Energy&power

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6.002

CIRCUITS AND ELECTRONICS

Energy and Power

6.002 Fall 2000

Lecture

22

1

Why worry about energy?

-

small batteries Æ good

Today: How long will the battery last? in standby mode in active use „ Will the chip overheat and self-destruct? „

6.002 Fall 2000

Lecture

22

2

Look at energy dissipation in MOSFET gates VS R + + vI –

C

vO –

C: wiring capacitance and CGS of following gate Let us determine standby power active use power Let’s work out a few related examples first. 6.002 Fall 2000

Lecture

22

3

Example 1:

I

V + –

Power

R

+ V –

V2 P = VI = R

Energy dissipated in time T

E = VIT

6.002 Fall 2000

Lecture

22

4

Example 1: for our gate

VS

VS RL

RL

vO

vI high

vI low

RON

RON

2

VS P= RL + RON

6.002 Fall 2000

Lecture

vO

P=0

22

5

Example 2: Consider R1 S1

VS + –

S2 R2

C

T T1

T2

S1 closed S1 open S 2 open

S 2 closed t

Find energy dissipated in each cycle. Find average power P. 6.002 Fall 2000

Lecture

22

6

T1 : S1 closed, S2 open i VS + –

R1

+ vC –

C

vC

i

VS R1

VS

t

6.002 Fall 2000

assume vC = 0 at t = 0

Lecture

VS e R1

−t R1C

t

22

7

Total energy provided by source during T1 T1

E = ∫ VS i dt 0

T1

2

VS e R1 0

=∫

−t R1C

dt

2

=−

VS R1C e R1

−t T1 R1C 0

−T1   2 R C 1 = C VS  1 − e      2

≈ C VS if T1 >> R1C

I.e., if we wait long enough

1 2 C VS stored on C , 2 1 2 E1 = C VS dissipated in R1 2 6.002 Fall 2000

Lecture

22

Independent of R!

8

T2 : S2 closed, S1 open + vC –

C

R2

Initially, vC = VS

(recall T1 >> R1C)

So, initially,

1 2 energy stored in capacitor = CVS 2 Assume T2 >> R2C So, capacitor discharges ~fully in T2 So, energy dissipated in R2 during T2

1 2 E2 = CVS 2 E1, E2 independent of R2 ! 6.002 Fall 2000

Lecture

22

9

Putting the two together: Energy dissipated in each cycle E = E1 + E2 1 1 2 2 = CVS + CVS 2 2

E = CVS

2

energy dissipated in charging & discharging C

Assumes C charges and discharges fully. Average power

P=

E T

CVS = T

2

2

= CVS f

frequency f = 6.002 Fall 2000

Lecture

22

1 T 10

Back to our inverter — VS RL

vO vIN

RON

C

What is P for the following input?

vIN T 2

T 2 T

6.002 Fall 2000

t 1 T= f

Lecture

22

11

Equivalent Circuit RL

VS + –

C RON

What is P for the following input? vIN

T 2

T 2 T

6.002 Fall 2000

t 1 T= f

Lecture

22

12

What is P for gate? We can show (see section 12.2 of A & L) 2

P=

2

VS RL 2 + CVS f 2( RL + RON ) (RL + RON )2

when RL >> RON 2

VS 2 P= + CVS f 2 RL

r e b m e m re

P STATIC independent of f. MOSFET ON half the time.

6.002 Fall 2000

e b m e rem

r

P DYNAMIC related to switching capacitor

Lecture

22

13

What is P for gate? when RL >> RON 2

VS 2 P= + CVS f 2 RL In standby mode, half the gates in a chip can be assumed to be on. So P STATIC per gate is still VS2 .

In standby mode, fÆ0, so dynamic power is 0

2RL Relates to standby power.

6.002 Fall 2000

Lecture

22

14

Some numbers… a chip with 106 gates clocking C =1f F at 100 MHZ RL = 10 kΩ f = 100 × 10 6 VS = 5 V 25 −15 6 P = 10 6  10 25 100 10 + × × × 4   2 × 10 = 10 6 [1.25 milliwatts + 2.5 microwatts ]

problem !

1.25KW!

must get rid of this

α VS 2 α f reduce VS

next lecture 6.002 Fall 2000

2.5W not bad

5 V → 1V 2.5 W → 150 mW Lecture

22

15