Analog Devices and Circuits:Operational Amplifier

Dr. Rand Basil Alhashimie

rand.basil@ishik.edu.iq

Tishk International University

Mechatronics Engineering Department

Analog Devices and Circuits ME227

Lecture 9: 19/20-05-2019

Introduction to Operational Amplifier

● Early operational amplifiers (op-amps) were used primarily to perform

mathematical operations such as addition, subtraction, integration, and

differentiation—thus the term operational. These early devices were

constructed with vacuum tubes and worked with high voltages. Today’s op-

amps are linear integrated circuits (ICs) that use relatively low dc supply

voltages and are reliable and inexpensive.

● The standard operational amplifier (op-amp) symbol is shown in Figure 1(a). It

has two input terminals, the inverting (-) input and the noninverting (+) input,

and one output terminal. Most op-amps operate with two dc supply voltages,

one positive and the other negative, as shown in Figure 1(b), although some

have a single dc supply.

Introduction to Operational Amplifier

Usually these dc voltage

terminals are left off the

schematic symbol for

simplicity but are understood

to be there. Some typical op-

amp IC packages are shown

in Figure 1(c).

Figure 1

Ideal Op-Amp

● To illustrate what an op-amp is, let’s consider its ideal characteristics. A

practical op-amp, of course, falls short of these ideal standards, but it is much

easier to understand and analyze the device from an ideal point of view.

● First, the ideal op-amp has infinite voltage gain and infinite bandwidth.

Also, it has an infinite input impedance (open) so that it does not load the

driving source. Finally, it has a zero output impedance. Op-amp

characteristics are illustrated in Figure 2(a).

● The input voltage, Vin, appears between the two input terminals, and the

output voltage is AvVin, as indicated by the internal voltage source symbol.

The concept of infinite input impedance is a particularly valuable analysis tool

for the various op-amp configurations.

Ideal Op-Amp

Figure 2

Practical Op-Amp

● Although integrated circuit (IC) op-amps approach parameter values that can

be treated as ideal in many cases, the ideal device can never be made. Any

device has limitations, and the IC op-amp is no exception.

● Op-amps have both voltage and current limitations. Peak-to-peak output

voltage, for example, is usually limited to slightly less than the two

supply voltages. Output current is also limited by internal restrictions such as

power dissipation and component ratings.

● Characteristics of a practical op-amp are very high voltage gain, very

high input impedance, and very low output impedance. These are labelled

in Figure 2(b).

Practical Op-Amp

● Another practical consideration is that there is always noise generated within

the op-amp. Noise is an undesired signal that affects the quality of a desired

signal. Today, circuit designers are using smaller voltages that require high

accuracy, so low-noise components are in greater demand.

● All circuits generate noise; op-amps are no exception, but the amount can be

minimized. Internal Block Diagram of an Op-Amp A typical op-amp is made up

of three types of amplifier circuits: a differential amplifier, a voltage amplifier,

and a push-pull amplifier, as shown in Figure 3.

Practical Op-Amp

● The differential amplifier is the input stage for the op-amp. It provides

amplification of the difference voltage between the two inputs.

● The second stage is usually a class A amplifier that provides additional gain.

Some op-amps may have more than one voltage amplifier stage.

● A push-pull class B amplifier is typically used for the output stage.

Figure 3

Input Signals Modes

● Recall that the input signal modes are determined by the differential amplifier

input stage of the op-amp.

● Differential Mode In the differential mode, either one signal is applied to an

input with the other input grounded or two opposite-polarity signals are applied to

the inputs.

● When an op-amp is operated in the single-ended differential mode, one input is

grounded and a signal voltage is applied to the other input, as shown in Figure 4.

● In the case where the signal voltage is applied to the inverting input as in part

(a), an inverted, amplified signal voltage appears at the output. In the case where

the signal is applied to the noninverting input with the inverting input grounded,

as in Figure 4(b), a non inverted, amplified signal voltage appears at the output.

Single Ended Differential Modes

Figure 4

Double Ended Differential Modes

Figure 5

In the double-ended differential mode, two opposite-polarity (out-of-phase)

signals are applied to the inputs, as shown in Figure 5(a). The amplified

difference between the two inputs appears on the output. Equivalently, the

double-ended differential mode can be represented by a single source connected

between the two inputs, as shown in Figure 5(b).

Common Mode

● In the common mode, two signal voltages of the same phase, frequency, and

amplitude are applied to the two inputs, as shown in Figure 6. When equal input

signals are applied to both inputs, they tend to cancel, resulting in a zero output

voltage.

Figure 6

Common Mode

● This action is called common-mode rejection. Its importance lies in the

situation where an unwanted signal appears commonly on both op-amp inputs.

Common-mode rejection means that this unwanted signal will not appear on

the output and distort the desired signal.

● Common-mode signals (noise) generally are the result of the pick-up of

radiated energy on the input lines, from adjacent lines, the 60 Hz power line, or

other sources.

Op-Amp Parameters

● Common-Mode Rejection Ratio Desired signals can appear on only one input or

with opposite polarities on both input lines. These desired signals are amplified

and appear on the output as previously discussed. Unwanted signals (noise)

appearing with the same polarity on both input lines are essentially

cancelled by the op-amp and do not appear on the output. The measure of

an amplifier’s ability to reject common-mode signals is a parameter called the

CMRR (common-mode rejection ratio).

Op-Amp Parameters

● Ideally, an op-amp provides a very high gain for differential-mode signals and

zero gain for common-mode signals. Practical op-amps, however, do exhibit a

very small common-mode gain (usually much less than 1), while providing a high

open-loop differential voltage gain (usually several thousand).

● The higher the open-loop gain with respect to the common-mode gain, the better

the performance of the op-amp in terms of rejection of common-mode signals.

Op-Amp Parameters

● This suggests that a good measure of the op-amp’s performance in rejecting

unwanted common-mode signals is the ratio of the open-loop differential

voltage gain, Aol, to the common-mode gain, Acm. This ratio is the common-

mode rejection ratio, CMRR.

● The higher the CMRR, the better. A very high value of CMRR means that the

open-loop gain, Aol , is high and the common-mode gain, Acm , is low. The

CMRR is often expressed in decibels (dB) as

Open Loop Voltage Gain (AoL)

● The open-loop voltage gain, Aol , of an op-amp is the internal voltage gain of the

device and represents the ratio of output voltage to input voltage when

there are no external components. The open-loop voltage gain is set entirely

by the internal design.

● Open-loop voltage gain can range up to 200,000 (106 dB) and is not a well-

controlled parameter.

● Datasheets often refer to the open-loop voltage gain as the large-signal voltage

gain.

● A CMRR of 100,000, for example, means that the desired input signal

(differential) is amplified 100,000 times more than the unwanted noise (common-

mode). If the amplitudes of the differential input signal and the common-mode

noise are equal, the desired signal will appear on the output 100,000 times

greater in amplitude than the noise. Thus, the noise or interference has been

essentially eliminated.

Example: A certain op-amp has an open-loop differential voltage gain of 100,000

and a common-mode gain of 0.2. Determine the CMRR and express it in decibels.

Solution:

Aol = 100,000 and Acm = 0.2

Then CMRR = Aol/Acm = 500,000

CMRR(dB) = 20 log(500,000) = 114dB

Maximum Output Voltage Swing (VO(p-p))

● With no input signal, the output of an opamp is ideally 0 V. This is called the

quiescent output voltage. When an input signal is applied, the ideal limits of

the peak-to-peak output signal are ±VCC.

● In practice, however, this ideal can be approached but never reached Vo(pp)

varies with the load connected to the op-amp and increases directly with load

resistance.

● Some op-amps do not use both positive and negative supply voltages. One

example is when a single dc voltage source is used to power an op-amp that

drives an analog-to-digital converter.

Maximum Output Voltage Swing (VO(p-p))

● In this case, the op-amp output is designed to operate between ground and a

full scale output that is near (or at) the positive supply voltage. Op-amps that

operate on a single supply use the terminology VOH and VOL to specify the

maximum and minimum output voltage. (Note that these are not the same as

the digital definitions of VOL and VOH .)

Input Offset Voltage

● The ideal op-amp produces zero volts out for zero volts in. In a practical op-amp,

however, a small dc voltage, VOUT(error), appears at the output when no

differential input voltage is applied. Its primary cause is a slight mismatch of the

base-emitter voltages of the differential amplifier input stage of an op-amp.

● As specified on an op-amp datasheet, the input offset voltage, VOS , is the

differential dc voltage required between the inputs to force the output to zero volts.

Typical values of input offset voltage are in the range of 2 mV or less. In the ideal

case, it is 0 V.

Input Offset Voltage

● The input offset voltage drift is a parameter related to VOS that specifies how

much change occurs in the input offset voltage for each degree change in

temperature. Typical values range anywhere from about 5µV per degree Celsius

to about 50µV per degree Celsius. Usually, an op-amp with a higher nominal

value of input offset voltage exhibits a higher drift.

Input Bias Current

You have seen that the input terminals of a bipolar differential amplifier are the

transistor bases and, therefore, the input currents are the base currents.

The input bias current is the dc current required by the inputs of the amplifier to

properly operate the first stage. By definition, the input bias current is the average

of both input currents and is calculated as follows:

Input Impedance

● Two basic ways of specifying the input impedance of an op-amp are the

differential and the common mode. The differential input impedance is the total

resistance between the inverting and the noninverting inputs, as illustrated in

Figure 7(a).

● Differential impedance is measured by determining the change in bias current

for a given change in differential input voltage. The common-mode input

impedance is the resistance between each input and ground and is measured

by determining the change in bias current for a given change in common-mode

input voltage. It is depicted in Figure 7(b).

Input Impedance

Figure 7

Input Offset Current

● Ideally, the two input bias currents are equal, and thus their difference is zero.

In a practical op-amp, however, the bias currents are not exactly equal.

● The input offset current, IOS , is the difference of the input bias currents,

expressed as an absolute value.

● Actual magnitudes of offset current are usually at least an order of magnitude

(ten times) less than the bias current. In many applications, the offset current

can be neglected.

● However, high-gain, high-input impedance amplifiers should have as little IOS

as possible because the difference in currents through large input resistances

develops a substantial offset voltage, as shown in Figure 8.

Input Offset Current

Output Impedance

The output impedance is the resistance viewed from the output terminal of the op-

amp, as indicated in Figure 9.

Figure 9

Slew Rate

● The maximum rate of change of the output voltage in response to a step input

voltage is the slew rate of an op-amp. The slew rate is dependent upon the

high-frequency response of the amplifier stages within the op-amp.

● Slew rate is measured with an op-amp connected as shown in Figure 10(a).

This particular op-amp connection is a unity-gain. It gives a worst-case

(slowest) slew rate. Recall that the high frequency components of a voltage

step are contained in the rising edge and that the upper critical frequency of

an amplifier limits its response to a step input. For a step input, the slope on

the output is inversely proportional to the upper critical frequency. Slope

increases as upper critical frequency decreases.

Slew Rate

Figure 10

Slew Rate

A pulse is applied to the input and the resulting ideal output voltage is indicated in

Figure 10(b). The width of the input pulse must be sufficient to allow the output to

“slew” from its lower limit to its upper limit. A certain time interval, is required for

the output voltage to go from its lower limit -Vmax to its upper limit +Vmax once

the input step is applied. The slew rate is expressed as

where ΔVout = +Vmax - (-Vmax ). The unit of slew rate is volts per microsecond

(V/µs).

Example: The output voltage of a certain op-amp appears as

shown in Figure below in response to a step input. Determine

the slew rate.

Solution:

Slew Rate = ΔVout/Δt

= (9 - (-9) ) / 1µs

= 18 V/µs

Op_Amp with Negative Feedback

An op-amp can be connected using negative feedback to stabilize the gain and

increase frequency response. Negative feedback takes a portion of the output and

applies it back out of phase with the input, creating an effective reduction in gain.

This closed loop gain is usually much less than the open-loop gain and

independent of it.

The closed-loop voltage gain is the voltage gain of an op-amp with external

feedback. The amplifier configuration consists of the op-amp and an external

negative feedback circuit that connects the output to the inverting input. The

closed-loop voltage gain is determined by the external component values and can

be precisely controlled by them.

Closed Loop Voltage Gain, Ad

Non Inverting Amplifier

An op-amp connected in a closed-loop configuration as a noninverting amplifier with a

controlled amount of voltage gain is shown in Figure 11. The input signal is applied to the

non inverting (+) input. The output is applied back to the inverting input through the

feedback circuit (closed loop) formed by the input resistor Ri and the feedback resistor Rf.

This creates negative feedback as follows. Resistors Ri and Rf form a voltage-divider

circuit, which reduces Vout and connects the reduced voltage Vf to the inverting input. The

feedback voltage is expressed as:

Non Inverting Amplifier

The difference of the input voltage, Vin , and the feedback voltage, Vf , is the

differential input to the op-amp, as shown in Figure 12. This differential voltage

is amplified by the open-loop voltage gain of the op-amp (Aol ) and produces an

output voltage expressed as

The attenuation, B , of the feedback circuit is

Substituting BVout for Vf in the Vout equation,

Non Inverting Amplifier

Non Inverting Amplifier

Example: Determine the closed-loop voltage gain of the

amplifier in Figure below.

Voltage Follower

The voltage-follower configuration is a special case of the noninverting amplifier where all

of the output voltage is fed back to the inverting (-) input by a straight connection, as

shown in Figure 13. As you can see, the straight feedback connection has a voltage gain

of 1 (which means there is no gain). The closed-loop voltage gain of a noninverting

amplifier is 1/B as previously derived. Since B = 1 for a voltage-follower, the closed-loop

voltage gain of the voltage-follower is:

The most important features of the voltage-follower

configuration are its very high input impedance and its very low

output impedance. These features make it a nearly ideal buffer

amplifier for interfacing high-impedance sources and low-

impedance loads.

Inverting Amplifier

An op-amp connected as an inverting amplifier with a controlled amount of

voltage gain is shown in Figure 14. The input signal is applied through a series

input resistor Ri to the inverting (-) input. Also, the output is fed back through Rf

to the same input. The noninverting (+ ) input is grounded.

Inverting Amplifier

At this point, the ideal op-amp parameters mentioned earlier are useful in

simplifying the analysis of this circuit. In particular, the concept of infinite input

impedance is of great value.

Inverting Amplifier

An infinite input impedance implies zero current at the inverting input. If there is

zero current through the input impedance, then there must be no voltage drop

between the inverting and noninverting inputs. This means that the voltage at the

inverting input (-) is zero because the noninverting (+) input is grounded. This zero

voltage at the inverting input terminal is referred to as virtual ground.

This condition is illustrated in Figure 15(a). Since there is no current at the

inverting input, the current through Ri and the current through Rf are equal, as

shown in Figure 15(b).

Iin = If

Example: Given the op-amp configuration in Figure below,

determine the value of Rf required to produce a closed-loop

voltage gain of - 100.

Summing Amplifier

● A summing amplifier has two or more

inputs, and its output voltage is proportional

to the negative of the algebraic sum of its

input voltages.

● A two-input summing amplifier is shown in

Figure 16, but any number of inputs can be

used. The operation of the circuit and

derivation of the output expression are as

follows. Two voltages, VIN1 and VIN2 , are

applied to the inputs and produce currents

I1 and I2 , as shown in Figure 16.

Summing Amplifier

● Using the concepts of infinite input impedance and virtual ground, you can

determine that the inverting input of the op-amp is approximately 0 V and has

no current through it. This means that the total current IT , which goes through

Rf divides into I1 and I2 at summing point A , as indicated in Figure 16.

Summing Amplifier

● The previous equation shows that the

output voltage has the same magnitude

as the sum of the two input voltages but

with a negative sign, indicating inversion.

● A general expression is given in equation

below for a unity-gain summing amplifier

with n inputs, as shown in Figure 17

where all resistors are equal in value.

Example: Determine the output voltage in Figure below

Solution:

VOUT = -( VIN1 + VIN2 + VIN3) = -(3 V + 1V + 8 V) = -12 V

Summing Amplifier with Gain Greater Than Unity

When Rf is larger than the input resistors, the amplifier has a gain of Rf /R, where

R is the value of each equal-value input resistor. The general expression for the

output is Equation below

As you can see, the output voltage has the same magnitude as the sum of all the

input voltages multiplied by a constant determined by the ratio - (Rf /R ).

Example: Determine the output voltage for the summing

amplifier in Figure below

Averaging Amplifier

A summing amplifier can be made to produce the mathematical average of the

input voltages. This is done by setting the ratio Rf /R equal to the reciprocal of the

number of inputs (n ).

You obtain the average of several numbers by first adding the numbers and then

dividing by the quantity of numbers you have.

Example: Show that the amplifier in

Figure below produces an output whose

magnitude is the mathematical average

of the input voltages.

Scaling Adder

A different weight can be assigned to each input of a summing amplifier by simply

adjusting the values of the input resistors. As you have seen, the output voltage

can be expressed as

The weight of a particular input is set by the ratio of Rf to the resistance, Rx , for

that input (Rx ! R1 , R2 , . . . Rn ). For example, if an input voltage is to have a

weight of 1, then Rx = Rf . Or, if a weight of 0.5 is required, Rx = 2Rf . The smaller

the value of input resistance Rx , the greater the weight, and vice versa.

Example: Determine the weight of each input voltage for the scaling adder in

Figure below and find the output voltage.

Application

● D/A conversion is an important interface process for converting digital signals to

analog (linear) signals. An example is a voice signal that is digitized for storage,

processing, or transmission and must be changed back into an approximation of

the original audio signal in order to drive a speaker.

● One method of D/A conversion uses a scaling adder with input resistor values

that represent the binary weights of the digital input code. Although this is not the

most widely used method, it serves to illustrate how a scaling adder can be

applied.

● A more common method for D/A conversion is known as the R/2R ladder

method. The R/2R ladder is introduced here for comparison although it does not

use a scaling adder.

● Figure 18 shows a four-digit digital-to-analog converter (DAC) of this type (called

a binary-weighted resistor DAC).

Application

● The switch symbols represent transistor

switches for applying each of the four

binary digits to the inputs. The inverting

input (-) is at virtual ground, and so the

output voltage is proportional to the current

through the feedback resistor Rf (sum of

input currents). The lowest-value resistor R

corresponds to the highest weighted binary

input (2^23). All of the other resistors are

multiples of R and correspond to the binary

weights 2^22, 2^21, and 2^20.

Example: Determine the output voltage of the DAC in Figure below. The sequence

of four digit binary codes represented by the waveforms in Figure (b) are applied

to the inputs. A high level is a binary 1, and a low level is a binary 0. The least

significant binary digit is D0 .

● From Figure (b), the first binary input code is

0000, which produces an output voltage of 0

V.

● The next input code is 0001 (it stands for

decimal 1). For this, the output voltage is -

0.25.

● The next code is 0010, which produces an

output voltage of -0.5V.

● The next code is 0011, which produces an

output voltage of -0.25 + (-0.5) = -0.75V

● Each successive binary code increases the

output voltage by -0.25V. So, for this

particular straight binary sequence on the

inputs, the output is a stair step waveform

● going from 0 V to -3.75V in -.025V steps, as

shown in this figure.

● If the steps are very small, the output

approximates a straight line (linear).

Integrators and Differentiators

The Ideal Integrator An ideal integrator is shown in Figure 19. Notice that the

feedback element is a capacitor that forms an RC circuit with the input resistor.

How a Capacitor Charges

To understand how an integrator works, it is important

to review how a capacitor charges. Recall that the

charge Q on a capacitor is proportional to the

charging current (IC) and the time (t).

Also, in terms of the voltage, the charge on a

capacitor is

From these two relationships, the capacitor voltage

can be expressed as

● This expression has the form of an equation for a straight line that begins at

zero with a constant slope of IC/C. Remember from algebra that the general

formula for a straight line is y = mx + b. In this case, y = VC, m = IC/C, x = t, and

b = 0.

● Recall that the capacitor voltage in a simple RC circuit is not linear but is

exponential.

● This is because the charging current continuously decreases as the capacitor

charges and causes the rate of change of the voltage to continuously decrease.

● The key thing about using an op-amp with an RC circuit to form an integrator is

that the capacitor charging current is made constant, thus producing a straight-

line (linear) voltage rather than an exponential voltage. Now let’s see why this is

true.

In Figure 20, the inverting input of the op-amp is at virtual ground (0 V), so the

voltage across Ri equals Vin. Therefore, the input current is

● If Vin is a constant voltage, then Iin is also a constant because the inverting

input always remains at 0 V, keeping a constant voltage across Ri . Because

of the very high input impedance of the op-amp, there is negligible current at

the inverting input. This makes all of the input current go through the

capacitor, as indicated in Figure 20, so IC = Iin

● The Capacitor Voltage Since Iin is constant, so is IC. The constant IC

charges the capacitor linearly and produces a linear voltage across C. The

positive side of the capacitor is held at 0 V by the virtual ground of the op-

amp. The voltage on the negative side of the capacitor, which is the op-amp

output voltage, decreases linearly from zero as the capacitor charges, as

shown in Figure 21. This voltage, VC, is called a negative ramp and is the

consequence of a constant positive input.

The Output Voltage Vout is the same as the voltage on the negative side of the

capacitor. When a constant positive input voltage in the form of a step or pulse (a

pulse has a constant amplitude when high) is applied, the output ramp decreases

negatively until the op-amp saturates at its maximum negative level. This is

indicated in Figure 21.

Rate of Change of the Output Voltage The rate at which the capacitor charges,

and therefore the slope of the output ramp, is set by the ratio IC/C, as you have

seen. Since IC = Vin/Ri, the rate of change or slope of the integrator’s output

voltage is ΔVout /Δt.

Example:

(a) Determine the rate of change

of the output voltage in response

to the input square wave, as

shown for the ideal integrator in

Figure (a). The output voltage is

initially zero. The pulse width is

100µs

(b) Describe the output and draw

the waveform.

Practical Integrator

● The ideal integrator uses a capacitor in the feedback path, which is open to

dc. This implies that the gain at dc is the open-loop gain of the op-amp.

● In a practical integrator, any dc error voltage due to offset error will cause the

output to produce a ramp that moves toward either positive or negative

saturation (depending on the offset), even when no signal is present.

● Practical integrators must have some means of overcoming the effects of

offset and bias current. Various solutions are available, such as chopper

stabilized amplifiers; however, the simplest solution is to use a resistor in

parallel with the capacitor in the feedback path, as shown in Figure 22.

Practical Integrator

● The feedback resistor, Rf, should be

large compared to the input resistor

Rin, in order to have a negligible

effect on the output waveform. In

addition, a compensating resistor,

Rc, may be added to the

noninverting input to balance the

effects of bias current.

The OP-AMP Differentiator

The Ideal Differentiator An ideal differentiator is shown in Figure 23. Notice how

the placement of the capacitor and resistor differ from the integrator. The

capacitor is now the input element, and the resistor is the feedback element. A

differentiator produces an output that is proportional to the rate of change of the

input voltage.

The OP-AMP Differentiator

To see how the differentiator works, apply a positive-going ramp voltage to the

input as indicated in Figure 24. In this case, IC ! Iin and the voltage across the

capacitor is equal to Vin at all times (VC = Vin ) because of virtual ground on the

inverting input. From the basic formula, VC = (IC /C )t , the capacitor current is

Since the current at the inverting input is negligible, IR = IC . Both currents are

constant because the slope of the capacitor voltage (VC/t ) is constant. The

output voltage is also constant and equal to the voltage across Rf because one

side of the feedback resistor is always 0 V (virtual ground).

Notice in Equation that the term VC/t is the slope of the input. If the slope increases, Vout

increases . If the slope decreases, Vout decreases. The output voltage is proportional to the

slope (rate of change) of the input. The constant of proportionality is the time constant, Rf *C.

Example: Determine the output voltage of the ideal op-amp differentiator in Figure

below for the triangular-wave input shown.

The Practical Differentiator

The ideal differentiator uses a capacitor in series with the inverting input. Because

a capacitor has very low impedance at high frequencies, the combination of Rf

and C form a very high gain amplifier at high frequencies. This means that a

differentiator circuit tends to be noisy because electrical noise mainly consists of

high frequencies.

The solution to this problem is simply to add a resistor, Rin , in series with the

capacitor to act as a low-pass filter and reduce the gain at high frequencies. The

resistor should be small compared to the feedback resistor in order to have a

negligible effect on the desired signal. Figure 25 shows a practical differentiator. A

bias compensating resistor may also be used on the noninverting input.

End of Analog Devices and Circuits ME227