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Ricky wrote:
Oh, BTW, you seem to ignore the need for an N bit state detector for whatever is making your sequence generator. That\'s even more logic.
On a sunny day (Sun, 18 Jun 2023 20:47:44 -0700 (PDT)) it happened Ricky
gnuarm.deletethisbit@gmail.com> wrote in
83110f6f-e327-443e-bd6e-3c36a3ccaec1n@googlegroups.com>:
Uh, how do you count down without using flipflops?
Use your fingers
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is minimal?
RV3028 needs 48nA wile being significantly more complex than a simple
2^15 divider. What lower value would make you happy?
Best regards, Piotr
Some flops use about the same power whether they change state or not.
Most, probably.
On Tue, 4 Jul 2023 22:51:24 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is minimal?
RV3028 needs 48nA wile being significantly more complex than a simple
2^15 divider. What lower value would make you happy?
Some flops use about the same power whether they change state or not.
Most, probably.
On Tue, 4 Jul 2023 22:45:10 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
Ricky wrote:
Oh, BTW, you seem to ignore the need for an N bit state detector for whatever is making your sequence generator. That\'s even more logic.
Three lines and only four insults. Ricky is off his form.
On Tue, 4 Jul 2023 22:51:24 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is minimal?
RV3028 needs 48nA wile being significantly more complex than a simple
2^15 divider. What lower value would make you happy?
Best regards, Piotr
Some flops use about the same power whether they change state or not.
Most, probably.
On Tuesday, July 4, 2023 at 3:03:02?PM UTC-7, John Larkin wrote:
On Tue, 4 Jul 2023 22:51:24 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is minimal?
RV3028 needs 48nA wile being significantly more complex than a simple
2^15 divider. What lower value would make you happy?
Some flops use about the same power whether they change state or not.
Most, probably.
None, actually; a flipflop that changes state, without holding a memory of its
previous state, takes an irreversible step. Irreversibility implies entropy, thus
energy is lost to heat.
https://en.wikipedia.org/wiki/Landauer%27s_principle
On Tuesday, July 4, 2023 at 6:03:02â¯PM UTC-4, John Larkin wrote:
On Tue, 4 Jul 2023 22:51:24 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is
minimal? RV3028 needs 48nA wile being significantly more complex
than a simple 2^15 divider. What lower value would make you
happy?
Best regards, Piotr
Some flops use about the same power whether they change state or
not. Most, probably.
Depending on what they are driving. You can\'t just count the power
in the gates themselves. The output loads are the majority of the
power consumption.
On 05/07/2023 07:42, Ricky wrote:
On Tuesday, July 4, 2023 at 6:03:02â¯PM UTC-4, John Larkin wrote:
On Tue, 4 Jul 2023 22:51:24 +0200, Piotr Wyderski
bom...@protonmail.com> wrote:
whit3rd wrote:
Capacitive energy loss thus favors a Gray code for counting
with minimal energy cost
That\'s not true, as pointed out by Rick, but how minimal is
minimal? RV3028 needs 48nA wile being significantly more complex
than a simple 2^15 divider. What lower value would make you
happy?
Best regards, Piotr
Some flops use about the same power whether they change state or
not. Most, probably.
That seems a bit unlikely in general, but in this application yes.
The average power used for changing states of a divide by 2^N chain is
typically only about twice what it would be for one of them. Peak
consumption N x when all ones to zero rolls over.
Plus N x whatever standing current each flip-flop requires to function.
At such very low clock rates I expect the quiescent current is quite a
bit bigger than the total current used to change states.
If they were clocked at >10MHz then it might be a different story.
Depending on what they are driving. You can\'t just count the power
in the gates themselves. The output loads are the majority of the
power consumption.
And that is part of the difference between a generic divide by 2^N chip
with every output available buffered to an external pin and a custom
watch chip where only 1Hz pulse has any connection to the outside world.
On Tue, 4 Jul 2023 20:20:58 -0700 (PDT), whit3rd <whi...@gmail.com
wrote:
On Tuesday, July 4, 2023 at 3:03:02?PM UTC-7, John Larkin wrote:
Some flops use about the same power whether they change state or not.
Most, probably.
None, actually; a flipflop that changes state, without holding a memory of its
previous state, takes an irreversible step. Irreversibility implies entropy, thus
energy is lost to heat.
https://en.wikipedia.org/wiki/Landauer%27s_principle
It wouldn\'t shock me if ECL, especially the eclips parts, use less
power when they are clocked hard.
But I did say \"about the same power\", which doesn\'t violate any basic
principles.
On Wednesday, July 5, 2023 at 3:38:44?AM UTC-7, John Larkin wrote:
On Tue, 4 Jul 2023 20:20:58 -0700 (PDT), whit3rd <whi...@gmail.com
wrote:
On Tuesday, July 4, 2023 at 3:03:02?PM UTC-7, John Larkin wrote:
Some flops use about the same power whether they change state or not.
Most, probably.
None, actually; a flipflop that changes state, without holding a memory of its
previous state, takes an irreversible step. Irreversibility implies entropy, thus
energy is lost to heat.
https://en.wikipedia.org/wiki/Landauer%27s_principle
It wouldn\'t shock me if ECL, especially the eclips parts, use less
power when they are clocked hard.
But I did say \"about the same power\", which doesn\'t violate any basic
principles.
Oh, I realize the Landauer effect isn\'t very large; still, entropy IS a basic principle.
On Wednesday, July 5, 2023 at 8:30:37â¯AM UTC-4, Martin Brown wrote:
Some flops use about the same power whether they change state or
not. Most, probably.
That seems a bit unlikely in general, but in this application yes.
The average power used for changing states of a divide by 2^N chain is
typically only about twice what it would be for one of them. Peak
consumption N x when all ones to zero rolls over.
How do you know this? Dynamic power dissipation is a property that is very sensitive to implementation.
Plus N x whatever standing current each flip-flop requires to function.
At such very low clock rates I expect the quiescent current is quite a
bit bigger than the total current used to change states.
Again, how do you know this? Again, static power is very sensitive to implementation.
If they were clocked at >10MHz then it might be a different story.
And that is part of the difference between a generic divide by 2^N chip
with every output available buffered to an external pin and a custom
watch chip where only 1Hz pulse has any connection to the outside world.
No one has said anything about whether this is being designed from a custom chip or relays.
This is all a thought project, which has never been defined, except for wanting to do something similar to some fancy watch.
I vote for relay logic and a 1 Hz crystal.
On 05/07/2023 14:05, Ricky wrote:
On Wednesday, July 5, 2023 at 8:30:37â¯AM UTC-4, Martin Brown wrote:
Some flops use about the same power whether they change state or
not. Most, probably.
That seems a bit unlikely in general, but in this application yes.
The average power used for changing states of a divide by 2^N chain is
typically only about twice what it would be for one of them. Peak
consumption N x when all ones to zero rolls over.
How do you know this? Dynamic power dissipation is a property that is very sensitive to implementation.
Because unless the chip designers are complete morons they won\'t allow
both the FETs to be biassed to conduct hard simultaneously. The typical
cost per 0-1-0 transition is the same although there might be some
slight asymmetry between 0-1 and 1-0 transitions.
The simple rule is that if it costs x to change state then each change
of state will slightly affect the current consumption by adding x.
The rest is simple mathematics of dividing by 2.
1 + 1/2 + 1/4 + 1/8 + ... < 2
Plus N x whatever standing current each flip-flop requires to function.
At such very low clock rates I expect the quiescent current is quite a
bit bigger than the total current used to change states.
Again, how do you know this? Again, static power is very sensitive to implementation.
But for any given implementation the current drawn by a D flip-flop is
going to be about the same value y. If there are N of them that is N*y.
I suppose there might be a slight difference due to packaging overhead
for any given N if they are available 1, 2, 4 or 8 to a chip.
If they were clocked at >10MHz then it might be a different story.
And that is part of the difference between a generic divide by 2^N chip
with every output available buffered to an external pin and a custom
watch chip where only 1Hz pulse has any connection to the outside world.
No one has said anything about whether this is being designed from a custom chip or relays.
This is all a thought project, which has never been defined, except for wanting to do something similar to some fancy watch.
I vote for relay logic and a 1 Hz crystal.
That might make it a little difficult to wear on a wrist.
I recall one of the world\'s top experimentalists showing off his brand
new Sinclair matchbox sized digital watch in the mid 70\'s - he was very
proud of it. (at the time it was quite amazing for the price)
I recall one of the world\'s top experimentalists showing off his brand
new Sinclair matchbox sized digital watch in the mid 70\'s - he was very
proud of it. (at the time it was quite amazing for the price)
--
Martin Brown
Martin Brown <\'\'\'newspam\'\'\'@nonad.co.uk> wrote:
I recall one of the world\'s top experimentalists showing off his brand
new Sinclair matchbox sized digital watch in the mid 70\'s - he was very
proud of it. (at the time it was quite amazing for the price)
--
Martin Brown
First Atomic Clock Wristwatch
http://www.leapsecond.com/pages/atomic-bill/
On 05/07/2023 14:05, Ricky wrote:
On Wednesday, July 5, 2023 at 8:30:37â¯AM UTC-4, Martin Brown wrote:
Some flops use about the same power whether they change state or
not. Most, probably.
That seems a bit unlikely in general, but in this application yes.
The average power used for changing states of a divide by 2^N chain is
typically only about twice what it would be for one of them. Peak
consumption N x when all ones to zero rolls over.
How do you know this? Dynamic power dissipation is a property that is very sensitive to implementation.
Because unless the chip designers are complete morons they won\'t allow
both the FETs to be biassed to conduct hard simultaneously. The typical
cost per 0-1-0 transition is the same although there might be some
slight asymmetry between 0-1 and 1-0 transitions.
The simple rule is that if it costs x to change state then each change
of state will slightly affect the current consumption by adding x.
The rest is simple mathematics of dividing by 2.
1 + 1/2 + 1/4 + 1/8 + ... < 2
Plus N x whatever standing current each flip-flop requires to function..
At such very low clock rates I expect the quiescent current is quite a
bit bigger than the total current used to change states.
Again, how do you know this? Again, static power is very sensitive to implementation.
But for any given implementation the current drawn by a D flip-flop is
going to be about the same value y. If there are N of them that is N*y.
I suppose there might be a slight difference due to packaging overhead
for any given N if they are available 1, 2, 4 or 8 to a chip.
I expect the quiescent current is quite a
bit bigger than the total current used to change states.
If they were clocked at >10MHz then it might be a different story.
And that is part of the difference between a generic divide by 2^N chip
with every output available buffered to an external pin and a custom
watch chip where only 1Hz pulse has any connection to the outside world.
No one has said anything about whether this is being designed from a custom chip or relays.
This is all a thought project, which has never been defined, except for wanting to do something similar to some fancy watch.
I vote for relay logic and a 1 Hz crystal.
That might make it a little difficult to wear on a wrist.
On Thursday, July 6, 2023 at 4:17:05?AM UTC-4, Martin Brown wrote:
On 05/07/2023 14:05, Ricky wrote:
On Wednesday, July 5, 2023 at 8:30:37?AM UTC-4, Martin Brown wrote:
Some flops use about the same power whether they change state or
not. Most, probably.
That seems a bit unlikely in general, but in this application yes.
The average power used for changing states of a divide by 2^N chain is
typically only about twice what it would be for one of them. Peak
consumption N x when all ones to zero rolls over.
How do you know this? Dynamic power dissipation is a property that is very sensitive to implementation.
Because unless the chip designers are complete morons they won\'t allow
both the FETs to be biassed to conduct hard simultaneously. The typical
cost per 0-1-0 transition is the same although there might be some
slight asymmetry between 0-1 and 1-0 transitions.
The simple rule is that if it costs x to change state then each change
of state will slightly affect the current consumption by adding x.
The rest is simple mathematics of dividing by 2.
1 + 1/2 + 1/4 + 1/8 + ... < 2
I recall we used Rb clocks disciplined by MSF Rugby time signatures for
intermediate length LBI back then and so noticed that when there was dew
on the ground in the early morning MSF signals arrived late.
--
Martin Brown