Lect01-6P-C

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    Linear Amplifiers (2)

    • Symbol for a single-ended

    input linear voltage amplifier.

    • Ideally provides linearvoltage gain regardless of

    the amplitude of the input

    signal

    • Real amplifiers have power 

    supplies that limit the

    amplitude of the output

    • If input is too large, output

    clamps

    Transfer Characteristics (1)

    Plot of amplifier output versus amplifier input

    Transfer Characteristics (2)

    • Gain = Slope

    • To operate amplifier in its linear region, the inputmust be kept small enough

    Real Transfer Characteristics

    • Each circle represents

    a different DC

    component for the

    input and output

    signals – called an

    operating point

    • Location of operating

    point has an effect on

    • input signal range

    • amplifier gain magnitude

    • amount of distortion

    Operating Point (1)

    Voltage gain, output DC voltage, allowable input

    magnitude range are affected

    Operating Point (2)

    Derivative of

    transfercharacteristic

    gives measure

    of amplifier gain

    linearity (and

    distortion)

    Input and output signal amplitude ranges are

    maximized when operating point is near middle of

    linear region

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    Signal Convention (1)

    • DC magnitudes in uppercase symbol and

    subscript

     – Example: ID, VD

    • Ac signal quantities in lowercase symbol &

    subscript

     – Example: id(t), vd(t)

    • Total DC + ac signal quantities in lowercase

    symbol, uppercase subscript

     – Example: iD(t), vD (t)

    Signal Convention (2)In general

    vD(t) = VD + vd(t)

    iC(t) = IC + ic(t)

    Superposition

    • If the amplifier is linear, superposition can be

    applied. Each component can be determined

    seperately.

    • Determine the DC magnitude of the output using

    DC model (i.e capacitors are open circuited,

    inductors are short circuited)

     – Example: IO, VO

    • Determine the ac signal component of the output

    using the ac model (i.e., DC sources are killed and

    capacitors are short circuited, inductors are open

    circuited) as

     – Example: vo(t) = Av vi(t)

    Amplifier Classification

    CurrentCurrentCurrent

    TransresistanceVoltageCurrent

    TransconductanceCurrentVoltage

    VoltageVoltageVoltage

    TypeOutputInput

    +

    vi

    -

    +

    vo

    -

    Ideal Voltage Amplifier 

    Avi

    +

    R oAvi

    +

    vi-

    +

    vo-

    R i

    +

    Real Voltage Amplifier 

    Loaded Ideal Voltage Amplifier 

    Avi

    +

    vi-

    +

    vo-

    +

    R LR s

    vs

    +

    Av

    v

    s

    o=

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    Loaded Voltage Amplifier 

    R o

    Avi

    +

    vi-

    +

    vo-

    R i

    +

    R L

    R s

    vs

    +

    Lo

    L

    is

    i

    s

    o

    R R 

    R R 

    R A

    v

    v

    ++=

    Ideal Current Amplifier 

    Real Current Amplifier 

    Aii

    ii io

    Aiiii io

    R i R o

    Loaded Ideal Current Amplifier 

    Ai

    i

    s

    o=

    Aii R Lii

    R sisio

    Loaded Current Amplifier 

    Aii R Lii

    R Sisio

    R i R o

    Lo

    o

    is

    s

    s

    o

    R R 

    R R 

    R A

    i

    i

    ++=

    To determine Input Resistance R i

    R oAvi

    +

    vi-

    +

    vo-

    R i

    +

    • Apply input voltage vi(or input current ii)

    • Determine the input

    current ii (or inputvoltage vi)

    Then

    R i = vi / iiAiiii io

    R i R o

    ii

    +

    vi-

    Voltage Amplifier 

    Current Amplifier 

    To determine Output Resistance R o

    R oAvi

    +

    vi-

    +

    vx-

    R i

    +

    • Kill the input signal

    (set vi or ii to zero)

    • Apply a test voltage

    vx to the output node• Determine the current

    ix that the source

    delivers to the circuit

    Then

    R o = vx / ix

    Aiiii ix

    R i R o

    ii

    +

    vi-

    Voltage Amplifier 

    Current Amplifier 

    +

    vx-

    ix