Digital Circuit Speed

6.002

CIRCUITS AND ELECTRONICS

Digital Circuit

6.002 Fall 2000

Lecture

13

1

Review vI

R

vI + –

VI

+ vC –

C

t

0

vC (0 ) = VO vC = VI + (VO − VI ) e

−t RC

1

vC VI

VO RC

6.002 Fall 2000

time constant RC t

Lecture

13

2

Let’s apply the result to an inverter. B

A X

First, rising delay tr at B

VS

VS

A vA 5V

0 1 Æ 0 at A

6.002 Fall 2000

B CGS

X t

Lecture

13

3

First, rising delay tr at B VS

A

VS

B vA 5V

CGS

X

0 1 Æ 0 at A

t

vB

5V

ideal observed

t

0

6.002 Fall 2000

Lecture

13

4

First, rising delay tr at B VS

A vA 5V

VS

B CGS

X

0 1 Æ 0 at A

t 5V VOH

rising delay of X

6.002 Fall 2000

Lecture

vB

0

tr

13

t

5

Equivalent circuit for 0Æ1 at B

vI = VS

RL

+ –

vI = VS vB (0 ) = 0

From

CGS

+ vB –

for t ≥ 0

1 vB = VS + (0 − VS ) e

−t RL CGS

Now, we need to find t for which vB = VOH .

6.002 Fall 2000

Lecture

13

6

Or vOH = VS − VS e

Find tr : VS e

−t r RL CGS

−t RL CGS

= VS − VOH

VS − VOH − tr = ln RL CGS VS

VS − VOH t r = − RL CGS ln VS

6.002 Fall 2000

Lecture

13

7

Or vOH = VS − VS e

Find tr : VS e

−t r RL CGS

−t RL CGS

= VS − VOH

VS − VOH − tr = ln RLCGS VS VS − VOH t r = − RL CGS ln VS e.g.

RL = 1K

VS = 5V

CGS = 0.1 pF

VOH = 4V

t r = −1 × 10 × 0.1 × 10 3

= 0.16 ns

−12

5−4 ln 5

RC = 0.1 ns ! 6.002 Fall 2000

Lecture

13

8

Falling Delay tf Falling delay tf is the t for which vB falls to VOL

Equivalent circuit for 1 Æ 0 at B vB (0 ) = VS (5V )

RL

VS + –

CGS

RON

+ vB –

X

6.002 Fall 2000

Lecture

13

9

Falling Delay tf

Equivalent circuit for 1 Æ 0 at B vB (0 ) = VS (5V )

RL

VS + – RON

CGS

+ vB –

CGS

+ vB –

X Thévenin replacement … RTH

VTH + –

RTH = RL || RON VTH 6.002 Fall 2000

RON = VS RON + RL Lecture

13

10

From

1 vB = VTH + (VS − VTH ) e

−t RTH CGS

Falling decay tf is the t for which vB falls to VOL −t f

VOL = VTH + (VS − VTH ) e RTH CGS or

VOL − VTH t f = − RTH CGS ln VS − VTH

6.002 Fall 2000

Lecture

13

11

t f = − RTH CGS ln

e.g.

RL = 1K

VS = 5V

CGS = 0.1 pF RTH ≈ 10Ω,

VOL − VTH VS − VTH RON = 10Ω

VOL = 1V

VTH ≈ 0V

t f = −10 ⋅ 0.1 ⋅10 = 1.6 ps

−12

1 ln 5

RC = 1 ps !

6.002 Fall 2000

Lecture

13

12

For recitation: Slow may be better

Problem

chip

pin 2 pin 1

v CL

v:

ideal

observed

slow!

So the engineers decided to speed it up…

RL RON

6.002 Fall 2000

made RL small made RON small

Lecture

13

13

For recitation: Slow may be better

Problem

chip

pin 2 pin 1

v CL

v:

ideal

…

observed

slow!

but, disaster!

v:

observed expected

6.002 Fall 2000

VIL

Lecture

13

14

Why? Consider Case

1

…

Demo

R1

pin1

R0 ok

6.002 Fall 2000

Lecture

13

15

Why? Consider Case 2

…

Demo

CP

R1

pin1

pin2

R0

R2

crosstalk! CP

R model for crosstalk:

+ v

+ –

–

6.002 Fall 2000

Lecture

13

16

Case 3

…

6.002 expert saw the solution R1

CP

R0

R2

+ –

slower transitions!

Detailed analysis in recitation.

6.002 Fall 2000

Lecture

13

17