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Operational Amplifiers Lecture 4 BME 372 New Schesser 94

Operational Amplifiersjoelsd/electronicsnew/Classnotes... · 2020. 9. 29. · Op-amp • Because we assumed that the Op-amp was ideal, we found that with negative feedback we can

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  • OperationalAmplifiers

    Lecture4

    BME372NewSchesser 94

  • 95

    Ideal Amplifiers •  Parameters:

    vs vo vi

    Ro

    Ri

    Rs

    RL Avovi

    io ii + +

    - -

    +

    -

    vovs

    =Ri

    Ri + RsAvo

    RLRL + RO

    This states the gain of the amplifier depends on the external components.This is BAD!!!!!

    However, if Ri →∞ and RO → 0; then Ri

    Ri + Rs→1 and

    RLRL + RO

    →1.

    Therefore, vovs

    → Avo and the gain of the amplifier is independent of the external components.

    BME372NewSchesser

  • 96

    Operational Amplifiers

    •  An operational Amplifier is an ideal differential with the following characteristics: –  Infinite input impedance –  Infinite gain for the differential signal – Zero gain for the common-mode signal – Zero output impedance –  Infinite Bandwidth vo-

    +v1v2

    Inverting input

    Non-Inverting input

    BME372NewSchesser

  • 97

    Operational Amplifier Feedback •  Operational Amplifiers are used with negative feedback •  Feedback is a way to return a portion of the output of an

    amplifier to the input –  Negative Feedback: returned output opposes the source signal –  Positive Feedback: returned output aids the source signal

    •  For Negative Feedback –  In an Op-amp, the negative feedback returns a fraction of the

    output to the inverting input terminal forcing the differential input to zero.

    –  Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit

    –  Since the input impedance is infinite the current flowing into the input is also zero.

    –  These latter two points are called the summing-point constraint.

    BME372NewSchesser

  • 98

    Operational Amplifier Analysis Using the Summing Point Constraint

    •  In order to analyze Op-amps, the following steps should be followed:

    1.  Verify that negative feedback is present 2.  Assume that the voltage and current at the input

    of the Op-amp are both zero (Summing-point Constraint

    3.  Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest.

    BME372NewSchesser

  • 99

    Example: Inverting Amplifier 1.  Verify Negative Feedback: Note that a

    portion of vo is fed back via R2 to the inverting input. So if vi increases and, therefore, increases vo, the portion of vo fed back will then have the affect of reducing vi (i.e., negative feedback).

    2.  Use the summing point constraint.

    3.  Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output

    RLvin

    R2

    R1

    vivo

    -+

    0 i1=vin/R1 i2

    inverted) :input the toopposite isoutput that the(note oft independen is which -

    zero is since 0constraintpoint -summing the todue

    constraintpoint -summing the todue zero is since 0

    1

    2

    220

    21

    11

    Lin

    i

    iin

    RvRR

    vRivii

    vRiv

    =

    +−==

    +=1

    1

    11

    1

    RiRi

    ivZ inin ===

    -+ ++

    --

    BME372NewSchesser

  • 100

    Op-amp

    •  Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is:

    1.  Independent of the load 2.  Dependent only on values of the circuit

    parameter 3.  We can choose the gain of our amplifier by

    proper selection of resistors.

    BME372NewSchesser

  • 101

    Non-inverting Amp 1.  First check: negative feedback? 2.  Next apply, summing point constraint 3.  Use circuit analysis

    +

    -- vo

    +

    --

    RL R2

    R1

    +

    --

    vi

    +

    --

    vf

    io

    Iin= 0

    +

    -

    vin

    1

    1 2

    2 1 2

    1 1

    0

    ;

    1

    Since 0;

    in i f f f

    f o in

    ov

    in

    inin in

    in

    v v v v vRv v v

    R Rv R R RAv R R

    vi Zi

    = + = + =

    = =+

    += = = +

    = = = ∞Note:1.  Thegainisalwaysgreaterthanone2.  Theoutputhasthesamesignasthe

    input

    BME372NewSchesser

  • 102

    Non-inverting Amp Special Case What happens if R2 = 0?

    +

    -- vo

    +

    --

    RL

    +

    --

    vi

    io

    Iin= 0

    +

    -

    vin

    ∞===

    =+==

    ==+

    =

    =+=+=

    in

    ininin

    in

    o

    inoof

    fffiin

    ivZi

    RR

    vvAv

    vvvRRv

    vvvvv

    ;0 Since

    10

    ;0

    0

    1

    1

    1

    1

    This is a unity gain amplifier and is also called a voltage follower.

    BME372NewSchesser

  • 103

    Special Amplifiers

    •  Summer (Homework Problem) •  Instrumentation Amplifier – Uses 3 Op-amps – One as a differential amplifier – Two Non-inverting Amps using for providing gain

    BME372NewSchesser

  • 104

    Medical Instrumentation Amplifier

    Differential Amplifier

    +

    --

    R2

    R1

    +

    -

    v2

    +

    --

    R2

    R1

    +

    -

    v1

    R3

    R4 +

    -

    vo

    R5

    R6

    Non-inverting Amplifier

    Non-inverting Amplifier

    BME372NewSchesser

  • 105

    Medical Instrumentation Amplifier

    +

    --

    R2

    R1

    +

    -

    v2

    +

    --

    R2

    R1

    +

    -

    v1

    R3

    R4

    vo

    R5

    R6

    Differential Amplifier

    Non-inverting Amplifier

    Non-inverting Amplifier

    v2D

    v1D

    +

    -

    BME372NewSchesser

  • 106

    Medical Instrumentation Amplifier 2

    6 5

    2

    6 6 5 5

    5 62

    6 6 5 5

    41

    3 4

    5 62 41

    6 3 4 6 5 5

    5 5 6 41 2

    6 5 3 4

    5 6 4

    5 3 4

    51 2

    6

    1 1( )

    ( )

    ( )

    ( )

    Chose 1

    ( )

    D x x o

    oDx

    oDx

    y D x i x

    oDD

    o D D

    o D D

    v v v vR R

    vv vR R R R

    R R vv vR R R R

    Rv v v v vR R

    R R vv R vR R R R R R

    R R R Rv v vR R R RR R RR R R

    Rv v vR

    − −=

    −− + =

    + −− =

    = = − =+

    + −− =+

    += −+

    + =+

    = −

    R3

    R4

    vo

    R5

    R6

    Differential Amplifier

    v2D

    v1D

    vx

    +

    -

    4

    3

    5

    6

    3564

    45356454

    43

    4

    5

    65

    1 Chose

    RR

    RR

    RRRRRRRRRRRR

    RRR

    RRR

    =

    =+=+

    =+

    +

    vi=0

    vy

    BME372NewSchesser

  • 107

    Medical Instrumentation Amplifier

    +

    --

    R2

    R1

    +

    -

    v2

    +

    --

    R2

    R1

    +

    -

    v1

    R3

    R4 +

    -

    vo

    R5

    R6

    Differential Amplifier

    Non-inverting Amplifier

    Non-inverting Amplifier

    v2D

    v1D

    BME372NewSchesser

  • 108

    Medical Instrumentation Amplifier

    +

    --

    R2

    R1

    +

    -

    v2

    +

    --

    R2

    R1

    +

    -

    v1 Non-inverting Amplifier

    Non-inverting Amplifier

    v2D

    v1D vA

    AD

    AD

    AD

    AD

    vRRv

    RRRv

    vRRv

    RRRv

    vRRv

    RRRv

    Rvv

    Rvv

    1

    21

    1

    211

    1

    22

    1

    212

    1

    22

    12

    22

    1

    2

    2

    22

    Likewise

    )11(

    −+=

    −+=

    −+=

    −=−

    BME372NewSchesser

  • 109

    Medical Instrumentation Amplifier

    ))(1())((

    )]([

    )(

    21

    1

    2

    6

    521

    1

    21

    6

    5

    1

    22

    1

    21

    1

    21

    1

    21

    6

    5

    1

    21

    1

    211

    1

    22

    1

    212

    21

    6

    5

    vvRR

    RRvv

    RRR

    RRv

    vRRv

    RRRv

    RRv

    RRR

    RRv

    vRRv

    RRRv

    vRRv

    RRRv

    vvRRv

    o

    AAo

    AD

    AD

    DDo

    −+=−+=

    −+−−+=

    −+=

    −+=

    −=

    +

    -- R2

    R1

    +

    -

    v2

    +

    --

    R2

    R1 + -

    v1

    R3

    R4

    vo

    R5

    R6

    Differential Amplifier

    Non-inverting Amplifier

    Non-inverting Amplifier

    +

    -

    -

    +

    BME372NewSchesser

  • 110

    Uses of the Differential Amplifier

    BME372NewSchesser

  • 111

    Integrators and Differentiators

    ∫∫ −=−=

    ==

    t

    in

    t

    o

    in

    dxxvRC

    dxxiC

    v

    tiRtvti

    002

    21

    )(1)(1

    )()()(

    vin

    C

    Rvi

    vo

    -+

    0 i2 i1

    vin

    R

    Cvi

    vo

    -+

    0 i2 i1

    dttdvRCRtiv

    tidttCdvti

    ino

    in

    )()(

    )()()(

    2

    21

    −=−=

    ==

    BME372NewSchesser

  • 112

    Frequency Analysis

    vinvi

    vo

    -+

    0 i1=vin/Z1

    i2

    1in

    o

    Lin1

    o

    1

    11in

    ZZ

    VV

    ZVZZ

    ZIVII

    ZIV

    2

    2

    22

    2

    - )()(

    oft independen is which )(-

    zero )(virtually is since 0)()(constraintpoint -summing the todue )()(

    zero )(virtually is since 0)()()(

    =

    =

    +−==

    +=

    ωω

    ω

    ωωωω

    ωωω

    jj

    j

    vjjjj

    vjjj

    i

    i

    ZLvin

    C

    Rvi vo

    -+

    0i1=vin/Z1 i2

    ZLvin

    R

    Cvi vo

    -+

    0i1=vin/Z1 i2

    integratoran 1- )()( 2

    RCjjj

    ωωω −==

    1in

    o

    ZZ

    VV tordifferenia a - )(

    )( 2 RCjjj ωωω −==

    1in

    o

    ZZ

    VV

    Z2

    Z1

    BME372NewSchesser

  • 113

    Frequency Response

    vin

    CRvi

    vo

    -+

    0 i2 i1

    vin

    RCvi

    vo

    -+

    0 i2 i1

    1- )()( 2

    RCjjj

    ωωω −==

    1in

    o

    ZZ

    VV

    RCjjj ωωω −==

    1in

    o

    ZZ

    VV 2-

    )()(

    ω

    ω

    BME372NewSchesser

  • 114

    Frequency Response

    vin

    C2R1vi

    vo

    -+

    0 i2 i1

    vin

    R2

    C1vi

    vo

    -+

    0 i2

    i1

    R2 VoutVin

    = −Z2Z1

    = −R2R1

    1(1+ jωC2R2 )

    =R2R1

    1

    1+ (ωC2R2 )2∠π − tan−1(ωC2R2 )

    0

    500

    1,000

    1,500

    2,000

    2,500

    1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

    R151C110mfR2100k

    R1

    12 2 1 1 2 1 11 12

    1 1 1 1 1 1 1

    tan ( )(1 ) 21 ( )

    out

    in

    V Z R j C R R C R C RV Z R j C R R C R

    ω ω π ωω ω

    −= − = − = ∠− −+ +

    0

    500

    1,000

    1,500

    2,000

    2,500

    1.E+00 1.E+02 1.E+04 1.E+06 1.E+08HZ

    R151C110mfR2100k

    BME372NewSchesser

  • 115

    Homework •  Probs 2.2, 2.5, 2.6, 2.10, 2.22, 2.24, 2.25,

    2.28 •  Calculate and plot the output vs frequency

    for these circuits. R1=1k, R2=3k, C=1µf Use Matlab to calculate the Bode plot

    vin

    C

    R1vi

    vo-+

    R2

    vinCR1

    vivo

    -+

    R2

    vin

    C

    R1vi

    vo-+

    R2R2

    vinC

    R1

    vivo

    -+

    BME372NewSchesser