Chapter 2 Operational Amps

8/11/2019 Chapter 2 Operational Amps

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Figure 2.1Circuit symbol for the op amp.

Op Amps

Op Amps are Operational Amplifiers Simple, standardized conceptual amplifier which you can buy for pennies

Wide range of specifications and specialties

Low cost, precision, fast, low power, high voltage, high impedance,

high current drive, multiple inputs, differential, etc. etc.

Manufacturers may sell hundreds of versions in multiple packages

Microelectronic Circuits Sedra/Smith Copyright 2010 Oxford University Press, Inc.ECE 3210 Fall 2013 John Lindsey


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Op Amps

Op Amps versatile to work withcan do a wide variety of tasks Feedback makes op amps reliable, manufacturable

Op amps considered here have three terminals

Inverting input

Noninverting input

Output

Dont forget op amps actually must have power connections, usually two in

this chapter

VCCpositive supply

VEEnegative supply

May have other inputs like clock signals, frequency compensation, offset

correction, etc.



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Op Amps

Inverting input

Noninverting input

Output



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Op Amps

Inverting input

Noninverting input

Output



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Op Amps

Inverting input

Noninverting input

Output

Remember, VCCand VEEneed to be connected to ground for real in theLab or insimulation!

Ground will be the circuit common nodeall signals referenced

to ground



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Op Amps

Inverting input

Noninverting input

Output

Whats the minimum number of terminals required for a single op amp?

Whats the minimum number of terminals required an op amp in a

package containing two op amps?

D

D



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Op Amps

Inverting input

Noninverting input

Output

Whats the minimum number of terminals required for a single op ampIC? 3+2=5

Whats the minimum number of terminals required an op amp in an IC

package containing two op amps?

-- 3+3+2=8

(power and ground internally routed)



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Figure 2.3Equivalent circuit of the ideal op amp.

Op Amps

Op amps sense the difference between the inputs and amplifiesthat difference, everything is referenced to the circuit ground

We should write

Inverting input=v1-ground=v1-0

Noninverting input=v2-ground=v2-0

These voltage differences appear at

the output amplified by the gain A

(v3-0)=A[(v2-0)-(v1-0)]

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Op Amps

Op amps sense the difference between the inputs and amplifiesthat difference, everything is referenced to the circuit ground

We should write

Inverting input=v1-ground=v1-0

Noninverting input=v2-ground=v2-0

These voltage differences appear at

the output amplified by the gain A

(v3-0)=A[(v2-0)-(v1-0)]

These zeros just make it confusing, so

leave them off and write:

v3=A(v2-v1)

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Note, the noninverting input and the out put have the same signthey

are in phase, while the inverting input and the output have opposite

sign, they are opposite phase



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Op Amps

Real op amps seem to approach some of the performance of ideal op amps,but not quite. Ideal characteristics:

Infinite input impedance

Zero output impedance

Zero common-mode gain

(infinite common-mode-

rejection)

Infinite open-loop gain A

Infinite bandwidth



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Op Amps


Infinite input impedance means

The ideal op amp draws no

current from the inputs:

i1=i2=0

No power drawn from the

inputs

The inputs behave as if they

are open circuits



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Op Amps


Zero output impedance means

The output acts as an ideal

voltage source, able to supply

any amount of current yet

maintain the voltage

The output always supplies

the voltage A(v2-v1)

The output impedance is zero



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Op Amps


Zero output impedance means

The output acts as an ideal

voltage source, able to supply

any amount of current yet

maintain the voltage

The output always supplies

the voltage A(v2-v1)

The output impedance is zero

No resistance

herea short



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Op Amps


Zero common-mode gain means

infinite common-mode-rejection

Only the difference is amplified

If v1=v2the output is 0

If v1=v2=1.5, the difference is

0, output will be 0, and the

common-mode signal is 1.5



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Op Amps


Zero common-mode gain means

infinite common-mode-rejection

Only the difference is amplified

If v1=v2the output is 0

If v1=v2=1.5, the difference is 0,

output will be 0, and the

common-mode signal is 1.5

If v1=1.75, v2=1.70, then

v1-v2=(1.75-1.70)=0.05

The difference is 0.05 (this

is amplified)

The common-mode signal

is the average signal,

1.725V (this is rejected)



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Op Amps

Differential and common-mode signals Differential input signal is the difference

between the inputs

= Common-mode input signal is

=12 ( )



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Op Amps

Differential and common-mode signals Differential input signal is the difference

between the inputs

= Common-mode input signal is

=12 ( )

vIdand vIcmcan be used to write the inputs in

a little different and very useful way

= 2

= 2 Half the difference between 1 and 2 is addedto the common signal to make input 2

Half the difference between 1 and 2 is

subtracted from the common signal to make

input 1



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Op Amps

Signals on the inputs v1and v2

Common mode and difference signal

= =12 ( )

Difference between v1

and v2=



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Op Amps

Ideally, only the difference signal is amplified, while the common signal is rejected



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Op Amps

Ideally, only the difference signal is amplified, while the common signal is rejected

Common mode signals are everywherepower supply ripple, instability, noise of

many kinds. Concentrating on the difference signal gives much cleaner results

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Op Amps


Infinite input impedance

Zero output impedance

Zero common-mode gain

(infinite common-mode-

rejection)

Infinite open-loop gain A

Infinite bandwidth



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Op Amps


Infinite open-loop gain A means

Ideally, the open-loop gain is

very large, might as well call it

infinite

Infinite gain is impractical of

course, but open loop gain of real

amplifiers can be very large



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Op Amps


Infinite bandwidth means

The gain does not depend on

frequency

Gain at dc levels is the same as

the gain of a time-varying signal

at any frequency

Of all the ideal specifications,

real op amps may diverge the

most on bandwidth



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Op Amps


Infinite open-loop gain A means

Open-loop is without feedback

(with feedback gain is closed-

loop gain)

The open loop gain of transistors

is not stable in manufacturing

processes, since it is influenced

by properties which naturally

vary, even if slightly

Adding feedback can stabilize the

gain and make amplifier systems

based on transistors much easier

to manufacture. Feedback has

other benefits as well



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Exercise 2.2

a, c; you do b, d

Ideal op amp except gain A=103

= =12 ( )

= 2

= 2 3=

v1=1.002V, v2=0.998V

= = 0.998 1.002 = 4=12 =12 0.998 1.002 = 13= = 1000 0.004 = 4



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Exercise 2.2

a, c; you do b, d

Ideal op amp except gain A=103

= =12 ( )

= 2

= 2 3= v2=0V, v3=2V

=3 = 21000= 2

=

2 = 0 0.002

2 = 1

= 2 = 0.001 0.0022 = 2



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Exercise 2.3

find v3 and Gain

3= = = = = 3=



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Exercise 2.3

find v3 and Gain

3= = = = = 3=

= =

= 3= = = /( ) =



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Figure 2.5The inverting closed-loop configuration.

Inverting Configuration

Finally a real op amp

The non-inverting input is at ground

The value of v1is referenced to ground at the inverting input



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Figure 2.5The inverting closed-loop configuration.

Review -- Inverting Configuration

R1and R2make a feedback network

The value at the inverting is a result the voltage

divider formed by R1and R2 Information (the voltage, in this case) is fed-back

from the output to the inputfeedback

R2makes a loop, it closes the loop


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Figure 2.6Analysis of the inverting configuration. The circled numbers indicate the order of the analysis steps.


The open-loop gain is A, now define the closed-loop gain as G

Assume the open-loop gain A is very very large

Output is = and A is very large The larger A is, the closer gets to zero, since approaches zero

=

= 0 Because of the feedback and the

huge open-loop gain A, the inputs

are forced to be almost the same;

in fact just say the voltage on the

inverting terminal, v1should be

nearly zero since= =ground



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2v1=0 (virtual ground)


Because = the two terminals track each other in potential There is a virtual short circuit between terminals 1 and 2

Since terminal 2 is at ground, terminal 1

is brought to a virtual ground as well

But, terminal 1 and 2 are not shorted

together! The huge open-loop gain of the

amplifier makes it seem as if they are



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Now can calculate the current i1since

The source voltage vIis known

Inverting input voltage v1is known (the virtual ground)

Know the voltage on both sides of R1, so i1is easy to find

= = 0

=



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Since the op amp has ideal infinite input resistance, terminal 1 will not accept the

current, and it all must flow on through R2

Assume load supplies a lower impedance path to ground

= =



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Since we know the voltage at all nodes, we can calculate the output voltage

= =



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Since we know the voltage at all nodes, we can calculate the output voltage

= = = =

And vO/vIis the closed loop, Gain G

= =

While the open loop gain A is

infinite, the closed-loop gain is

finite, and controlled by easily

manufactured and controlled passiveresistors.

The uncontrolled gain A is tamed by

negative feedback

0



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Inverting ConfigurationEffect of Finite Gain

Assume the open-loop gain A is not necessarily large

No longer does

approach zero, instead v1is not so close to ground

This changes the current through the feedback resistors

Still assuming infinite input resistance

= = (0 ) =

0

= = ( )

=


=


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As before, all of i1goes through R2 Assume load supplies a lower impedance path to ground

Output voltage=v1-drop in R2

= =

After some math its possible

to find the closed loop gain

=

/1 1 / /



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As A goes to infinity, G goes to the ideal value = =

/1 1 / / 0

To make the infinite open loop gain approximation, and

have the virtual ground at the inverting input, its not

necessary for A to go to infinity, it just must be largeenough so that the quantity 1 / /becomesvery small compared to one.

This happens when 1 /



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ECE 3210 Fall 2013 John Lindsey

Example 2.1

Calculate closed loop gain for several values of open loop gain for the

inverting configuration with = 1 = 100compare to the idealvalue of g and find the percentage error. Find v1when vI=0.1V

= /

1 1 / /

Substitute these values into the equations for closed loop gain

and try with a variety of values for A with Excel or MATLAB

= 100% = ( /)(/) 100%

=

v1is not virtual ground, it is changed by the open loop gain:

Output voltage is related to input voltage by close loop gain:

Substitute in to get a relation for v1in terms of vI: =

Microelectronic Circuits Sedra/Smith Copyright 2010 Oxford University Press, Inc.


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ECE 3210 Fall 2013 John Lindsey

Example 2.1

Calculate closed loop gain for several values of open loop gain for the

inverting configuration with = 1 = 100compare to the idealvalue of g and find the percentage error. Find v1when vI=0.1V

As open lop gain becomes 10,000 and

above, the closed loop gain changes verylittle for further changes in open loop gain;

the error becomes small for large A; tiny

changes in v1for large changes of large A

A G error v1

1E+00 0.98 -99% -9.80E-02

1E+01 9.01 -91% -9.01E-02

1E+02

49.75

-50%

-4.98E-02

1E+03 90.83 -9.2% -9.08E-03

1E+04 99.00 -1.0% -9.90E-04

1E+05 99.90 -0.10% -9.99E-05

1E+06 99.99 -0.01% -1.00E-05

Microelectronic Circuits Sedra/Smith Copyright 2010 Oxford University Press, Inc.


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Inverting Configuration Input Resistance

Input resistance of the ideal op amp is infinite; for the inverting configuration,

however, the feedback network provides a path from input to ground through the

resistors Input resistance is defined as / The input current provided by the source is exactly the current flowing through R1

this makes finding the input resistance simple since already know i1from the source

voltage:

This is a problem, actually, since the

closed loop gain G is proportional to

high input impedance makes for low

gain (or an impractically high R2)

Example 2.2 shows one work around


= = = = /=


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Inverting Configuration Output Resistance

Output resistance of the ideal op amp is a shortthe voltage source inside the op

amp can supply any amount of current required by the load to keep the outputvoltage steady


=

=

= 0

For the inverting configuration,

the feedback network does not

influence the output resistance


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Figure 2.10A weighted summer.


Weighted Summer

Adds or sums the inputs

Gives a weighta multiplierindividually for each input Based on the inverting configurationrelies on two concepts

Currents at a node sum, so can add currents

Voltages through resistors in series a multiplied by the resistor divider

The current through any of the

inputs is given by Ohms law

= And the currents sum at the

inverting input

=


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Weighted Summer

The summed current flows through the feedback resistor and the virtual ground at

the inverting input simplifies finding the output voltage= = 0 = Substitute in for the summed

currents, and get the output in

terms of the weighted and summedcurrents

= =

= +

The values for R1Rncan be

individually adjusted


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Weighted Summer

Example D2.7 = 5 max output 10V current 1mA Since we are given maximum output currents and voltage, using the virtual

ground Rf is specified = = .= 10,000

v2weighted 5x

=

Picking a value for R1 then sets a

value for R2

=

5

Or=

if pick R1=10,000R2=2,000


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Figure 2.11A weighted summer capable of implementing summing coefficients of both signs.


Weighted Summer when some inputs have a different sign

By adding a second op amp, inputs of either sign can be weighted and added

Possible since each stage produces the inverse of the input


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Figure 2.12The noninverting configuration.


Noninverting Configuration

A non-inverting closed-loop amplifier can be made by

Maintaining the feedback network Moving the source to the non-inverting input

To analyze

Assume ideal op amp with infinite gain

Use the difference equation for the input

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Figure 2.13Analysis of the noninverting circuit. The sequence of the steps in the analysis is

indicated by the circled numbers.



Virtual short between the inputs means the source voltage appears at both inputs

Assuming there is an open at the output (and infinite output impedance of the opamp), the input voltage seen at the inverting input will flow through R1 to ground;

so we can get the current through R1 as

The virtual short is not a real short; any current in R1 must come from R2.

Fortunately, the op amp can supply current, since the output is open.


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Figure 2.13Analysis of the noninverting circuit. The sequence of the steps in the analysis is

indicated by the circled numbers.



Can get the output voltage as the sum of

Voltage drop in R2, which is known since the current is known Voltage at the noninverting node, which is forced to be the same as the input

voltage by the virtual short (which is due to the huge open-loop gain A)

= Collect the voltage terms to get closed loop gain:

= = 1


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Noninverting Configuration with Finite Gain

In the case where A is finite, the closed loop gain can be calculated as

= 1 1

1

When A is infinite, the closed loop gain becomes 1

As with the inverting case,

the bottom of the fraction

becomes close to one when

the A 1 Its the same result because

the feedback network is the

same (short the sources and

they are identical)


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Noninverting Configuration with Finite Gain

Input resistance is infinite

because there are no paths to

ground on the noninverting

terminal

Output resistance is again ashort because of the internal

ideal voltage source

As with the inverting case, the bottom of the fraction becomes close to one

when the A 1 Its the same result as the inverting configuration because the feedback

network is the same (short the sources and they are identical)

=

1 1 1


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Noninverting Configuration as a Voltage Follower

To make the noninvertingconfiguration into a voltage

follower want the closed loop gain

to become 1

The infinite input impedance and zero output impedance make the

noninverting configuration perfect as a voltage follower, also known as a

buffer

A good buffer circuit wont interfere with the source (high input

impedance for a voltage source) and will be able to drive any load (output

impedance a short)


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Noninverting Configuration as a Voltage Follower

This circuit has 100% negative feedback

G1

R1 becomes open

R2 becomes a short

= 1

0


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Example 2.14

A transducer with an open-circuit voltage of 1V and source resistance of

1M is connected to a 1k load. Find the load voltage for the case of

direction connection and with a unity gain voltage follower

For the direct connection, its a voltage divider

= += 1 3

.=1mV


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Example 2.14

A transducer with an open-circuit voltage of 1V and source resistance of

1M is connected to a 1k load. Find the load voltage for the case of

direction connection and with a unity gain voltage follower

For the voltage follower

= = = 1

vs

1V

Rs

1megaRL

1k

follower

1vo


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Difference Amplifiersin more detail

Difference amplifies amplify the difference between the two inputs

Rejects all other signals

Real circuits, even reasonable simulations of imaginary circuits will have some

common-mode gain

=

= = = =12 ( )

A useful figure of merit for real circuits is the common mode rejection ratio

how well a circuit amplifies the difference signal divided by how poorly the

amplifier amplifies the common signal

= 20 ||||


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Figure 2.15 Representing the input signals to a differential amplifier in terms of their

differential and common-mode components.




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Figure 2.16A difference amplifier.


A difference amplifier using feedback to improve stability and performance

Essentially a inverting and noninverting configuration together

The voltage divider of R3 and R4 is used to attenuate the noninvertingsignal from 1 down to the level of the inverting signal



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Figure 2.17Application of superposition to the analysis of the circuit of Fig. 2.16.


Difference AmplifiersAnalyze by Superposition

Analyzer by superpositionset one input to ground, find the output, then set the

other input to ground, find the output, and add the outputs together

Works because the system is linear


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First ground the noninverting input vI2 Call the resulting output vO1 R3 and R4 on the noninvering input have no effect on the circuit output, since

they are at circuit ground on one side, and the infinite impedance input on the

otherno current flows, no voltage exists at input 2

Replacing those with a short to ground then exactly the circuit is the inverting

configuration


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First ground the noninverting input vI2 Call the resulting output vO1 R3 and R4 on the noninvering input have no effect on the circuit output, since

they are at circuit ground on one side, and the infinite impedance input on the

otherno current flows, no voltage exists at input 2

Replacing those with a short to ground then exactly the circuit is the inverting

configuration

The gain is:

=


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Second ground the inverting input vI1 Call the resulting output vO2

This is the noninverting configuration, with the voltage divider R3 and R4 on theinput to deal with

= 3 So the output v

O2is given by

= 1 =

3 1

But we want the gain of the two

inputs to be the same magnitude


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Set the gain portions equal to each other:

= 3 1 = =


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= 3 1 = = Solving only for the gains

3 1

=


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= 3 1 = = Solving only for the gains

3 1

=

Rearrange

3 =

1

1 =


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R3 and R4 should be like R2 and R1 to match the gains on the inputs

+=

+

=

So back to the output from the non-inverting input

= 1 =

3 1

=


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Now finish the superposition by adding the two outputs to find the output

= =

=

= This is exactly the form for a difference amplifier, where the gain is

=

Now must find the common mode gain


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Difference Amplifierscommon mode response

Apply only a common mode signal

Tie the inputs together with Vicm

Then proceed as usual

Find the current in the feedback loop

Find the output voltage

Gain is Vout/VIcm


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Find the current in R1

To do that, find the voltage drop across R1

On the left side of R1 there is the common mode voltage

On the right side, there is a virtual short between the two inputs

The feedback network is the thing making the virtual short through the

huge open loop gain Ainput 1 is driven to be a virtual short with

input 2 The voltage on input 2 is from

the R3 R4 divider network

= +


71/121



The drop in R1 is the difference between the source and input 1

= =

= 1

3

= 3 1

Now can find the output voltage

Know the drop in R2 due to i2

Know the voltage at input 1

due to the virtual short

ff f


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= = + -

+

R

Am= = 3 1 3

= =

3 = =

3

1

Diff A lifi d


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Am= = 3 1 3

The common mode gain will be zero when

= 1

which is the condition established for getting the gain on

both inputs the same:=

In reality there will be some common

mode gainCMRR will not be infinite

because its impossible to matchresistors perfectly

Diff A lifi i t i t


74/121

Figure 2.19Finding the input resistance of the difference

amplifier for the case R3= R1and R4= R2.


Difference Amplifiersinput resistance

=

While this difference amplifier can amplify the difference and reject the

common signal, it suffers the low input resistance of the inverting

configurations topology

Find the differential input resistance Rid given that=

The two inputs each have iIflowing to their resistors The virtual short connects the resistors, making a loop

= 2This topology suffers from low input

resistance if high gain is required

I t t ti A lifi


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Figure 2.20A popular circuit for an instrumentation

amplifier. (a)Initial approach to the circuit


Instrumentation Amplifier

Adding a high input impedance amplifier on the input side improves the

low input resistance issue of the difference amplifier

The voltage follower is just the circuit The voltage follower can have voltage gain if there is some resistance in

the feedback network

Adding some gain before the difference amplifier can be donegive the

difference circuit a larger difference to amplify

A1 and A2 are noninverting with gain

= 1

I t t ti A lifi


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The difference signal at the inputs is

=

Each input is amplified by 1 At the output of the voltage followers (the input of the difference amplifier A3)

= 1 1 or

(1

)

The difference amplifier gain is R4/R3 so

= =

3 1

RR

I t t ti A lifi


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However

This configuration will amplify the common mode signal as well as the

differential mode signal Mismatch between the set of R1/R2 will cause a change in the differential

signal

To change the differential gain, a pair of resistors must be changed instead

of just one resistor

Removing the ground at node X

resolves the issues by relying on the

virtual short circuits between the inputs

created by the infinite open loop gain

I t t ti A lifi


78/121


amplifier.



The virtual short between the terminals causes vI1and vI2to be on either side of

2RI The voltage drop in RIis then the difference between the inputsjust what

should be amplified

The current in RIis then

= =

This current then flows through the R2 resistors, producing the voltage at the

input of the difference stage of the amplifier

Instr mentation Amplifier


79/121


amplifier.



Each R2 has a voltage drop of

= = R = R = = =

The difference between these outputs is the difference amplifiers (A3) input

= =

= 1 RR



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amplifier.



The difference amplifier A3 has a gain of R4/R3, so the overall output is

vO= ()(/3) = 1 and the overall gain is = =

1



81/121


amplifier.



Gains independence from variations in R2 is removedif there are two

different resistors R2: R2 and R2 the differential gain may change, but the

difference signal is preserved since current flows through the resistors in series

==RR3 1

R R

Overall CMRR of the system is improved since

Common signals at vI1and vI2will pass through the first stage, but they will

not be amplified

The difference signal is amplified in the first stage

Changing the gain can be done

by adjusting the value of RIalone

Integrators and Differentiators


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Figure 2.22The inverting configuration with general impedances in the feedback and the feed-in paths.


Integrators and Differentiators

Redraw the inverting configuration with impedances instead of resistors

The closed loop transfer function is now =

Inverting Integrator


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Figure 2.24(a)The Miller or inverting integrator. (b)Frequency

response of the integrator.



Replace resistor R2 with a capacitor

There still is a virtual ground at the inverting input due to the high open loop gain All of the input voltage drop appears across R, and the current in the

capacitor can then be found

= = After the clock starts at t=o, current through R leaves charge on the plates of

capacitor C, where it accumulates (assuming no leakage) The charge will be:

=



84/121





Recall that V=Q/C, so the voltage accumulating will be, and we should include an

initial condition VC

=1

V

And Ohms law relates current and voltage with resistance

= 1



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= 1

This is the voltage on the input side of the capacitor; the output side is the

opposite sign

= So the output is given by the integral:

=1



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A different way to analyze starts with the general closed loop transfer function

=

Then substitute in the impedances Z1=R and Z2=1/sC

= 1 = 1sCRAnd = =

1CR



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The circuit has a transfer function magnitude

= 1CR

And phase of

= 90 As frequency doubles (an octave) magnitude

decreases by half, 20Log0.5=6dB

(20dB/decade)

Where gain becomes 0dB, the frequency is

the integrator frequency, or the inverse of the

time constant

=1= 1



88/121



The circuit has a transfer function magnitude

= 1CR

At frequency=0, the capacitor passes no signalit

is open. The closed-loop transfer function goes to

infinity, which means the system is now running

without feedbackopen loop gain dc signals can be amplified greatly by this

circuita problemfix it by adding a

resistor in parallel with the capacitor to allow

some dc feedback gain ofRF/R

Figure 2.25The Miller integrator with a large resistance RF

connected in parallel with Cin order to provide negative

feedback and hence finite gain at dc.

Vo sV(s)= RF/R1 s C R

Op Amp Differentiator


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Microelectronic Circuits, Sixth Edition Sedra/Smith Copyright 2010 by Oxford University Press, Inc.

Figure 2.27 (a)A differentiator. (b)Frequency response of a differentiator with a time-constant CR.

The virtual ground causes the input to appear across the capacitor

Q=VC, and current is Q/time=CV/time

() =()

= () =

Op Amp Differentiator


90/121


Figure 2.27 (a)A differentiator. (b)Frequency response of a differentiator with a time-constant CR.

In the frequency domain, the transfer function

=

=

Tend to be unstable

Tend to magnify noise spikes

Offset Voltage


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Figure 2.28Circuit model for an op amp with input offset voltageVOS.

Real op amps never perfectly amplify

There is always some extra voltage on v1 or v2, so one or both inputs are slightly

different from the actual input value, and this wrong difference is amplified

= Inside the op amp there are transistors, capacitances, resistances, inductances all

with non-ideal values, all with slightly different values for each op amp

manufactured

A tiny difference might onlymake a V difference, but

many of those can add and

multiplycan easily become a

mV difference

mV still seems small, but

multiply by a large open loopgain Adand its a big factor

Offset Voltage


92/121


Figure 2.28Circuit model for an op amp with input offset voltageVOS.

Real op amps never perfectly amplify

There is always some extra voltage on v1 or v2, so one or both inputs are slightly

different from the actual input value, and this wrong difference is amplified

= Inside the op amp there are transistors, capacitances, resistances, inductances all

with non-ideal values, all with slightly different values for each op amp

manufactured, and under use also vary

A tiny difference might onlymake a V difference, but

many of those can add and

multiplycan easily become a

mV difference

mV still seems small, but

multiply by a large open loopgain Adand its a big factor

1mV5mV is typical VOS

Offset Voltagean example


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A simple single-input transistor

amplifier

Single input and output

Single power supply 5V

M2 nchan

Vdd

5V

pchanM3pchan

M4

Vbias1

2.5

M1 nchan

Ibias

100A

Vin

1.083

Vout

V_in

*0.8 micron mod.MODEL nchan N

+ LD=1.2E-7 NS+PHI=0.70 PB=0.+MJ=0.5 CGSO=3

Offset Voltagean example


94/121

Sweep input from 0V5V

Output falls from 5V to 0.5V

Linear region

of operation

Offset Voltagezoom in even more


95/121

A good region to operate might

be with the input centered and

near Vin

=1.083V

Change transistor M1 gate width to

12.1 microns from 12.0 microns and

get a very different plotnow need

1.080V Vinfor best operation

Offset VoltageAn offset voltage


96/121


Figure E2.21Transfer characteristic of an op amp with VOS= 5 mV.

Offset voltages

Caused by slight differences in manufacturing

May be made better or worse through design

Usually are sensitive to use

Temp effects

Voltage supply effects

Wear-out

The designer

may expect for

an input=0V, theoutput will be 0V

An offset voltage

may cause the

output instead to

rail at 10V

In this case 0V

output occurs

for -5mV input

due to the offset

voltage

Offset Voltage: what is the gain


97/121



= =

Question: what

is the gain here

Question: what

is the gain here

Offset Voltage: what is the gain


98/121



= =

Is it

= ?

Is it

.

= 0?

Gain: think about the slope of the transfer characteristic


99/121



= =

10 100 0.001= 0

10 0.006 0.005=

100.001

= 10,000/

Offset Voltage


100/121


Figure 2.29Evaluating the output dc offset voltage due to VOSin a closed-loop amplifier.

Offset voltage in the op amp causes the output to be influenced by the

offset voltage just as the non-inverting configurations input

= 1

The offsets effect can be removed, in principle, if there is a VOSsupply

which can be adjusted to the oppositevoltage

In a closed-loopamplifier, the offset

voltage can be

thought of as a shift

in the circuit ground

at the non-inverting

input

Offset VoltageCommercial Op Amps Allow Correction


101/121


Figure 2.30The output dc offset voltage of an op amp can be trimmed to zero by connecting a potentiometer to the

two offset-nulling terminals. The wiper of the potentiometer is connected to the negative supply of the op amp.

Commercial Op Amps may have inputs to allow adjustment to remove

the offset voltage

The circuit may still drift due to temperature, supply, age

Offset Voltage DC coupled input


102/121


Figure 2.31(a)A capacitively coupled inverting amplifier. (b)The equivalent circuit for determining its dc output offset voltage VO.

If the inverting configuration is coupled at the input by a capacitor

Low frequency (and dc) signals are blocked

VOS

has 100% negative feedback, so passes only at the low level of a

few mVit will be a small error on the amplified AC signal

Considering DC input, gain

for VOSis 1

Input Bias and Offset Current


103/121


Figure 2.32The op-amp input bias currents represented by two current sources IB1and IB2.

Op Amps have high but non-zero input resistance

Op Amps made from BJTs need some current to runthe bipolar

transistor amplifies a small current in the base

Op Amps made from MOS transistors have very little input current atlow frequencies, but capacitance on the gate allows current to flow at

higher frequencies

Input Bias and Offset Current


104/121


Figure 2.32The op-amp input bias currents represented by two current sources IB1and IB2.

Input bias current is the average of the two input currents

=

2

The input offset current is the difference in the currents

= 100 Typcially

10 Typically

Input Bias and Offset Current Effect


105/121


Figure 2.33Analysis of the closed-loop amplifier, taking into account the input bias currents.

Ground the input signal sources to see the effect of the input currents on the

output voltage

There are currents in R1 and R2, but the non-inverting input is at ground

The virtual short between inputs due to the high open loop gain brings the

inverting input to a virtual short

The output voltage is due to the current in R2, IB1 The current IB2has no effect

=

If R2is large, the output will be

significantly affected by the input

bias current

Input Bias Current Effect Reduction


106/121


Figure 2.34 Reducing the effect of the input bias currents by introducing a resistor R3.

Add a resistor R3 to the non-inverting input

Changes the voltage level at the non-inverting input

Through the virtual short between inputs also at the inverting input The voltage level at the inputs is then:

= 3 The current in R1 is defined by the input voltage, since R1 is connected to

ground

= = 3


3


107/121



= = 3

The current in R2 is the sum of the current in R1 and the current supplied by the

input, IB1

3 This gives the output voltage, which is the drop in R2 and the voltage on the inputs

= 3 3


If the bias currents are equal I =I =I then output is simplified to


108/121



If the bias currents are equal, IB=IB1=IB2then output is simplified to

= 3(1

In order to minimize the effects of the bias currents, VOshould be zero when there

are no input signals as in this case. Minimizing VOmeans adjusting R2 so that

when multiplied by 1 the result equals R2

3= 1 = RRR R


Finally check the value of the offset current in this configuration


109/121



Finally check the value of the offset current in this configuration

= 2 = 2

Substitute into the output relation

= 3 3

= This value can be much less

than the effect of the offset

current without the feedback

and R3

AC coupled amplifier


110/121


Figure 2.36 Illustrating the need for a continuous dc path for each of the op-amp input terminals. Specifically, note

that the amplifier will notwork without resistor R3.

For AC coupling, there must be a path to ground for the amplifier to work

R3 should be the same as R2

Input offset effect on inverting integrator


111/121


Figure 2.37 Determining the effect of the op-amp input offset voltage VOSon the Miller integrator circuit.

Note that since the output rises with time, the op amp eventually saturates.

For the inverting integrator, input offset current will add linearly to the output,

eventually saturating the outputthe term increases linearly with time

Open Loop Gain and Frequency Response

h fi i l i f l d i h f b


112/121


Figure 2.39 Open-loop gain of a typical general-purpose internally compensated op amp.

The finite open loop gain of a real Op Amp decreases with frequency above a

certain point

The -3dB frequency fb break or corner frequency may be quite low

ftis the unity gain frequency, where gain falls to 0db (1V/V) Gain often falls at -20dB /decade due to internal capacitance making the op

amp a STC circuit

frequency compensation


113/121

Open Loop Gain and Frequency Response


114/121


Figure 2.39 Open-loop gain of a typical general-purpose internally compensated op amp.

For frequencies an order of

magnitude or so beyond the

-3dB frequency, an

approximation for gain is

= Gain reaches 0dB at

= =1v/v=0dB

= Unity gain bandwidth,

ft=t/2is an important

figure of merit given by

manufacturers

Once you know ft, you canwork backwards up the

curve to find gain at any

frequency

Gain bandwidth trade-off


115/121

The finite-gain closed loop performance of an op amp was found earlier in the

chapter as

= 1

1

This can be re-written in terms of frequency as

()()= 1 1 (1

) 1

Usually open loop gain is very large, or

1



116/121

The term in the middle then goes away and closed loop gain then reduces to

()()= 1 1

This is the form of a low-pass STC where the inverting configuration has

DC gain magnitude is R2/R1 Gain rolls off at -20dB/decade

-3dB point is given by

3= +/ Similar math for the non-inverting configuration gives the form of a low-pass

STC where

DC gain magnitude is 1+R2/R1

Gain rolls off at -20dB/decade

-3dB point is given by

3= +/


Example 2 6 find 3dB frequency and gains of closed loop amplifiers


117/121

Example 2.6, find -3dB frequency and gains of closed loop amplifiers

were ft=1MHz

The governing rules

are

DC gain

magnitude is

R2/R1

Gain rolls off at -20dB/decade

-3dB point is

given by

3=

1 /

Operating Limits

O A l h ithi b t 1V f th l lt


118/121


Figure 2.42 (a)A noninverting amplifier with a nominal gain of 10 V/V designed using an op amp that saturates at

13-V output voltage and has 20-mA output current limits.

(b) When the input sine wave has a peak of 1.5 V, the output is clipped off at 13 V.

Op Amps can only reach within about 1V or so of the power supply voltage

For example, if the power supplies are +-5V, output may be maximum of about

+-4V

Current is also limited with similar results

Example where

the amplifier

saturates at +-13V

Operating Limits


119/121

Op Amps cant respond quickly enough in a second way, different from frequency

response

A large load requires a large current, for instance charging a large capacitance.If the current required is large enough, the output voltage will lag a change in

the input voltage

The maximum rate of change is the Slew Rate

=

Typically specified as volts/microsecond=V/s

Operating Limits


120/121


Slew rate limitedwhile an amplifier is slewing the output voltage

rises at a fixed rate

Result is a non-linear distortion in the output

Sharply rising input

Slew-rate exceeded, the

output rises at the SR

Operating Limits


121/121

Slew rate limitedwhile an

amplifier is slewing the output

voltage rises at a fixed rate Result is a non-linear distortion in

the output

Documents

Chapter 2 Operational Amps