Current Mirror Application

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    EECE488 Set 5 - Current Mirrors 1SM

    EECE488: Analog CMOS Integrated Circuit Design

    Set 5

    Current Mirrors

    Shahriar Mirabbasi

    Department of Electrical and Computer Engineering

    University of British Columbia

    [email protected]

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    Applications of Current Sources

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    Simple Resistive Biasing for Current Source

    2

    12

    2 )(2

    THDDoxn

    OUT VVRR

    RLWCI

    +

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    Problems

    Output current depends on:

    Supply

    Process

    Temperature

    What if the bias voltage is independent of supply voltage?

    Is there a way of generating reliable currents?

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    Basic Idea

    Iout is a function of gate-source voltage

    Typically we assume that one precisely defined current source isavailable and other current sources copy their current from this precise

    source.

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    Basic Idea

    This structure is called current mirror

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    Question

    What happens if the two transistors in the basic current mirrorhave different sizes?

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    Example

    Assuming all the transistors are in saturation region, find Iout:

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    Current Mirrors: Amplifier Bias Example

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    Board Notes

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    Current Mirrors: Signal Amplification Example

    Find the small signal voltage gain of the following circuit.

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    Effect of Channel Length Modulation

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    Cascode Current Mirror

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    Board Notes

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    Cascode Current Mirror

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    Cascode Current Mirror

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    Cascode Current Mirror

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    Board Notes

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    Cascode Current Mirror Biasing

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    Cascode Current Mirror Biasing

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    Current Mirror Biasing

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    Basic Circuit to Generate Supply Independent Current

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    Supply Independent Current

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    Board Notes

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    Supply Independent Current

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    Start-up Problem

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    Start-up Problem

    DDGSTHGSDDTHTHTH VVVVandVVVV >++

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    Active Current Mirrors

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    Active Current Mirrors in Differential to Single-Ended

    Amplifiers

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    Differential to Single-Ended Amplifiers

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    Calculation of Gm

    ID1 = ID3 = ID4 = gm1,2Vin / 2 ID2 = gm1,2Vin / 2

    Iout = ID2 ID4 = gm1,2Vin , Gm = gm1, 2

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    Small-Signal Gain

    Av gm1,2 (ro2 || ro4 )

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    Common Mode Characteristics

    ACM =Vout

    Vin ,CM

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    Common Mode

    ACM

    1

    2gm3,4||ro3,4

    2

    12gm1,2

    +RSS

    =1

    1 + 2gm1,2RSS

    gm1,2

    gm3, 4

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    Common Mode

    CMRR =ADM

    ACM

    = gm1, 2(ro1,2 || ro3,4 )gm3,4 (1 + 2gm1,2RSS )

    gm1,2

    = gm3,4 (ro1, 2 || ro3,4 )(1 + 2gm1, 2RSS)