6.002
CIRCUITS AND ELECTRONICS
Energy and Power
6.002 Fall 2000
Lecture
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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?
6.002 Fall 2000
Lecture
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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. 6.002 Fall 2000
Lecture
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Example 1:
I
V + –
Power
R
+ V –
V2 P = VI = R
Energy dissipated in time T
E = VIT
6.002 Fall 2000
Lecture
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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
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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
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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
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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
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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
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
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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
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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.
6.002 Fall 2000
e b m e rem
r
P DYNAMIC related to switching capacitor
Lecture
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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.
6.002 Fall 2000
Lecture
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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
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