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Electronic Devices, 9th edition Thomas L. Floyd

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 Vs = 1 mV and frequency ωs with 0 phase (signal is used as reference): Amplifier output is sinusoidal with same frequency but different amplitude Vo and phase θ:



Amplification Introduction (cont.)

Amplifier output power is: Here, we desire PO = 100 W with RL = 8 Ω and Vs = 1 mV Output power also requires output current which is: Input current is given by Output phase is zero because circuit is purely resistive.



Amplification Voltage Gain & Current Gain

• Voltage Gain:

Magnitude and phase of voltage gain are given by

and

For our example,

• Current Gain:

Magnitude of current gain is given by



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

For our 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 Rin >> Rs and Rout<< RL, then

In an ideal voltage amplifier, and Rout = 0

For the voltage amplifier shown



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

Denominator = desired component



Differential Amplifier Basic Model

vo= A vid A = open-circuit voltage gain vid = (v+- v-) = differential input

signal voltage Rid = amplifier input resistance Ro = 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 180o out of phase with that applied at - input (inverting) terminal.



Differential Amplifier Model Impact of Source and Load

RL = load resistance RS = Thevenin equivalent

resistance of signal source vs = Thevenin equivalent voltage

of signal source

•Op amp circuits are mostly dc-coupled amplifiers. Signals vo and vs 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, Rid = 100kΩ, Ro = 100Ω, RS = 10kΩ, RL = 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 (Rid >> RS or infinite Rid, and Ro << RL or zero Ro ). 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.

1201

8

DIP1

8

DIP SMT

18

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.

–

+

Zin = ‘Vin VoutZout = 0

AvVin

Av = ‘

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.

–

+

ZinVin VoutZout

AvVin

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.

Differentialamplifier

input stage

Voltageamplifier(s)gain stage

Push-pullamplifier

outputstage

Vin Vout

+

–



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

Vin

–

+

Vout

Vin

–

+

Vout

Single-ended differential amplifier

Double-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.

Vin

Vin

–

+

Vout

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

AA

=CMRRwhere Aol is the open-loop differential-gain and Acm is the common-mode gain.

CMRR can also be expressed in decibels as 20 log ol

cm

AA

=

CMRR

Acm is zero in ideal op-amp and much less than 1 is practical op-amps.

Aol ranges up to 200,000 (106dB)

CMRR = 100,000 means that desired signal is amplified 100,000 times more than un wanted noise signal.

Op-Amp parametrs



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

20log ol

cm

AA

=

CMRR

200,00020log6.3

= =

(The minimum specified CMRR is 70 dB.)



Maximum Output Voltage Swing

VO(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 ± VCC .

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 VOUT(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 VOS .

Typical value of VOS is about 2mV.



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 2BIAS 2

I II +=

Input Bias Current (IBIAS )



The input impedance of an op-amp is specified in two ways:

Differential input impedance and common-mode input impedance.

Differential input impedance, ZIN(d), is the total resistance between inverting and noninverting input.

Common-mode input impedance, ZIN(c), is the resistance between each input and ground.

Input Impedance

ZIN(d)

–

+

ZIN(cm)

–

+



The offset voltage developed by the input offset current is 𝑉𝑉𝑂𝑂𝑂𝑂 = 𝐼𝐼𝑂𝑂𝑂𝑂𝑅𝑅𝑖𝑖𝑖𝑖

The output error volt is 𝑉𝑉𝑂𝑂𝑈𝑈𝑈𝑈(𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒) = 𝐴𝐴𝑣𝑣𝐼𝐼𝑂𝑂𝑂𝑂𝑅𝑅𝑖𝑖𝑖𝑖

Input offset Current IOS

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 𝐼𝐼𝑒𝑒𝑜𝑜 = 𝐼𝐼1 − 𝐼𝐼2



Zout : The output impedance is the resistance viewed from the output of the circuit.

Output Impedance

Zout

–

+



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.

Vout

+

–

Vin

Vf

Internal inversion makes Vf180° out of phase with Vin.

Negativefeedback

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 Acl) 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 (Acl ) is usually much less than the open-loop gain (Aol).

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 Ri and feedback resistance Rf .

Rf

Ri

Vf

Vin

+

–

Feedbackcircuit

Vout

This feedback creates a voltage divider circuit which reduces Vout and connects the reduced voltage Vf to the inverting input and can be expressed as:




The differential input is amplified by the open-loop gain and produces the output voltage as:

The attenuation, B, of the feedback circuit is




(NI) 1 fcl

i

RA

R= +

Substituting BVout for Vf , we get

The overall voltage gain of the amplifier can be expressed as

The product AolB is typically much greater than 1, so the equation simplifies to

Which means




(NI) 1 fcl

i

RA

R= +

Determine the gain of the noninverting amplifier shown.

Rf82 kΩ

Vin +

–

Vout

Ri3.3 kΩ

82 k13.3 k

Ω= +

Ω

= 25.8



Voltage Follower

A special case of the inverting amplifier is when Rf =0 and Ri = ∞. This forms a voltage follower or unity gain buffer with a gain of 1.

Rf82 kΩ

Vin +

–

Vout

Ri3.3 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

Vin +

–

Vout

It produces an excellent circuit for isolating one circuit stage from another, which avoids "loading" effects.



Inverting Amplifier

We have: and

Since Iin = If , then

The overall gain is



Example

Determine the gain of the inverting amplifier shown.

82 k3.3 k

Ω= −

Ω

= −24.8

–

+

Rf

Vout

Ri

Vin

82 kΩ

3.3 kΩ

(I)f

cli

RA

R= −

The minus sign indicates inversion.

Inverting Amplifier



Oxford University Publishing Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)

2.2.1. Closed-Loop Gain

• question: how will we… – step #4: define vOut in terms of

current flowing across R2 – step #5: substitute vin / R1 for

i1.

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

closed-loop gain

G = -R2/R1



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

previous, result is presented below…




if then the previousgain expression is

2 1

2

yielded

2

11

/1 ( / )1

AG

A

OutA

In

v R RGR RvA

RR

=∞

=∞

<∞−

= =+ +

≠ −

ideal gain non-ideal gain

Collecting terms, the closed-loop gain G is found as




• Q: Under what condition can G = -R2 / R1 be employed over the more complex expression? – A: If 1 + (R2/R1) << A, then simpler expression

may be used.

2 2 2 1

2 11 1

if then el /1 1 ( /1

se)A A

R R R RA G GR RR RA

=∞ <∞−

+ << = − =+ +

ideal gain non-ideal gain



Oxford University Publishing Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)



From the Figure we get

Impedances of the Noninverting Amplifier

Input Impedance

Substituting IinZin = Vd , where Zin is open loop input impedance

( )(NI) 1in ol inZ A B Z= +

The input impedance of the noninverting amplifier with negative feedback is much greater than the internal open-loop input impedance of the op-amp.



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) 1out

outol

ZZA 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 Ri and that is connected to virtual ground so that is the only resistance seen by input.

Input Impedance



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..

Output Impedance



Noninverting amplifier:

Impedances

( )(NI) 1in ol inZ A B Z= +

( )(NI) 1out

outol

ZZA B

=+

Generally, assumed to be ∞

Generally, assumed to be 0

(I) in iZ R≅

( )(I) 1out

outol

ZZA B

=+

Generally, assumed to be Ri

Generally, assumed to be 0

Inverting amplifier:

Note that the output impedance has the same form for both amplifiers.



Example



Solution



Example



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

(a) 11 (b) 101 (c) 47.8 (d) 23



If a signal voltage of 10 mVrms is applied to each amplifier in Figure, what are the output voltages and what is there phase relationship with inputs?.

(a) Vout ≅ Vin = 10 mV, in phase (b) Vout = AclVin = – 10 mV, 180º out of phase (c) Vout = 233 mV, in phase (d) Vout = – 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 Ri is zero because the input voltage is zero and the voltage at the inverting terminal is zero.

The small input bias current, I1, is through Rf from the output terminal. This produces an output error voltage I1Rf when it should be zero



A voltage-follower with zero input voltage and a source resistance, Rs is shown.

In this case, an input bias current, I1, produces a drop across Rs and creates an output voltage error -I1 Rs as shown.

Effect of Input Bias Current – Voltage Follower



Consider a noninverting amplifier with zero input voltage.

The input bias current, I1, produces a voltage drop across Rf and thus creates an output error voltage of I1Rf

Effect of Input Bias Current - Noninverting Amplifier



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, Rf , equal to the source resistance, Rs , in the feedback path.

The voltage drop created by I1 across the added resistor subtracts from the output error voltage (if I1 = I2 ).



Bias Current Compensation – Inverting and Noninverting

To compensate for the effect of bias current in Inverting and Noninverting Amplifiers, a resistor Rc is added at the noninverting terminal.

The compensating resistor value equals the parallel combination of Ri and Rf.

The input current creates a voltage drop across Rc that offsets the voltage across the combination of Ri and Rf , thus sufficiently reducing the output error voltage.



–

+

Rf

Vout

Ri

Vin

Rc = Ri || Rf

–

+

Rf

Vout

Ri

Vin

Rc = Ri || Rf

Noninverting amplifier Inverting

amplifier

Bias Current Compensation – Inverting and Noninverting



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.




–20 dB/decade roll-off

Unity-gain frequency (fT)Critical frequency

101 100 1k 10k 100k 1Mf (Hz)

106100

75

50

25

0

Aol (dB)Midrange




An open-loop response curve (Bode plot) for a certain op-amp is shown.

The differential open-loop gain Aol 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 low-pass response shown.



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:

The critical frequency of an RC circuit is:



Gain-Versus-Frequency Analysis

If an op-amp is represented by a voltage gain element with a gain of Aol(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, Aol(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: 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) = 𝑓𝑓𝑐𝑐 𝑒𝑒𝑐𝑐 (1 + 𝐵𝐵𝐴𝐴𝑒𝑒𝑐𝑐 𝑚𝑚𝑖𝑖𝑚𝑚 )

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

The above expression shows that the closed-loop critical frequency, 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐), is higher than the open-loop critical frequency 𝑓𝑓𝑐𝑐 𝑒𝑒𝑐𝑐 by a factor of (1 + 𝐵𝐵𝐴𝐴𝑒𝑒𝑐𝑐 𝑚𝑚𝑖𝑖𝑚𝑚 )

Since 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) equals bandwidth therefore the closed-loop bandwidth, 𝐵𝐵𝐵𝐵𝑐𝑐𝑐𝑐, is also increased:

𝐵𝐵𝐵𝐵𝑐𝑐𝑐𝑐 = 𝐵𝐵𝐵𝐵𝑒𝑒𝑐𝑐(1 + 𝐵𝐵𝐴𝐴𝑒𝑒𝑐𝑐 𝑚𝑚𝑖𝑖𝑚𝑚 )



Summary


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.

Av

f

Aol(mid )

0 fc(ol) fc(cl )

Acl(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 𝐴𝐴𝑐𝑐𝑐𝑐 is the gain of an op-amp with 𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) bandwidth then: 𝐴𝐴𝑐𝑐𝑐𝑐𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐) = 𝐴𝐴𝑒𝑒𝑐𝑐𝑓𝑓𝑐𝑐(𝑒𝑒𝑐𝑐)

The equation, Acl fc(cl) = Aol fc(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, 𝑓𝑓𝑈𝑈)

𝑓𝑓𝑈𝑈 = 𝐴𝐴𝑐𝑐𝑐𝑐𝑓𝑓𝑐𝑐(𝑐𝑐𝑐𝑐)



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 (fT ) of 3 MHz.



Solution



Selected Key Terms

Operational 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 impedance



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

Vin

–

+

Vout



Quiz

3. CMRR can be expressed in

a. amps

b. volts

c. ohms

d. none of the above



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




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 Ri

c. approximately equal to Rf

d. extremely large



Quiz

7. The noninverting amplifier has a gain of 11. Assume that Vin = 1.0 V. The approximate value of Vf is

a. 0 V

b. 100 mV

c. 1.0 V

d. 11 V

Rf10 kΩ

V

V

in

f

+

–

Vout

Ri1.0 kΩ



Quiz

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

–

+

Rf

Vout

Ri

Vin

1.0 kΩ

10 kΩ



Quiz

9. To compensate for bias current, the value of Rc should be equal to

a. Ri

b. Rf

c. Ri||Rf

d. Ri + Rf

–

+

Rf

Vout

Ri

Vin

Rc



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

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