Download EE 174 Spring 2016

EE 174
Spring 2016
Operational Amplifiers
Contents
• Introduction
• Brief of History
• Fundamentals of Op-Amps
• Basic operation
• Gain
• Offset
• Applications
Introduction
• Operational Amplifier (Op-Amp) name comes from the fact that it was originally used to perform
mathematical operations.
• Op-Amp is an active circuit element which is basic component used to build analog circuits.
• Op-Amp is a low cost integrating circuit consisting of transistors, resistors, diodes and capacitors.
• Op-Amp amplify an input signal produces an output voltage equal to the difference between the two
input terminals multiplied by the gain A.
• Op-Amps are two-port networks in which the output voltage or current is directly proportional to
either input voltage or current. Four different kind of ampliers exits:
•
•
•
•
•
Voltage amplifier: Av = Vo / Vi
Current amplifier: Ai = Io / Ii
Transconductance amplifier: Gm = Io / Vi
Transresistance amplifier: Rm = Vo / Ii
Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting
Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters (High, Low, Band
Pass and Notch Filters), Schmitt trigger, Comparators, A/D converters.
Brief History of Op-Amp
Monolithic IC Op-Amp
Vacuum Tube Op-Amps (1930’s-1940’s) Solid State Discrete Op-Amps
Dual-supply voltage of +300/-300 V
(1960’s)
Output swing +/- 50 volts
Open-loop voltage gain of 15,000 to 20,000,
Slew rate of +/- 12 volts/µsecond
Maximum output current of 1 mA
George Philbrick
Dual-supply voltage of +15/-15 V
Output swing +/- 11 volts
Open-loop voltage gain of 40,000,
Slew rate of +/- 1.5 volts/µsecond
Maximum output current of 2.2 mA
• First created in 1963 μA702
by Fairchild Semiconductor
• μA741 created in 1968,
became widely used due to
its ease of use 8 pin, dual
in-line package (DIP)
• Further advancements
include use of field effects
transistors (FET), greater
precision, faster response,
and smaller packaging
Internal Op-Amp Circuits
Op-Amp Ideal, Equivalent Circuit, Characteristics and Features
Ideal Op-Amp
Op-Amp Equivalent
Circuit
Op-Amp Characteristics
Op-Amp Symbol
741 Op-Amp Features
+V2: Non-inverting input
-V1: Inverting input
+Vs: Positive source PS
-Vs: Negative source PS
Vout: Output voltage
ON: Offset Null
NC: Not Connected
The Ideal Op-Amp Assumptions
vd = v+ - vvo = Avd = A(v+ – v-)
Note: v+ = v2, v- = v1
1) The input impedance Ri is infinite - i.e. no current flows into either input.
2) The output impedance Ro is zero - i.e. the op-amp can drive any load impedance to
any voltage.
3) The open-loop gain (A) is infinite.
4) The bandwidth is infinite.
5) The output voltage is zero when the input voltage difference is zero.
6) The slew rate is infinite
Op-Amp two basic operations
The Non-inverting Op Amp
The Inverting Op Amp
Op-Amp Gain
Open loop gain: This form of gain is measured when
no feedback is applied to the op amp.
Closed loop gain: This form of gain is measured when
the feedback loop is closed and the overall gain of the
circuit is much reduced. It has two forms, signal gain
and noise gain.
The expression for the gain of a closed-loop amplifier
involves the open-loop gain. If G is the actual gain, NG
is the noise gain, and AVOL is the open-loop gain of the
amplifier, then:
Since the open-loop gain is very high, the closed-loop gain of the circuit is simply the noise gain.
Loop gain: The difference between the open-loop gain and the closed-loop gain is known as the loop gain.
This is useful information because it gives you the amount of negative feedback that can apply to the
amplifier system.
Signal Gain and Noise Gain
Signal gain: This is the gain applied to the input signal, with the feedback loop connected. It can be
inverting or non-inverting. It can even be less than unity for the inverting case. Signal gain is the gain that
we are primarily interested in when designing circuits.
Noise gain: This is the gain applied to a noise source in series with an op amp input. It is also the gain applied
to an offset voltage. Noise gain is equal to the signal gain of a non-inverting amp. It is the same for either an
inverting or non-inverting stage. It is the noise gain that is used to determine stability.
Summary of Signal Gain and Noise Gain
Op-Amp Gain and Bandwidth
The Voltage Gain (A) of the operational amplifier
can be found using the following formula:
and in Decibels or (dB) is given as:
Gain versus Bandwidth
Applying feedback will
reduce the gain but increase
the bandwidth.
Gain Bandwidth Product (GBP)
Gain-bandwidth (GBW) product is defined as the
open-loop gain multiplied by the bandwidth. The
GBW product can be used to calculate the
closed-loop gain bandwidth.
GBW = A x BW
where A is in ratio (not in dB)
GBW/closed-loop gain = closed-loop BW
Closed-loop BW = GBW / A
Example 1: If the GBW is 10 MHz @ unity gain
and the closed-loop gain is 10.
Closed-loop BW = 10 MHz/10 = 1 MHz
Example 2: GBW 1 MHz @ unity gain, closed
loop gain 70 dB. Closed-loop BW?
Gain: 20 log A = 70 dB  log A = 70/20 = 3.5
 A = 103.5 = 3162
Closed-loop BW = 1 MHz/3162 = 316 Hz
Gain Bandwidth Product Examples
Given the open loop gain graph for an op-amp as shown.
a) Find the closed loop BW for gain of 88 dB.
b) Find the closed loop gain in dB for a BW of 200 Hz
c) Find the closed loop gain in dB for a BW of 250K Hz.
Solutions:
a) 88 dB = 20 log A  A = 104.4 = 25,119
Closed loop BW = 1 M / 25,119 ≈ 40 Hz
b) Gain A = 1M / 200 = 5,000 or 70.45 dB
c) Gain A = 1M / 250K = 4 or 12 dB
12dB
250K
Bode Plot Example
Given a transfer function:
H(s) =
=
100
100
=
𝑠
𝑠+30
30 (30 +1)
100
1
1
= 3.3 𝑠
𝑠
30 ( +1)
( +1)
30
30
Constant gain = 3.3
Gain in dB = 20 log (3.3) ≈ 10 dB
Pole at 30 rad/sec
Phase = 45o at 30 rad/sec
Op-Amp Saturation
The op amp has three distinct regions of operation: Linear region (−Vsat
< Vo < Vsat), positive (Vo > Vsat) and negative saturation Vo < −Vsat .
The output voltage of a practical amplifier cannot exceed certain
threshold value is called saturation. A voltage amplifier behaves
linearly, i.e., Vo/Vi = Av = constant as long as the output voltage remains
below the “saturation” voltage,
−Vsat < Vo < Vsat
Note that the saturation voltage, in general, is not symmetric.
For an amplifier with a given gain, Av, the above range of Vo translate
into a certain range for Vi
− Vsat < Vo < Vsat
− Vsat < Av Vi < Vsat
− Vsat / Av < Vi < Vsat / Av
Any amplifier will enter its saturation region if Vi is raised above certain
limit. The figure shows how the amplifier output clips when amplifier is
not in the linear region.
Example: For Av = 105, Vsat,1 = L- = -12V, Vsat,2 = L+ = +15V, find range of input Vi to prevent saturation.
Solution: -12V / 105 < Vi < 15V / 105 or -0.12mV < Vi < 0.15mV
Op-Amp slew rate basics
Square Wave: The slew rate (SR)of an op amp is the rate of
change in the output voltage caused by a step change on the
input. It is measured as a voltage change in a given time - typically
V / µs. The slew rate may be very important in many applications.
For the given graph, it takes 2 µs to swing 10 V so SR = 5V/ µs.
Sine Waves: The maximum rate of change for a sine wave occurs
at the zero crossing and may derived as follows:
Given vo = VP sin(2π f t)
Slew Rate (SR) =
𝑑vo
𝑑𝑡
V/µs
max
= 2π f VP cos(2π f t)
= 2π f VP
t=0
SR = 2π fmax VP
Example:
Given an application where an op amp
is required to amplify a signal with a
peak amplitude of 5 volts at a frequency
of 25kHz. What is the slew rate
requirement?
Solution:
SR = 2 π x 25 000 x 5 = 785,000V/s or
0.785V/µs or higher would be required
Op-Amp slew rate Example
Given: Slew rate S0 = 1 V/μs; input voltage vin (t) = sin(2π × 105t); closed loop gain AV =10.
1) Sketch the theoretically correct output and the actual output of the amplifier in the same graph.
2) What is the maximum frequency that will not violating the given slew rate?
Solutions:
1) With the closed-loop voltage gain of 10  vo (t) = 10 sin(2π × 105t)
= Aω = A x 2π x f = 10 × 2π × 105 = 6.28 V/μs
max
2) fmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHz
Op-Amp Offset Parameters
Input Offset Voltage:
Input offset voltage is defined as the voltage that must be applied between
the two input terminals of an OPAMP to null or zero the output. fig. 2, shows
that two dc voltages are applied to input terminals to make the output zero.
Vio = Vdc1 – Vdc2
Vdc1 and Vdc2 are dc voltages and RS represents the source resistance. Vio is the
difference of Vdc1 and Vdc2. It may be positive or negative. For a 741C OPAMP
the maximum value of Vio is 6mV. It means a voltage ± 6 mV is required to
one of the input to reduce the output offset voltage to zero. The smaller the
input offset voltage the better the differential amplifier, because its
transistors are more closely matched.
fig. 2
Input Offset Current:
The input offset current Iio is the difference between the currents into
inverting and non-inverting terminals of a balanced amplifier.
Iio = | IB1 – IB2 |
The Iio for the 741C is 200nA maximum. As the matching between two input
terminals is improved, the difference between IB1 and IB2 becomes smaller,
i.e. the Iio value decreases further. For a precision OPAMP 741C, Iio is 6 nA
Op-Amp Offset Parameters
Input Bias Current:
The input bias current IB is the average of the current entering the input terminals
of a balanced amplifier i.e.
IB = (IB1 + IB2 ) / 2
For 741C IB(max) = 700 nA and for precision 741C IB = ± 7 nA
Offset Voltage Adjustment Range:
741 OPAMP have offset voltage null capability. Pins 1 and 5 are marked offset null
for this purpose. It can be done by connecting 10 K ohm pot between 1 and 5 as
shown in fig. 3.
By varying the potentiometer, output offset voltage (with inputs grounded) can be
reduced to zero volts. Thus the offset voltage adjustment range is the range
through which the input offset voltage can be adjusted by varying 10 K pot. For the
741C the offset voltage adjustment range is ± 15 mV.
fig. 3
Op-Amp Offset Example
Example: Determine the output voltage in each of the following cases for the open
loop differential amplifier of fig. 4:
a.vin 1 = 5 m V dc, vin 2 = -7 µVdc
b.vin 1 = 10 mV rms, vin 2= 20 mV rms
Specifications of the OPAMP are given below:
A = 200,000, Ri = 2 M Ω , R O = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage
swing = ± 14V.
Solution:
(a). The output voltage of an OPAMP is given by
Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero
when the input signals are zero.
(b). The output voltage equation is valid for both ac and dc input signals. The output
voltage is given by
Thus the theoretical value of output voltage vo = -2000 V rms. However, the OPAMP
saturates at ± 14 V. Therefore, the actual output waveform will be clipped as shown
fig. 5. This non-sinusoidal waveform is unacceptable in amplifier applications.
Op-Amp Applications
Differential Amplifier
If R1 = R2 and R3 = R4
Op-Amp Applications
Find i, if and vo.
i=0A
For the ideal op amp shown below, what should be the
value of resistor Rf to obtain a gain of 5?
Op-Amp Applications
1) Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity
markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (AV), both as a ratio and
as a figure in units of decibels (dB):
Solutions: Gain = AV = 1.468 = 3.335 dB
2) Given gain RF/RS = 100, GBW product = A0 ω0 = K = 4π × 106 .
Determine the overall 3-dB bandwidth of the cascade amplifier Solutions: The 3-dB bandwidth for each amplifier is:
below:
Thus, the bandwidth of the cascade amplifier is 4π × 104,
the cascade amplifier gain is A1A1= 100 × 100 = 104.
To achieve the same gain with a single-stage amplifier
having the same K, the bandwidth would be
Op-Amp Applications
Problem: Suppose that we need an amplifier with input resistance of 500 kΩ or greater and a voltage gain of -10. The
feedback resistors are to be implemented in integrated form and have values of 10 kΩ or less to conserve chip area.
Choose a suitable circuit configuration and specify the resistance values.
Solution:
Gain of -10  Use inverting Op-Amp where gain
A = - R2/R1 = -10
To attain desired input resistance R1 = 500 kΩ  R2 = 5MΩ.
These values exceed the maximum values allowed.
To attain large input resistance with moderate resistances
for an inverting amplifier, we cascade a voltage follower
with an inverter.
To achieve the desired gain = - 10.
Choose R1 = 1 kΩ  R2 = 10 kΩ.
R2
The Op-Amp Comparators
The comparator is an electronic decision making circuit that makes use of an operational amplifiers very high gain
in its open-loop state, that is, there is no feedback resistor. The Op-amp comparator compares one analogue
voltage level with another analogue voltage level, or some preset reference voltage, VREF and produces an output
signal based on this voltage comparison. In other words, the op-amp voltage comparator compares the magnitudes
of two voltage inputs and determines which is the largest of the two.
Then an op-amp comparator can be configured to operate in what is called an inverting or a non-inverting
configuration.
The Op-Amp Comparators
The Schmitt Trigger
The Schmitt trigger is a comparator application which switches the output negative when the input passes upward
through a positive reference voltage. It then uses positive feedback of a negative voltage to prevent switching back to
the other state until the input passes through a lower threshold voltage, thus stabilizing the switching against rapid
triggering by noise as it passes the trigger point. That is, it provides feedback which is not reversed in phase, but in this
case the signal that is being fed back is a negative signal and keeps the output driven to the negative supply voltage until
the input drops below the lower design threshold.
The Schmitt Trigger
A variation of the previous circuit is the single power supply op-amp Schmitt trigger circuit. This circuit does not
require positive and negative voltage for the op-amp to operate. The negative voltage of the op-amp is connected
directly to the ground (0V) of the circuit.
The calculation of the two threshold points is different. To calculate
the High Threshold Level, you solve the following formula:
RTOT = (R1 x RFB) / (R1 + RFB)
VTHRESHOLD_HIGH = V * R2 / (R2 + RTOT)
For the Low Threshold Level, you solve the following formula:
RTOT = (R2 x RFB) / (R2 + RFB)
VTHRESHOLD_LOW = V x RTOT / (R1 + RTOT)
with Schmitt Trigger
Let V = 5V, R1 = R2 = RFB = 10 kΩ.
3.4 V
RTOT = 10 kΩ // 10 kΩ = 5 kΩ
1.7 V
VTH = 5 V x (10 kΩ / 15 kΩ) = 3.4 V
5.0 V
VTL = 5 V x (5 kΩ / 15 kΩ) = 1.7 V
0.0 V
without Schmitt Trigger
www.isu.edu.tw/upload/52/.../dept_34_lv_3_17785.ppt
References:
• http://www.ume.gatech.edu/mechatronics_course/OpAmp_F08.ppt
• http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/operational-amplifier-models/
• http://www.ti.com/lit/an/slaa068a/slaa068a.pdf
• http://www.radio-electronics.com/info/circuits/opamp_basics/operational-amplifier-slew-rate.php
• http://www.planetanalog.com/author.asp?section_id=483&doc_id=562347
• http://www-ferp.ucsd.edu/najmabadi/CLASS/ECE60L/02-S/NOTES/opamp.pdf
• http://www-inst.eecs.berkeley.edu/~ee105/fa14/lectures/Lecture06-Non-ideal%20Op%20Amps%20(Offset-Slew%20rate).pdf
• http://nptel.ac.in/courses/117107094//lecturers/lecture_6/lecture6_page1.htm
• http://www.ti.com/ww/en/bobpease/assets/AN-31.pdf
• http://www.cs.tut.fi/kurssit/TLT-8016/Chapter2.pdf
• http://www.electronics-tutorials.ws/opamp/op-amp-comparator.html
• http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/schmitt.html
• http://lpsa.swarthmore.edu/Bode/BodeExamples.html
• http://www.allaboutcircuits.com/worksheets/inverting-and-noninverting-opamp-voltage-amplifier-circuits/
• Fundamentals of Electrical Engineering, Giorgio Rizzoni, McGraw-Hill Higher Education