note 7: For military specifications see RETS741X for LM741 and RETS741AX for LM741A.
Note 8: Human body model, 1.5 kΩ in series with 100 pF.
Schematic Diagram
00934101
www.national.com
Appendix D
4
16
LF411 Low Offset, Low Drift
JFET Input Operational Amplifier
General Description
Features
These devices are low cost, high speed, JFET input operational amplifiers with very low input offset voltage and guaranteed input offset voltage drift. They require low supply
current yet maintain a large gain bandwidth product and fast
slew rate. In addition, well matched high voltage JFET input
devices provide very low input bias and offset currents. The
LF411 is pin compatible with the standard LM741 allowing
designers to immediately upgrade the overall performance
of existing designs.
These amplifiers may be used in applications such as high
speed integrators, fast D/A converters, sample and hold
circuits and many other circuits requiring low input offset
voltage and drift, low input bias current, high input impedance, high slew rate and wide bandwidth.
Y
Typical Connection
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Internally trimmed offset voltage
0.5 mV(max)
Input offset voltage drift
10 mV/§ C(max)
Low input bias current
50 pA
Low input noise current
0.01 pA/0Hz
Wide gain bandwidth
3 MHz(min)
High slew rate
10V/ms(min)
Low supply current
1.8 mA
High input impedance
1012X
k 0.02%
Low total harmonic distortion AV e 10,
RL e 10k, VO e 20 Vp-p, BW e 20 Hzb20 kHz
Low 1/f noise corner
50 Hz
Fast settling time to 0.01%
2 ms
Ordering Information
X
Y
Z
LF411XYZ
indicates electrical grade
indicates temperature range
‘‘M’’ for military
‘‘C’’ for commercial
indicates package type
‘‘H’’ or ‘‘N’’
Connection Diagrams
Metal Can Package
TL/H/5655 – 5
Top View
Note: Pin 4 connected to case.
Order Number LF411ACH
or LF411MH/883*
See NS Package Number H08A
TL/H/5655–1
Simplified Schematic
Dual-In-Line Package
TL/H/5655 – 7
TL/H/5655 – 6
BI-FET IITM is a trademark of National Semiconductor Corporation.
C1995 National Semiconductor Corporation
Top View
Order Number LF411ACN,
LF411CN or LF411MJ/883*
See NS Package Number
N08E or J08A
*Available per JM38510/11904
RRD-B30M115/Printed in U. S. A.
TL/H/5655
Appendix D
17
LF411 Low Offset, Low Drift JFET Input Operational Amplifier
February 1995
Absolute Maximum Ratings
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales
Office/Distributors for availability and specifications.
(Note 8)
LF411A
LF411
g 22V
g 18V
Supply Voltage
g 38V
g 30V
Differential Input Voltage
Input Voltage Range
g 19V
g 15V
(Note 1)
Output Short Circuit
Duration
Continuous
Continuous
H Package
N Package
670 mW
670 mW
150§ C
162§ C/W (Still Air)
65§ C/W (400 LF/min
Air Flow)
120§ C/W
Power Dissipation
(Notes 2 and 9)
Tjmax
ijA
ijC
Operating Temp.
Range
Storage Temp.
Range
115§ C
20§ C/W
(Note 3)
(Note 3)
b 65§ C s TA s 150§ C
b 65§ C s TA s 150§ C
Lead Temp.
(Soldering, 10 sec.)
ESD Tolerance
260§ C
260§ C
Rating to be determined.
DC Electrical Characteristics (Note 4)
Symbol
Parameter
LF411A
Conditions
Min
VOS
Input Offset Voltage
RS e 10 kX, TA e 25§ C
DVOS/DT
Average TC of Input
Offset Voltage
RS e 10 kX (Note 5)
IOS
Input Offset Current
VS e g 15V
(Notes 4, 6)
IB
Input Bias Current
Typ
Max
0.3
0.5
Tj e 25§ C
VS e g 15V
(Notes 4, 6)
LF411
Min
Tj e 25§ C
VS e g 15V, VO e g 10V,
RL e 2k, TA e 25§ C
VO
Output Voltage Swing
VCM
Input Common-Mode
Voltage Range
25
100
25
100
pA
2
nA
Tj e 125§ C
25
25
nA
50
200
50
PSRR
Supply Voltage
Rejection Ratio
(Note 7)
IS
Supply Current
pA
4
nA
50
nA
50
50
200
25
1012
X
200
V/mV
25
200
15
200
V/mV
g 12
g 13.5
g 12
g 13.5
V
g 16
a 19.5
g 11
a 14.5
V
b 11.5
V
b 16.5
RSs10k
200
4
VS e g 15V, RL e 10k
Common-Mode
Rejection Ratio
mV/§ C
7
2
Over Temperature
CMRR
mV
10
1012
Large Signal Voltage
Gain
2.0
7
Tj e 125§ C
Input Resistance
0.8
Tj e 70§ C
Tj e 25§ C
AVOL
Max
20
(Note 5)
Tj e 70§ C
RIN
Units
Typ
80
100
70
100
dB
80
100
70
100
dB
1.8
2.8
1.8
3.4
mA
AC Electrical Characteristics (Note 4)
Symbol
Parameter
LF411A
Conditions
Min
Typ
LF411
Max
Min
Typ
Units
Max
SR
Slew Rate
VS e g 15V, TA e 25§ C
10
15
8
15
V/ms
GBW
Gain-Bandwidth Product
VS e g 15V, TA e 25§ C
3
4
2.7
4
MHz
en
Equivalent Input Noise Voltage
TA e 25§ C, RS e 100X,
f e 1 kHz
25
25
nV/ S0Hz
in
Equivalent Input Noise Current
TA e 25§ C, f e 1 kHz
0.01
0.01
pA/ S0Hz
2
Appendix D
18
Note 1: Unless otherwise specified the absolute maximum negative input voltage is equal to the negative power supply voltage.
Note 2: For operating at elevated temperature, these devices must be derated based on a thermal resistance of ijA.
Note 3: These devices are available in both the commercial temperature range 0§ C s TA s 70§ C and the military temperature range b 55§ C s TA s 125§ C. The
temperature range is designated by the position just before the package type in the device number. A ‘‘C’’ indicates the commercial temperature range and an ‘‘M’’
indicates the military temperature range. The military temperature range is available in ‘‘H’’ package only.
Note 4: Unless otherwise specified, the specifications apply over the full temperature range and for VS e g 20V for the LF411A and for VS e g 15V for the LF411.
VOS, IB, and IOS are measured at VCM e 0.
Note 5: The LF411A is 100% tested to this specification. The LF411 is sample tested to insure at least 90% of the units meet this specification.
Note 6: The input bias currents are junction leakage currents which approximately double for every 10§ C increase in the junction temperature, Tj. Due to limited
production test time, the input bias currents measured are correlated to junction temperature. In normal operation the junction temperature rises above the ambient
temperature as a result of internal power dissipation, PD. Tj e TA a ijA PD where ijA is the thermal resistance from junction to ambient. Use of a heat sink is
recommended if input bias current is to be kept to a minimum.
Note 7: Supply voltage rejection ratio is measured for both supply magnitudes increasing or decreasing simultaneously in accordance with common practice, from
g 15V to g 5V for the LF411 and from g 20V to g 5V for the LF411A.
Note 8: RETS 411X for LF411MH and LF411MJ military specifications.
Note 9: Max. Power Dissipation is defined by the package characteristics. Operating the part near the Max. Power Dissipation may cause the part to operate
outside guaranteed limits.
Typical Performance Characteristics
Input Bias Current
Input Bias Current
Supply Current
Positive Common-Mode
Input Voltage Limit
Negative Common-Mode
Input Voltage Limit
Positive Current Limit
Negative Current Limit
Output Voltage Swing
Output Voltage Swing
TL/H/5655 – 2
3
Appendix D
19
Typical Performance Characteristics
(Continued)
Gain Bandwidth
Bode Plot
Slew Rate
Distortion vs Frequency
Undistorted Output
Voltage Swing
Open Loop Frequency
Response
Common-Mode Rejection
Ratio
Power Supply
Rejection Ratio
Equivalent Input Noise
Voltage
Open Loop Voltage Gain
Output Impedance
Inverter Settling Time
TL/H/5655 – 3
4
Appendix D
20
Ultraprecision
Operational Amplifier
OP177
PIN CONFIGURATION
Ultralow offset voltage
TA = 25°C, 25 μV maximum
Outstanding offset voltage drift 0.1 μV/°C maximum
Excellent open-loop gain and gain linearity
12 V/μV typical
CMRR: 130 dB minimum
PSRR: 115 dB minimum
Low supply current 2.0 mA maximum
Fits industry-standard precision op amp sockets
VOS TRIM 1
–IN
+IN 3
V–
OP177
2
4
8
VOS TRIM
7
V+
6 OUT
TOP VIEW
5 NC
(Not to Scale)
NC = NO CONNECT
00289-001
FEATURES
Figure 1. 8-Lead PDIP (P-Suffix),
8-Lead SOIC (S-Suffix)
GENERAL DESCRIPTION
operational amplifier. The combination of outstanding
specifications of the OP177 ensures accurate performance in
high closed-loop gain applications.
The OP177 features one of the highest precision performance of
any op amp currently available. Offset voltage of the OP177 is
only 25 μV maximum at room temperature. The ultralow VOS of
the OP177 combines with its exceptional offset voltage drift
(TCVOS) of 0.1 μV/°C maximum to eliminate the need for
external VOS adjustment and increases system accuracy over
temperature.
This low noise, bipolar input op amp is also a cost effective
alternative to chopper-stabilized amplifiers. The OP177
provides chopper-type performance without the usual problems
of high noise, low frequency chopper spikes, large physical size,
limited common-mode input voltage range, and bulky external
storage capacitors.
The OP177 open-loop gain of 12 V/μV is maintained over the
full ±10 V output range. CMRR of 130 dB minimum, PSRR of
120 dB minimum, and maximum supply current of 2 mA are
just a few examples of the excellent performance of this
The OP177 is offered in the −40°C to +85°C extended industrial
temperature ranges. This product is available in 8-lead PDIP, as
well as the space saving 8-lead SOIC.
FUNCTIONAL BLOCK DIAGRAM
V+
R2A*
(OPTIONAL NULL)
R2B*
C1
R7
R1B
R1A
Q19
2B
Q10
Q9
Q7
NONINVERTING
INPUT
INVERTING
INPUT
R3
Q3
Q5
Q11
Q8
Q6
Q4
Q1
R4
OUTPUT
Q27
Q21
Q23
Q22
Q24
R9
Q12
Q26
C3
C2
Q17
R10
Q16
R5
Q20
Q25
Q15
Q2
Q18
Q14
*R2A AND R2B ARE ELECTRONICALLY ADJUSTED ON CHIP AT FACTORY.
R6
R8
00289-002
Q13
V–
Figure 2. Simplified Schematic
Rev. E
Information furnished by Analog Devices is believed to be accurate and reliable. However, no
responsibility is assumed by Analog Devices for its use, nor for any infringements of patents or other
rights of third parties that may result from its use. Specifications subject to change without notice. No
license is granted by implication or otherwise under any patent or patent rights of Analog Devices.
Trademarks and registered trademarks are the property of their respective owners.
One Technology Way, P.O. Box 9106, Norwood, MA 02062-9106, U.S.A.
Appendix DTel: 781.329.4700
21
Fax: 781.461.3113
www.analog.com
©2006 Analog Devices, Inc. All rights reserved.
OP177
SPECIFICATIONS
ELECTRICAL CHARACTERISTICS
@ VS = ±15 V, TA = 25°C, unless otherwise noted.
Table 1.
Parameter
INPUT OFFSET VOLTAGE
LONG-TERM INPUT OFFSET 1
Voltage Stability
INPUT OFFSET CURRENT
INPUT BIAS CURRENT
INPUT NOISE VOLTAGE
INPUT NOISE CURRENT
INPUT RESISTANCE
Differential Mode 3
INPUT RESISTANCE COMMON MODE
INPUT VOLTAGE RANGE 4
COMMON-MODE REJECTION RATIO
POWER SUPPLY REJECTION RATIO
LARGE SIGNAL VOLTAGE GAIN
OUTPUT VOLTAGE SWING
Symbol
VOS
SLEW RATE2
CLOSED-LOOP BANDWIDTH2
OPEN-LOOP OUTPUT RESISTANCE
POWER CONSUMPTION
SR
BW
RO
PD
SUPPLY CURRENT
OFFSET ADJUSTMENT RANGE
ISY
Conditions
Min
OP177F
Typ
10
Max
25
−0.2
0.3
0.3
+1.2
118
3
1.5
+2
150
8
Min
OP177G
Typ
Max
20
60
Unit
μV
T
ΔVOS/time
IOS
IB
en
in
RIN
RINCM
IVR
CMRR
PSRR
AVO
VO
2
fO = 1 Hz to 100 Hz
fO = 1 Hz to 100 Hz2
26
VCM = ±13 V
VS = ±3 V to ±18 V
RL ≥ 2 kΩ, VO = ±10 V 5
RL ≥ 10 kΩ
RL ≥ 2 kΩ
RL ≥ 1 kΩ
RL ≥ 2 kΩ
AVCL = 1
VS = ±15 V, no load
VS = ±3 V, no load
VS = ±15 V, no load
RP = 20 kΩ
1
±13
130
115
5000
±13.5
±12.5
±12.0
0.1
0.4
45
200
±14
140
125
12,000
±14.0
±13.0
±12.5
0.3
0.6
60
50
3.5
1.6
±3
−0.2
18.5
±13
115
110
2000
±13.5
±12.5
±12.0
0.1
0.4
60
4.5
2
0.4
0.3
+1.2
118
3
45
200
±14
140
120
6000
±14.0
±13.0
±12.5
0.3
0.6
60
50
3.5
1.6
±3
2.8
+2.8
150
8
60
4.5
2
μV/mo
nA
nA
nV rms
pA rms
MΩ
GΩ
V
dB
dB
V/mV
V
V
V
V/μs
MHz
Ω
mW
mW
mA
mV
Long-term input offset voltage stability refers to the averaged trend line of VOS vs. time over extended periods after the first 30 days of operation. Excluding the initial
hour of operation, changes in VOS during the first 30 operating days are typically less than 2.0 μV.
2
Sample tested.
3
Guaranteed by design.
4
Guaranteed by CMRR test condition.
5
To ensure high open-loop gain throughout the ±10 V output range, AVO is tested at −10 V ≤ VO ≤ 0 V, 0 V ≤ VO ≤ +10 V, and –10 V ≤ VO ≤ +10 V.
Appendix D
Rev. E22
| Page 3 of 16
OP177
@ VS = ±15 V, −40°C ≤ TA ≤ +85°C, unless otherwise noted.
Table 2.
Parameter
INPUT
Input Offset Voltage
Average Input Offset Voltage Drift 1
Input Offset Current
Average Input Offset Current Drift 2
Input Bias Current
Average Input Bias Current Drift2
Input Voltage Range 3
COMMON-MODE REJECTION RATIO
POWER SUPPLY REJECTION RATIO
LARGE-SIGNAL VOLTAGE GAIN 4
OUTPUT VOLTAGE SWING
POWER CONSUMPTION
SUPPLY CURRENT
Symbol
Conditions
VOS
TCVOS
IOS
TCIOS
IB
TCIB
IVR
CMRR
PSRR
AVO
VO
PD
ISY
OP177F
Typ
Min
15
0.1
0.5
1.5
+2.4
8
±13.5
140
120
6000
±13
60
20
−0.2
VCM = ±13 V
VS = ±3 V to ±18 V
RL ≥ 2 kΩ, VO = ±10 V
RL ≥ 2 kΩ
VS = ±15 V, no load
VS = ±15 V, no load
±13
120
110
2000
±12
1
Max
Min
40
0.3
2.2
40
+4
40
±13
110
106
1000
±12
75
2.5
OP177G
Typ
20
0.7
0.5
1.5
+2.4
15
±13.5
140
115
4000
±13
60
2
TCVOS is sample tested.
Guaranteed by endpoint limits.
3
Guaranteed by CMRR test condition.
4
To ensure high open-loop gain throughout the ±10 V output range, AVO is tested at −10 V ≤ VO ≤ 0 V, 0 V ≤ VO ≤ +10 V, and −10 V ≤ VO ≤ +10 V.
2
TEST CIRCUITS
200kΩ
50Ω
–
OP177
VOS =
VO
00289-003
+
VO
4000
Figure 3. Typical Offset Voltage Test Circuit
20kΩ
V+
–
–
INPUT
OUTPUT
OP177
VOS TRIM RANGE IS
TYPICALLY ±3.0mV
V–
Figure 4. Optional Offset Nulling Circuit
20kΩ
+20V
–
OP177
+
–20V
Figure 5. Burn-In Circuit
Appendix D
Rev. E | Page
234 of 16
00289-005
PINOUTS SHOWN FOR
P AND Z PACKAGES
00289-004
+
+
Max
Unit
100
1.2
4.5
85
±6
60
μV
μV/°C
nA
pA/°C
nA
pA/°C
V
dB
dB
V/mV
V
mW
mA
75
2.5
OP177
ABSOLUTE MAXIMUM RATINGS
Table 3.
Parameter
Supply Voltage
Internal Power Dissipation1
Differential Input Voltage
Input Voltage
Output Short-Circuit Duration
Storage Temperature Range
Operating Temperature Range
Lead Temperature (Soldering, 60 sec)
DICE Junction Temperature (TJ)
1
Ratings
±22 V
500 mW
±30 V
±22 V
Indefinite
−65°C to +125°C
−40°C to +85°C
300°C
−65°C to +150°C
For supply voltages less than ±22 V, the absolute maximum input voltage is
equal to the supply voltage.
Stresses above those listed under Absolute Maximum Ratings
may cause permanent damage to the device. This is a stress
rating only; functional operation of the device at these or any
other conditions above those indicated in the operational
section of this specification is not implied. Exposure to absolute
maximum rating conditions for extended periods may affect
device reliability.
THERMAL RESISTANCE
θJA is specified for worst-case mounting conditions, that is, θJA
is specified for device in socket for PDIP; θJA is specified for
device soldered to printed circuit board for SOIC package.
Table 4. Thermal Resistance
Package Type
8-Lead PDIP (P-Suffix)
8-Lead SOIC (S-Suffix)
ESD CAUTION
ESD (electrostatic discharge) sensitive device. Electrostatic charges as high as 4000 V readily accumulate on
the human body and test equipment and can discharge without detection. Although this product features
proprietary ESD protection circuitry, permanent damage may occur on devices subjected to high energy
electrostatic discharges. Therefore, proper ESD precautions are recommended to avoid performance
degradation or loss of functionality.
Appendix D
Rev. E24
| Page 5 of 16
θJA
103
158
θJC
43
43
Unit
°C/W
°C/W
OP177
TYPICAL PERFORMANCE CHARACTERISTICS
20
TA = 25°C
VS = ±15V
RL = 10kΩ
VS = ±15V
ABSOLUTE CHANGE IN
INPUT OFFSET VOLTAGE (µV)
25
1
0
–1
DEVICE IMMERSED IN
70° OIL BATH (20 UNITS)
30
35
40
–2
–10
–5
0
OUTPUT VOLTAGE (V)
5
00289-009
45
00289-006
INPUT VOLTAGE (µV)
(NULLED TO 0mV @ VOUT = 0V)
2
50
10
0
50
60
70
25
100
VS = ±15V
TA = 25°C
OPEN-LOOP GAIN (V/µV)
20
10
15
10
1
0
10
20
30
TOTAL SUPPLY VOLTAGE, V+ TO V– (V)
0
–55
40
00289-010
5
00289-007
POWER CONSUMPTION (mW)
20
30
40
TIME (Seconds)
Figure 9. Offset Voltage Change Due to Thermal Shock
Figure 6. Gain Linearity (Input Voltage vs. Output Voltage)
–35
–15
5
25
45
65
TEMPERATURE (°C)
85
105
125
Figure 10. Open-Loop Gain vs. Temperature
Figure 7. Power Consumption vs. Power Supply
16
5
TA = 25°C
RL = 2kΩ
4
3
1
OPEN-LOOP GAIN (V/µV)
LOT A
LOT B
LOT C
LOT D
2
0
–1
–2
12
8
4
–4
–5
0
20
40
60
80
100
120
TIME (Seconds)
140
160
0
180
00289-011
–3
00289-008
VOS (µV)
10
0
±5
±10
±15
POWER SUPPLY VOLTAGE (V)
Figure 11. Open-Loop Gain vs. Power Supply Voltage
Figure 8. Warm-Up VOS Drift (Normalized) Z Package
Appendix D
Rev. E | Page
256 of 16
±20
OP177
4
160
VS = ±15V
TA = 25°C
VS = ±15V
OPEN-LOOP GAIN (dB)
3
2
1
–50
0
50
TEMPERATURE (°C)
100
80
60
40
0
0.01
100
00289-015
0
120
20
00289-012
INPUT BIAS CURRENT (nA)
140
0.1
1
10
100
1k
FREQUENCY (Hz)
10k
100k
1M
Figure 15. Open-Loop Frequency Response
Figure 12. Input Bias Current vs. Temperature
2.0
150
VS = ±15V
TA = 25°C
1.5
CMRR (dB)
130
1.0
110
90
–50
0
50
TEMPERATURE (°C)
80
100
00289-016
0
120
100
0.5
00289-013
1
100
1k
FREQUENCY (Hz)
10k
100k
Figure 16. CMRR vs. Frequency
Figure 13. Input Offset Current vs. Temperature
130
100
TA = 25°C
VS = ±15V
TA = 25°C
120
80
110
60
PSRR (dB)
CLOSED-LOOP GAIN (dB)
10
40
20
90
80
00289-014
0
–20
100
10
100
1k
10k
100k
FREQUENCY (Hz)
1M
10M
Figure 14. Closed-Loop Response for Various Gain Configurations
Appendix D
Rev. E26
| Page 7 of 16
70
60
0.1
00289-017
INPUT OFFSET CURRENT (nA)
140
1
10
100
FREQUENCY (Hz)
Figure 17. PSRR vs. Frequency
1k
10k
Low Cost Low Power
Instrumentation Amplifier
AD620
FEATURES
CONNECTION DIAGRAM
1
8
–IN
2
7 +VS
+IN
3
6
–VS
4
AD620
RG
OUTPUT
5 REF
TOP VIEW
Figure 1. 8-Lead PDIP (N), CERDIP (Q), and SOIC (R) Packages
PRODUCT DESCRIPTION
The AD620 is a low cost, high accuracy instrumentation
amplifier that requires only one external resistor to set gains of
1 to 10,000. Furthermore, the AD620 features 8-lead SOIC and
DIP packaging that is smaller than discrete designs and offers
lower power (only 1.3 mA max supply current), making it a
good fit for battery-powered, portable (or remote) applications.
The AD620, with its high accuracy of 40 ppm maximum
nonlinearity, low offset voltage of 50 µV max, and offset drift of
0.6 µV/°C max, is ideal for use in precision data acquisition
systems, such as weigh scales and transducer interfaces.
Furthermore, the low noise, low input bias current, and low power
of the AD620 make it well suited for medical applications, such
as ECG and noninvasive blood pressure monitors.
APPLICATIONS
Weigh scales
ECG and medical instrumentation
Transducer interface
Data acquisition systems
Industrial process controls
Battery-powered and portable equipment
The low input bias current of 1.0 nA max is made possible with
the use of Superϐeta processing in the input stage. The AD620
works well as a preamplifier due to its low input voltage noise of
9 nV/√Hz at 1 kHz, 0.28 µV p-p in the 0.1 Hz to 10 Hz band,
and 0.1 pA/√Hz input current noise. Also, the AD620 is well
suited for multiplexed applications with its settling time of 15 µs
to 0.01%, and its cost is low enough to enable designs with one
in-amp per channel.
30,000
10,000
3 OP AMP
IN-AMP
(3 OP-07s)
RTI VOLTAGE NOISE
(0.1 – 10Hz) (µV p-p)
1,000
20,000
15,000
AD620A
10,000
RG
TYPICAL STANDARD
BIPOLAR INPUT
IN-AMP
100
G = 100
10
AD620 SUPERβETA
BIPOLAR INPUT
IN-AMP
1
5,000
0
0
5
10
SUPPLY CURRENT (mA)
15
20
0.1
1k
Figure 2. Three Op Amp IA Designs vs. AD620
10k
100k
1M
SOURCE RESISTANCE (Ω)
10M
100M
00775-0-003
25,000
00775-0-002
TOTAL ERROR, PPM OF FULL SCALE
RG
00775-0-001
Easy to use
Gain set with one external resistor
(Gain range 1 to 10,000)
Wide power supply range (±2.3 V to ±18 V)
Higher performance than 3 op amp IA designs
Available in 8-lead DIP and SOIC packaging
Low power, 1.3 mA max supply current
Excellent dc performance (B grade)
50 µV max, input offset voltage
0.6 µV/°C max, input offset drift
1.0 nA max, input bias current
100 dB min common-mode rejection ratio (G = 10)
Low noise
9 nV/√Hz @ 1 kHz, input voltage noise
0.28 µV p-p noise (0.1 Hz to 10 Hz)
Excellent ac specifications
120 kHz bandwidth (G = 100)
15 µs settling time to 0.01%
Figure 3. Total Voltage Noise vs. Source Resistance
Rev. G
Information furnished by Analog Devices is believed to be accurate and reliable.
However, no responsibility is assumed by Analog Devices for its use, nor for any
infringements of patents or other rights of third parties that may result from its use.
Specifications subject to change without notice. No license is granted by implication
or otherwise under any patent or patent rights of Analog Devices. Trademarks and
registered trademarks are the property of their respective owners.
One Technology Way, P.O. Box 9106, Norwood, MA 02062-9106, U.S.A.
www.analog.com
© 2004 Analog Devices, Inc. All rights reserved.
Appendix DTel: 781.329.4700
Fax: 781.326.8703
27
AD620
SPECIFICATIONS
Typical @ 25°C, VS = ±15 V, and RL = 2 kΩ, unless otherwise noted.
Table 1.
AD620A
Parameter
GAIN
Gain Range
Gain Error2
G=1
G = 10
G = 100
G = 1000
Nonlinearity
G = 1–1000
G = 1–100
Gain vs. Temperature
VOLTAGE OFFSET
Input Offset, VOSI
Overtemperature
Average TC
Output Offset, VOSO
Overtemperature
Average TC
Offset Referred to the
Input vs. Supply (PSR)
G=1
G = 10
G = 100
G = 1000
INPUT CURRENT
Input Bias Current
Overtemperature
Average TC
Input Offset Current
Overtemperature
Average TC
INPUT
Input Impedance
Differential
Common-Mode
Input Voltage Range3
Overtemperature
Overtemperature
Conditions Min
G = 1 + (49.4 kΩ/RG)
1
VOUT = ±10 V
VOUT = −10 V to +10 V
RL = 10 kΩ
RL = 2 kΩ
Typ
AD620B
Max
Min
10,000
1
Typ
Max
Min
10,000
1
AD620S1
Typ Max
Unit
10,000
0.03
0.15
0.15
0.40
0.10
0.30
0.30
0.70
0.01
0.10
0.10
0.35
0.02
0.15
0.15
0.50
0.03
0.15
0.15
0.40
0.10
0.30
0.30
0.70
%
%
%
%
10
10
40
95
10
10
40
95
10
10
40
95
ppm
ppm
10
−50
ppm/°C
ppm/°C
125
µV
225
µV
G=1
Gain >12
(Total RTI Error = VOSI + VOSO/G)
VS = ±5 V
30
to ± 15 V
VS = ±5 V
to ± 15 V
VS = ±5 V
0.3
to ± 15 V
VS = ±15 V
400
VS = ± 5 V
VS = ±5 V
to ± 15 V
VS = ±5 V
5.0
to ± 15 V
10
−50
10
−50
125
15
185
50
30
85
1.0
0.1
0.6
0.3
1.0
µV/°C
1000
1500
2000
200
500
750
1000
400
1000
1500
2000
µV
µV
µV
15
2.5
7.0
5.0
15
µV/°C
VS = ±2.3 V
to ±18 V
80
95
110
110
100
120
140
140
0.5
3.0
0.3
VS = ±2.3 V −VS + 1.9
to ±5 V
−VS + 2.1
VS = ± 5 V
−VS + 1.9
to ±18 V
−VS + 2.1
80
100
120
120
2.0
2.5
100
120
140
140
0.5
3.0
0.3
1.0
1.5
80
95
110
110
1.0
1.5
100
120
140
140
0.5
8.0
0.3
0.5
0.75
1.5
1.5
8.0
10||2
10||2
10||2
10||2
10||2
10||2
dB
dB
dB
dB
2
4
1.0
2.0
nA
nA
pA/°C
nA
nA
pA/°C
+VS − 1.2
−VS + 1.9
+VS − 1.2
−VS + 1.9
+VS − 1.2
GΩ_pF
GΩ_pF
V
+VS − 1.3
+VS − 1.4
−VS + 2.1
−VS + 1.9
+VS − 1.3
+VS − 1.4
−VS + 2.1
−VS + 1.9
+VS − 1.3
+VS − 1.4
V
V
+VS − 1.4
−VS + 2.1
+VS + 2.1
−VS + 2.3
+VS − 1.4
V
Appendix D
Rev. G28
| Page 3 of 20
AD620
AD620A
AD620B
Parameter
Conditions Min
Typ Max
Common-Mode Rejection
Ratio DC to 60 Hz with
1 kΩ Source Imbalance VCM = 0 V to ± 10 V
G=1
73
90
G = 10
93
110
G = 100
110
130
G = 1000
110
130
OUTPUT
Output Swing
RL = 10 kΩ
VS = ±2.3 V −VS +
+VS − 1.2
to ± 5 V
1.1
Overtemperature
−VS + 1.4
+VS − 1.3
VS = ±5 V
−VS + 1.2
+VS − 1.4
to ± 18 V
Overtemperature
−VS + 1.6
+VS – 1.5
Short Circuit Current
±18
DYNAMIC RESPONSE
Small Signal –3 dB Bandwidth
G=1
1000
G = 10
800
G = 100
120
G = 1000
12
Slew Rate
0.75
1.2
Settling Time to 0.01%
10 V Step
G = 1–100
15
G = 1000
150
NOISE
Voltage Noise, 1 kHz
Total RTI Noise = (e 2ni ) + (e / G)2
Min
Typ
80
100
120
120
90
110
130
130
Max
Min
AD620S1
Typ Max
73
93
110
110
90
110
130
130
Unit
dB
dB
dB
dB
−VS + 1.1
+VS − 1.2
−VS + 1.1
+VS − 1.2
V
−VS + 1.4
−VS + 1.2
+VS − 1.3
+VS − 1.4
−VS + 1.6
−VS + 1.2
+VS − 1.3
+VS − 1.4
V
V
+VS – 1.5
–VS + 2.3
+VS – 1.5
−VS + 1.6
0.75
±18
±18
V
mA
1000
800
120
12
1.2
1000
800
120
12
1.2
kHz
kHz
kHz
kHz
V/µs
15
150
µs
µs
0.75
15
150
no
Input, Voltage Noise, eni
Output, Voltage Noise, eno
RTI, 0.1 Hz to 10 Hz
G=1
G = 10
G = 100–1000
Current Noise
0.1 Hz to 10 Hz
REFERENCE INPUT
RIN
IIN
Voltage Range
Gain to Output
POWER SUPPLY
Operating Range4
Quiescent Current
9
72
3.0
0.55
0.28
100
10
f = 1 kHz
20
50
VIN+, VREF = 0
−VS + 1.6
1 ± 0.0001
±2.3
VS = ±2.3 V
to ±18 V
Overtemperature
TEMPERATURE RANGE
For Specified Performance
13
100
60
+VS − 1.6
0.9
±18
1.3
1.1
1.6
−40 to +85
9
72
13
100
9
72
13
100
nV/√Hz
nV/√Hz
3.0
0.55
0.28
100
10
6.0
0.8
0.4
3.0
0.55
0.28
100
10
6.0
0.8
0.4
µV p-p
µV p-p
µV p-p
fA/√Hz
pA p-p
20
50
−VS + 1.6
1 ± 0.0001
±2.3
−40 to +85
1
See Analog Devices military data sheet for 883B tested specifications.
Does not include effects of external resistor RG.
3
One input grounded. G = 1.
4
This is defined as the same supply range that is used to specify PSR.
2
Appendix D
Rev. G29
| Page 4 of 20
60
+VS − 1.6
0.9
±18
1.3
1.1
1.6
20
50
60
+VS − 1.6
kΩ
µA
V
0.9
±18
1.3
V
mA
1.1
1.6
mA
−VS + 1.6
1 ± 0.0001
±2.3
−55 to +125
°C
AD620
ABSOLUTE MAXIMUM RATINGS
Table 2.
Parameter
Supply Voltage
Internal Power Dissipation1
Input Voltage (Common-Mode)
Differential Input Voltage
Output Short-Circuit Duration
Storage Temperature Range (Q)
Storage Temperature Range (N, R)
Operating Temperature Range
AD620 (A, B)
AD620 (S)
Lead Temperature Range
(Soldering 10 seconds)
1
Rating
±18 V
650 mW
±VS
25 V
Indefinite
−65°C to +150°C
−65°C to +125°C
Stresses above those listed under Absolute Maximum Ratings
may cause permanent damage to the device. This is a stress
rating only; functional operation of the device at these or any
other condition s above those indicated in the operational
section of this specification is not implied. Exposure to absolute
maximum rating conditions for extended periods may affect
device reliability.
−40°C to +85°C
−55°C to +125°C
300°C
Specification is for device in free air:
8-Lead Plastic Package: θJA = 95°C
8-Lead CERDIP Package: θJA = 110°C
8-Lead SOIC Package: θJA = 155°C
ESD CAUTION
ESD (electrostatic discharge) sensitive device. Electrostatic charges as high as 4000 V readily accumulate on the
human body and test equipment and can discharge without detection. Although this product features
proprietary ESD protection circuitry, permanent damage may occur on devices subjected to high energy
electrostatic discharges. Therefore, proper ESD precautions are recommended to avoid performance
degradation or loss of functionality.
Appendix D
Rev. G30
| Page 5 of 20
AD620
TYPICAL PERFORMANCE CHARACTERISTICS
(@ 25°C, VS = ±15 V, RL = 2 kΩ, unless otherwise noted.)
2.0
50
SAMPLE SIZE = 360
1.5
INPUT BIAS CURRENT (nA)
PERCENTAGE OF UNITS
40
30
20
10
1.0
+IB
–I B
0.5
0
–0.5
–1.0
–40
0
40
80
INPUT OFFSET VOLTAGE (µV)
175
00775-0-008
–80
5
00775-0-009
0
00775-0-005
–1.5
–2.0
–75
–25
25
75
TEMPERATURE (°C)
125
Figure 8. Input Bias Current vs. Temperature
Figure 5. Typical Distribution of Input Offset Voltage
2.0
50
CHANGE IN OFFSET VOLTAGE (µV)
SAMPLE SIZE = 850
PERCENTAGE OF UNITS
40
30
20
0
–1200
–600
0
600
1200
INPUT BIAS CURRENT (pA)
00775-0-006
10
1.5
1.0
0.5
0
0
1
2
3
WARM-UP TIME (Minutes)
4
Figure 9. Change in Input Offset Voltage vs. Warm-Up Time
Figure 6. Typical Distribution of Input Bias Current
1000
50
SAMPLE SIZE = 850
GAIN = 1
VOLTAGE NOISE (nV/ Hz)
30
20
10
100
GAIN = 10
10
GAIN = 100, 1,000
0
–400
–200
0
200
400
INPUT OFFSET CURRENT (pA)
Figure 7. Typical Distribution of Input Offset Current
1
1
10
100
1k
FREQUENCY (Hz)
10k
100k
00775-0-010
GAIN = 1000
BW LIMIT
00775-0-007
PERCENTAGE OF UNITS
40
Figure 10. Voltage Noise Spectral Density vs. Frequency (G = 1−1000)
Appendix D
Rev. G31
| Page 7 of 20
AD620
THEORY OF OPERATION
I1
20µA
VB
I2
20µA
A1
The input transistors Q1 and Q2 provide a single differentialpair bipolar input for high precision (Figure 38), yet offer 10×
lower input bias current thanks to Superϐeta processing.
Feedback through the Q1-A1-R1 loop and the Q2-A2-R2 loop
maintains constant collector current of the input devices Q1
and Q2, thereby impressing the input voltage across the external
gain setting resistor RG. This creates a differential gain from the
inputs to the A1/A2 outputs given by G = (R1 + R2)/RG + 1. The
unity-gain subtractor, A3, removes any common-mode signal,
yielding a single-ended output referred to the REF pin potential.
A2
10kΩ
C2
C1
10kΩ
A3
R3
400Ω
R1
R2
Q1
Q2
R4
400Ω
RG
GAIN
SENSE
OUTPUT
10kΩ
REF
+IN
GAIN
SENSE
–VS
00775-0-038
– IN
10kΩ
Figure 38. Simplified Schematic of AD620
The AD620 is a monolithic instrumentation amplifier based on
a modification of the classic three op amp approach. Absolute
value trimming allows the user to program gain accurately
(to 0.15% at G = 100) with only one resistor. Monolithic
construction and laser wafer trimming allow the tight matching
and tracking of circuit components, thus ensuring the high level
of performance inherent in this circuit.
The value of RG also determines the transconductance of the
preamp stage. As RG is reduced for larger gains, the
transconductance increases asymptotically to that of the input
transistors. This has three important advantages: (a) Open-loop
gain is boosted for increasing programmed gain, thus reducing
gain related errors. (b) The gain-bandwidth product
(determined by C1 and C2 and the preamp transconductance)
increases with programmed gain, thus optimizing frequency
response. (c) The input voltage noise is reduced to a value of
9 nV/√Hz, determined mainly by the collector current and base
resistance of the input devices.
The internal gain resistors, R1 and R2, are trimmed to an
absolute value of 24.7 kΩ, allowing the gain to be programmed
accurately with a single external resistor.
The gain equation is then
G=
49.4 kΩ
RG =
RG
+1
49.4 kΩ
G−1
Make vs. Buy: a Typical Bridge Application Error Budget
The AD620 offers improved performance over “homebrew”
three op amp IA designs, along with smaller size, fewer
components, and 10× lower supply current. In the typical
application, shown in Figure 39, a gain of 100 is required to
amplify a bridge output of 20 mV full-scale over the industrial
temperature range of −40°C to +85°C. Table 3 shows how to
calculate the effect various error sources have on circuit
accuracy.
Appendix D
Rev. G32
| Page 13 of 20
AD620
Note that for the homebrew circuit, the OP07 specifications for
input voltage offset and noise have been multiplied by √2. This
is because a three op amp type in-amp has two op amps at its
inputs, both contributing to the overall input error.
Regardless of the system in which it is being used, the AD620
provides greater accuracy at low power and price. In simple
systems, absolute accuracy and drift errors are by far the most
significant contributors to error. In more complex systems
with an intelligent processor, an autogain/autozero cycle will
remove all absolute accuracy and drift errors, leaving only the
resolution errors of gain, nonlinearity, and noise, thus allowing
full 14-bit accuracy.
10V
10kΩ *
R = 350Ω
10kΩ **
REFERENCE
PRECISION BRIDGE TRANSDUCER
AD620A MONOLITHIC
INSTRUMENTATION
AMPLIFIER, G = 100
SUPPLY CURRENT = 1.3mA MAX
100Ω **
10kΩ **
OP07D
00775-0-040
R = 350Ω
00775-0-039
R = 350Ω
10kΩ *
OP07D
AD620A
OP07D
10kΩ *
10kΩ*
"HOMEBREW" IN-AMP, G = 100
*0.02% RESISTOR MATCH, 3ppm/°C TRACKING
**DISCRETE 1% RESISTOR, 100ppm/ °C TRACKING
SUPPLY CURRENT = 15mA MAX
Figure 39. Make vs. Buy
Table 3. Make vs. Buy Error Budget
Error Source
ABSOLUTE ACCURACY at TA = 25°C
Input Offset Voltage, µV
Output Offset Voltage, µV
Input Offset Current, nA
CMR, dB
DRIFT TO 85°C
Gain Drift, ppm/°C
Input Offset Voltage Drift, µV/°C
Output Offset Voltage Drift, µV/°C
RESOLUTION
Gain Nonlinearity, ppm of Full Scale
Typ 0.1 Hz to 10 Hz Voltage Noise, µV p-p
Error, ppm of Full Scale
AD620
Homebrew
AD620 Circuit Calculation
“Homebrew” Circuit Calculation
125 µV/20 mV
1000 µV/100 mV/20 mV
2 nA ×350 Ω/20 mV
110 dB(3.16 ppm) ×5 V/20 mV
(150 µV × √2)/20 mV
((150 µV × 2)/100)/20 mV
(6 nA ×350 Ω)/20 mV
(0.02% Match × 5 V)/20 mV/100
6,250
500
18
791
10,607
150
53
500
Total Absolute Error
7,559
11,310
100 ppm/°C Track × 60°C
(2.5 µV/°C × √2 × 60°C)/20 mV
(2.5 µV/°C × 2 × 60°C)/100 mV/20 mV
3,600
3,000
450
6,000
10,607
150
Total Drift Error
7,050
16,757
40
14
54
14,663
40
27
67
28,134
(50 ppm + 10 ppm) ×60°C
1 µV/°C × 60°C/20 mV
15 µV/°C × 60°C/100 mV/20 mV
40 ppm
0.28 µV p-p/20 mV
40 ppm
(0.38 µV p-p × √2)/20 mV
Total Resolution Error
Grand Total Error
G = 100, VS = ±15 V.
(All errors are min/max and referred to input.)
Appendix D
Rev. G33
| Page 14 of 20
00775-0-041
R = 350Ω
RG
499Ω
AD620
5V
3kΩ
3kΩ
3kΩ
3kΩ
20kΩ
7
3
REF
8
AD620B
G = 100
499Ω
6
IN
10kΩ
5
1
ADC
4
2
20kΩ
1.7mA
DIGITAL
DATA
OUTPUT
AD705
0.10mA
00775-0-042
1.3mA
MAX
AGND
0.6mA
MAX
Figure 40. A Pressure Monitor Circuit that Operates on a 5 V Single Supply
Pressure Measurement
Medical ECG
Although useful in many bridge applications, such as weigh
scales, the AD620 is especially suitable for higher resistance
pressure sensors powered at lower voltages where small size and
low power become more significant.
The low current noise of the AD620 allows its use in ECG
monitors (Figure 41) where high source resistances of 1 MΩ or
higher are not uncommon. The AD620’s low power, low supply
voltage requirements, and space-saving 8-lead mini-DIP and
SOIC package offerings make it an excellent choice for batterypowered data recorders.
Figure 40 shows a 3 kΩ pressure transducer bridge powered
from 5 V. In such a circuit, the bridge consumes only 1.7 mA.
Adding the AD620 and a buffered voltage divider allows the
signal to be conditioned for only 3.8 mA of total supply current.
Small size and low cost make the AD620 especially attractive for
voltage output pressure transducers. Since it delivers low noise
and drift, it will also serve applications such as diagnostic
noninvasive blood pressure measurement.
R1
10kΩ
R4
1MΩ
The value of capacitor C1 is chosen to maintain stability of
the right leg drive loop. Proper safeguards, such as isolation,
must be added to this circuit to protect the patient from
possible harm.
+3V
PATIENT/CIRCUIT
PROTECTION/ISOLATION
C1
Furthermore, the low bias currents and low current noise,
coupled with the low voltage noise of the AD620, improve the
dynamic range for better performance.
R3
24.9kΩ
R2
24.9kΩ
RG
8.25kΩ
AD620A
G=7
0.03Hz
HIGHPASS
FILTER
G = 143
OUTPUT
1V/mV
OUTPUT
AMPLIFIER
–3V
Figure 41. A Medical ECG Monitor Circuit
Appendix D
Rev. G34
| Page 15 of 20
00775-0-043
AD705J
UMT4401 / SST4401 / MMST4401 / 2N4401
Transistors
NPN Medium Power Transistor
(Switching)
UMT4401 / SST4401 / MMST4401 / 2N4401
!External dimensions (Units : mm)
UMT4401
2.0±0.2
0.65 0.65
SST4401
MMST4401
2N4401
Packaging type
UMT3
R2X
SST3
R2X
SMT3
R2X
TO-92
-
T106
T116
T146
T93
3000
MMST4401
V
V
V
A
Tstg
2N4401
W
0.625
150
All terminals have the same
+0.1
dimensions
4.8±0.2
˚C
-55~+150
0.15 −0.06
0.4 +0.1
−0.05
(1) Emitter
(2) Base
(3) Collector
3.7±0.2
2.5Min.
Tj
Storage temperature
0~0.1
(3)
ROHM : SMT3
EIAJ : SC-59
0.2
PC
Junction temperature
1.6+0.2
−0.1
Unit
60
40
6
0.6
2N4401
0.8±0.1
(2)
(1)
4.8±0.2
UMT4401
SST4401
MMST4401
1.1+0.2
−0.1
(12.7Min.)
Collector power
dissipation
Limits
VCBO
VCEO
VEBO
IC
2.1±0.1
2.9±0.2
1.9±0.2
0.95 0.95
Symbol
(1) Emitter
(2) Base
(3) Collector
0.15 −0.06
0.4 +0.1
−0.05
!Absolute maximum ratings (Ta=25°C)
Parameter
0.2Min.
All terminals have the same
dimensions
+0.1
ROHM : SST3
Collector-base voltage
Collector-emitter voltage
Emitter-base voltage
Collector current
0~0.1
(3)
0.3~0.6
3000
0.45±0.1
(2)
(1)
2.8±0.2
3000
(1) Emitter
(2) Base
(3) Collector
0.95 +0.2
−0.1
0.95 0.95
1.3+0.2
−0.1
SST4401
2.4±0.2
UMT4401
3000
2.9±0.2
1.9±0.2
Part No.
Code
Basic ordering unit (pieces)
0~0.1
0.3+0.1
0.15±0.05
−0
All terminals have the same
dimensions
ROHM : UMT3
EIAJ : SC-70
!Package, marking, and packaging specifications
0.7±0.1
0.2
(2)
1.25±0.1
(1)
(3)
Marking
0.9±0.1
1.3±0.1
0.1~0.4
!Features
1) BVCEO>40V (IC=1mA)
2) Complements the UMT4403 / SST4403 / MMST4403
/ PN4403.
˚C
0.5±0.1.
ROHM : TO-92
EIAJ : SC-43
(1)
(2)
(3)
+0.3
2.5 −
0.1
5
0.45±0.1
!Electrical characteristics (Ta=25°C)
Symbol
Min.
Typ.
Max.
Unit
Collector-base breakdown voltage
Collector-emitter breakdown voltage
Parameter
BVCBO
BVCEO
60
40
-
-
V
V
IC=100µA
IC=1mA
Emitter-base breakdown voltage
BVEBO
6
-
-
0.1
0.1
V
µA
IE=100µA
-
-
0.4
-
-
0.75
-
-
0.95
-
-
1.2
20
40
-
-
Collector cutoff current
ICBO
Emitter cutoff current
IEBO
Collector-emitter saturation voltage
Base-emitter saturation voltage
DC current transfer ratio
VCE(sat)
VBE(sat)
hFE
-
µA
V
V
VCB=35V
VEB=5V
IC/IB=150mA/15mA
IC/IB=500mA/50mA
IC/IB=150mA/15mA
IC/IB=500mA/50mA
VCE=1V, IC=0.1mA
VCE=1V, IC=1mA
80
-
-
100
-
300
-
40
250
-
-
6.5
MHz
pF
pF
VCE=1V, IC=10mA
VCE=1V, IC=150mA
VCE=2V, IC=500mA
VCE=10V, IE=-20mA, f=100MHz
VCB=10V, f=100kHz
VEB=0.5V, f=100kHz
Transition frequency
Collector output capacitance
Cob
Emitter input capacitance
Cib
-
-
30
td
-
-
15
ns
VCC=30V, VEB(OFF)=2V, IC=150mA, IB1=15mA
tr
tstg
-
-
20
225
ns
ns
VCC=30V, VEB(OFF)=2V, IC=150mA, IB1=15mA
VCC=30V, IC=150mA, IB1=-IB2=15mA
tf
-
-
30
ns
VCC=30V, IC=150mA, IB1=-IB2=15mA
Delay time
Rise time
Storage time
Fall time
fT
-
Conditions
Appendix D
35
2.3
(1) Emitter
(2) Base
(3) Collector
UMT4401 / SST4401 / MMST4401 / 2N4401
Transistors
!Electrical characteristic curves
100
COLLECTOR CURRENT : Ic(mA)
Ta=25°C
1000
600
DC CURRENT GAIN : hFE
500
400
50
Ta=25°C
VCE=10V
100
300
200
1V
100
IB=0µA
10
0.1
0
0
10
5
COLLECTOR-EMITTER VOLTAGE : VCE(V)
COLLECTOR EMITTER SATURATION VOLTAGE : VCE(sat)(V)
10
COLLECTOR CURRENT : Ic(mA)
100
1000
Fig.3 DC current gain vs. collector current(Ι)
Fig.1 Grounded emitter output
characteristics
Ta=25°C
IC / IB=10
1.0
1000
VCE=10V
DC CURRENT GAIN : hFE
0.3
0.1
0
1.0
25°C
100
10
100
1000
COLLECTOR CURRENT : Ic(mA)
−55°C
10
0.1
1.0
1000
AC CURRENT GAIN : hFE
Ta=25°C
VCE=10V
f=1kHz
100
1.0
10
COLLECTOR CURRENT : Ic(mA)
100
1000
Fig.4 DC current gain vs. collector current(ΙΙ)
Fig.2 Collector-emitter saturation
voltage vs. collector current
10
0.1
10
COLLECTOR CURRENT : Ic(mA)
100
1000
Fig.5 AC current gain vs. collector current
Appendix D
36
BASE EMITTER SATURATION VOLTAGE : VBE(sat)(V)
0.2
Ta=125°C
1.8
1.6
Ta=25°C
IC / IB=10
1.2
0.8
0.4
0
1.0
10
100
1000
COLLECTOR CURRENT : Ic(mA)
Fig.6 Base-emitter saturation
voltage vs. collector current
A
Appendix D
37
A
Appendix D
38
LM111/LM211/LM311
Voltage Comparator
1.0 General Description
The LM111, LM211 and LM311 are voltage comparators that
have input currents nearly a thousand times lower than
devices like the LM106 or LM710. They are also designed to
operate over a wider range of supply voltages: from standard
± 15V op amp supplies down to the single 5V supply used for
IC logic. Their output is compatible with RTL, DTL and TTL
as well as MOS circuits. Further, they can drive lamps or
relays, switching voltages up to 50V at currents as high as
50 mA.
Both the inputs and the outputs of the LM111, LM211 or the
LM311 can be isolated from system ground, and the output
can drive loads referred to ground, the positive supply or the
negative supply. Offset balancing and strobe capability are
provided and outputs can be wire OR’ed. Although slower
than the LM106 and LM710 (200 ns response time vs 40 ns)
3.0 Typical Applications
the devices are also much less prone to spurious oscillations. The LM111 has the same pin configuration as the
LM106 and LM710.
The LM211 is identical to the LM111, except that its performance is specified over a −25˚C to +85˚C temperature range
instead of −55˚C to +125˚C. The LM311 has a temperature
range of 0˚C to +70˚C.
2.0 Features
n
n
n
n
n
Operates from single 5V supply
Input current: 150 nA max. over temperature
Offset current: 20 nA max. over temperature
Differential input voltage range: ± 30V
Power consumption: 135 mW at ± 15V
(Note 3)
Offset Balancing
Strobing
00570436
00570437
Note: Do Not Ground Strobe Pin. Output is turned off when current is
pulled from Strobe Pin.
Increasing Input Stage Current (Note 1)
Detector for Magnetic Transducer
00570438
Note 1: Increases typical common mode slew from 7.0V/µs to 18V/µs.
00570439
© 2004 National Semiconductor Corporation
DS005704
Appendix D
39
www.national.com
LM111/LM211/LM311 Voltage Comparator
January 2001
LM111/LM211/LM311
3.0 Typical Applications
(Note 3)
(Continued)
Digital Transmission Isolator
Relay Driver with Strobe
00570440
00570441
*Absorbs inductive kickback of relay and protects IC from severe voltage
transients on V++ line.
Note: Do Not Ground Strobe Pin.
Strobing off Both Input and Output Stages (Note 2)
00570442
Note: Do Not Ground Strobe Pin.
Note 2: Typical input current is 50 pA with inputs strobed off.
Note 3: Pin connections shown on schematic diagram and typical applications are for H08 metal can package.
Positive Peak Detector
Zero Crossing Detector Driving MOS Logic
00570424
00570423
*Solid tantalum
www.national.com
Appendix D
2
40
LM111/LM211/LM311
5.0 Absolute Maximum Ratings for
the LM311(Note 12)
Output Short Circuit Duration
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales Office/
Distributors for availability and specifications.
Storage Temperature Range
Total Supply Voltage (V84)
36V
Soldering Information
Output to Negative Supply Voltage
(V74)
40V
Ground to Negative Supply Voltage
(V14)
30V
10 sec
Operating Temperature Range
0˚ to 70˚C
−65˚C to 150˚C
Lead Temperature (soldering, 10 sec)
260˚C
Voltage at Strobe Pin
V+−5V
Dual-In-Line Package
Soldering (10 seconds)
260˚C
Small Outline Package
± 30V
± 15V
Differential Input Voltage
Input Voltage (Note 13)
Power Dissipation (Note 14)
500 mW
ESD Rating (Note 19)
300V
Electrical Characteristics
(Note 15)
Parameter
Vapor Phase (60 seconds)
215˚C
Infrared (15 seconds)
220˚C
See AN-450 “Surface Mounting Methods and Their Effect
on Product Reliability” for other methods of soldering
surface mount devices.
for the LM311
Typ
Max
Units
Input Offset Voltage (Note 16)
TA=25˚C, RS≤50k
Conditions
Min
2.0
7.5
mV
Input Offset Current(Note 16)
TA=25˚C
6.0
50
nA
Input Bias Current
TA=25˚C
100
250
Voltage Gain
TA=25˚C
Response Time (Note 17)
Saturation Voltage
40
nA
200
V/mV
TA=25˚C
200
ns
VIN≤−10 mV, IOUT=50 mA
0.75
1.5
V
2.0
5.0
mA
0.2
50
nA
10
mV
Input Offset Current (Note 16)
70
nA
Input Bias Current
300
nA
TA=25˚C
Strobe ON Current (Note 18)
TA=25˚C
Output Leakage Current
VIN≥10 mV, VOUT=35V
TA=25˚C, ISTROBE=3 mA
V− = Pin 1 = −5V
Input Offset Voltage (Note 16)
RS≤50K
Input Voltage Range
−14.5
13.8,−14.7
13.0
V
0.23
0.4
V
Saturation Voltage
V+≥4.5V, V−=0
Positive Supply Current
TA=25˚C
5.1
7.5
mA
Negative Supply Current
TA=25˚C
4.1
5.0
mA
VIN≤−10 mV, IOUT≤8 mA
Note 12: “Absolute Maximum Ratings indicate limits beyond which damage to the device may occur. Operating Ratings indicate conditions for which the device is
functional, but do not guarantee specific performance limits.”
Note 13: This rating applies for ± 15V supplies. The positive input voltage limit is 30V above the negative supply. The negative input voltage limit is equal to the
negative supply voltage or 30V below the positive supply, whichever is less.
Note 14: The maximum junction temperature of the LM311 is 110˚C. For operating at elevated temperature, devices in the H08 package must be derated based
on a thermal resistance of 165˚C/W, junction to ambient, or 20˚C/W, junction to case. The thermal resistance of the dual-in-line package is 100˚C/W, junction to
ambient.
Note 15: These specifications apply for VS= ± 15V and Pin 1 at ground, and 0˚C < TA < +70˚C, unless otherwise specified. The offset voltage, offset current and
bias current specifications apply for any supply voltage from a single 5V supply up to ± 15V supplies.
Note 16: The offset voltages and offset currents given are the maximum values required to drive the output within a volt of either supply with 1 mA load. Thus, these
parameters define an error band and take into account the worst-case effects of voltage gain and RS.
Note 17: The response time specified (see definitions) is for a 100 mV input step with 5 mV overdrive.
Note 18: This specification gives the range of current which must be drawn from the strobe pin to ensure the output is properly disabled. Do not short the strobe
pin to ground; it should be current driven at 3 to 5 mA.
Note 19: Human body model, 1.5 kΩ in series with 100 pF.
www.national.com
Appendix D
4
41
Supply Current
LM111/LM211/LM311
6.0 LM111/LM211 Typical Performance Characteristics
(Continued)
Supply Current
00570455
00570456
Leakage Currents
00570457
7.0 LM311 Typical Performance Characteristics
Input Bias Current
Input Offset Current
00570458
00570459
Appendix
D
7
42
www.national.com
LM111/LM211/LM311
7.0 LM311 Typical Performance Characteristics
Offset Error
(Continued)
Input Characteristics
00570461
00570460
Common Mode Limits
Transfer Function
00570462
Response Time for Various
Input Overdrives
Response Time for Various
Input Overdrives
00570464
www.national.com
00570463
Appendix D
8
43
00570465
LM111/LM211/LM311
7.0 LM311 Typical Performance Characteristics
(Continued)
Response Time for Various
Input Overdrives
Output Saturation Voltage
00570466
00570467
Response Time for Various
Input Overdrives
Output Limiting Characteristics
00570469
00570468
Supply Current
Supply Current
00570470
00570471
Appendix
D
9
44
www.national.com
LM555
Timer
General Description
Features
The LM555 is a highly stable device for generating accurate
time delays or oscillation. Additional terminals are provided
for triggering or resetting if desired. In the time delay mode of
operation, the time is precisely controlled by one external resistor and capacitor. For astable operation as an oscillator,
the free running frequency and duty cycle are accurately
controlled with two external resistors and one capacitor. The
circuit may be triggered and reset on falling waveforms, and
the output circuit can source or sink up to 200mA or drive
TTL circuits.
n
n
n
n
n
n
n
n
n
Direct replacement for SE555/NE555
Timing from microseconds through hours
Operates in both astable and monostable modes
Adjustable duty cycle
Output can source or sink 200 mA
Output and supply TTL compatible
Temperature stability better than 0.005% per ˚C
Normally on and normally off output
Available in 8-pin MSOP package
Applications
n
n
n
n
n
n
n
Precision timing
Pulse generation
Sequential timing
Time delay generation
Pulse width modulation
Pulse position modulation
Linear ramp generator
Schematic Diagram
DS007851-1
© 2000 National Semiconductor Corporation
DS007851
Appendix D
45
www.national.com
LM555 Timer
February 2000
LM555
Connection Diagram
Dual-In-Line, Small Outline
and Molded Mini Small Outline Packages
DS007851-3
Top View
Ordering Information
Package
8-Pin SOIC
8-Pin MSOP
8-Pin MDIP
www.national.com
Part Number
Package Marking
Media Transport
LM555CM
LM555CM
Rails
LM555CMX
LM555CM
2.5k Units Tape and Reel
LM555CMM
Z55
1k Units Tape and Reel
LM555CMMX
Z55
3.5k Units Tape and Reel
LM555CN
LM555CN
Rails
Appendix D
2
46
NSC Drawing
M08A
MUA08A
N08E
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales Office/
Distributors for availability and specifications.
Supply Voltage
Power Dissipation (Note 3)
LM555CM, LM555CN
LM555CMM
Operating Temperature Ranges
LM555C
Storage Temperature Range
+18V
1180 mW
613 mW
Soldering Information
Dual-In-Line Package
Soldering (10 Seconds)
260˚C
Small Outline Packages
(SOIC and MSOP)
Vapor Phase (60 Seconds)
215˚C
Infrared (15 Seconds)
220˚C
See AN-450 “Surface Mounting Methods and Their Effect
on Product Reliability” for other methods of soldering
surface mount devices.
0˚C to +70˚C
−65˚C to +150˚C
Electrical Characteristics (Notes 1, 2)
(TA = 25˚C, VCC = +5V to +15V, unless othewise specified)
Parameter
Conditions
Limits
Units
LM555C
Min
Supply Voltage
Supply Current
Typ
4.5
Max
16
V
6
15
mA
VCC = 5V, RL = ∞
VCC = 15V, RL = ∞
(Low State) (Note 4)
3
10
1
%
RA = 1k to 100kΩ,
50
ppm/˚C
Timing Error, Monostable
Initial Accuracy
Drift with Temperature
C = 0.1µF, (Note 5)
Accuracy over Temperature
1.5
%
Drift with Supply
0.1
%/V
Timing Error, Astable
Initial Accuracy
Drift with Temperature
RA, RB = 1k to 100kΩ,
2.25
%
150
ppm/˚C
C = 0.1µF, (Note 5)
Accuracy over Temperature
3.0
%
Drift with Supply
0.30
%/V
Threshold Voltage
Trigger Voltage
0.667
x VCC
VCC = 15V
5
V
VCC = 5V
1.67
Trigger Current
Reset Voltage
0.4
Reset Current
Threshold Current
Control Voltage Level
(Note 6)
VCC = 15V
VCC = 5V
9
2.6
Pin 7 Leakage Output High
V
0.5
0.9
µA
0.5
1
V
0.1
0.4
mA
0.1
0.25
µA
10
3.33
11
4
V
1
100
nA
200
mV
Pin 7 Sat (Note 7)
Output Low
VCC = 15V, I7 = 15mA
180
Output Low
VCC = 4.5V, I7 = 4.5mA
80
Appendix
D
3
47
mV
www.national.com
LM555
Absolute Maximum Ratings (Note 2)
LM555
Electrical Characteristics (Notes 1, 2)
(Continued)
(TA = 25˚C, VCC = +5V to +15V, unless othewise specified)
Parameter
Conditions
Limits
Units
LM555C
Min
Output Voltage Drop (Low)
Typ
Max
ISINK = 10mA
0.1
0.25
ISINK = 50mA
0.4
0.75
V
ISINK = 100mA
2
2.5
V
ISINK = 200mA
2.5
VCC = 15V
V
V
VCC = 5V
ISINK = 8mA
Output Voltage Drop (High)
V
ISINK = 5mA
0.25
ISOURCE = 200mA, VCC = 15V
12.5
ISOURCE = 100mA, VCC = 15V
0.35
V
V
12.75
13.3
V
2.75
3.3
V
Rise Time of Output
100
ns
Fall Time of Output
100
ns
VCC = 5V
Note 1: All voltages are measured with respect to the ground pin, unless otherwise specified.
Note 2: Absolute Maximum Ratings indicate limits beyond which damage to the device may occur. Operating Ratings indicate conditions for which the device is functional, but do not guarantee specific performance limits. Electrical Characteristics state DC and AC electrical specifications under particular test conditions which guarantee specific performance limits. This assumes that the device is within the Operating Ratings. Specifications are not guaranteed for parameters where no limit is
given, however, the typical value is a good indication of device performance.
Note 3: For operating at elevated temperatures the device must be derated above 25˚C based on a +150˚C maximum junction temperature and a thermal resistance
of 106˚C/W (DIP), 170˚C/W (S0-8), and 204˚C/W (MSOP) junction to ambient.
Note 4: Supply current when output high typically 1 mA less at VCC = 5V.
Note 5: Tested at VCC = 5V and VCC = 15V.
Note 6: This will determine the maximum value of RA + RB for 15V operation. The maximum total (RA + RB) is 20MΩ.
Note 7: No protection against excessive pin 7 current is necessary providing the package dissipation rating will not be exceeded.
Note 8: Refer to RETS555X drawing of military LM555H and LM555J versions for specifications.
www.national.com
Appendix D
4
48
LM555
Typical Performance Characteristics
Minimuim Pulse Width
Required for Triggering
Supply Current vs.
Supply Voltage
DS007851-4
High Output Voltage vs.
Output Source Current
DS007851-19
Low Output Voltage vs.
Output Sink Current
DS007851-20
Low Output Voltage vs.
Output Sink Current
DS007851-21
Low Output Voltage vs.
Output Sink Current
DS007851-22
DS007851-23
Appendix
D
5
49
www.national.com
LM555
Typical Performance Characteristics
(Continued)
Output Propagation Delay vs.
Voltage Level of Trigger Pulse
Output Propagation Delay vs.
Voltage Level of Trigger Pulse
DS007851-24
Discharge Transistor (Pin 7)
Voltage vs. Sink Current
DS007851-25
Discharge Transistor (Pin 7)
Voltage vs. Sink Current
DS007851-26
www.national.com
Appendix D
6
50
DS007851-27
MONOSTABLE OPERATION
In this mode of operation, the timer functions as a one-shot
(Figure 1). The external capacitor is initially held discharged
by a transistor inside the timer. Upon application of a negative trigger pulse of less than 1/3 VCC to pin 2, the flip-flop is
set which both releases the short circuit across the capacitor
and drives the output high.
NOTE: In monostable operation, the trigger should be driven
high before the end of timing cycle.
DS007851-7
FIGURE 3. Time Delay
DS007851-5
FIGURE 1. Monostable
ASTABLE OPERATION
If the circuit is connected as shown in Figure 4 (pins 2 and 6
connected) it will trigger itself and free run as a multivibrator.
The external capacitor charges through RA + RB and discharges through RB. Thus the duty cycle may be precisely
set by the ratio of these two resistors.
The voltage across the capacitor then increases exponentially for a period of t = 1.1 RA C, at the end of which time the
voltage equals 2/3 VCC. The comparator then resets the
flip-flop which in turn discharges the capacitor and drives the
output to its low state. Figure 2 shows the waveforms generated in this mode of operation. Since the charge and the
threshold level of the comparator are both directly proportional to supply voltage, the timing internal is independent of
supply.
DS007851-8
FIGURE 4. Astable
DS007851-6
VCC = 5V
TIME = 0.1 ms/DIV.
RA = 9.1kΩ
C = 0.01µF
Top Trace: Input 5V/Div.
Middle Trace: Output 5V/Div.
Bottom Trace: Capacitor Voltage 2V/Div.
In this mode of operation, the capacitor charges and discharges between 1/3 VCC and 2/3 VCC. As in the triggered
mode, the charge and discharge times, and therefore the frequency are independent of the supply voltage.
FIGURE 2. Monostable Waveforms
During the timing cycle when the output is high, the further
application of a trigger pulse will not effect the circuit so long
as the trigger input is returned high at least 10µs before the
end of the timing interval. However the circuit can be reset
during this time by the application of a negative pulse to the
reset terminal (pin 4). The output will then remain in the low
state until a trigger pulse is again applied.
When the reset function is not in use, it is recommended that
it be connected to VCC to avoid any possibility of false triggering.
Figure 3 is a nomograph for easy determination of R, C values for various time delays.
Appendix
D
7
51
www.national.com
LM555
Applications Information
LM555
Applications Information
(Continued)
Figure 5 shows the waveforms generated in this mode of
operation.
DS007851-11
VCC = 5V
TIME = 20µs/DIV.
RA = 9.1kΩ
C = 0.01µF
DS007851-9
VCC = 5V
TIME = 20µs/DIV.
RA = 3.9kΩ
RB = 3kΩ
C = 0.01µF
Top Trace: Output 5V/Div.
Bottom Trace: Capacitor Voltage 1V/Div.
Top Trace: Input 4V/Div.
Middle Trace: Output 2V/Div.
Bottom Trace: Capacitor 2V/Div.
FIGURE 7. Frequency Divider
PULSE WIDTH MODULATOR
When the timer is connected in the monostable mode and
triggered with a continuous pulse train, the output pulse
width can be modulated by a signal applied to pin 5. Figure
8 shows the circuit, and in Figure 9 are some waveform
examples.
FIGURE 5. Astable Waveforms
The charge time (output high) is given by:
t1 = 0.693 (RA + RB) C
And the discharge time (output low) by:
t2 = 0.693 (RB) C
Thus the total period is:
T = t1 + t2 = 0.693 (RA +2RB) C
The frequency of oscillation is:
Figure 6 may be used for quick determination of these RC
values.
The duty cycle is:
DS007851-12
FIGURE 8. Pulse Width Modulator
DS007851-13
DS007851-10
FIGURE 6. Free Running Frequency
FREQUENCY DIVIDER
The monostable circuit of Figure 1 can be used as a frequency divider by adjusting the length of the timing cycle.
Figure 7 shows the waveforms generated in a divide by three
circuit.
www.national.com
Appendix D
8
52
VCC = 5V
Top Trace: Modulation 1V/Div.
TIME = 0.2 ms/DIV. Bottom Trace: Output Voltage 2V/Div.
RA = 9.1kΩ
C = 0.01µF
FIGURE 9. Pulse Width Modulator
DAC0808
8-Bit D/A Converter
General Description
Features
The DAC0808 is an 8-bit monolithic digital-to-analog converter (DAC) featuring a full scale output current settling time
of 150 ns while dissipating only 33 mW with ± 5V supplies.
No reference current (IREF) trimming is required for most
applications since the full scale output current is typically ± 1
LSB of 255 IREF/256. Relative accuracies of better than
± 0.19% assure 8-bit monotonicity and linearity while zero
level output current of less than 4 µA provides 8-bit zero
accuracy for IREF≥2 mA. The power supply currents of the
DAC0808 is independent of bit codes, and exhibits essentially constant device characteristics over the entire supply
voltage range.
The DAC0808 will interface directly with popular TTL, DTL or
CMOS logic levels, and is a direct replacement for the
MC1508/MC1408. For higher speed applications, see
DAC0800 data sheet.
n
n
n
n
Relative accuracy: ± 0.19% error maximum
Full scale current match: ± 1 LSB typ
Fast settling time: 150 ns typ
Noninverting digital inputs are TTL and CMOS
compatible
n High speed multiplying input slew rate: 8 mA/µs
n Power supply voltage range: ± 4.5V to ± 18V
n Low power consumption: 33 mW @ ± 5V
Block and Connection Diagrams
DS005687-1
Dual-In-Line Package
DS005687-2
Top View
Order Number DAC0808
See NS Package M16A or N16A
© 2001 National Semiconductor Corporation
DS005687
Appendix D
53
www.national.com
DAC0808 8-Bit D/A Converter
May 1999
DAC0808
Block and Connection Diagrams
(Continued)
Small-Outline Package
DS005687-13
Ordering Information
ACCURACY
OPERATING
TEMPERATURE RANGE
8-bit
N PACKAGE (N16A)
(Note 1)
0˚C≤TA≤+75˚C
DAC0808LCN
Note 1: Devices may be ordered by using either order number.
www.national.com
Appendix D
2
54
MC1408P8
SO PACKAGE
(M16A)
DAC0808LCM
Storage Temperature Range
Lead Temp. (Soldering, 10 seconds)
Dual-In-Line Package (Plastic)
Dual-In-Line Package (Ceramic)
Surface Mount Package
Vapor Phase (60 seconds)
Infrared (15 seconds)
If Military/Aerospace specified devices are required,
please contact the National Semiconductor Sales Office/
Distributors for availability and specifications.
Power Supply Voltage
VCC
VEE
Digital Input Voltage, V5–V12
Applied Output Voltage, VO
Reference Current, I14
Reference Amplifier Inputs, V14, V15
Power Dissipation (Note 4)
ESD Susceptibility (Note 5)
−10 VDC
−11 VDC
+18 VDC
−18 VDC
to +18 VDC
to +18 VDC
5 mA
VCC, VEE
1000 mW
TBD
−65˚C to +150˚C
260˚C
300˚C
215˚C
220˚C
Operating Ratings
TMIN ≤ TA ≤ TMAX
0 ≤TA ≤ +75˚C
Temperature Range
DAC0808
Electrical Characteristics
(VCC = 5V, VEE = −15 VDC, VREF/R14 = 2 mA, and all digital inputs at high logic level unless otherwise noted.)
Symbol
Er
Parameter
Relative Accuracy (Error Relative
Conditions
Min
Typ
Max
Units
(Figure 4)
%
to Full Scale IO)
± 0.19
DAC0808LC (LM1408-8)
Settling Time to Within ⁄ LSB
TA =25˚C (Note 7),
(Includes tPLH)
(Figure 5)
TA = 25˚C, (Figure 5)
12
tPLH, tPHL
Propagation Delay Time
TCIO
Output Full Scale Current Drift
MSB
Digital Input Logic Levels
VIH
High Level, Logic “1”
VIL
Low Level, Logic “0”
MSB
I15
IO
Digital Input Current
%
150
30
ns
100
ns
± 20
ppm/˚C
(Figure 3)
2
VDC
0.8
VDC
(Figure 3)
High Level
VIH = 5V
0
0.040
mA
Low Level
VIL = 0.8V
−0.003
−0.8
mA
Reference Input Bias Current
(Figure 3)
−1
−3
µA
Output Current Range
(Figure 3)
Output Current
VEE = −5V
0
2.0
2.1
mA
VEE = −15V, TA = 25˚C
0
2.0
4.2
mA
1.9
1.99
2.1
mA
0
4
µA
VREF = 2.000V,
R14 = 1000Ω,
(Figure 3)
SRIREF
Output Current, All Bits Low
(Figure 3)
Output Voltage Compliance (Note 3)
Er ≤ 0.19%, TA = 25˚C
VEE =−5V, IREF =1 mA
−0.55, +0.4
VDC
VEE Below −10V
−5.0, +0.4
VDC
Reference Current Slew Rate
(Figure 6)
Output Current Power Supply
−5V ≤ VEE ≤ −16.5V
4
8
mA/µs
0.05
2.7
µA/V
2.3
22
mA
−4.3
−13
mA
Sensitivity
Power Supply Current (All Bits
(Figure 3)
Low)
ICC
IEE
Power Supply Voltage Range
TA = 25˚C, (Figure 3)
VCC
4.5
5.0
5.5
VDC
VEE
−4.5
−15
−16.5
VDC
Power Dissipation
Appendix
D
3
55
www.national.com
DAC0808
Absolute Maximum Ratings (Note 2)
DAC0808
Electrical Characteristics
(Continued)
(VCC = 5V, VEE = −15 VDC, VREF/R14 = 2 mA, and all digital inputs at high logic level unless otherwise noted.)
Symbol
Parameter
All Bits Low
All Bits High
Typ
Max
VCC = 5V, VEE = −5V
Conditions
Min
33
170
Units
mW
VCC = 5V, VEE = −15V
106
305
mW
VCC = 15V, VEE = −5V
90
mW
VCC = 15V, VEE = −15V
160
mW
Note 2: Absolute Maximum Ratings indicate limits beyond which damage to the device may occur. DC and AC electrical specifications do not apply when operating
the device beyond its specified operating conditions.
Note 3: Range control is not required.
Note 4: The maximum power dissipation must be derated at elevated temperatures and is dictated by TJMAX, θJA, and the ambient temperature, TA. The maximum
allowable power dissipation at any temperature is PD = (TJMAX − TA)/θJA or the number given in the Absolute Maixmum Ratings, whichever is lower. For this device,
TJMAX = 125˚C, and the typical junction-to-ambient thermal resistance of the dual-in-line J package when the board mounted is 100˚C/W. For the dual-in-line N
package, this number increases to 175˚C/W and for the small outline M package this number is 100˚C/W.
Note 5: Human body model, 100 pF discharged through a 1.5 kΩ resistor.
Note 6: All current switches are tested to guarantee at least 50% of rated current.
Note 7: All bits switched.
Note 8: Pin-out numbers for the DAL080X represent the dual-in-line package. The small outline package pinout differs from the dual-in-line package.
Typical Application
DS005687-23
DS005687-3
FIGURE 1. +10V Output Digital to Analog Converter (Note 8)
Typical Performance Characteristics
Logic Input Current vs
Input Voltage
VCC = 5V, VEE = −15V, TA = 25˚C, unless otherwise noted
Logic Threshold Voltage vs
Temperature
Bit Transfer Characteristics
DS005687-14
DS005687-15
DS005687-16
www.national.com
Appendix D
4
56
DAC0808
Typical Performance Characteristics
VCC = 5V, VEE = −15V, TA = 25˚C, unless otherwise
noted (Continued)
Output Current vs Output
Voltage (Output Voltage
Compliance)
Output Voltage Compliance
vs Temperature
Typical Power Supply
Current vs Temperature
DS005687-18
DS005687-19
DS005687-17
Typical Power Supply
Current vs VEE
Typical Power Supply
Current vs VCC
DS005687-20
Reference Input
Frequency Response
DS005687-21
DS005687-22
Unless otherwise specified: R14 = R15 = 1 kΩ, C = 15 pF, pin 16 to VEE; RL = 50Ω, pin 4 to ground.
Curve A: Large Signal Bandwidth Method of Figure 7, VREF = 2 Vp-p offset 1V above ground.
Curve B: Small Signal Bandwidth Method of Figure 7, RL = 250Ω, VREF = 50 mVp-p offset 200 mV above ground.
Curve C: Large and Small Signal Bandwidth Method of Figure 9 (no op amp, RL = 50Ω), RS = 50Ω, VREF = 2V, VS = 100 mVp-p
centered at 0V.
Appendix
D
5
57
www.national.com
www.national.com
Appendix D
6
58
FIGURE 2. Equivalent Circuit of the DAC0808 Series (Note 8)
DS005687-4
DAC0808
Low-Cost Multifunction DAQ for USB
NEW
NI USB-6008, NI USB-6009
• Small, portable multifunction data
acquisition devices
• 12 or 14-bit input resolution,
at up to 48 kS/s
• Built-in, removable connectors
for easier and more
cost-effective connectivity
• 2 true DAC analog outputs
for accurate output signals
• 12 digital I/O lines
(TTL/LVTTL/CMOS)
• 32-bit event counter
• Student kits available
Product
Bus
Analog Inputs1
USB-6009
USB
8 SE/4 DI
USB-6008
USB
8 SE/4 DI
1 SE = single ended, DI = differential
Operating Systems
• Windows 2000/XP
• Mac OS X
• Linux
Recommended Software
• LabVIEW
• LabWindows/CVI
Measurement Services
Software (included)
• NI-DAQmx Base
• Ready-to-Run Data Logger
Input Resolution Max Sampling
(bits)
Rate (kS/s)
14
48
12
10
Input Range
(V)
±1 to ±20
±1 to ±20
Analog Outputs
2
2
Output Resolution
(bits)
12
12
Output Rate
(Hz)
150
150
Output Range
(V)
0 to 5
0 to 5
Digital I/O
Lines
12
12
32-bit Counter
1
1
Trigger
Digital
Digital
Hardware Description
Information for Student Ownership
The National Instruments USB-6008 and USB-6009 multifunction
data acquisition devices provide reliable data acquisition at a low
price. With plug-and-play USB connectivity, these devices are simple
enough for quick measurements, but versatile enough for more
complex measurement applications.
To supplement simulation, measurement, and automation theory
courses with practical experiments, NI has developed the USB-6008
and USB-6009 student kits that include LabVIEW Student Edition
and a ready-to-run data logger application. These kits are exclusively
for students, giving them a powerful, low-cost hands-on learning
tool. Visit ni.com/academic for more details.
Software Description
The NI USB-6008 and USB-6009 include a ready-to-run data
logger application that acquires and logs up to eight channels of
analog data. For more functionality, NI-DAQmx Base software is a
multiplatform driver with a subset of the NI-DAQmx programming
interface. Use it to develop customized DAQ applications with
NI LabVIEW or C-based development environments.
Recommended Accessories
The USB-6008 and USB-6009 have built-in connectivity, so no
additional accessories are required.
Common Applications
Information for OEM Customers
For information on special configurations and pricing, please
visit ni.com/oem.
Ordering Information
NI USB-60081 ................................................................779051-01
NI USB-60091 ................................................................779026-01
NI USB-6008 Student-kit1,2 ..........................................779320-22
NI USB-6009 Student-kit1,2 ..........................................779321-22
1Includes NI-DAQmx Base Software, NI-Ready-to-Run
Data Logger Software, and a USB cable.
2Includes LabVIEW Student Edition
The USB-6008 and USB-6009 are ideal for a number of applications
where economy, small size, and simplicity are essential, such as:
• Data logging – Log environmental or voltage data quickly and easily
• Academic lab use – The low price facilitates student ownership
of DAQ hardware for completely interactive lab-based courses.
Academic pricing available. Visit ni.com/academic for details.
• Embedded OEM applications
Appendix D
59
Low-Cost Multifunction DAQ for USB
Specifications
Typical at 25 °C unless otherwise noted.
Digital I/O
Analog Input
Number of channels......................................... 12 total
8 (P0.<0..7>)
4 (P1.<0..3>)
Direction control............................................... Each channel individually programmable as input or output
Output driver type
USB-6008................................................... Open-drain
USB-6009................................................... Each channel individually programmable as push-pull or
open-drain.
Compatibility .................................................... CMOS, TTL, LVTTL
Internal pull-up resistor.................................... 4.7 kΩ to +5 V
Power-on state ................................................. Input (high impedance)
Absolute maximum voltage range ................... -0.5 to +5.8 V
Absolute accuracy, single-ended
Range
±10
Typical at 25 ˚C (mV)
14.7
Maximum (0 to 55 ˚C) (mV)
138
Absolute accuracy at full scale, differential1
Range
±20
±10
±5
±4
±2.5
±2
±1.25
±1
1
Typical at 25 ˚C (mV)
14.7
7.73
4.28
3.59
2.56
2.21
1.70
1.53
Maximum (0 to 55 ˚C) (mV)
138
84.8
58.4
53.1
45.1
42.5
38.9
37.5
Input voltages may not exceed the working voltage range
Number of channels......................................... 8 single-ended / 4 differential
Type of ADC...................................................... Successive approximation
ADC resolution (bits)
Device
USB-6008
USB-6009
Differential
12
14
Single-Ended
11
13
Maximum sampling rate (system dependent)
Device
USB-6008
USB-6009
Maximum Sampling Rate (kS/s)
10
48
Input range, single-ended ................................
Input range, differential ...................................
Maximum working voltage ..............................
Overvoltage protection.....................................
FIFO buffer size.................................................
Timing resolution..............................................
Timing accuracy................................................
Input Impedance...............................................
Trigger source...................................................
System noise ....................................................
±10 V
±20, ±10, ±5, ±4, ±2.5, ±2, ±1.25, ±1 V
±10 V
±35 V
512 B
41.67 ns (24 MHz timebase)
100 ppm of actual sample rate
144 kΩ
Software or external digital trigger
0.3 LSBrms (±10 V range)
Digital logic levels
Level
Input low voltage
Input high voltage
Input leakage current
Output low voltage (I = 8.5 mA)
Output high voltage (Push-pull, I = -8.5 mA)
Output high voltage (Open-drain, I = -0.6 mA, nominal)
Output high voltage (Open-drain, I = -8.5 mA,
with external pull-up resistor)
Min
-0.3
2.0
–
–
2.0
2.0
Max
0.8
5.8
50
0.8
3.5
5.0
Units
V
V
µA
V
V
V
2.0
–
V
Counter
Number of counters .........................................
Resolution.........................................................
Counter measurements ....................................
Pull-up Resistor ................................................
Maximum input frequency ...............................
Minimum high pulse width ..............................
Minimum low pulse width ...............................
Input high voltage ............................................
Input low voltage .............................................
1
32 bits
Edge counting (falling edge)
4.7 kΩ to 5 V
5 MHz
100 ns
100 ns
2.0 V
0.8 V
Power Available at I/O Connector
+5 V output (200 mA maximum) ...................... +5 V typical
+4.85 V minimum
+2.5 V output (1 mA maximum) ....................... +2.5 V typical
+2.5 V output accuracy..................................... 0.25 % max
Voltage reference temperature drift................ 50 ppm/°C max
Physical Characteristics
If you need to clean the module, wipe it with a dry towel.
Analog Output
Absolute accuracy (no load).............................
Number of channels.........................................
Type of DAC......................................................
DAC resolution .................................................
Maximum update rate......................................
Output range ....................................................
Output impedance ............................................
Output current drive .........................................
Power-on state .................................................
Slew rate ..........................................................
Short-circuit current .........................................
2
7 mV typical, 36.4 mV maximum at full scale
2
Successive approximation
12 bits
150 Hz, software-timed
0 to +5 V
50 Ω
5 mA
0V
1 V/µs
50 mA
Dimensions (without connectors) .................... 6.35 by 8.51 by 2.31 cm
(2.50 by 3.35 by 0.91 in.)
Dimensions (with connectors).......................... 8.18 by 8.51 by 2.31 cm
(3.22 by 3.35 by 0.91 in.)
Weight (without connectors) ........................... 59 g (2.1 oz.)
Weight (with connectors)................................. 84 g (3 oz.)
I/O Connectors ................................................. USB series B receptacle
(2) 16-position (screw-terminal) plug headers
Screw-terminal wiring ..................................... 16 to 28 AWG
Screw-terminal torque ..................................... 0.22 to 0.25 N•m
(2.0 to 2.2 lb•in.)
Appendix D
60
National Instruments • Tel: (800) 813-3693 • info@ni.com • ni.com
Low-Cost Multifunction DAQ for USB
Bus Interface
Voltages
USB specification ............................................. USB 2.0 full-speed
USB bus speed ................................................. 12 Mb/s
Connect only voltages that are within the absolution maximum limits of the connection point.
See pertinent specification section for appropriate limits.
Power Requirement
Hazardous Locations
USB (4.10 to 5.25 VDC) .................................... 80 mA typical
500 mA maximum
USB Suspend.................................................... 300 µA typical
500 µA maximum
The USB-6008 and USB-6009 are not certified for use in hazardous locations.
Electromagnetic Compatibility
Environmental
The USB-6008 and USB-6009 are intended for indoor use only.
Operating Environment
Ambient temperature range...................... 0 to 55 °C (tested in accordance with IEC-60068-2-1
and IEC-60068-2-2.)
Relative humidity range ............................ 10% to 90%, non-condensing (tested in accordance
with IEC-60068-2-56.)
Storage Environment
Ambient temperature range...................... -40 to 85 °C (tested in accordance with IEC-60068-2-1
and IEC-60068-2-2.)
Relative humidity range ............................ 5% to 90%, non-condensing (tested in accordance
with IEC-60068-2-56.)
Maximum altitude ............................................ 2,000 m (at 25 °C ambient temperature)
Pollution Degree............................................... 2
Certifications and Compliances
Emissions.......................................................... EN 55011 Class A at 10 m
FCC Part 15A above 1 GHz
Immunity........................................................... Industrial levels per EN 61326:1997 + A2:2001, Table 1
EMC/EMI .......................................................... CE, C-Tick, and FCC Part 15 (Class A) Compliant
Note: The USB-6008 and USB-6009 may experience temporary variations in analog input
readings when exposed to radiated and conducted RF noise. Device returns to normal operation
after RF exposure is removed.
CE Compliance
This product meets the essential requirements of applicable European Directives, as amended for
CE marking, as follows:
Low-Voltage Directive (safety)......................... 73/23/EEC
Electromagnetic Compatibility
Directive (EMC).......................................... 89/336/EEC
Note Refer to the Declaration of Conformity (DoC) for this product for any additional regulatory compliance
information. To obtain the DoC for this product, visit ni.com/certification, search by model number or
product line, and click the appropriate link in the Certification column.
The USB-6008 and USB-6009 are designed to meet the requirements of the following standards of safety
for electrical equipment for measurement, control, and laboratory use:
• IEC 61010-1, EN 61010-1
• UL 61010-1
• CAN/CSA C22.2 No. 61010-1
Note For UL and other safety certifications, refer to the product label, or visit ni.com/certification,
search by model number or product line, and click the appropriate link in the Certification column.
Appendix D
61
National Instruments • Tel: (800) 813-3693 • info@ni.com • ni.com
3
NI Services and Support
NI has the services and support to meet your
needs around the globe and through the
application life cycle – from planning
and development through deployment
and ongoing maintenance. We offer
services and service levels to meet
customer requirements in research,
design, validation, and manufacturing.
Visit ni.com/services for more information.
Local Sales and Technical Support
In offices worldwide, NI staff is local to the country, giving you access
to engineers who speak your language. NI delivers industry-leading
technical support through an online KnowledgeBase, applications
engineers, and access to 14,000 measurement and automation
professionals within NI Developer Exchange forums. Find
immediate answers to your questions at ni.com/support.
We also offer service programs that provide automatic upgrades to
your application development environment and higher levels of
technical support. Visit ni.com/ssp.
SERVICE
NEEDS
Training and Certification
NI training is the fastest, most certain route to productivity with our
products. NI training can shorten your learning curve, save
development time, and reduce maintenance costs over the
application life cycle. NI schedules instructor-led courses in cities
worldwide, or can hold a course at your facility. NI also offers a
professional certification program that identifies individuals who
have high levels of skill and knowledge on using NI products.
Visit ni.com/training.
Hardware Services
NI Factory Installation Services
NI Factory Installation Services (FIS) is the fastest and easiest way to
use your PXI or PXI/SCXI combination systems right out of the box.
Trained NI technicians install the software and hardware and
configure the system to your specifications. NI extends the standard
warranty by one year on hardware components (controllers, chassis,
modules) purchased with FIS. To use FIS, simply configure your
system online with ni.com/pxiadvisor.
Professional Services
The NI Professional Services Team is comprised of NI applications
engineers, NI consulting services, and a worldwide National Instruments
Alliance Partner Program of more than 600 independent consultants
and integrators. Services range from
start-up assistance to turnkey system
integration. Visit ni.com/alliance for
more information.
Calibration Services
NI recognizes the need to maintain properly calibrated devices for
high-accuracy measurements. We provide manual calibration
procedures, services to recalibrate your products, and automated
calibration software specifically designed for use by metrology
laboratories. Visit ni.com/calibration.
Repair and Extended Warranty
We offer design-in consulting and product integration assistance
if you want to use our products for OEM applications. For
information about special pricing and services for OEM customers,
visit ni.com/oem for more information.
NI provides complete repair services for our products. Express repair
and advance replacement services are also available. We offer
extended warranties to help you meet project life-cycle requirements.
Visit ni.com/services.
ni.com • (800) 813-3693
National Instruments • info@ni.com
Appendix D
62
© 2005 National Instruments Corporation. All rights reserved. CVI, LabVIEW, National Instruments Alliance Partner, ni.com, NI-DAQ, and SCXI are trademarks or trade names of National Instruments.
Other products and company names listed are trademarks or trade names of their respective companies.
National Instruments Alliance Partner Program Members are business entities independent from National Instruments and have no agency, partnership or joint-venture relationship with National Instruments.
2004_4947_301_101_D
OEM Support