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## ECE 331 – Digital System Design

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ECE 331 – Digital System Design

Power Dissipation

and

Additional Design Constraints

(Lecture #14)

The slides included herein were taken from the materials accompanying

Fundamentals of Logic Design, 6th Edition, by Roth and Kinney,

and were used with permission from Cengage Learning.

ECE 331 - Digital System Design

Power Dissipation

ECE 331 - Digital System DesignPower Dissipation

Each integrated circuit (IC) dissipates power

PT = PS + PD

PT = total power dissipated by IC

PS = static or quiescent power dissipation

PD = dynamic power dissipation

ECE 331 - Digital System DesignStatic Power Dissipation

PS = VCC * ICC

VCC = supply voltage

ICC = quiescent supply current

PS = static power consumption

ICC and VCC are specified in the datasheet for the integrated circuit (IC).

For CMOS devices, PS is very small.

ECE 331 - Digital System DesignStatic Power Dissipation

Example: 74LS00 (Quad 2-input NAND)

Supply voltage

4.75 V <= VCC <= 5.25 V

Supply current

High output: ICCmax = 1.6 mA

Low output: ICCmax = 4.4 mA

Maximum static power dissipation

High output: PS = 8.4 mW

Low output: PS = 23.1 mW

ECE 331 - Digital System DesignStatic Power Dissipation

Duty Cycle

Clock signal typically has 50% duty cycle

PS = PS_high * thigh + PS_low * tlow

PS_high = 8.4 mW

PS_low = 23.1 mW

Assume 50% duty cycle (high / low half the time)

PS = 8.4 mW * 0.5 + 23.1 mW * 0.5 = 15.8 mW

Assume 60% duty cycle (high 60% of the time)

PS = 8.4 mW * 0.6 + 23.1 mW * 0.4 = 14.28 mW

ECE 331 - Digital System DesignDynamic Power Dissipation

For TTL devices, PD is negligible compared to PS.

Assume PS = 0

For CMOS devices, PD dominates PT.

PD >> PS

PD in CMOS circuits arises from the movement of charge into and out of the device capacitance.

ECE 331 - Digital System DesignDynamic Power Dissipation

In CMOS devices, charge is stored in the

CPD = power dissipation capacitance (internal)

CL = capacitance of the load and wires (external)

These capacitors are in parallel

CT = CPD + CL

The stored charge (on these capacitors) is

QT = CT * VDD = (CPD + CL) * VDD

ECE 331 - Digital System DesignDynamic Power Dissipation

The charge moves into and out of the capacitors on every transition of the output.

Low → High

High → Low

Current = movement of charge

IAVG = (CPD + CL) * VDD * fT

Where fT = output frequency

PD = IAVG * VDD = (CPD + CL) * V2DD * fT

ECE 331 - Digital System DesignDynamic Power Dissipation

Example: 74HC00 (Quad 2-input NAND)

VDD = 5V

CPD = 20 pF, CL = 50 pF

PD = (20 + 50 pF) * (5V)2 * fT

fT (Hz) PD

1K 1.8 mW

1M 1.8 mW

100M 180 mW

ECE 331 - Digital System DesignTotal Power Dissipation

For the 74HC00, PS is determined as follows

VCC = 5V

ICC = 20 mA

PS = VCC * ICC = 5V * 20 mA = 100 mA

The PT is then determined from

PT = PS + PD

where PD is a function of fT

ECE 331 - Digital System DesignTotal Power Dissipation

PT = PS + PD

Compare PT for Quad 2-input NAND (74xx00)

0 Hz 1 MHz 100 MHz

TTL 15.8 mW 15.8 mW 15.8 mW

CMOS 100 W 1.805 mW 180 mW

Compare TTL and CMOS

TTL CMOS

PS VCC * ICC VDD * IDD

PD ~ 0 W (CPD + CL) * V2DD * fT

ECE 331 - Digital System DesignHazards

When the input to a combinational logic circuit changes, unwanted switching transients may appear on the output.

These transients occur when different paths from input to output have different propagation delays.

ECE 331 - Digital System DesignHazards

When analyzing combinational logic circuits for hazards we will consider the case where only one input changes at a time.

Under this condition, a static 1-hazard occurs when the input change causes one product term (in a SOP expression) to transition from 1 to 0 and another product term to transition from 0 to 1.

Both product terms can be transiently 0, resulting in the static 1-hazard.

ECE 331 - Digital System DesignHazards

Under the same condition, a static 0-hazard occurs when the input change causes one sum term (in a POS expression) to transition from 0 to 1 and another sum term to transition from 1 to 0.

Both sum terms can be transiently 1, resulting in the static 0-hazard.

ECE 331 - Digital System DesignDetecting Static 1-Hazards

We can detect hazards in a two-level AND-OR circuit using the following procedure:

Write down the sum-of-products expression for the circuit.

Plot each term on the map and loop it.

If any two adjacent 1′s are not covered by the same loop, a 1-hazard exists for the transition between the two 1′s. For an n-variable map, this transition occurs when one variable changes and the other n – 1 variables are held constant.

ECE 331 - Digital System DesignDetecting Static 0-Hazards

We can detect hazards in a two-level OR-AND circuit using the following procedure:

Write down the product-of-sums expression for the circuit.

Plot each sum term on the map and loop the zeros.

If any two adjacent 0′s are not covered by the same loop, a 0-hazard exists for the transition between the two 0′s. For an n-variable map, this transition occurs when one variable changes and the other n – 1 variables are held constant.

ECE 331 - Digital System DesignRemoving Static 0-Hazards

How many redundant gates are necessary to remove the 0-hazards?

ECE 331 - Digital System DesignHazards

Exercise:

Design a hazard-free combinational logic circuit to implement the following logic function

F(A,B,C) = A'.C' + A.D + B.C.D'

ECE 331 - Digital System DesignHazards

Exercise:

Design a hazard-free combinational logic circuit to implement the following logic function

F(A,B,C) = (A'+C').(A+D).(B+C+D')

ECE 331 - Digital System DesignHazards

Two-level AND-OR circuits (SOP) cannot have static 1-Hazards.

Why?

Two-level OR-AND circuits (POS) cannot have static 0-Hazards.

Why?

ECE 331 - Digital System Design

Questions?

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