KNL4343 lecture4

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  • KNL4343 VLSI Design And Technology

    Lecture 04: CMOS Inverter (static view)

    Norhuzaimin Julai

    [Adapted from Rabaeys Digital Integrated Circuits, 2002, J. Rabaey et al.]

    Irwin&Vijay, PSU, 2002

  • Review: Design Abstraction Levels

    SYSTEM

    GATE

    CIRCUIT

    Vout Vin

    CIRCUIT

    Vout Vin

    MODULE

    +

    DEVICE

    n+

    S D

    n+

    G

  • Review: The MOS Transistor

    Gate oxide

    n+

    Source Drain

    p substrate

    Bulk (Body)

    p+ stopper

    Field-Oxide

    (SiO2) n+

    Polysilicon

    Gate

    L

    W

  • CMOS Inverter: A First Look

    VDD

    Vout

    CL

    Vin

  • CMOS Inverter: Steady State Response

    VDD

    Rn

    Vout = 0

    Vin = V DD

    VDD

    Rp

    Vout = 1

    Vin = 0

    VOL = 0

    VOH = VDD VM = f(Rn, Rp)

  • CMOS Properties

    Full rail-to-rail swing high noise margins

    Logic levels not dependent upon the relative device sizes transistors can be minimum size ratioless

    Always a path to Vdd or GND in steady state low output impedance (output resistance in k range) large fan-out (albeit with degraded performance)

    Extremely high input resistance (gate of MOS transistor is near perfect insulator) nearly zero steady-state input current

    No direct path steady-state between power and ground no static power dissipation

    Propagation delay function of load capacitance and resistance of transistors

  • Review: Short Channel I-V Plot (NMOS)

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5VDS (V)

    X 10-4

    VGS = 1.0V

    VGS = 1.5V

    VGS = 2.0V

    VGS = 2.5V

    NMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = 0.4V

  • Review: Short Channel I-V Plot (PMOS)

    -1

    -0.8

    -0.6

    -0.4

    -0.2

    0

    0-1-2 VDS (V)

    X 10-4

    VGS = -1.0V

    VGS = -1.5V

    VGS = -2.0V

    VGS = -2.5V

    PMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = -0.4V

    All polarities of all voltages and currents are reversed

  • Transforming PMOS I-V Lines

    IDSp = -IDSn

    VGSn = Vin ; VGSp = Vin - VDD VDSn = Vout ; VDSp = Vout - VDD

    Vout

    IDn

    VGSp = -2.5

    VGSp = -1

    Mirror around x-axis

    Vin = VDD + VGSp IDn = -IDp

    Vin = 1.5

    Vin = 0

    Vin = 1.5

    Vin = 0

    Horiz. shift over VDD Vout = VDD + VDSp

    Want common coordinate set Vin, Vout, and IDn

  • CMOS Inverter Load Lines

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5Vout (V)

    X 10-4

    Vin = 1.0V

    Vin = 1.5V

    Vin = 2.0V

    Vin = 2.5V

    0.25um, W/Ln = 1.5, W/Lp = 4.5, VDD = 2.5V, VTn = 0.4V, VTp = -0.4V

    Vin = 0V

    Vin = 0.5V

    Vin = 1.0V

    Vin = 1.5V

    Vin = 0.5V Vin = 2.0V

    Vin = 2.5V

    Vin = 2V Vin = 1.5V

    Vin = 1V

    Vin = 0.5V

    Vin = 0V

    PMOS NMOS

  • CMOS Inverter VTC

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5

    Vin (V)

    Vo

    ut (

    V)

    NMOS off

    PMOS res

    NMOS sat

    PMOS res

    NMOS sat

    PMOS sat

    NMOS res

    PMOS sat NMOS res

    PMOS off

  • CMOS Inverter: Switch Model of Dynamic Behavior

    VDD

    Rn

    Vout

    CL

    Vin = V DD

    VDD

    Rp

    Vout

    CL

    Vin = 0

    Gate response time is determined by the time to charge CL

    through Rp (discharge CL through Rn)

  • Relative Transistor Sizing

    When designing static CMOS circuits, balance the driving strengths of the transistors by making the PMOS section wider than the NMOS section to

    maximize the noise margins and

    obtain symmetrical characteristics

  • Switching Threshold

    VM where Vin = Vout (both PMOS and NMOS in saturation since VDS = VGS)

    VM rVDD/(1 + r) where r = kpVDSATp/knVDSATn

    Switching threshold set by the ratio r, which compares the relative driving strengths of the PMOS and NMOS transistors

    Want VM = VDD/2 (to have comparable high and low noise margins), so want r 1

    (W/L)p knVDSATn(VM-VTn-VDSATn/2)

    (W/L)n kpVDSATp(VDD-VM+VTp+VDSATp/2) =

  • Switch Threshold Example

    In our generic 0.25 micron CMOS process, using the process parameters from slide L03.25, a VDD = 2.5V, and a minimum size NMOS device ((W/L)n of 1.5)

    VT0(V) (V0.5) VDSAT(V) k(A/V

    2) (V-1)

    NMOS 0.43 0.4 0.63 115 x 10-6 0.06

    PMOS -0.4 -0.4 -1 -30 x 10-6 -0.1

    (W/L)p 115 x 10-6 0.63 (1.25 0.43 0.63/2)

    (W/L)n -30 x 10-6 -1.0 (1.25 0.4 1.0/2)

    = x x = 3.5

    (W/L)p = 3.5 x 1.5 = 5.25 for a VM of 1.25V

  • Simulated Inverter VM

    0.8

    0.9

    1

    1.1

    1.2

    1.3

    1.4

    1.5

    0 1 10(W/L)p/(W/L)n

    VM is relatively

    insensitive to variations in

    device ratio setting the ratio to 3, 2.5

    and 2 gives VMs of 1.22V, 1.18V, and 1.13V

    Increasing the width of

    the PMOS moves VM

    towards VDD

    Increasing the width of

    the NMOS moves VM

    toward GND

    .1

    Note: x-axis is semilog

    ~3.4

  • Noise Margins Determining VIH and VIL

    0

    1

    2

    3

    VIL VIHVin

    VOH = VDD

    VM

    By definition, VIH and VIL are

    where dVout/dVin = -1 (= gain)

    VOL = GND

    A piece-wise linear

    approximation of VTC

    NMH = VDD - VIH NML = VIL - GND

    Approximating:

    VIH = VM - VM /g

    VIL = VM + (VDD - VM )/g

    So high gain in the transition

    region is very desirable

  • CMOS Inverter VTC from Simulation

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5

    Vin (V)

    Vo

    ut (

    V)

    0.25um, (W/L)p/(W/L)n = 3.4

    (W/L)n = 1.5 (min size)

    VDD = 2.5V

    VM 1.25V, g = -27.5

    VIL = 1.2V, VIH = 1.3V

    NML = NMH = 1.2

    (actual values are

    VIL = 1.03V, VIH = 1.45V

    NML = 1.03V & NMH = 1.05V)

    Output resistance

    low-output = 2.4k

    high-output = 3.3k

  • Gain Determinates

    -18

    -16

    -14

    -12

    -10

    -8

    -6

    -4

    -2

    0

    0 0.5 1 1.5 2

    Vin Gain is a strong function of the

    slopes of the currents in the

    saturation region, for Vin = VM (1+r) g ---------------------------------- (VM-VTn-VDSATn/2)(n - p )

    Determined by technology

    parameters, especially channel

    length modulation (). Only

    designer influence through

    supply voltage and VM (transistor

    sizing).

  • Impact of Process Variation on VTC Curve

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5

    Vin (V)

    Vo

    ut (

    V)

    Nominal

    Good PMOS

    Bad NMOS

    Bad PMOS

    Good NMOS

    Process variations (mostly) cause a shift in the switching threshold

  • Scaling the Supply Voltage

    0

    0.5

    1

    1.5

    2

    2.5

    0 0.5 1 1.5 2 2.5Vin (V)

    Vo

    ut (

    V)

    Device threshold voltages are

    kept (virtually) constant

    0

    0.05

    0.1

    0.15

    0.2

    0 0.05 0.1 0.15 0.2

    Vin (V)

    Vo

    ut (

    V)

    Gain=-1

    Device threshold voltages are

    kept (virtually) constant

  • Next Time: CMOS Inverter max Layout

    VDD

    GND

    NMOS (2/.24 = 8/1)

    PMOS (4/.24 = 16/1)

    metal2

    metal1 polysilicon

    In Out

    metal1-poly via

    metal2-metal1 via

    metal1-diff via

    pdiff

    ndiff

  • Next Lecture and Reminders

    Next lecture

    IC manufacturing

    - Reading assignment Rabaey, et al, 2.1-2.3

    Reminders