CMOS capacitive loads, transition probabilities and FPGAs

>  I think the OP was just
> confused about the variations in the formulae.  It is just a matter of
> how you figure the constant.  Normally the constant is measured rather
> than calculated, so you just need to use the formulae that you are given
> with the constant.

On the subject of variations and locations of 0.5 and 2 factors:

if we have

C as the node capacitance
'a' as the transition density (0 to 1 and 1 to 0) at the node
then
P = 0.5aCV^2F
holds I believe?

Alternatively,

If 'a' is only counting "power consuming" 0 to 1 transitions then we have:

P = 0.5(a*2)CV^2F
= aCV^2F

Correct?

Or is the factor of 2/0.5 worked into the C constant?

Ken

Rick,

Thanks for clearing that up.  Good way to think about it.  The 1/2CV^2 
comment was confusing, I agree.

Austin

rickman wrote:

> Austin Lesea wrote:
> 
>>Ken,
>>
>>A commont confusion is what dissipates power?
>>
>>When you charge a node, you waste power in the resistance of the
>>charging transistor.
>>
>>When you discharge a node, you then waste power in the resistance of the
>>discharging transistors (both generate heat).
>>
>>If both edges generate heat, then you have to count them both.
>>
>>The energy stored in a capacitor is 1/2 CV^2, but don't let that confuse
>>you:  you have to put it in, and then take it out!  The devices that do
>>the work dissipate the power.
> 
> 
> This is a bit misleading and irrelevant.  The fact that power is only
> dissipated in the resistance has nothing to do with the total amount of
> energy expended in charging and discharging a capacitor.  Regardless of
> what value resistance, even if it is not constant, the energy drawn from
> the supply is the same as long as the capacitor is charged to the same
> voltage.  The energy may be dissipated in the transistor or in the poly
> track or a metal track or the bond wire or the external pin or even in
> the wire from the power supply to the board.  But add them all up and
> you will get the same value each time you charge a capacitor to a given
> voltage.  
> 
> So the amount of power consumed is related to the frequency of
> transitions and the size of the capacitance.  Whether you calculate it
> from the rate of the rising edges or both edges is not relevant, that
> just changes the constant that you use by a factor of 2.  As the OP
> said, when you charge the cap from 0 volts to Vcc, half goes into the
> cap and half is wasted.  But the total always comes from the PSU and is
> always the same amount.  
> 
> You can't have a 0 to 1 transition without a 1 to 0 transition, so why
> is this even an issue?  
> 

Ken wrote:
> 
> >  I think the OP was just
> > confused about the variations in the formulae.  It is just a matter of
> > how you figure the constant.  Normally the constant is measured rather
> > than calculated, so you just need to use the formulae that you are given
> > with the constant.
> 
> On the subject of variations and locations of 0.5 and 2 factors:
> 
> if we have
> 
> C as the node capacitance
> 'a' as the transition density (0 to 1 and 1 to 0) at the node
> then
> P = 0.5aCV^2F
> holds I believe?
> 
> Alternatively,
> 
> If 'a' is only counting "power consuming" 0 to 1 transitions then we have:
> 
> P = 0.5(a*2)CV^2F
> = aCV^2F
> 
> Correct?
> 
> Or is the factor of 2/0.5 worked into the C constant?

Actually, I don't recall ever seeing a formula with the "a"
coefficient.  I have always just worked with a variable F to account for
differences in the number of transitions.  

The energy required to move charge onto a capacitor is Q*V.  As you move
more charge onto the capacitor the voltage rises linearly with the
amount of charge so that V = Q/C.  The total amount of energy stored on
the capacitor at any voltage is the integral of the V vs. Q curve which
is Int(V dQ) = Int(C*V dV) = 0.5 * C * V^2.  If the capacitor is charged
fully from a constant voltage source via a resistor, the curve to
integrate is just a constant Vcc vs. Q which gives you an integral of C
* Vcc^2.  If you want a power level you must multiply by F, the rate of
charging the cap (not discharging since that consumes no extra power, it
just dissapates the power stored in the cap).  That gives you P =
F*C*V^2.   So the whole "where did the 0.5 come from" question has no
answer.  There is *no* 0.5 inherent in the equation.  If someone wants
to change it by counting *both* the positive and negative transitions,
then you will have to multiply by 0.5 to compensate.  

-- 

Rick "rickman" Collins

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>Actually, I don't recall ever seeing a formula with the "a"
>coefficient.  I have always just worked with a variable F to account for
>differences in the number of transitions.  

There is another factor of 2 to consider.  If the clock is running
at F, a FF can only toggle at F/2.

Is your F the clock frequency or the frequency of a FF toggling
on each clock tick?  (bottom bit of a counter)

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Hi Rick,

> Actually, I don't recall ever seeing a formula with the "a"
> coefficient.  I have always just worked with a variable F to account for
> differences in the number of transitions.

I guess the 'a' is just a suitable modifer for F so that readily available 
numbers can just be plugged in (i.e. fmax for FPGA).

One example of many I have found that define 'a' is at:
http://www.lowpower.de/charter/introduction.php
(see section 3.2)

Also, "Minimizing Power Consumption in Digital CMOS Circuits" by 
Chandrakasan seems be often referenced and defines such a term (alpha, on 
page 2, right-hand column).


> The energy required to move charge onto a capacitor is Q*V.  As you move
> more charge onto the capacitor the voltage rises linearly with the
> amount of charge so that V = Q/C.  The total amount of energy stored on
> the capacitor at any voltage is the integral of the V vs. Q curve which
> is Int(V dQ) = Int(C*V dV) = 0.5 * C * V^2.  If the capacitor is charged
> fully from a constant voltage source via a resistor, the curve to
> integrate is just a constant Vcc vs. Q which gives you an integral of C
> * Vcc^2.  If you want a power level you must multiply by F, the rate of
> charging the cap (not discharging since that consumes no extra power, it
> just dissapates the power stored in the cap).  That gives you P =
> F*C*V^2.   So the whole "where did the 0.5 come from" question has no
> answer.  There is *no* 0.5 inherent in the equation.  If someone wants
> to change it by counting *both* the positive and negative transitions,
> then you will have to multiply by 0.5 to compensate.

Ok, in that case, my 1st option holds?

'a' is the transition density (0 to 1 and 1 to 0) at the node
F is the clock freq
C is the node capacitance
V is supply voltage
then
P = 0.5aCV^2F

where the 0.5 is modifying 'a' since 'a' is counting all transitions (0 to 1 
and 1 to 0) instead of just the power consuming ones (0 to 1)?

Cheers,

Ken

Hi Hal,

(snip)

> There is another factor of 2 to consider.  If the clock is running
> at F, a FF can only toggle at F/2.
>
> Is your F the clock frequency or the frequency of a FF toggling
> on each clock tick?  (bottom bit of a counter)

I think that this must be taken care of in the 'a' term so that the clock 
freq can just be plugged in to F.

In
http://bwrc.eecs.berkeley.edu/Classes/Icbook/SLIDES/slides4a.pdf
they actually modify the C term with transition probability and call it 
"effective capacitance"!

One example they give for the probability of a 0 to 1 transition for a 2 
input NOR gates comes out to 3/16 which is multiplied with C but is 
effectively the 'a' term we have been discussing I believe.

Ken