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OpAmp pratik

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Rachit Agarwal
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    Table of ContentsTable of ContentsThe Operational Amplifier______________________________slides 3The Operational Amplifier______________________________slides 3--44

    The Four Amplifier Types______________________________slide 5The Four Amplifier Types______________________________slide 5

    VCVS(Voltage Amplifier) Summary:VCVS(Voltage Amplifier) Summary:

    Noninverting Configuration____________slides 6Noninverting Configuration____________slides 6--99

    Inverting Configuration________________slides 10Inverting Configuration________________slides 10--1212

    ICIC(Current Amplifier) Summary________________________slide 13ICIC(Current Amplifier) Summary________________________slide 13

    VCIS (Transconductance Amplifier) Summary_____________slides 14VCIS (Transconductance Amplifier) Summary_____________slides 14--1515

    ICVS (Transresistance Amplifier) Summary_______________slides 16ICVS (Transresistance Amplifier) Summary_______________slides 16--1818Power Bandwidth_____________________________________slide 19Power Bandwidth_____________________________________slide 19

    Slew Rate____________________________________________slide 20Slew Rate____________________________________________slide 20

    Slew Rate Output Distortion____________________________slide 21Slew Rate Output Distortion____________________________slide 21

    Noise Gain___________________________________________slide 22Noise Gain___________________________________________slide 22

    GainGain--Bandwidth Product_______________________________slide 23Bandwidth Product_______________________________slide 23Cascaded AmplifiersCascaded Amplifiers -- Bandwidth________________________slide 24Bandwidth________________________slide 24

    Common Mode Rejection Ratio__________________________slides 25Common Mode Rejection Ratio__________________________slides 25--2626

    Power Supply Rejection Ratio___________________________slide 27Power Supply Rejection Ratio___________________________slide 27

    Sources_____________________________________________slide 28Sources_____________________________________________slide 28

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    The Operational AmplifierThe Operational Amplifier

    Usually Called Op AmpsUsually Called Op Amps

    An amplifier is a device that accepts a varying input signal andAn amplifier is a device that accepts a varying input signal andproduces a similar output signal with a larger amplitude.produces a similar output signal with a larger amplitude.

    Usually connected so part of the output is fed back to the input.Usually connected so part of the output is fed back to the input.

    (Feedback Loop)(Feedback Loop)

    Most Op Amps behave like voltage amplifiers. They take an inputMost Op Amps behave like voltage amplifiers. They take an input

    voltage and output a scaled version.voltage and output a scaled version.

    They are the basic components used to build analog circuits.They are the basic components used to build analog circuits.

    The name operational amplifier comes from the fact that they wereThe name operational amplifier comes from the fact that they were

    originally used to perform mathematical operations such asoriginally used to perform mathematical operations such as

    integration and differentiation.integration and differentiation.

    Integrated circuit fabrication techniques have made highIntegrated circuit fabrication techniques have made high--

    performance operational amplifiers very inexpensive in comparisonperformance operational amplifiers very inexpensive in comparison

    to older discrete devices.to older discrete devices.

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    ii(+)(+), i, i((--)) : Currents into the amplifier on the inverting and noninverting lines: Currents into the amplifier on the inverting and noninverting lines

    respectivelyrespectively

    vvidid : The input voltage from inverting to non: The input voltage from inverting to non--inverting inputsinverting inputs +V+VSS ,, --VVSS : DC source voltages, usually +15V and: DC source voltages, usually +15V and 15V15V

    RRii : The input resistance, ideally infinity: The input resistance, ideally infinity

    A : The gain of the amplifier. Ideally very high, in the 1x10A : The gain of the amplifier. Ideally very high, in the 1x101010 range.range.

    RROO: The output resistance, ideally zero: The output resistance, ideally zero

    vvOO: The output voltage; v: The output voltage; vOO = A= AOLOLvvidid where Awhere AOLOL is the openis the open--loop voltage gainloop voltage gain

    The Operational AmplifierThe Operational Amplifier+V+VSS

    --VVSS

    vvidid

    InvertingInverting

    NoninvertingNoninverting

    OutputOutput

    ++

    __ii((--))

    ii(+)(+)

    vvOO = A= Addvvidid

    RROOAA

    RRii

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    The Four Amplifier TypesThe Four Amplifier Types

    Description

    Gain

    Symbol

    Transfer

    Function

    Voltage Amplifier

    or

    Voltage Controlled Voltage Source (VCVS)

    Av vo/vin

    Current Amplifier

    or

    Current Controlled Current Source (ICIS)

    Ai io/iin

    Transconductance Amplifier

    or

    Voltage Controlled Current Source (VCIS)

    gm

    (siemens)io

    /vin

    Transresistance Amplifier

    or

    Current Controlled Voltage Source (ICVS)

    rm(ohms)

    vo/iin

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    VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting ConfigurationNoninverting Configuration

    ++

    __

    vvinin

    ++++

    --

    vvOO

    vvidid

    ii(+)(+)

    ii((--))

    iiOO

    iiFF

    RRFF RRLL

    RR11

    ii11

    vvidid

    = v= voo/A/A

    OLOLAssuming AAssuming AOLOL

    vvidid =0=0

    Also, with theAlso, with the

    assumption that Rassumption that Rinin ==

    ii(+)(+) = i= i((--)) = 0= 0

    __vvFF

    ++

    __

    vv11

    ++

    __

    vvLL

    ++

    __

    iiLL

    Applying KVL theApplying KVL the

    following equationsfollowing equations

    can be found:can be found:

    vv11 = v= vinin

    vvOO = v= v11 + v+ vFF = v= vinin+ i+ iFFRRFF

    This means that,This means that,

    iiFF = i= i11

    Therefore: iTherefore: iFF = v= vinin/R/R11

    Using the equation to the left the outputUsing the equation to the left the output

    voltage becomes:voltage becomes:

    vvoo = v= vinin + v+ vininRRFF = v= vinin RRFF + 1+ 1

    RR11 RR11

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    VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued

    The closedThe closed--loop voltage gain is symbolized by Aloop voltage gain is symbolized by Avv and is found to be:and is found to be:AAvv = v= voo = R= RFF + 1+ 1

    vvinin RR11

    The original closed loop gain equation is:The original closed loop gain equation is:

    AAvv = A= AFF = A= AOLOL

    1 + A1 + AOLOLFF

    Ideally AIdeally AOLOL , Therefore A, Therefore Avv = 1= 1

    FF

    Note: The actual value of ANote: The actual value of AOLOL is given for the specific device andis given for the specific device and

    usually ranges from 50kusually ranges from 50k 500k.500k.

    FF is the feedback factor and by assuming openis the feedback factor and by assuming open--loop gain is infinite:loop gain is infinite:

    FF = R= R11

    RR11 + R+ RFF

    AAFF is the amplifieris the amplifier

    gain withgain withfeedbackfeedback

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    VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued

    Input and Output ResistanceInput and Output ResistanceIdeally, the input resistance for this configuration is infinity, but the aIdeally, the input resistance for this configuration is infinity, but the a

    closer prediction of the actual input resistance can be found with thecloser prediction of the actual input resistance can be found with the

    following formula:following formula:

    RRinFinF = R= Rinin (1 +(1 +FFAAOLOL)) Where RWhere Rinin is given for theis given for the

    specified device. Usually Rspecified device. Usually Rinin isisin the Min the M;; range.range.

    Ideally, the output resistance is zero, but the formula below gives aIdeally, the output resistance is zero, but the formula below gives a

    more accurate value:more accurate value:

    RRoFoF = R= Roo Where RWhere Roo is given for theis given for theFFAAOLOL + 1+ 1 specified device. Usually Rspecified device. Usually Roo is inis in

    the 10the 10ss ofof;;ss range.range.

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    VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Noninverting Configuration ExampleNoninverting Configuration Example

    ++

    __

    vvinin++

    ++

    --

    vvOO

    vvidid

    ii(+)(+)

    ii((--))

    iiOO

    iiFF

    RRFF RRLL

    RR11

    ii11

    __vvFF

    ++

    __

    vv11

    ++

    __

    vvLL++

    __

    iiLL

    Given:Given: vvinin = 0.6V, R= 0.6V, RFF = 200 k= 200 k;;RR11 = 2 k= 2 k;; , A, AOLOL = 400k= 400k

    RRinin = 8 M= 8 M ;; , R, Roo = 60= 60 ;;

    Find: vFind: voo , i, iFF , A, Avv ,, FF , R, RinFinF and Rand RoFoF

    Solution:Solution:

    vvoo = v= vinin + v+ vininRRFF = 0.6 + 0.6*2x10= 0.6 + 0.6*2x1055 == 60.6 V60.6 V iiFF = v= vinin = 0.6 == 0.6 = 0.3 mA0.3 mA

    RR11 20002000 RR11 20002000

    AAvv = R= RFF + 1 = 2x10+ 1 = 2x1055

    + 1 =+ 1 = 101101 FF = 1 = 1 == 1 = 1 = 9.9x109.9x10--33

    RR11 20002000 AAOLOL 101101

    RRinFinF = R= Rinin (1 +(1 +FFAAOLOL) = 8x10) = 8x1066 (1 + 9.9x10(1 + 9.9x10--33*4x10*4x1055) =) = 3.1688x103.1688x101010 ;;

    RRoFoF = R= Roo = 60= 60 == 0.0150.015 ;;

    FFAAOLOL + 1 9.9x10+ 1 9.9x10--33

    *4x10*4x1055

    + 1+ 1

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    VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting ConfigurationInverting Configuration

    ++

    __

    RRLL

    vvOO

    ++

    --

    vvinin

    ++

    __

    RR11ii11

    RRFFiiFF

    The sameThe same

    assumptions used toassumptions used tofind the equations forfind the equations for

    the noninvertingthe noninverting

    configuration areconfiguration are

    also used for thealso used for the

    invertinginvertingconfiguration.configuration.

    General Equations:General Equations:

    ii11 = v= vinin/R/R11

    iiFF = i= i11

    vvoo == --iiFFRRFF == --vvininRRFF/R/R11

    AAvv = R= RFF/R/R11 FF = R= R11/R/RFF

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    Input and Output ResistanceInput and Output ResistanceIdeally, the input resistance for this configuration is equivalent to RIdeally, the input resistance for this configuration is equivalent to R11..

    However, the actual value of the input resistance is given by theHowever, the actual value of the input resistance is given by the

    following formula:following formula:

    RRinin = R= R11 + R+ RFF

    1 + A1 + AOLOL

    Ideally, the output resistance is zero, but the formula below gives aIdeally, the output resistance is zero, but the formula below gives a

    more accurate value:more accurate value:

    RRoFoF = R= Roo

    1 +1 +FFAAOLOL

    Note:Note: FF = R= R11 This is different from the equation usedThis is different from the equation used

    RR11 + R+ RFF on the previous slide, which can be confusing.on the previous slide, which can be confusing.

    VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting Configuration ContinuedInverting Configuration Continued

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    VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Inverting Configuration ExampleInverting Configuration Example

    ++

    __

    RRLL

    ++

    --

    vvinin

    ++

    __

    RR11ii11

    RRFFiiFF

    Given:Given: vvinin = 0.6 V, R= 0.6 V, RFF = 20 k= 20 k;;RR11 = 2 k= 2 k;; , A, AOLOL = 400k= 400k

    RRinin = 8 M= 8 M ;; , R, Roo = 60= 60 ;;

    Find: vFind: voo , i, iFF , A, Avv ,, FF , R, RinFinF and Rand RoFoFvvOO

    Solution:Solution:

    vvoo == --iiFFRRFF == --vvininRRFF/R/R11 == --(0.6*20,000)/2000 =(0.6*20,000)/2000 = 12 V12 V

    iiFF = i= i11 == vvinin/R/R11 = 1 / 2000 == 1 / 2000 = 0.50.5 mAmA

    AAvv = R= RFF/R/R11 = 20,000 / 2000 == 20,000 / 2000 = 1010 FF = R= R11/R/RFF = 2000 / 20,000 == 2000 / 20,000 = 0.10.1

    RRinin = R= R11 + R+ RFF = 2000 + 20,000= 2000 + 20,000 == 2,000.052,000.05 ;;

    1 + A1 + AOLOL 1 + 400,0001 + 400,000

    RRoFoF = R= Roo = 60= 60 == 1.67 m1.67 m ;;

    1 +1 +FFAAOLOL 1 + 0.09*400,0001 + 0.09*400,000 Note:Note: FF is 0.09 because usingis 0.09 because usingdifferent formula than abovedifferent formula than above

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    ICIS (Current Amplifier) SummaryICIS (Current Amplifier) Summary Not commonly done using operational amplifiersNot commonly done using operational amplifiers

    ++

    __LoadLoad

    iiinin

    iiLL

    Similar to the voltageSimilar to the voltage

    follower shown below:follower shown below:

    Both these amplifiers haveBoth these amplifiers have

    unity gain:unity gain:

    AAvv = A= Aii = 1= 1

    ++

    __

    iiinin = i= iLL

    vvinin = v= voovvinin++

    __ ++

    --

    vvOO

    Voltage FollowerVoltage Follower

    1 Possible1 Possible

    ICISICIS

    OperationalOperational

    AmplifierAmplifier

    ApplicationApplication

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    VCIS (Transconductance Amplifier) SummaryVCIS (Transconductance Amplifier) SummaryVoltage to Current ConverterVoltage to Current Converter

    ++

    __

    LoadLoad

    iiLL

    RR11ii11

    vvinin

    ++

    __

    OROR++

    __

    LoadLoad

    iiLL

    RR11ii11

    vvinin

    ++

    __

    vvinin++

    __

    General Equations:General Equations:

    iiLL = i= i11 = v= v11/R/R11

    vv11 = v= vinin

    The transconductance, gThe transconductance, gmm = i= ioo/v/vinin = 1/R= 1/R11

    Therefore, iTherefore, iLL = i= i11 = v= vinin/R/R11 = g= gmmvvinin

    The maximum load resistance is determined by:The maximum load resistance is determined by:

    RRL(max)L(max) = v= vo(max)o(max)/i/iLL

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    VCIS (Transconductance Amplifier)VCIS (Transconductance Amplifier)Voltage to Current Converter ExampleVoltage to Current Converter Example

    ++

    __

    LoadLoad

    iiLL

    RR11ii11

    vvinin

    ++

    __

    Given: vGiven: vinin = 2 V, R= 2 V, R11 = 2 k= 2 k;;vvo(max)o(max) == 10 V10 V

    Find: iFind: iLL , g, gmm and Rand RL(max)L(max)

    Solution:Solution:

    iiLL = i= i11 = v= vinin/R/R11 = 2 / 2000 == 2 / 2000 = 1 mA1 mA

    ggmm = i= ioo/v/vinin = 1/R= 1/R11 = 1 / 2000 == 1 / 2000 = 0.5 mS0.5 mS

    RRL(max)L(max) = v= vo(max)o(max)/i/iLL = 10 V / 1 mA= 10 V / 1 mA

    == 10 k10 k ;;

    Note:Note:

    If RIf RLL > R> RL(max)L(max) the op ampthe op ampwill saturatewill saturate

    The output current, iThe output current, iLL isis

    independent of the loadindependent of the load

    resistance.resistance.

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    VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter

    General Equations:General Equations:

    iiFF = i= iinin

    vvoo == --iiFFRRFF

    rrmm = v= voo/i/iinin = R= RFF

    ++

    __

    iiFF

    iiinin

    RRFF

    vvOO

    ++

    --

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    VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter

    Transresistance Amplifiers are used for lowTransresistance Amplifiers are used for low--powerpowerapplications to produce an output voltage proportional toapplications to produce an output voltage proportional to

    the input current.the input current.

    Photodiodes and Phototransistors, which are used in thePhotodiodes and Phototransistors, which are used in the

    production of solar power are commonly modeled asproduction of solar power are commonly modeled as

    current sources.current sources.

    Current to Voltage Converters can be used to convert theseCurrent to Voltage Converters can be used to convert these

    current sources to more commonly used voltage sources.current sources to more commonly used voltage sources.

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    VCIS (Transresistance Amplifier)VCIS (Transresistance Amplifier)Current to Voltage Converter ExampleCurrent to Voltage Converter Example

    ++

    __

    iiFF

    iiinin

    RRFF

    vvOO

    ++

    --

    Given: iGiven: iinin = 10 mA= 10 mA

    RRFF = 200= 200 ;;

    Find: iFind: iFF , v, voo and rand rmm

    Solution:Solution:

    iiFF = i= iinin == 10 mA10 mA

    vvoo == --iiFFRRFF = 10 mA * 200= 10 mA * 200 ;; == 2 V2 V

    rrmm = v= voo/i/iinin = R= RFF == 200200

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    Power BandwidthPower BandwidthThe maximum frequency at which a sinusoidal output signal can beThe maximum frequency at which a sinusoidal output signal can be

    produced without causing distortion in the signal.produced without causing distortion in the signal.

    The power bandwidth, BWThe power bandwidth, BWpp is determined using the desiredis determined using the desired

    output signal amplitude and the the slew rate (output signal amplitude and the the slew rate (see next slidesee next slide))

    specifications of the op amp.specifications of the op amp.

    BWBWpp = SR= SR

    22TTVVo(max)o(max)

    SR = 2SR = 2TTffVVo(max)o(max) where SR is the slew ratewhere SR is the slew rate

    Example:Example:

    Given: VGiven: Vo(max)o(max) = 12 V and SR = 500 kV/s= 12 V and SR = 500 kV/s

    Find: BWFind: BWpp

    Solution:Solution: BWBWpp == 500 kV/s500 kV/s = 6.63 kHz= 6.63 kHz

    22TT * 12 V* 12 V

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    Slew RateSlew RateA limitation of the maximum possible rate of change of theA limitation of the maximum possible rate of change of the

    output of an operational amplifier.output of an operational amplifier.

    As seen on the previous slide,As seen on the previous slide, This is derived from:This is derived from:

    SR = 2SR = 2TTffVVo(max)o(max) SR =SR = vvoo//ttmaxmax

    Slew Rate is independent of theSlew Rate is independent of the

    closedclosed--loop gain of the op amp.loop gain of the op amp.

    Example:Example:

    Given: SR = 500 kV/s andGiven: SR = 500 kV/s and vvoo = 12 V (Vo(max) = 12V)= 12 V (Vo(max) = 12V)

    Find: TheFind: The t and f.t and f.

    Solution:Solution: t =t = vo / SR = (10 V) / (5x10vo / SR = (10 V) / (5x1055 V/s) = 2x10V/s) = 2x10--55 ss

    f = SR /f = SR / 22TTVVo(max)o(max) == (5x10(5x1055 V/s) / (V/s) / (22TT * 12) = 6,630 Hz* 12) = 6,630 Hz

    f is thef is the

    frequency infrequency in

    HzHz

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    Slew Rate DistortionSlew Rate Distortion

    vv

    tt

    desired outputdesired output

    waveformwaveform

    actual outputactual output

    because ofbecause of

    slew rateslew rate

    limitationlimitation

    tt

    vv

    The picture above shows exactly what happens when theThe picture above shows exactly what happens when the

    slew rate limitations are not met and the output of theslew rate limitations are not met and the output of the

    operational amplifier is distorted.operational amplifier is distorted.

    SR =SR = v/v/t = m (slope)t = m (slope)

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    Noise GainNoise GainThe noise gain of an amplifier is independent of the amplifiersThe noise gain of an amplifier is independent of the amplifiers

    configuration (inverting or noninverting)configuration (inverting or noninverting)

    The noise gain is given by the formula:The noise gain is given by the formula:

    AANN = R= R11 + R+ RFF

    RR11

    Example 1: Given a noninverting amplifier with the resistanceExample 1: Given a noninverting amplifier with the resistance

    values, Rvalues, R11 = 2 k= 2 k;; and Rand RFF = 200 k= 200 k;;Find: The noise gain.Find: The noise gain.

    AANN == 2 k2 k;; + 200 k+ 200 k;; = 101= 101 Note: For theNote: For the

    2 k2 k;; noninverting amplifier Anoninverting amplifier ANN = A= AVV

    Example 2: Given an inverting amplifier with the resistanceExample 2: Given an inverting amplifier with the resistancevalues, Rvalues, R11 = 2 k= 2 k;; and Rand RFF = 20 k= 20 k;;

    Find: The noise gain.Find: The noise gain.

    AANN == 2 k2 k;; + 20 k+ 20 k;; = 12= 12 Note: For theNote: For the

    2 k2 k;; inverting amplifier Ainverting amplifier ANN > A> AVV

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    GainGain--Bandwidth ProductBandwidth Product

    In most operational amplifiers, the openIn most operational amplifiers, the open--loop gain beginsloop gain begins

    dropping off at very low frequencies. Therefore, to make thedropping off at very low frequencies. Therefore, to make theop amp useful at higher frequencies, gain is traded forop amp useful at higher frequencies, gain is traded for

    bandwidth.bandwidth.

    The GainThe Gain--Bandwidth Product (GBW) is given by:Bandwidth Product (GBW) is given by:

    GBW = AGBW = ANNBWBW

    Example: For a 741 op amp, a noise gain of 10 k correspondsExample: For a 741 op amp, a noise gain of 10 k corresponds

    to a bandwidth of ~200 Hzto a bandwidth of ~200 Hz

    Find: The GBWFind: The GBW

    GBW = 10 k * 200 Hz = 2 MHzGBW = 10 k * 200 Hz = 2 MHz

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    Cascaded AmplifiersCascaded Amplifiers -- BandwidthBandwidth

    Quite often, one amplifier does not increase the signal enoughQuite often, one amplifier does not increase the signal enough

    and amplifiers are cascaded so the output of one amplifier is theand amplifiers are cascaded so the output of one amplifier is theinput to the next.input to the next.

    The amplifiers are matched so:The amplifiers are matched so:

    BWBWSS = BW= BW11 = BW= BW22 == GBWGBW where, BWwhere, BWSS is the bandwidth of allis the bandwidth of all

    AANN the cascaded amplifiers and Athe cascaded amplifiers and ANN isis

    the noise gainthe noise gain

    The Total Bandwidth of the Cascaded Amplifiers is:The Total Bandwidth of the Cascaded Amplifiers is:

    BWBWTT = BW= BWss(2(21/n1/n 1)1)1/21/2 where n is the number of amplifierswhere n is the number of amplifiers

    that are being cascadedthat are being cascaded

    Example: Cascading 3 Amplifiers with GBW = 1 MHz and AExample: Cascading 3 Amplifiers with GBW = 1 MHz and ANN = 15,= 15,Find: The Total Bandwidth, BWFind: The Total Bandwidth, BWTT

    BWBWSS = 1 MHz / 15 = 66.7 kHz= 1 MHz / 15 = 66.7 kHz

    BWBWTT = 66.7 kHz (2= 66.7 kHz (21/31/3 1)1)1/21/2 = 34 kHz= 34 kHz

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    CommonCommon--Mode Rejection RatioMode Rejection Ratio

    The commonThe common--mode rejection ratio (CMRR) relates to the ability ofmode rejection ratio (CMRR) relates to the ability of

    the op amp to reject commonthe op amp to reject common--mode input voltage. This is verymode input voltage. This is very

    important because commonimportant because common--mode signals are frequentlymode signals are frequentlyencountered in op amp applications.encountered in op amp applications.

    CMRR = 20 log|ACMRR = 20 log|ANN/ A/ Acmcm||

    AAcmcm == AANN

    loglog--11

    (CMRR / 20)(CMRR / 20)We solve for AWe solve for Acmcm because Op Amp data sheets list the CMRR value.because Op Amp data sheets list the CMRR value.

    The commonThe common--mode input voltage is an average of the voltages thatmode input voltage is an average of the voltages that

    are present at the nonare present at the non--inverting and inverting terminals of theinverting and inverting terminals of the

    amplifier.amplifier.vvicmicm = v= v(+)(+) + v+ v((--))

    22

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    CommonCommon--Mode Rejection RatioMode Rejection RatioExampleExample

    Given: A 741 op amp with CMRR = 90 dB and a noise gain,Given: A 741 op amp with CMRR = 90 dB and a noise gain,AANN = 1 k= 1 k

    Find: The common mode gain, AFind: The common mode gain, Acmcm

    AAcmcm == AANN = 1000= 1000

    loglog--11 (CMRR / 20)(CMRR / 20) loglog--11 (90 / 20)(90 / 20)

    = 0.0316= 0.0316

    It is very desirable for the commonIt is very desirable for the common--mode gain to be small.mode gain to be small.

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    Power Supply Rejection RatioPower Supply Rejection Ratio

    One of the reasons op amps are so useful, is that they canOne of the reasons op amps are so useful, is that they can

    be operated from a wide variety of power supply voltages.be operated from a wide variety of power supply voltages.

    The 741 op amp can be operated from bipolar suppliesThe 741 op amp can be operated from bipolar supplies

    ranging fromranging from 5V to5V to 18V with out too many changes to the18V with out too many changes to the

    parameters of the op amp.parameters of the op amp.

    The power supply rejection ratio (SVRR) refers to the slightThe power supply rejection ratio (SVRR) refers to the slight

    change in output voltage that occurs when the powerchange in output voltage that occurs when the power

    supply of the op amp changes during operation.supply of the op amp changes during operation.

    SVRR = 20 log (SVRR = 20 log (VVss// VVoo))

    The SVRR value is given for a specified op amp. For theThe SVRR value is given for a specified op amp. For the

    741 op amp, SVRR = 96 dB over the range741 op amp, SVRR = 96 dB over the range 5V to5V to 18V.18V.

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    OpenOpen--Loop Op Amp CharacteristicsLoop Op Amp CharacteristicsTable 12.1Table 12.111

    Device LM741C LF351 OP-07 LH0003 AD549K

    Technology BJT BiFET BJTHybrid

    BJTBiFET

    AOL(typ)

    200 k 100 k 400 k 40 k 100 k

    Rin 2 M; 1012 ; 8 M; 100 k; 1013; || 1 pF

    Ro 50 ; 30 ; 60 ; 50 ; ~100 ;

    SR 0.5 V/Qs 13 V/Qs 0.3 V/Qs 70 V/Qs 3 V/Qs

    CMRR 90 dB 100 dB 110 dB 90 dB 90 dB

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    SourcesSources

    Dailey, Denton.Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated.Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, NewPrentice Hall, New

    Jersey: 2001. (pp 456Jersey: 2001. (pp 456--509)509)

    11Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457Table 12.1: Selected Op Amps and Their Open Loop Characteristics, pg 457

    Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press,Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press,

    New York: 1998.New York: 1998.

    Neamen, Donald.Neamen, Donald. Semiconductor Physics & Devices. Basic Principles.Semiconductor Physics & Devices. Basic Principles. McGrawMcGraw--Hill,Hill,

    Boston: 1997. (pp 351Boston: 1997. (pp 351--357)357)

    Web SourcesWeb Sources

    www.infoplease.com/ce6/sci/A0803814.htmlwww.infoplease.com/ce6/sci/A0803814.html

    http://www.infoplease.com/ce6/sci/A0836717.htmlhttp://www.infoplease.com/ce6/sci/A0836717.html

    http://people.msoe.edu/~saadat/PSpice230Part3.htmhttp://people.msoe.edu/~saadat/PSpice230Part3.htm