Energy And Power

6.002

CIRCUITS AND ELECTRONICS

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Energy and Power

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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? „

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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.

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Example 1:

I

V + –

Power

R

+ V –

V2 P = VI = R

Energy dissipated in time T

E = VIT

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Example 1: for our gate

VS

VS RL

RL

vO

v I high

RON 2

VS P= RL + RON

vO

v I low

RON

P=0

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Example 2: Consider R1 S1

VS + –

S2

C

R2

T T1

T2

S1 closed S1 open S 2 open

S 2 closed t

Find energy dissipated in each cycle. Find average power P.

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T1 : S1 closed, S2 open i VS + –

assume vC = 0 at t = 0

R1

C

+ vC –

vC

i

VS R1

VS

t

VS e R1

−t R1C

t

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Total energy provided by source during T1 T1

E = ∫ VS i dt 0

T1

2

−t

VS R1C =∫ e dt R1 0 2

=−

VS R1C e R1

−t T1 R1C 0

−T1 ⎛ ⎞ 2 R C 1 = C VS ⎜ 1 − e ⎟ ⎟ ⎜ ⎠ ⎝

≈ C VS

2

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

Independent of R!

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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 !

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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

E P= T CVS = T

2

= CVS f 2

1 frequency f = T

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Back to our inverter — VS RL

vO vIN

RON

C

What is P for the following input?

vIN T 2

T 2 T

t 1 T= f

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Equivalent Circuit RL

VS + –

C RON

What is P for the following input? vIN

T 2

T 2 T

t 1 T= f

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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.

e b m e rem

r

P DYNAMIC related to switching capacitor

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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.

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Some numbers… a chip with 106 gates clocking C =1f F at 100 MHZ RL = 10 kΩ

25 ⎡ P = 10 ⎢ 4 ⎣ 2 × 10 6

f = 100 × 10 6 VS = 5 V + 10 −15 × 25 × 100 × 10 6 ⎤⎥ ⎦

= 10 6 [1.25 milliwatts + 2.5 microwatts ]

problem !

1.25KW!

must get rid of this

next lecture

2.5W not bad

α VS 2 α f reduce VS 5 V → 1V 2.5 W → 150 mW

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