Delay of a Driver with Load · synthesis

Part 2 · Logic Optimisation and Mapping · Intermediate

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The delay-table form

A library delay table gives cell delay against input slew and load capacitance. This lesson fixes the slew and reads delay against load only.

The EXAMPLE table below is linear: delay equals a base plus a slope times load. Real tables curve, so this form is a teaching stand-in.

D = d_{0} + k \cdot C_{l o a d}

Load (EXAMPLE fF)	Delay (EXAMPLE ns)
1	0.5
2	0.7
4	1.1
8	1.9

A worked loaded driver

Take 2 fF of wire plus 1 fF per sink. The calc block compares 1 sink against 8 sinks.

Diagram

EXAMPLE table at fixed input slew: d = 0.3 + 0.2 x C, EXAMPLE ns and EXAMPLE fF
1 sink:  C = 2 + 1 x 1 = 3 fF,   d = 0.3 + 0.2 x 3 = 0.9 ns
8 sinks: C = 2 + 8 x 1 = 10 fF,  d = 0.3 + 0.2 x 10 = 2.3 ns
extra delay from 7 more sinks = 2.3 - 0.9 = 1.4 ns

A runnable delay model

The program below implements the EXAMPLE table. Its five self-checks replay the rows and the worked driver.

example.py

import sys

D0 = 0.3
K = 0.2

def driver_delay(load_ff):
    return D0 + K * load_ff

def close(got, want):
    return abs(got - want) < 1e-9

checks = [
    (driver_delay(1), 0.5),
    (driver_delay(3), 0.9),
    (driver_delay(4), 1.1),
    (driver_delay(8), 1.9),
    (driver_delay(10), 2.3),
]

for i in range(len(checks)):
    got, want = checks[i]
    if not close(got, want):
        print("FAIL: check " + str(i + 1) + " got " + str(got) + " want " + str(want))
        sys.exit(1)

print("PASS: delay model matches the table on 5 loads")

Common mistakes

Check yourself The EXAMPLE table gives d = 0.3 + 0.2 x C. What is the delay at 4 fF?

1.1 EXAMPLE ns: 0.3 + 0.8. That row sits in both the table and the program checks.

Check yourself All five program checks pass. What does that prove, and what stays open?

It proves the code agrees with the hand-worked rows. It leaves open whether the EXAMPLE slope matches any real library.

Related topics

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Practice this lesson Report an error

The delay-table form

A library delay table gives cell delay against input slew and load capacitance. This lesson fixes the slew and reads delay against load only.

The EXAMPLE table below is linear: delay equals a base plus a slope times load. Real tables curve, so this form is a teaching stand-in.

D = d_{0} + k \cdot C_{l o a d}

Load (EXAMPLE fF)	Delay (EXAMPLE ns)
1	0.5
2	0.7
4	1.1
8	1.9

A worked loaded driver

Take 2 fF of wire plus 1 fF per sink. The calc block compares 1 sink against 8 sinks.

Diagram

EXAMPLE table at fixed input slew: d = 0.3 + 0.2 x C, EXAMPLE ns and EXAMPLE fF
1 sink:  C = 2 + 1 x 1 = 3 fF,   d = 0.3 + 0.2 x 3 = 0.9 ns
8 sinks: C = 2 + 8 x 1 = 10 fF,  d = 0.3 + 0.2 x 10 = 2.3 ns
extra delay from 7 more sinks = 2.3 - 0.9 = 1.4 ns

A runnable delay model

The program below implements the EXAMPLE table. Its five self-checks replay the rows and the worked driver.

example.py

import sys

D0 = 0.3
K = 0.2

def driver_delay(load_ff):
    return D0 + K * load_ff

def close(got, want):
    return abs(got - want) < 1e-9

checks = [
    (driver_delay(1), 0.5),
    (driver_delay(3), 0.9),
    (driver_delay(4), 1.1),
    (driver_delay(8), 1.9),
    (driver_delay(10), 2.3),
]

for i in range(len(checks)):
    got, want = checks[i]
    if not close(got, want):
        print("FAIL: check " + str(i + 1) + " got " + str(got) + " want " + str(want))
        sys.exit(1)

print("PASS: delay model matches the table on 5 loads")

Common mistakes

Check yourself The EXAMPLE table gives d = 0.3 + 0.2 x C. What is the delay at 4 fF?

1.1 EXAMPLE ns: 0.3 + 0.8. That row sits in both the table and the program checks.

Check yourself All five program checks pass. What does that prove, and what stays open?

It proves the code agrees with the hand-worked rows. It leaves open whether the EXAMPLE slope matches any real library.