Tài liệu Circuits & Electronics P22 pdf

6.002 CIRCUITS ANDELECTRONICS Energy and Power... Why worry about energy?small batteries Æ good Today: „ How long will the battery last?. in standby mode in active use... -Look at energy

6.002 CIRCUITS AND

ELECTRONICS

Energy and Power

Why worry about energy?

small batteries

Æ good

Today:

„ How long will the battery last?

in standby mode

in active use

-Look at energy dissipation in

MOSFET gates

Let us determine

standby power active use power Let’s work out a few related examples first

C: wiring capacitance and

C GS of following gate

V S

C

R

O

v

+

–

I

v

+ –

Example 1:

Power

Energy dissipated in time T

R

V VI

P

2

=

=

VIT

E =

+

I

V

+ –

V

Example 1:

for our gate

ON L

S

R R

V P

+

O

v

S

V

ON

R

L

R

I

v high

S

V

ON

R

L

R

O

v

I

v low

Example 2:

Consider

Find energy dissipated in each cycle

Find average power P

S

V +–

1

R

1

open

S

closed

S

2

1

t

closed S

open S

2 1

1

T

T1 : S1 closed, S2 open

t

C

v

S

1

S e 1

R

V −

t

i

1

S

R V

+ –C

v

+ –

1

R

C

S

V

v C = 0 at t = 0

Total energy provided by source during T 1

dt i V E

1

T

0 S

∫

=

dt

e R

V R C t

T

0 1

2

S 1

∫

=

1 1

T

0

C R

t

1 1

2

S R C e R

−

=

















−

=

−

C R

T 2

1

e 1 V

C

1

2 S 1

2 S

R in dissipated V

C

1 E

, C on stored V

C 2 1

=

C R T

if V

C S 2 1 >> 1

≈

I.e., if we wait long enough

Independent

of R!

T2 : S2 closed, S1 open

+ –C

v

2

R

C

So, initially,

2 S

CV 2

1

= energy stored in capacitor

Assume T 2 >> R 2 C

So, capacitor discharges ~fully in T 2

So, energy dissipated in R 2 during T 2

2 S

2 CV

2

1

Initially, v C = V S (recall T 1 >> R 1 C)

Putting the two together:

Energy dissipated in each cycle

2 S

2

S CV

2

1 CV

2

=

2

1 E E

E = +

C g dischargin

&

charging

in dissipated energy

CV

Assumes C charges and discharges fully.

1

T

E

P =

T

CV S2

=

f

CV S2

=

Average power

Back to our inverter —

O

v

IN

S

V

L

R

ON

R

t 2

T

T

2 T

IN

v

f

T = 1

What is for the following input?P

Equivalent Circuit

S

V +–

L

R

C

ON

R

t

2

T

T

2

T

IN

v

f

T = 1

What is for the following input?P

We can show (see section 12.2 of A & L)

ON L

2 L

2 S ON

L

2 S

R R

R f

CV R

R 2

V P

+

+ +

=

f

CV R

2

V

L

2

S +

=

when R L >> R ON

What is for gate? P

ber

remem

ber

STATIC

P P DYNAMIC

related to switching capacitor

independent of f

MOSFET ON half

the time

CV R

V

L

2

=

when R L >> R ON

In standby mode,

half the gates in a

chip can be

assumed to be on

So per

gate is still

Relates to standby

power.

STATIC

P

L

2 S

R 2 V

What is for gate? P

In standby mode,

f Æ 0 ,

so dynamic power

is 0

Some numbers…

a chip with 106 gates clocking

at 100 MHZ

V 5 V

10 100

f

k 10 R

F f 1 C

S

6 L

=

×

=

Ω

=

=







×

4

6 10 25 100 10

10 2

25 10

P

[1 25 milliwatts 2 5 microwatts]

=

problem! 1.25KW! 2.5Wnot bad

V 1 V

5

V reduce f

V

S

2 S

→

α α

next lecture must get rid of this