The Operational Amplifier

The Operational Amplifier

(Op-Amp)

Chapter Goals

• Develop understanding of linear amplification

concepts such as:

– Voltage gain, current gain, and power gain

– Gain conversion to decibel representation

– Input and output resistances

– Biasing for linear amplification

– Distortion in amplifiers

– Two-port representations of amplifiers

– Understand behavior and characteristics of ideal differential

and op amps.

– Demonstrate circuit analysis techniques for ideal op amps.

– Characterize inverting, non-inverting amplifiers.

Amplification

Introduction

A complex periodic signal can be represented as the sum of many

individual sine waves. We consider only one component with

amplitude

V

_s

= 1 mV and frequency

ω

_s

with 0 phase (signal is

used as reference):

Amplifier output is sinusoidal with same frequency but different

amplitude

V

_o

and phase

θ:

Amplification

Introduction (cont.)

Amplifier output power is:

Here, we desire

P

_O

= 100 W with

R

_L

= 8 Ω and

V

_s

= 1 mV

Output power also requires output current which is:

Input current is given by

Amplification

Voltage Gain & Current Gain

• Voltage Gain:

Magnitude and phase of voltage gain are given by

and

For our example,

• Current Gain:

Amplification

Power Gain

• Power Gain:

For our example,

Amplification

Expressing Gain in

Decibels (dB)

The logarithmic decibel or dB scale compresses the huge numeric range of

gains encountered in real systems.

Amplification

Expressing

Gain in dB - Example

Mismatched Source and Load

Resistances



In introductory circuit theory, the maximum power transfer

theorem is usually discussed.



Maximum power transfer occurs when the source and load

resistances are matched (equal in value).



In most amplifier applications, however, the opposite situation is

desired.



A completely mismatched condition is used at both the input and

output ports of the amplifier.

Mismatched Source and Load

Resistances

If

R

_in

>>

R

_s

and

R

_out

<< R

_L

, then

In an ideal voltage amplifier,

and

R

_out

= 0

Distortion in Amplifiers

•

In this graph,

different gains for

positive and

negative values of

the input cause

distortion in the

output.

•

Total Harmonic

Distortion (THD) is

a measure of signal

distortion that

compares

undesired harmonic

content of a signal

to the desired

component.

Total Harmonic Distortion

dc

desired

output

2nd harmonic

distortion

3rd harmonic

distortion

Numerator = rms amplitude of distortion terms

Differential Amplifier

Basic Model

v

_o

= A v

_id

A

= open-circuit voltage gain

v

_id

= (

v

₊

- v

_-

) = differential input

signal voltage

R

_id

= amplifier input resistance

R

_o

= amplifier output

resistance

An ideal differential amplifier produces an

output that depends on the voltage

difference between its two input terminals.

Signal developed at amplifier output is in

phase with the voltage applied at + input

(non-inverting) terminal and 180

o

_{out of}

phase with that applied at - input (inverting)

terminal.

Differential Amplifier Model

Impact of Source and Load

R

_L

= load resistance

R

_S

= Thevenin equivalent

resistance of signal source

v

_s

= Thevenin equivalent voltage

of signal source

•Op amp circuits are mostly dc-coupled amplifiers. Signals

v

_o

and

v

_s

may have

a dc component representing a dc shift of the input away from the Q-point.

•Op-amp amplifies both dc and ac components.

Differential Amplifier Model

Example including Source and Load Resistances

•

Problem: Calculate voltage gain for an amplifier

•

Given Data:

A

= 100,

R

_id

= 100kΩ,

R

_o

= 100Ω,

R

_S

= 10kΩ,

R

_L

=

1000Ω

Differential Amplifier Model

Example including Source and Load Resistances

•

Analysis:

•

An ideal amplifier’s output depends only on the input voltage

difference and not on the source and load resistances. This can

be achieved by using a fully mismatched resistance condition

(

R

_id

>>

R

_S

or infinite

R

_id

, and

R

_o

<<

R

_L

or zero

R

_o

). Then:

•

A = open-loop gain (maximum voltage gain available from the

device)

Operational Amplifers

Op-amp is an electronic device that amplify the difference of voltage

at its two inputs.

–

+

+V

–V

Very high gain dc coupled amplifiers with differential inputs.

One of the inputs is called the inverting input (−); the other is called the

non-inverting input. Usually there is a single output.

Most op-amps operate from plus and minus supply voltages, which

may or may not be shown on the schematic symbol.

1 20 1 8 DIP 1 8 DIP SMT 1 8 SMT

The Ideal Op-Amp

Ideally, op-amps have characteristics (used in circuit analysis):



Infinite voltage gain



Infinite input impedance (does not load the driving sources)



Zero output impedance (drive any load)



Infinite bandwidth (flat magnitude response, zero phase shift)



Zero input offset voltage.

–

+

The ideal op-amp has characteristics that

simplify analysis of op-amp circuits.

The concept of infinite input impedance is

particularly a valuable analysis tool for

several op-amp configurations.

The Practical Op-Amp

Real op-amps differ from the ideal model in various respects.

In addition to finite gain, bandwidth, and input impedance,

they have other limitations.

–

+

 Finite open loop gain.

 Finite input impedance.

 Non-zero output impedance.

 Input current.

 Input offset voltage.

 Temperature effects.

Internal Block Diagram of an Op-Amp

Internally, the typical op-amp has a differential input, a

voltage amplifier, and a push-pull output. Recall from the

discussion in Section 6-7 of the text that the differential

amplifier amplifies the

difference

in the two inputs.

Differential

amplifier

input stage

Voltage

amplifier(s)

gain stage

Push-pull

amplifier

output

stage

V

V

+

–

Input Signal modes

 As the input stage of an op-amp is a differential amplifier,

there are two input modes possible: differential mode

and common mode.

 In

differential mode

any one of the two scenarios can

occur.

 Either one input is applied to one input while the other

input is grounded (single-ended).

 Or opposite polarity signals are applied to the inputs

(double-ended).

Differential Mode Operation

V

–

+

V

V

–

+

V

Single-ended differential amplifier

Common Mode Operation

 The same input tend to cancel

each other and the output is

zero.

 This is called common-mode

rejection.

V

V

–

+

V

 In common mode, two signals voltages of the same

amplitude, frequency and phase are applied to the two

inputs.

 This is useful to reject unwanted signal that appears to both

inputs. It is cancelled and does not appear at the output

Common-Mode Rejection Ratio

The ability of an amplifier to amplify differential signals and

reject common-mode signals is called the common-mode

rejection ratio (CMRR).

CMRR is defined as

ol

cm

A

A

=

CMRR

where

A

_ol

is the open-loop

differential-gain and

A

_cm

is

the common-mode gain.

CMRR can also be expressed in decibels as

20 log

ol

cm

A

A





=

_

_





CMRR

A

_cm

is zero in ideal op-amp and much less than 1

is practical op-amps.

A

_ol

ranges up to 200,000 (106dB)

CMRR = 100,000 means that desired signal is amplified 100,000 times

more than un wanted noise signal.

Common-Mode Rejection Ratio

Example

What is CMRR in decibels for a typical 741C op-amp?

The typical open-loop differential gain for the 741C is 200,000 and the

typical common-mode gain is 6.3.

90 dB

20 log

ol

cm

A

A





=

_

_





CMRR

200, 000

20 log

6.3

=

=

Maximum Output Voltage Swing

V

_O(p-p)

: The maximum output voltage swing is determined

by the op-amp and the power supply voltages

With no input signal, the output of an op-amp 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 ± V

_CC

.

In practice, however, this ideal can be approached but never

reached (varies with load resistance).

Input Offset Voltage

 Ideal op-amp produces zero output voltage if the

differential input is zero

 But practical op-amp produces a non-zero output voltage

when there is no differential input applied. This output

voltage is termed V

_OUT(error)

.

 It is due to unavoidable mismatches in the differential

stage of the op amp.

 The amount of differential input voltage required

between the inputs to force the output to zero volts is the

input offset voltage V

_OS

.

The input bias current is the

average of the two dc currents

required to bias the differential

amplifier

Ideally, input bias current is zero.

1

2

BIAS

2

I

I

I

=

+

 The input impedance of an op-amp is

specified in two ways:

 Differential input impedance and

common-mode input impedance.

 Differential input impedance,

Z

_IN(d)

,

is the total resistance between

inverting and noninverting input.

 Common-mode input impedance,

Z

_IN(c)

, is the resistance between each

input and ground.

Input Impedance

Z

–

+

Z

–

+

The offset voltage developed by the input offset current is

𝑉𝑉

_{𝑂𝑂𝑂𝑂}

= 𝐼𝐼

_{𝑂𝑂𝑂𝑂}

𝑅𝑅

_{𝑖𝑖𝑖𝑖}

The output error volt is

𝑉𝑉

𝑂𝑂𝑈𝑈𝑈𝑈(𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒)

= 𝐴𝐴

𝑣𝑣

𝐼𝐼

𝑂𝑂𝑂𝑂

𝑅𝑅

𝑖𝑖𝑖𝑖

Input offset Current

I

_OS

Ideally, the two input bias currents are equal, and thus their

difference is zero

The input offset current is

the difference of the input

bias currents

𝐼𝐼

_{𝑒𝑒𝑜𝑜}

= 𝐼𝐼

₁

− 𝐼𝐼

₂

Z

_out

:

The output impedance is the resistance viewed from the

output of the circuit.

Output Impedance

Z

–

Slew Rate

The slew rate is the maximum

rate of change of the output

voltage in response to a step input

voltage (V/µs)

Slew rate is measured with an

op-amp connected as shown and is

given as

The slew rate is dependent upon

the high-frequency response of

the amplifier stages within the

op-amp.

Example

Determine the slew rate for the

output response to a step input

shown.

Since this response is not

ideal, the limits are taken

at the 90% points.

Negative Feedback

Negative feedback is the process of returning a portion of

the output signal of an amplifier to the input with a phase

angle that is opposite to the input signal.

V

+

–

V

V

Internal inversion makes V

180

° out of phase with V

.

Negative

feedback

circuit

The advantage of negative

feedback is that precise

values of amplifier gain

can be set. In addition,

bandwidth and input and

output impedances can be

controlled.

Importance of Negative Feedback

 The usefulness of an

op-amp operated without

negative feedback is

generally limited to

comparator applications

 With negative feedback the gain of op-amp (called

close-loop gain

A

_cl

) can be reduced and controlled so that an

op-amp can function as a linear amplifier.

 The inherent open-loop voltage gain of a typical op-amp is

very high. Therefore, an extremely small input voltage (even

the input offset voltage) drives the op-amp into saturation.

Op-Amps with Negative Feedback

 An op-amp can be connected using negative feedback to

stabilize the gain and increase frequency response.

 This close-loop gain (

A

_cl

) is usually much less than the

open-loop gain (

A

_ol

).

 The close-loop voltage gain is the voltage of an op-amp with

external feedback.

 The amplifier circuit consists of an op-amp and an external

negative feedback circuit.

 The feedback from the output is connected to the inverting input

of the op-amp.

 The negative feedback is determined and controlled by external

components.

Noninverting Amplifier

A noninverting amplifier is a configuration in which the

signal is on the noninverting input and a portion of the

output is returned to the inverting input.

The feedback circuit is formed by input resistance

R

_i

and

feedback resistance

R

_f

.

R

R

V

V

+

–

Feedback

circuit

V

This feedback creates a

voltage divider circuit

which reduces

V

_out

and

connects the reduced

voltage

V

_f

to the

inverting input and can

be expressed as:

Noninverting Amplifier

The differential input is

amplified by the

open-loop gain and produces

the output voltage as:

The attenuation, B, of

the feedback circuit is

Noninverting Amplifier

(NI)

1

f

cl

i

R

A

R

= +

Substituting

BV

_out

for

V

_f

, we get

The overall voltage gain of the amplifier can be expressed

as

The product

A

_ol

B

is typically much greater than 1, so the

equation simplifies to

Noninverting Amplifier

(NI)

1

f

cl

i

R

A

R

= +

Determine the gain of the noninverting amplifier shown.

R

82 kΩ

V

+

–

V

R

3.3 kΩ

82 k

1

3.3 k

Ω

= +

Ω

=

25.8

Voltage Follower

A special case of the inverting amplifier is when

R

_f

=0 and

R

_i

=

∞. This forms a voltage follower or unity gain buffer

with a gain of 1.

R

82 kΩ

V

+

–

V

R

3.3 kΩ

This configuration offers 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

V

+

–

V

It produces an excellent circuit for isolating one circuit stage from another,

which avoids "loading" effects.

Inverting Amplifier

We have:

_and

Since

I

_in

=

I

_f

,

then

Example

Determine the gain of the inverting amplifier shown.

82 k

3.3 k

Ω

= −

Ω

=

−24.8

–

+

R

V

R

V

82 kΩ

3.3 kΩ

(I)

f

cl

i

R

A

R

= −

The minus sign

indicates inversion.

2.2.1.

Closed-Loop Gain

•

question:

how will we…

– step #4: define

v

_Out

in terms of

current flowing across

R

₂

– step #5: substitute

v

_in

/

R

₁

for

i

₁

.

Analysis of the inverting configuration. The circled numbers indicate

the order of the analysis steps.

closed-loop

gain

Effect of Finite Open-Loop Gain

• Q:

How does the gain expression change if

open

loop gain (

A

) is not assumed to be infinite?

– A:

One must employ analysis similar to the

Effect of Finite Open-Loop Gain



if

then the previous

gain expression is

2

1

2

yielded

2

1

1

/

1 ( / )

1

A

G

A

Out

A

In

v

R R

G

R R

v

A

R

R

=∞

=∞

<∞

−

=

=

+





+ 



_

≠ −





ideal gain

non-ideal gain

Collecting terms, the closed-loop

Effect of Finite Open-Loop Gain

• Q:

Under what condition can

G

= -

R

₂

/

R

₁

be

employed over the more complex expression?

– A:

If

1 + (

R

₂

/

R

₁

) <<

A,

then simpler expression

may be used.

2

2

2

1

2

1

1

1

if

1

then

el

/

1 ( /

1

se

)

A

A

R

_A

_G

R

_G

R R

R R

R

R

A

=∞

<∞

−

+

<<

= −

=

+





+ 

_



_

From the Figure we get

Impedances of the Noninverting Amplifier

Input Impedance

Substituting

I

_in

Z

_in

=

V

_d

, where

Z

_in

is open loop input impedance

(

)

(NI)

1

in

ol

in

Z

= +

A B Z

The input impedance of the noninverting amplifier with negative

feedback is much greater than the internal open-loop input

Referring to the shown Figure,

after some mathematical

manipulations (Floyd 620) it

can be shown that the output

impedance of a noninverting

(NI) amplifier can be given as

Impedances of the Noninverting Amplifier

Output Impedance

The output impedance of the noninverting amplifier with

negative feedback is much less than the internal open-loop input

impedance of the op-amp.

(

)

(NI)

1

out

out

ol

Z

Z

A B

=

+

Since voltage-follower is a special case on noninverting

amplifier, the same formulas are used with

B

= 1, therfore

Impedances of the Voltage-Follower

The input impedance of voltage-follower is very high, it can be

seen that it is extremely high as compared to the noninverting

amplifier as the feedback attenuation

B

= 1.

The same is true for the output impedance where the factor of

is removed so it becomes even less than the output impedance of

the noninverting amplifier.

The input impedance of the inverting (I) amplifier is

Impedances of the Inverting Amplifier

This is because the input is in series with

R

_i

and that is connected

to virtual ground so that is the only resistance seen by input.

As with the noninverting amplifier, the output impedance of an

inverting (I) amplifier is decreased by the negative feedback.

In fact the the expression is the same as for the noninverting case.

Impedances of the Inverting Amplifier

The output impedance of the both noninverting and inverting

amplifier is low. In fact, practically it can be considered zero..

Noninverting amplifier:

Impedances

(

)

(NI)

1

in

ol

in

Z

= +

A B Z

(

)

(NI)

1

out

out

ol

Z

Z

A B

=

+

Generally, assumed to be

∞

Generally, assumed to be 0

(I)

in

i

Z

≅

R

(

)

(I)

1

out

out

ol

Z

Z

A B

=

+

Generally, assumed to be

R

_i

Generally, assumed to be 0

Inverting amplifier:

Determine the closed-loop gain of each amplifier in Figure.

If a signal voltage of 10 mV

_rms

is applied to each amplifier in

Figure, what are the output voltages and what is there phase

relationship with inputs?.

(a) V

_out

≅ V

_in

= 10 mV, in phase (b) V

_out

= A

_cl

V

_in

= – 10 mV, 180º out of

phase (c) V

_out

= 233 mV, in phase (d) V

_out

= – 100 mV, 180º out of phase

Effect of Input Bias Current - Inverting Amplifier

An inverting amplifier with zero input voltage is shown.

Ideally, the current through

R

_i

is zero because the input voltage

is zero and the voltage at the inverting terminal is zero.

The small input bias current,

I

₁

, is through

R

_f

from the output

terminal. This produces an output error voltage

I

₁

R

_f

when it

should be zero

A voltage-follower with zero input voltage and a source

resistance,

R

_s

is shown.

In this case, an input bias current,

I

₁

, produces a drop across

R

_s

and creates an output voltage error -

I

₁

R

_s

as shown.

Consider a noninverting amplifier with zero input voltage.

The input bias current,

I

₁

, produces a voltage drop across

R

_f

and thus creates an output error voltage of

I

₁

R

_f

Bias Current Compensation – Voltage Follower

The output error voltage due to bias currents in a

voltage-follower can be sufficiently reduced by adding a resistor,

R

_f

, equal to the source resistance,

R

_s

, in the feedback

path.

The voltage drop created by

I

₁

across the added resistor

subtracts from the output error voltage (if

I

₁

=

I

₂

).

Bias Current Compensation – Inverting and Noninverting

To compensate for the effect of bias current in Inverting

and Noninverting Amplifiers, a resistor

R

_c

is added at the

noninverting terminal.

The compensating resistor value equals the parallel

combination of

R

_i

and

R

_f

.

The input current creates a voltage drop across

R

_c

that

offsets the voltage across the combination of

R

_i

and

R

_f

,

thus sufficiently reducing the output error voltage.

–

+

R

V

R

V

R

= R

|| R

–

+

R

V

R

V

R

= R

|| R

Noninverting

amplifier

Inverting

amplifier

Effect of Input Offset Voltage

 The output of an op-amp should be zero when the

differential input is zero.

 Practically this is not the case. There is always an output

error voltage present whose range is typically in

microvolts to millivolts.

 This is due to the imbalances within the internal op-amp

transistors

 This output error voltage is aside from the one produced

by the input bias

Input Offset Voltage Compensation

Most ICs provide a mean of compensating for offset voltage.

An external potentiometer to the

offset null

pins of IC package

Bandwidth Limitations

 Frequency response of amplifiers is shown in a plot called Bode

Plot.

 In Bode plot, the frequency is on the horizontal axis and is in

logarithmic scale. It means that the frequency change is not

linear but ten-times. This ten-time change in frequency is called

a decade.

 The vertical axis shows the voltage gain in decibel (dB).

 The maximum gain on the plot is called the midrange gain.

 The point in the frequency response of amplifiers where the

gain is 3dB less than the midrange gain is called the critical

frequency.

Bandwidth Limitations

–20 dB/decade roll-off

Unity-gain frequency (f

)

Critical frequency

10

1

100

1k

10k

100k

1M

f (Hz)

106

100

75

50

25

0

A

(dB)

Midrange

Bandwidth Limitations

 An open-loop response curve (Bode plot) for a certain op-amp

is shown.

 The differential open-loop gain A

_ol

of an op amp is not

infinite; rather, it is finite and decreases with frequency.

 Note that although the gain is quite high at dc and low

frequencies, it starts to fall off at a rather low frequency.

 The process of modifying the open-loop gain is termed

frequency compensation, and its purpose is to ensure that

op-amp circuits will be stable (as opposed to oscillatory).

 These are units that have a network (usually a single

capacitor) included within the same IC chip whose function is

to cause the op-amp gain to have the single-time-constant

Gain-Versus-Frequency Analysis

The RC lag (low-pass) circuits within an

op-amp are responsible for the roll-off in

gain as the frequency increases, just as for

the discrete amplifiers. The attenuation of

an RC lag circuit shown is expressed as:

Gain-Versus-Frequency Analysis

If an op-amp is represented by a voltage gain element with a

gain of

A

_ol(mid

₎

plus a single RC lag circuit, as shown, it is

known as a compensated op-amp. The total open-loop gain of

the op-amp is the product of the midrange open-loop gain,

A

_ol(mid)

, and the attenuation of the RC circuit as:

Phase Shift

An RC circuit causes a propagation delay from input to

output, thus creating a phase shift between the input signal

and the output signal. An RC lag circuit such as found in an

op-amp stage causes the output signal voltage to lag the input.

The phase shift,

θ

, is given by:

Overall Frequency Response

 Most op-amps have a constant roll-off of -20 dB/decade

above its critical frequency. The more complex IC

operational amplifier may consist of two or more

cascaded amplifier stages.

 The gain of each stage is frequency dependent and rolls

off at above its critical frequency. Therefore, the total

response of an op-amp is a composite of the individual

responses of the internal stages. dB gains are added and

phase lags of the stages are added as shown in next slide

for a three stage op-amp.

closed-loop frequency response

 Op-amps are usually used in a closed-loop configuration

with negative feedback in order to achieve precise

control of gain and bandwidth.

 Midrange gain of an op-amp is reduced by negative

feedback as we have already seen in previous sections.

 Now we will see its effects on bandwidth.

Effect of Negative Feedback on Bandwidth

 The closed-loop critical frequency of an op-amp is given by:

𝑓𝑓

_{𝑐𝑐(𝑐𝑐𝑐𝑐)}

= 𝑓𝑓

_{𝑐𝑐 𝑒𝑒𝑐𝑐}

(1 + 𝐵𝐵𝐴𝐴

_{𝑒𝑒𝑐𝑐 𝑚𝑚𝑖𝑖𝑚𝑚}

)

Where

B

is the feedback attenuation of the closed-loop op-amp.

 The above expression shows that the closed-loop critical

frequency,

𝑓𝑓

_{𝑐𝑐(𝑐𝑐𝑐𝑐)}

, is higher than the open-loop critical

frequency

𝑓𝑓

_{𝑐𝑐 𝑒𝑒𝑐𝑐}

by a factor of

(1 + 𝐵𝐵𝐴𝐴

_{𝑒𝑒𝑐𝑐 𝑚𝑚𝑖𝑖𝑚𝑚}

)

 Since

_𝑓𝑓

_{𝑐𝑐(𝑐𝑐𝑐𝑐)}

equals bandwidth therefore the closed-loop

bandwidth,

𝐵𝐵𝐵𝐵

_{𝑐𝑐𝑐𝑐}

, is also increased:

Summary

Bandwidth Limitations

The Figure shows the concept of closed-loop response. When the

open-loop gain is reduced due to negative feedback, the bandwidth is

increased.

This means that you can

achieve a higher

BW

by

accepting less gain.

A_v f Aol(mid ) 0 fc(ol) fc (cl ) A_{cl(mid )} Closed-loop gain Open-loop gain

Gain-Bandwidth Product

 An increase in the closed-loop gain causes a decrease in the

bandwidth and vice versa, such that the product of gain and

bandwidth is constant.

 If

_𝐴𝐴

_{𝑐𝑐𝑐𝑐}

is the gain of an op-amp with

𝑓𝑓

_{𝑐𝑐(𝑐𝑐𝑐𝑐)}

bandwidth then:

𝐴𝐴

_{𝑐𝑐𝑐𝑐}

𝑓𝑓

_{𝑐𝑐(𝑐𝑐𝑐𝑐)}

= 𝐴𝐴

_{𝑒𝑒𝑐𝑐}

𝑓𝑓

_{𝑐𝑐(𝑒𝑒𝑐𝑐)}

 The equation,

A

_cl

f

_c(cl

₎

=

A

_ol

f

_c

₍

_ol

₎

shows that the product of the

gain and bandwidth are constant.

 The gain-bandwidth product is always equal to the frequency

at which the op-amp’s open-loop gain is unity or 0 dB (unity

gain bandwidth,

𝑓𝑓

_𝑈𝑈

)

Example

Determine the bandwidth of each of the amplifiers shown.

Both op-amps have an open-loop gain of 100 dB and a

unity-gain bandwidth (

f

_T

) of 3 MHz.

Selected Key Terms

Oper ational

amplifier

Differential

mode

Common mode

A type of amplifier that has very high voltage

gain, very high input impedance, very low

output impedance and good rejection of

common-mode signals.

A mode of op-amp operation in which two

opposite-polarity signals voltages are applied to

the two inputs (double-ended) or in which a

signal is applied to one input and ground to the

other input (single-ended).

A condition characterized by the presence of

the same signal on both inputs

Selected Key Terms

Open-loop

voltage gain

Negative

feedback

Closed-loop

voltage gain

Gain-bandwidth

product

The voltage gain of an op-amp without external

feedback.

The process of returning a portion of the output

signal to the input of an amplifier such that it is

out of phase with the input.

The voltage gain of an op-amp with external

feedback.

A constant parameter which is always equal to

the frequency at which the op-amp’s open-loop

gain is unity (1).

Quiz

1. The ideal op-amp has

a. zero input impedance and zero output impedance

b. zero input impedance and infinite output impedance

c. infinite input impedance and zero output impedance

d. infinite input impedance and infinite output

Quiz

2. The type of signal represented in the figure is a

a. single-ended common-mode signal

b. single-ended differential signal

c. double-ended common-mode signal

d. double-ended differential signal

V

–

+

Quiz

3. CMRR can be expressed in

a. amps

b. volts

c. ohms

Quiz

4. The difference in the two dc currents required to bias the

differential amplifier in an op-amp is called the

a. differential bias current

b. input offset current

c. input bias current

d. none of the above

Quiz

5. To measure the slew rate of an op-amp, the input signal is a

a. pulse

b. triangle wave

c. sine wave

Quiz

6. The input impedance of a noninverting amplifier is

a. nearly 0 ohms

b. approximately equal to

R

_i

c. approximately equal to

R

_f

d. extremely large

Quiz

7. The noninverting amplifier has a gain of 11. Assume that

V

_in

= 1.0 V. The approximate value of

V

_f

is

a. 0 V

b. 100 mV

c. 1.0 V

d. 11 V

R

10 k

Ω

V

V

+

–

V

R

1.0 kΩ

Quiz

8. The inverting amplifier has a gain of −10. Assume that

V

_in

= 1.0 V. The approximate value of the voltage at the

inverting terminal of the op-amp is

a. 0 V

b. 100 mV

c. 1.0 V

d. 10 V

–

+

R

V

R

V

1.0 kΩ

10 k

Ω

Quiz

9. To compensate for bias current, the value of

R

_c

should

be equal to

a.

R

_i

b.

R

_f

c.

R

_i

||

R

_f

d.

R

_i

+

R

_f

–

+

R

V

R

V

R

Quiz

10. Given a noninverting amplifier with a gain of 10 and a

gain-bandwidth product of 1.0 MHz, the expected high

critical frequency is

a. 100 Hz

b. 1.0 kHz

c. 10 kHz

d. 100 kHz

Quiz

Answers:

1. c

2. d

3. d

4. b

5. a

6. d

7. c

8. a

9. c

10. d