LINEAR APPLICATIONS OF OP-

AMPS

SYLLABUS

Applications of Op-Amps: Addition,

Subtraction, Multiplication, Current to

Voltage and Voltage to current conversion

operations , Integration, Differentiation,

Active Filters, Comparators, Schmitt

trigger, Instrumentation Amplifier, f to V

converter.

LINEAR APPLICATIONS

• Adder

• Subtractor

• Voltage follower

• Current to voltage converter

• Voltage to current converter

• Integrator

• Differentiator

• Active filters

Non-linear applications

• Comparators

• Logarithmic amplifiers

• Exponential amplifiers

• Peak detectors

• Precision rectifiers

• Waveform generators

• Clippers & clampers

If Rf = R, then Vout = v1+v2+v3

Non-inverting

Summing

Amplifier

Therefore, using the

superposition theorem, the

voltage V2 = V1

Vb & Vc = 0. Net resistance

= R+R/2

If RF=2R1, 1+RF/R1=3

Vo= Va+Vb+Vc

Determine the output voltage

=12

Averaging Amplifier

• A summing amplifier can be made to produce the mathematical average of the input voltages. The amplifier has a gain of Rf/R, where R is the value of each input resistor. The general expression for the output of an averaging amplifier is

• VOUT=-(Rf /R)(VIN1+VIN2+…+VINn)

• Averaging is done by setting the ratio Rf/R equal to the reciprocal of the number of inputs (n); that is , Rf/R=1/n.

ProblemShow that the amplifier in produces an output whose magnitude is the

mathematical average of the input voltages.

ADDER-SUBTRACTOR

It is possible to perform addition and subtraction simultaneously with a single

op-amp using the circuit shown above.

The output voltage Vo can be obtained by using superposition theorem. To find

output voltage Vo1 due to V1 alone, make all other input voltages V2, V3 and V4

equal to zero. The simplified circuit is shown in Fig.b. This is the circuit of an

inverting amplifier and its output voltage is,

4.3 (b) Op-amp adder- subtractor, (c)

Simplifier circuit for V2 = V3 = V4 = 0,

(d) Simplified circuit for V1 = V2 = V4 = 0

Similarly V02 = -V2

Now, the output voltage

Vo3 due to the input

voltage signal V3 alone

applied at the (+) input

terminal can be found by

setting V1=V2=V4=0. The

circuit now becomes a

non-inverting amplifier as

shown in Fig. 4.3 (d). The

voltage Va at the non-

inverting terminal is

Applying Thevinons theorem at inverting terminal, we get

=R/2

=1/3 R

=V3

So, the output voltage Vo3 due to V3 alone is

Similarly, it can be shown that the output

voltage V04: due to V4 alone is

Vo4 = V4 (4.14)

Thus, the output voltage Vo due to all four

input voltages is given by

Scaling Adder

• A different weight can be assigned to each input of a summing amplifier by simply adjusting the values of the input resistors. The output voltage can be expressed as

• VOUT=((Rf/R1)VIN1+(Rf/R2)VIN2+…+(Rf/Rn)VINn)

Problem

Determine the weight of each input voltage

for the scaling adder in Figure 7-16 and find

the output voltage.

Voltage to Current Converter (Transconductance Amplifier)

In many applications, one may have to convert a voltage signal to a

proportional output current. For this, there are two types of circuits

possible.

•V-I Converter with floating load

•V-I Converter with grounded load

Figure 4.8 (a) shows a voltage to current converter in which load ZL is

floating. Since voltage at node 'a' is vi therefore,

Fig. 4.8 Voltage to current converter

Vi

iL proportional to Vi

is controlled by R1

(b) Grounded load

(a) Floating load

iL = vi /R

Vi = iLR1

That is the input voltage vi is converted into an output current of vi /R1, it may be seen that

the same current flows through the signal source and load and,

therefore, signal source should be capable of providing this load current.

Since the op-amp is used in non-inverting

mode, the gain of the circuit is 1 + R/R= 2. The

output voltage is,

vo = 2v1 = vi + vo – iLR

Vi = iLR

iL = vi /R

A voltage-to-current converter with grounded load is shown in Fig. 4.8 (b). Let vi

be the voltage at node 'a'. Writing KVL, we get

As the input impedance of a non-inverting amplifier is very high, this circuit has the

advantage of drawing very little current from the source.

A voltage to current converter is used for low voltage dc and ac voltmeter, LED and

zenar diode tester.

Current to Voltage Converter (Trans-resistance Amplifier)

Photocell, photodiode and photovoltaic cell give an output

current that is proportional to an incident radiant energy or light.

The current through these devices can be converted to voltage by

using a current-to-voltage converter and thereby the amount of

light or radiant energy incident on the photo-device can be

measured.

Figure 4.9. shows an op-amp used as I to V converter. Since the

(-) input terminal is at virtual ground, current ii flows through the

feedback resistor Rf. Thus the output voltage v0 = - iiRf. It may be

pointed out that the lowest current that this circuit can measure

will depend upon the bias current of the op-amp. This means that

for 741 (bias current is 3 nA) can be used to detect lower

currents. The resistor Rf is sometimes shunted with a capacitor Cf

to reduce high frequency noise and the possibility of oscillations.

Inverting Current to Voltage Converter

Non-inverting Current to Voltage Converter.

INTEGRATORS AND

DIFFERENTIATORSIntegrators produce output voltages that are

proportional to the running time integral of the input

voltages. In a running time integral, the upper limit of

integration is t. Passive Integrator Gain is Less than

Unity. Attenuation is more. For perfect integration,

large values of RC are required.

Gain decreases with increase of frequency.

where C is the integration constant and is proportional to the value

of the output voltage vo at time t = 0 seconds.

For example, if the input is a sine wave, the output will be a cosine

wave.

If the input is a square wave, the output will be a triangular wave,

as shown in Figure 7-23(c) and (b), respectively.

Equation-7-23

When vin = 0, the integrator of Figure 7-23(a)

works as an open-loop amplifier. This is because

the capacitor CF acts as an open circuit (XCF = ∞)

to the input offset voltage Vio. In other words, the

input offset voltage Vio and the part of the input

current charging capacitor CF produce the error

voltage at the output of the integrator. Therefore, in

the practical integrator shown in Figure 7-25, to

reduce the error voltage at the output,

a resistor RF is connected across the feedback

capacitor CF. Thus, RF limits the low-frequency

gain and hence minimizes the variations in the

output voltage.

A practical Integrator

The frequency response of the basic integrator is shown in

Figure 7-24. In this figure, fb, is the frequency at which the

gain is 0 dB and is given by

Both the stability and the low-frequency roll-off problems can be

corrected by the addition of a resistor RF as shown in the

practical integrator of Figure 7-25. The frequency response of

the practical integrator is shown in Figure 7-24 by a dashed line.

In this figure, f is some relative operating frequency, and for

frequencies up to fa the gain RF/Ri is constant. However, after

fa, the gain decreases at a rate of 20 dB/decade. In other

words, between fa, and fb, the circuit of Figure 7-25 acts as an

integrator. The gain-limiting frequency fa is given by

Frequency Response

If the op-amp was ideal, an integrator as shown in above Figure would require just one

resistor, R, and one capacitor, C, and the relation between the output and input voltages

would be given by

fa < fb. If fa = fb/10, then RF = 10R1. The input signal will be integrated properly, if

time period T of the input signal

=fa =fb

The Integrating range is in between fa & fb

Frequency Response of an Ideal & Lossy Integrator

=fa =fb

Integrating Range

20db/decade

Square waves are integrated to triangles, triangles to parabolas etc.

EXAMPLE 7-15

In the circuit of Figure 7-23, R1CF = 1 second, and the input is a step (dc)

voltage, as shown in Figure 7-26(a). Determine the output voltage and

sketch it. Assume that the op-amp is initially nulled.

SOLUTION The input function is constant beginning at t = 0 seconds. That

is, Vm = 2 V for 0 < / =s 4. Therefore, using Equation (7-23),

The output voltage waveform is drawn in Figure 7-26(b); the waveform is

called a ramp function. The slope of the ramp is -2 V/s. Thus, with a

constant voltage applied at the input, the integrator gives a ramp at the

output.

Determine the rate of change of the output voltage in response to

the input square wave, as shown for the ideas integrator in Figure 7-

22(a). The output voltage is initially zero. The pulse width is 100μs.

Describe the output and draw the waveform.

Differentiator Circuit

Figure 7-27 Basic

differentiator, (a) Circuit,

(b) Frequency response.

fc = Unity Gain BW

Gain increases with frequency

Practical

Idealfc = Unity Gain BW

Figure 7-28

Figure 7-27b

Since IB = 0,

ic = iF

Since v1= v2 = 0 V,

fa is the frequency at which the gain is 0db and is given by

At fa gain is 0db

fC is the unity gain-bandwidth of the op-amp, and f is some relative operating

frequency.

Both the stability and the high-frequency noise problems can be corrected by

the addition of two components: R1 and CF, as shown in Figure 7-28(a), This

circuit is a practical differentiator, the frequency response of which is shown in

Figure 7-27(b) by a dashed line. From frequency f to fb, the gain increases at 20

dB/decade. However, after fb the gain decreases at 20 dB/decade. This 40-dB/

decade change in gain is caused by the R1C1 and RFCF combinations. The gain-

limiting frequency fb is given by (7-29)

where R1C1= RFCF , help to reduce significantly the effect of high-frequency

input, amplifier noise, and offsets. Above all, it makes the circuit more stable by

preventing the increase in gain with frequency. Generally, the value of fb and in

turn R1C1 and RFCF values should be selected such that

fa< fb< fc (7-30)

But v1= v2 = 0 V, because A is very large. Therefore,

Thus the output v0 is equal to the RFC times the negative

instantaneous rate of change of the input voltage vin with time.

Since the differentiator performs the reverse of the integrator's

function, a cosine wave input will produce a sine wave output, or a

triangular input will produce a square wave output. However, the

differentiator of Figure 7-27(a) will not do this because it has some

practical problems. The gain of the circuit

(RF /XC1) increases with increase in frequency at a rate of 20

dB/decade. This makes the circuit unstable. Also, the input

impedance XC decreases with increase in frequency, which makes

the circuit very susceptible to high-frequency noise. When

amplified, this noise can completely override the differentiated

output signal. The frequency response of the basic differentiator is

shown in Figure 7-27(b). In this figure, fa is the frequency at which

the gain is 0 dB and is given by

Figure 7-28 Practical differentiator, (a) Circuit, (b) Sine wave input and resulting cosine

wave output, (c) Square wave input and resulting spike output.

Steps For the Design of Practical Differentiator:

1. Select fa equal to the highest frequency of the input

signal to be differentiated. Then, assuming a value of C1

< 1 μF, calculate the value of RF .

2. Choose fb = 20 fa and calculate the values of R1, and CF

so that R1C1 = RFCF .

The differentiator is most commonly used in wave

shaping circuits to detect high-frequency components

in an input signal and also as a rate-of-change detector

in FM modulators.

EXAMPLE 7-16

Design a differentiator to differentiate an input signal that varies in frequency from

10 Hz to about 1 kHz. If a sine wave of 1 V peak at 1000 Hz is applied to the

differentiator of part (a), draw its output waveform.

SOLUTION (a) To design a differentiator, we simply follow the steps outlined

previously:

Problem

Determine the output voltage of the ideal op-amp

differentiator in Figure 7-26 for the triangular-wave input

shown.

Comparison Between Integrator & Differentiator.

The process of integration involves the

accumulation of signal over time and hence sudden

changes in the signal are suppressed. Therefore an

effective smoothing of signal is achieved and hence,

integration can be viewed as low-pass filtering.

The process of differentiation involves

identification of sudden changes in the input signal.

Constant and slowly changing signals are supressed

by a differentiator. It can be viewed as high-pass

filtering.

INSTRUMENTATION AMPLIFIER

In a number of industrial and consumer applications, one is required to measure

and control physical quantities. Some typical examples are measurement and

control of temperature, humidity, light intensity, water flow etc. These physical

quantities are usually measured with the help of transducers. The output of

transducer has to be amplified so that it can drive the indicator or display system.

This function is performed by an instrumentation amplifier. The important features

of an instrumentation amplifier are:

(i) high gain accuracy

(ii) high CMRR

(iii) high gain stability with low temperature coefficient (iv) low dc offset

(v) low output impedance

There are specially designed op-amps such as (µA725 to meet the above stated

requirements of a good instrumentation amplifier. Monolithic (single chip)

instrumentation amplifier are also available commercially such as AD521, AD524,

AD620, AD624 by Analog Devices, LM-363.XX (XX -»10,100,500) by National

Semiconductor and INA1O1,1O4, 3626, 3629 by Burr-Brown.

Non Linear Applications:

Precision rectifiers.•The major limitation of ordinary diode is that it cannot rectify

voltages below vγ (~ 0.6 V), the cut-in voltage of the diode.

• A circuit that acts like an ideal diode can be designed by placing

a diode in the feedback loop of an op-amp as in Fig. 4.10 (a).

Here the cut-in voltage is divided by the open loop gain A0L (~

104) of the op- amp so that vγ is virtually eliminated. When the

input Vi > Vγ /AOL then voA, the output of the op-amp exceeds Vγ

and the diode D conducts. Thus the circuit acts like a voltage

follower for input vi > Vγ /A0L (i.e., 0.6/104 = 60μv) and the

output v0 follows the input voltage vi during the positive half

cycle as shown in Fig. 4.10 (b).

•When vi is negative or less than Vγ /A0L, the diode D is off and

no current is delivered to the load RL except for small bias

current of the op-amp and the reverse saturation current of the

diode. This circuit is called the precision diode and is capable of

rectifying input signals of the order of mill volt. Some typical

applications of a precision diode discussed are:

•Half-wave Rectifier

•Full-Wave Rectifier

•Peak-Value Detector.

•Clipper.

•Clamper.

Fig. 4.10 (a) Precision diode, (b) Input and output

waveforms

When Vi is +ve, Voa is –ve. D2 is reverse biased and output is zero. When Vi is -ve,

Rf=R1, Voa is +ve, and D2 conducts even when the input is < 0.7V. The op-amp

should be a high speed version as it alternates between open loop & closed loop

operations. High Slew rate is required as the input passes through zero, Voa must

change from 0.6V to -0.6V. If the diodes are reversed, -ve output occurs.

An inverting amplifier can be converted into an ideal half-wave rectifier by adding

two diodes as shown in Fig. 4.11 (a). When vi is positive, diode D1 conducts

causing v0A to go to negative by one diode drop (~ 0.6 V). Hence diode D2 is

reverse biased. The output voltage v0 is zero, because, for all practical purposes,

no current flows through Rf and the input current flows through D1.

For negative input, i.e., vi < 0, diode D2 conducts and D1 is off. The negative input

vi forces the op-amp output v0A positive and causes D2 to conduct. The circuit

then acts like an inverter for Rf =R1 and output v0 becomes positive.

The input, output waveforms are shown in Fig. 4.11 (b). The op-amp in the circuit

of Fig. 4.11 (a) must be a high speed op-amp since it alternates between open

loop and closed loop operations. The principal limitation of this circuit is the slew

rate of the op-amp. As the input passes through zero, the op-amp output voA must

change from 0.6 V to - 0.6 V or vice-versa as quickly as possible in order to

switch over the conduction from one diode to the other. The circuit of Fig. 4.11(a)

provides a positive output. However, if both the diodes are reversed, then only

positive input signal is transmitted and gets inverted. The circuit, then provides a

negative output.

Full-wave Rectifier

Fig. 4.12 (a) Precision full wave rectifier, (b) Equivalent circuit for vi > 0; D1 is on

and D2 is OFF; op-amp A1 and A2 operate as inverting amplifier

A full wave rectifier or absolute value circuit is shown in Fig. 4.12 (a). For positive

input, i.e. vi > 0, diode D1 is on and D2 is off. Both the op-amps A1 and A2 act as

inverter as shown in equivalent circuit in Fig. 4.12 (b). It can be seen that v0 = vi

For negative input, i.e. vi< 0, diode D1 is off and D2 is on. The equivalent circuit

is shown in Fig. 4.12 (c). Let the output voltage of op-amp A1 be v. Since the

differential input to A2 is zero, the inverting input terminal is also at voltage v.

KCL at node 'a' gives

The equivalent circuit of Fig. 4.12 (c) is a non-inverting amplifier as shown in

Fig. 4.12 (d). The output v0 is,

(4.31)

(4.32)

Hence for vi < 0, the output is positive. The input and output waveforms are

shown in Fig. 4.12 (e). The circuit is also called an absolute value circuit as

output is positive even when input is negative. For example, the absolute

value of | +2 | and | -2 | is +2 only. It is possible to obtain negative outputs for

either polarity of input simply by reversing the diodes.

Fig. 4.12 (c) Equivalent circuit for v, < 0, (d) Equivalent circuit of (c)

Fig. 4.12 (e) Input and output waveforms

LOG -AMPLIFIER

Log Amplifiers

The basic log amplifier produces an output voltage as a

function of the logarithm of the input voltage; i.e.,

Vout = -K ln(Vin), where K is a constant.

Recall that the a diode has an exponential characteristic up to

a forward voltage of approximately 0.7 V. Hence, the

semiconductor pn junction in the form of a diode or the base

emitter junction of a BJT can be used to provide a logarithm

characteristic.

There are several applications of log and antilog amplifiers.

•Antilog computation may require functions such as In x,

log x or sinh x. These can be performed continuously with

log-amps.

•One would like to have direct dB display on digital

voltmeter and spectrum analyzer. Log-amp can easily

perform this function.

•Log-amp can also be used to compress the dynamic

range of a signal.

Log Amplifier

The fundamental log-amp circuit is shown in Fig. 4.34 (a)

where a grounded base transistor is placed in the feedback

path. Since the collector is held at virtual ground and the base

is also grounded, the transistor's voltage-current relationship

becomes that of a diode and is given by,

Since, IC = IE for a grounded base transistor

IS = emitter saturation current = 10-13 A

k = Boltzmann's Constant

T = absolute temperature (in 0K)

Since IE = IC

Taking natural Log on both sides,

we get

From Fig.4.18(a)

The circuit, however, has one problem.

•The emitter saturation current IS varies from transistor to

transistor and with temperature. Thus a stable reference

voltage Vref cannot be obtained.

•This is eliminated by the circuit given in Fig. 4.18 (b). The

input is applied to one log-amp, while a reference voltage is

applied to another log-amp. The two transistors are

integrated close together in the same silicon wafer. This

provides a close match of saturation currents and ensures

good thermal tracking.

V1

V2 Thermistor

Assume, Is1 = Is2 = Is (4.39)

and then, V1=

4.41

4.42

4.43

4.44

Thus reference level is now set with a single external voltage source. Its dependence on

device and temperature has been removed. The voltage Vo is still dependent upon

temperature and is directly proportional to T. This is compensated by the last op-amp

stage A4 which provides a non-inverting gain of (1 + R2/RTC ). NOW, the output voltage is,

Where RTC IS A PTC THERMISTOR.

Vin is converted in to current Ic = IEBOeVin/K

Antilog Amplifier

The circuit is shown in Fig. 4.19. The input V; for the antilog-amp is fed into the

temperature compensating voltage divider R2 and RTC and then to the base of

Q2. The output Vo of the antilog-amp is fed back to the inverting input of A1

through the resistor R1. The base to emitter voltage of transistors Ql and Q2 can

be written as

OP-AMP COMPARATORS

Non-Inverting Comparator.

(b) Input and Output wave-

forms when Vref is +ve

(c) Input and Output wave-

forms when Vref is -ve

Inverting Comparator.

b) Input and Output Wave

Forms when Vref is +ve and

c) Input and Output Wave

Forms when Vref is –ve

WINDOW COMPERATOR

Used in A.C Voltage Stabilizers.

Input(Volts) LED3 LED2 LED1

Less than 2V ON OFF OFF

Less than 4V & More than 2V

OFF ON OFF

More than 4V OFF OFF ON

If Vcc = 6V

Fig. 5.5 (a) Zero crossing detector and (b) Input and output waveforms

(a)

(b)

(a) Inverting Schmitt Trigger circuit (b)} (c) and (d) Transfer Characteristics of

Schmitt Trigger

INVERTING SCHMITT TRIGGER

The input voltage vi triggers the output vo every

time it exceeds certain voltage levels, VLT & VUT

If Vref = 0, then the voltage at the junction of R1

& R2 will form will determine VUT & VLT .

If Vi < VLT, Vo = +Vsat

Vi > VLT, Vo = -Vsat

ZERO

if Vref

is

Zero.

(a) Input and Output waveforms of Schmitt Trigger and (b) Output v0 versus Vi

plot of the hysteresis voltage.

If a sine wave frequency f=1/T is applied, a symmetrical square wave is obtained

at the output. The vertical edge is shifted in phase by from zero crossover

Where sin = VUT/Vm and Vm is the peak sinusoidal voltage.

NON-INVERTING SCHMITT TRIGGER

The input is applied to the non-inverting input terminal of the op-amp. To understand the

working of the circuit, let us assume that the output is positively saturated i.e. at +Vsat.

This is fedback to the non-inverting input through R1. This is a positive feedback.

Now though Vin is decreased, the output Continues its positive saturation level unless and

until the input becomes more negative than VLT. At lower threshold, the output changes

its state from positive saturation + Vsat to negative saturation - Vsat. It remains in negative

saturation till Vin increases beyond its upper threshold level VUT.

Now VA = voltage at point A =IinR2 = VUT

As op-amp input current is zero, I in entirely passes through R1.

Chattering can be defined as production of multiple output transitions the input

signal swings through the threshold region of a comparator. This is because of

the noise.

Eliminates Comparator Chatter.

S.No. Schmitt Trigger. Comparator.

1. The feedback is used. No feedback is used.

2. Op-amp is used in closed loop mode. Used in open loop mode.

3. No false triggering. False Triggering.

4. Two different threshold voltages exists as VUT & VLT

Single reference voltage Vref or –Vref.

5. Hysteresis exists. No Hysteresis exists.

Comparison.

Square & Triangular

waveform generation

Square Wave Generator

Vref = β Vsat

Let V0 initially be + Vsat. The capacitor charges

through R to + β Vsat. Then V0 goes to – Vsat . The

cycle repeats and output will be a Square Wave.

Where β = R2/(R1+R2)

VSat

+_

Triangular/rectangular wave generator.

Operation of the CircuitLet the output of the Schmitt trigger is + Vsat. This forces current + Vsat/R1

through C1, charging C1 with polarity positive to left and negative to right.

This produces negative going ramp at its output, for the time interval t1 to t2.

At t2 when ramp voltage attains a value equal to LTP of Schmitt trigger, the

output of Schmitt trigger changes its stage from

+ Vsat to -Vsat,

Now direction of current through C reverses. It discharges and recharges in

opposite direction with polarity positive to right and negative to left. This

produces positive going ramp at its output, for the time interval t2 to t3. At t3

when ramp voltage attains a value equal to UTP of Schmitt trigger, the

output of Schmitt trigger changes its state from - Vsat to + Vsat and cycle

continues.

The circuit acts as free running waveform generator producing

triangular and rectangular output waveforms.

Vo’ =+Vsat = Vin

-Vramp

+ -

0V

Vo(PP) = 2Vramp

Vin = VSat

Substitute V(pp) from Eqn. (4)