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EE534 VLSI Design System Summer 2004 Lecture 7: Static Dynamic CMOS inverter (CHAPTER 6)

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EE534VLSI Design SystemSummer 2004 Lecture 7: Static Dynamic CMOS inverter (CHAPTER 6)

The average current during high to low transition can be calculated by using the current values at the beginning and the end of the transition.

The average current during low to high transition can be calculated by using the current values at the beginning and the end of the transition.

- Total fall delay =(t1-t0) + (t2-t1)

- Similar calculation as for falling delay
- Separate into regions where PMOS is in linear, saturation

- What if input has finite rise/fall time?not a step pulse
- Both transistors are on for some amount of time
- Capacitor charge/discharge current is reduced

Empirical equations:

Review: Propagation delay simulation results

At very short channel width, the delay approaches a limit value of about 0.2nsec

, which is mainly determined by technology-specific parameters, independent

of extrinsic capacitance component.

CMOS Ring Oscillator Circuit

CMOS Ring Oscillator Circuit (cont.)

T=PHL1+PLH1+ PHL2+PLH2+ PHL3+PLH3

=2 P+ 2P+ 2P

=6P

- Model transistors as switches and resistances
- Resistance Ron = average resistance for a transition
- For NMOS tphl:

RP

A

Rn

CL

A

Delay estimation using switch-level model (for general RC circuit):

Rn

CL

- For fall delay tphl, V0=Vcc, V1=Vcc/2

Standard RC-delay equations

- For long interconnect lines, RC must be distributed to obtain accurate simulation results
- Elmore Delay – first order time constant
- simple, close approximation of delay

- Resistance is proportional to cross-sectional area
- As interconnect scales, increase aspect ratio to maintain same area.

- Reduce CL
- internal diffusion capacitance of the gate itself
- interconnect capacitance
- fanout

- Increase W/L ratio of the transistor
- the most powerful and effective performance optimization tool in the hands of the designer

- Increase VDD
- only minimal improvement in performance at the cost of increased energy dissipation

- Slope engineering - keeping signal rise and fall times smaller than or equal to the gate propagation delays and of approximately equal values
- good for performance
- good for power consumption

CMOS inverter Power Dissipation

Lead microprocessors power continues to increase

100

P6

Pentium ®

10

486

286

8086

Power (Watts)

386

8085

1

8080

8008

4004

0.1

1971

1974

1978

1985

1992

2000

Year

Power delivery and dissipation will be prohibitive

Source: Borkar, De Intel

10000

1000

Rocket

Sun’s

…chips might become hot…

Nozzle

Surface

100

Nuclear

Power Density (W/cm2)

Reactor

8086

10

4004

P6

Hot Plate

8008

Pentium®

8085

386

286

486

8080

1

1970

1980

1990

2000

2010

Year

Source: Borkar, De Intel

On-Die Temperature

Power Map

- Power density is not uniformly distributed across the chip
- Silicon is not a good heat conductor
- Max junction temperature is determined by hot-spots
- Impact on packaging, w.r.t. cooling

- Power has three components
- Static power: when input isn’t switching
- Dynamic capacitive power: due to charging and discharging of load capacitance
- Dynamic short-circuit power: direct current from VDD to Gnd when both transistors are on

- Static power consumption:
- Static current: in CMOS there is no static current as long as Vin < VTN or Vin > VDD+VTP
- Leakage current: determined by “off” transistor
- Influenced by transistor width, supply voltage, transistor threshold voltages

VDD

VDD

Ileak,p

VDD

VI<VTN

Vo(low)

Vcc

Ileak,n

Drain junction leakage

Sub-threshold current

Gate leakage

VDD Ileakage

Vout

Sub-threshold current is the dominant factor.

All increase exponentially with temperature!

- Continued scaling of supply voltage and the subsequent scaling of threshold voltage will make subthreshold conduction a dominate component of power dissipation.

10-2

- An 90mV/decade VT roll-off - so each 255mV increase in VT gives 3 orders of magnitude reduction in leakage (but adversely affects performance)

10-7

10-12

Ileakage(nA/m)

Temp(C)

From De,1999

Vdd

Vin

Vout

CL

Case I: When the input is at logic 0: Under this condition the PMOS is conducting and NMOS is in cutoff mode and the load capacitor must be charged through the PMOS device.

Power dissipation in the PMOS transistor is given by,

PP=iLVSD= iL(VDD-VO)

The current and output voltages are related by,

iL=CLdvO/dt

Similarly the energy dissipation in the PMOS device can be written as the output switches from low to high ,

Above equation showed the energy stored in the capacitor CL when the output is high.

Case II: when the input is high and out put is low:

During switching all the energy stored in the load capacitor is dissipated in the NMOS device because NMOS is conducting and PMOS is in cutoff mode. The energy dissipated in the NMOS inverter can be written as,

The total energy dissipated during one switching cycle is,

The power dissipated in terms of frequency can be written as

This implied that the power dissipation in the CMOS inverter is directly proportional to switching frequency and VDD2

- Formula for dynamic power:
- Observations
- Does not (directly) depend on device sizes
- Does not depend on switching delay
- Applies to general CMOS gate in which:
- Switched capacitances are lumped into CL
- Output swings from Gnd to VDD
- Input signal approximated as step function
- Gate switches with frequency f

Not a function of transistor sizes!

Data dependent - a function of switching activity!

Capacitance:

Function of fan-out, wire length, transistor sizes

Supply Voltage:

Has been dropping with successive generations

Clock frequency:

Increasing…

Pdyn = CL VDD2f

Vin

Isc

Vout

CL

Finite slope of the input signal causes a direct current path between VDD and GND for a short period of time during switching when both the NMOS and PMOS transistors are conducting.

- Short-circuit current flows from VDD to Gnd when both transistors are on
- Plot on VTC curve:

Imax: depends on saturation current of devices

VCC

Imax

Vout

ID

Vin

VCC

- Approximate short-circuit current as a triangular wave
- Energy per cycle:

Imax

- Duration and slope of the input signal, tsc
- Ipeak determined by
- the saturation current of the P and N transistors which depend on their sizes, process technology, temperature, etc.
- strong function of the ratio between input and output slopes
- a function of CL

Psc = tsc VDD Ipeak f01

Isc 0

Isc Imax

Vin

Vout

Vin

Vout

CL

CL

Large capacitive load

Output fall time significantly larger than input rise time.

Small capacitive load

Output fall time substantially smaller than the input rise time.

x 10-4

When load capacitance is small, Ipeak is large.

CL = 20 fF

CL = 100 fF

Ipeak (A)

Short circuit dissipation is minimized by matching the rise/fall times of the input and output signals - slope engineering.

CL = 500 fF

x 10-10

time (sec)

500 psec input slope

When load capacitance is small (tsin/tsout > 2 for VDD > 2V) the power is dominated by Psc

VDD= 3.3 V

P normalized

VDD = 2.5 V

If VDD < VTn + |VTp| then Psc is eliminated since both devices are never on at the same time.

VDD = 1.5V

tsin/tsout

W/Lp = 1.125 m/0.25 m

W/Ln = 0.375 m/0.25 m

CL = 30 fF

normalized wrt zero input rise-time dissipation

- Total power consumption

- Reducing dynamic capacitive power:
- Lower the voltage (Vdd)!
- Quadratic effect on dynamic power

- Reduce capacitance
- Short interconnect lengths
- Drive small gate load (small gates, small fan-out)

- Reduce frequency
- Lower clock frequency -
- Lower signal activity

- Lower the voltage (Vdd)!

- Reducing dynamic capacitive power:
- Lower the voltage (Vdd)!
- Quadratic effect on dynamic power

- Reduce capacitance
- Short interconnect lengths
- Drive small gate load (small gates, small fan-out)

- Reduce frequency
- Lower clock frequency -
- Lower signal activity

- Lower the voltage (Vdd)!

- Reducing dynamic capacitive power:
- Lower the voltage (VDD)
- Quadratic effect on dynamic power

- Reduce capacitance
- Short interconnect lengths
- Drive small gate load (small gates, small fan-out)

- Reduce frequency
- Lower clock frequency -
- Lower signal activity

- Lower the voltage (VDD)

f=500MHz

CL=15fF/gate

VDD=2.5V

Pdyn=50W

For a design with I million gate

Pdyn=50W!

Is this possible in reality? If not why?

Pdyn=Edyn/2tp=580W for tp=32.5ps

Pdyn=Edyn/2tp=155W for f=4GHz(250ps)

- Reducing short-circuit current:
- Fast rise/fall times on input signal
- Reduce input capacitance
- Insert small buffers to “clean up” slow input signals before sending to large gate

- Reducing leakage current:
- Small transistors (leakage proportional to width)
- Lower voltage

(Lower-power and Robust)

Good

Fast

Cheap

(Short Delay)

(Small Layout)

- Design trade-offs dance around the triangle, but still important
- Fundamental improvement that shrinks the triangle:
- Scaling in technology (lithography improvement)
- New functionality
- New architecture
- New algorithms