On 5/12/2016 5:36 PM, clipka wrote:
> Am 11.05.2016 um 08:09 schrieb Stephen:
>> On 5/11/2016 12:52 AM, clipka wrote:
>>> Am 11.05.2016 um 00:31 schrieb Stephen:
>>>
>>>>> Definitely too much petrol fumes.
>>>>
>>>> BTW Where do you think we got the petrol? The nearest petrol station was
>>>> either in Bergen or Lerwick and helicopters don't use petrol.
>>>
>>> Last time I checked, petrol is a component of crude oil, right?
>>>
>>
>> Yes, you get petrol from crude after it has been processed at a refinery.
>> It doesn't come out of the ground as petrol.
>
> Ha -- but it does come out of the ground /with/ petrol being part of it!
> So where there's crude oil, there /are/ petrol fumes. QED.
>

Yes, that's right. My job was to run about with a butterfly net and 
collect the petrol vapour.


-- 

Regards
     Stephen

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On 5/12/2016 6:24 PM, Stephen wrote:
>
> Ha -- but it does come out of the ground /with/ petrol being part of it!
> So where there's crude oil, there /are/ petrol fumes. QED.

Nurse!

-- 

Regards
     Stephen

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Am 2016-05-10 08:02, also sprach Thomas de Groot:

> Of course we have our anthem too:
>
> You know that it would be untrue
> You know that I would be a liar
> If I was to say to you
> Girl, we couldn't get much higher
>
> Come on baby, light my fire
> Come on baby, light my fire
> Try to set the night on fire
>
Oh Romeo and Juliet
Samson and Delilah
Baby you can bet
Their love they couldn't deny
Your words say split
But your words, they lie.
'Cause when we kiss...
oooo
Fire.

-- 
dik

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Am 2016-05-11 02:53, also sprach Thomas de Groot:

>
> So, what are you waiting for to implement this in your next alpha?
> Should have been done yesterday.
>

It is already done, and not done.

-- 
dik

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On 13-5-2016 0:51, dick balaska wrote:
> Am 2016-05-10 08:02, also sprach Thomas de Groot:
>
>> Of course we have our anthem too:
>>
>> You know that it would be untrue
>> You know that I would be a liar
>> If I was to say to you
>> Girl, we couldn't get much higher
>>
>> Come on baby, light my fire
>> Come on baby, light my fire
>> Try to set the night on fire
>>
> Oh Romeo and Juliet
> Samson and Delilah
> Baby you can bet
> Their love they couldn't deny
> Your words say split
> But your words, they lie.
> 'Cause when we kiss...
> oooo
> Fire.
>

Thank you Brother Balaska.

-- 
Thomas

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On 13-5-2016 0:52, dick balaska wrote:
> Am 2016-05-11 02:53, also sprach Thomas de Groot:
>
>>
>> So, what are you waiting for to implement this in your next alpha?
>> Should have been done yesterday.
>>
>
> It is already done, and not done.
>

Somehow I always get the dead cat.

-- 
Thomas

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On 5/13/2016 7:52 AM, Thomas de Groot wrote:
> On 13-5-2016 0:52, dick balaska wrote:
>> Am 2016-05-11 02:53, also sprach Thomas de Groot:
>>
>>>
>>> So, what are you waiting for to implement this in your next alpha?
>>> Should have been done yesterday.
>>>
>>
>> It is already done, and not done.
>>
>
> Somehow I always get the dead cat.
>

You are doing something wrong, then.
Try going to the world where it's living. See, easy.

-- 

Regards
     Stephen

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Le 12/05/2016 à 18:30, nemesis a écrit :
> Stephen <mca### [at] aolcom> wrote:
>> http://www.bbc.co.uk/news/technology-36203043
>>
>>
>> --
>>
>> Regards
>>      Stephen
>
>> General-purpose machines, which IBM calls "universal" quantum computers, will
eventually use more than 100,000 qubits
> .
>
> funny name for the measure:  quantum bits, qubits.  homophonous to the biblical
> cubits :)
>
> povray for quantum computing?  expect more 20 years, after they get into GPUs
>

By that time, the reflective sphere on a checkered plane meme will be 
replaced by a render made with a quantum computer of the Schrödinger's 
cat in a semi-transparent-polarising box.

"What is the colour of the cat in the box ?" is the input to trigger the 
start of the computation. (White, Black, .... )

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>> As clipka mentioned, a transistor used in digital circuits is either in
>> a state where the current flow is zero ("off") OR the voltage drop is
>> zero ("on"), so heat output is usually zero. It's switching between
>> those two states, when both voltage and current are non-zero, that
>> significant heat is dissipated within the transistor.
>
> Um... no?

It was a while ago I did my electronics course, but I was under the 
impression a FET worked a bit like a voltage controlled resistor. So 
with zero volts on the gate the resistance between drain and source was 
effectively infinite (so no current would flow), and as you increase the 
gate voltage the resistance went down to almost zero at a high enough 
gate voltage (so there would be current flowing, determined by whatever 
the output was connected to, but the voltage drop across the transistor 
would be near-zero).

For digital circuits you switch between fully on and fully off, whereas 
in an analog design (eg audio amplifier) you work in the "inbetween" 
region, controlling the output current based on the gate voltage. In 
this inbetween region, you have a non-zero voltage drop, and a non-zero 
current flow, so this generates heat inside the device according to P=VI.

Feel free to correct me where my understanding :-)

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On 5/13/2016 10:20 AM, scott wrote:
> Feel free to correct me where my understanding :-)

Are you forgetting about the constant heater current? ;-)

-- 

Regards
     Stephen

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>> Feel free to correct me where my understanding :-)
>
> Are you forgetting about the constant heater current? ;-)

My course was this millennium (just)...

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On 13-5-2016 9:28, Stephen wrote:
> On 5/13/2016 7:52 AM, Thomas de Groot wrote:
>> On 13-5-2016 0:52, dick balaska wrote:
>>> Am 2016-05-11 02:53, also sprach Thomas de Groot:
>>>
>>>>
>>>> So, what are you waiting for to implement this in your next alpha?
>>>> Should have been done yesterday.
>>>>
>>>
>>> It is already done, and not done.
>>>
>>
>> Somehow I always get the dead cat.
>>
>
> You are doing something wrong, then.
> Try going to the world where it's living. See, easy.
>

It is living but I burn my fingers instead.

-- 
Thomas

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On 5/13/2016 11:52 AM, scott wrote:
>>> Feel free to correct me where my understanding :-)
>>
>> Are you forgetting about the constant heater current? ;-)
>
> My course was this millennium (just)...

Just going back to basics.

There is a look of the quantum about thermionic emission.

-- 

Regards
     Stephen

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On 5/13/2016 1:06 PM, Thomas de Groot wrote:
> On 13-5-2016 9:28, Stephen wrote:
>> On 5/13/2016 7:52 AM, Thomas de Groot wrote:
>>> On 13-5-2016 0:52, dick balaska wrote:
>>>> Am 2016-05-11 02:53, also sprach Thomas de Groot:
>>>>
>>>>>
>>>>> So, what are you waiting for to implement this in your next alpha?
>>>>> Should have been done yesterday.
>>>>>
>>>>
>>>> It is already done, and not done.
>>>>
>>>
>>> Somehow I always get the dead cat.
>>>
>>
>> You are doing something wrong, then.
>> Try going to the world where it's living. See, easy.
>>
>
> It is living but I burn my fingers instead.
>

On consideration, I still think that you are doing something wrong.


-- 

Regards
     Stephen

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On 5/13/2016 1:14 PM, Stephen wrote:
>> It is living but I burn my fingers instead.
>>
>
> On consideration, I still think that you are doing something wrong.


Just a shot in the dark. Are you keeping your eye on Maxwell's Demon?


-- 

Regards
     Stephen

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On 13-5-2016 14:17, Stephen wrote:
> On 5/13/2016 1:14 PM, Stephen wrote:
>>> It is living but I burn my fingers instead.
>>>
>>
>> On consideration, I still think that you are doing something wrong.
>
>
> Just a shot in the dark. Are you keeping your eye on Maxwell's Demon?
>
>

Argh! Forgot about that one!

-- 
Thomas

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Am 13.05.2016 um 11:20 schrieb scott:
>>> As clipka mentioned, a transistor used in digital circuits is either in
>>> a state where the current flow is zero ("off") OR the voltage drop is
>>> zero ("on"), so heat output is usually zero. It's switching between
>>> those two states, when both voltage and current are non-zero, that
>>> significant heat is dissipated within the transistor.
>>
>> Um... no?
> 
> It was a while ago I did my electronics course, but I was under the
> impression a FET worked a bit like a voltage controlled resistor. So
> with zero volts on the gate the resistance between drain and source was
> effectively infinite (so no current would flow), and as you increase the
> gate voltage the resistance went down to almost zero at a high enough
> gate voltage (so there would be current flowing, determined by whatever
> the output was connected to, but the voltage drop across the transistor
> would be near-zero).
> 
> For digital circuits you switch between fully on and fully off, whereas
> in an analog design (eg audio amplifier) you work in the "inbetween"
> region, controlling the output current based on the gate voltage. In
> this inbetween region, you have a non-zero voltage drop, and a non-zero
> current flow, so this generates heat inside the device according to P=VI.
> 
> Feel free to correct me where my understanding :-)

There is a fallacy here: In contrast to a bipolar transistor, where the
voltage drop is more or less constant, in a FET it is the resistance
that is more or less constant (and low but non-zero), while the voltage
drop may actually vary wildly, depending on the current flowing; if
there is /any/ external mechanism that serves to stabilize the voltage
across the FET's source-drain channel, it /will/ remain high, and
according to R=V/I will instead drag the current up with it.

Note that R=V/I can be rewritten as I=V/R, so that P=VI = V^2/R --
meaning that, given a sufficiently stabilized voltage, a constant
near-zero resistance will actually lead to a *near-infinite* power
consumption (and consequently heat generation).

You can easily check this by connecting the terminals of a sufficiently
strong power source, such as a car battery, with a sufficiently constant
low-ohmic resistor, such as a piece of thick copper wire. (Spoiler
alert: *Don't!*)

And voltage-stabilizing elements are abundant in any modern digital
circuitry -- you want the power rails' voltage to be as stabile as
possible, after all; also, all the transistor gates to be switched
inevitably act as capacitances themselves, and you have parasitic
capacitances between the "wires"; at the same time you want to minimize
any current-stabilizing effects, such as parasitic inductivity, as they
stand in the way of increasing operating speed.


So again, no -- the fact that in modern digital circuitry switched-on
FETs don't draw any power is not because of any inherent near-zero
voltage drop in the on-state (as demonstrated above such an effect does
not exist), but because due to the design of the circuitry the current
is, for all purposes and intents, actually zero (which happens to have
the side effect that the voltage drop is indeed also zero) when the
circuitry is in a steady state. (As a matter of fact, switched-on FETs
/do/ consume power whenever they participate in the charging or
discharging of another FET's gate.)

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Am 12.05.2016 um 19:24 schrieb Stephen:

> Yes, that's right. My job was to run about with a butterfly net and
> collect the petrol vapour.

That explains a lot. :)

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On 5/17/2016 9:29 PM, clipka wrote:
> Am 12.05.2016 um 19:24 schrieb Stephen:
>
>> Yes, that's right. My job was to run about with a butterfly net and
>> collect the petrol vapour.
>
> That explains a lot. :)
>

It does. :P

-- 

Regards
     Stephen

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> There is a fallacy here: In contrast to a bipolar transistor, where the
> voltage drop is more or less constant, in a FET it is the resistance
> that is more or less constant (and low but non-zero), while the voltage
> drop may actually vary wildly, depending on the current flowing;

I thought the resistance (between source and drain) changes 
significantly as you change the gate voltage though? Otherwise how does 
a FET amplifier work? Like this one:

http://newton.ex.ac.uk/teaching/CDHW/Electronics2/PHY2028-C16.1.gif

That's the bit I'm missing/not understanding. If the effective 
resistance (Vds/Id) is always a constant, near zero (much less than 10K 
I assume), then wouldn't that mean the outputs in that circuit (Vd and 
Vs) were always 7.5V or thereabouts?

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Am 18.05.2016 um 10:18 schrieb scott:
>> There is a fallacy here: In contrast to a bipolar transistor, where the
>> voltage drop is more or less constant, in a FET it is the resistance
>> that is more or less constant (and low but non-zero), while the voltage
>> drop may actually vary wildly, depending on the current flowing;
> 
> I thought the resistance (between source and drain) changes
> significantly as you change the gate voltage though? Otherwise how does
> a FET amplifier work? Like this one:

Yes, of course -- the resistance does indeed vary with gate voltage.
However, given a constant gate voltage (which is kind of a prerequisite
for the "digital on" state), the resistance across the drain-source
channel remains pretty much constant.


> http://newton.ex.ac.uk/teaching/CDHW/Electronics2/PHY2028-C16.1.gif
> 
> That's the bit I'm missing/not understanding. If the effective
> resistance (Vds/Id) is always a constant, near zero (much less than 10K
> I assume), then wouldn't that mean the outputs in that circuit (Vd and
> Vs) were always 7.5V or thereabouts?

No, that's a misunderstanding; when I said the resistance was constant
near-zero, I was referring to the "digital on" state. In the "digital
off" state, the resistance is also constant, but near-infinite, and if
the gate is driven with an alternating voltage as is customary in analog
circuitry, the resistance will vary with time.

But varying the drain-source voltage or current will not change the
resistance, so in that sense the resistance can be considered "constant"
(across the drain-source parameter space, rather than time) even in
analog operating modes.

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On 5/18/2016 2:48 PM, clipka wrote:
> Yes, of course -- the resistance does indeed vary with gate voltage.
> However, given a constant gate voltage (which is kind of a prerequisite
> for the "digital on" state), the resistance across the drain-source
> channel remains pretty much constant.

I thought Scott was talking about the linear amplification state.

Which is where this part of the thread started. Transistors have three 
states. On, off and in between. The in between state is always there 
even is the transition between on and off is very small. You even get 
ringing in FET circuits.



-- 

Regards
     Stephen

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On 5/18/2016 2:48 PM, clipka wrote:
 > Yes, of course -- the resistance does indeed vary with gate voltage.
 > However, given a constant gate voltage (which is kind of a prerequisite
 > for the "digital on" state), the resistance across the drain-source
 > channel remains pretty much constant.

I thought Scott was talking about the linear amplification state.

Which is where this part of the thread started. Transistors have three 
states. On, off and in between. The in between state is always there 
even is the transition between on and off is very small. You even get 
ringing in FET circuits.

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>> Yes, of course -- the resistance does indeed vary with gate voltage.
>> However, given a constant gate voltage (which is kind of a prerequisite
>> for the "digital on" state), the resistance across the drain-source
>> channel remains pretty much constant.
>
> I thought Scott was talking about the linear amplification state.
>
> Which is where this part of the thread started. Transistors have three
> states. On, off and in between. The in between state is always there
> even is the transition between on and off is very small. You even get
> ringing in FET circuits.

I *think* I understand clipka's point now though. In digital circuits, 
because the output of a FET is usually connected to something with 
extremely high impedance (eg the gate of another FET, or through another 
FET that is off), even in the "in between" state still almost no current 
will flow and hence no power will be dissipated.

This is in contrast to if you had a load (eg 1K) connected to the 
output. In this case, there will be almost no power dissipation in the 
FET if it is fully "on" (no voltage drop across the FET) or "off" (no 
current flow), but there will be in the linear amplification state (eg 
when the FET effective resistance matches the output load).

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Am 18.05.2016 um 16:51 schrieb scott:
>>> Yes, of course -- the resistance does indeed vary with gate voltage.
>>> However, given a constant gate voltage (which is kind of a prerequisite
>>> for the "digital on" state), the resistance across the drain-source
>>> channel remains pretty much constant.
>>
>> I thought Scott was talking about the linear amplification state.
>>
>> Which is where this part of the thread started. Transistors have three
>> states. On, off and in between. The in between state is always there
>> even is the transition between on and off is very small. You even get
>> ringing in FET circuits.
> 
> I *think* I understand clipka's point now though. In digital circuits,
> because the output of a FET is usually connected to something with
> extremely high impedance (eg the gate of another FET, or through another
> FET that is off), even in the "in between" state still almost no current
> will flow and hence no power will be dissipated.

Um... no. I was contesting the following statement of yours:

"[...] a transistor used in digital circuits is either in a state where
the current flow is zero ("off") OR the voltage drop is zero ("on"), so
heat output is usually zero."

More specifically, I contested the claim that the normally-zero heat
output in the "on" state was due to the magnitude of the voltage drop. I
made no claims about the "in between" state of the FET.

I may also like to point out that I did not address the case when
/another/ FET, with the drain-source channel wired in series with the
drain-source channel of the FET in question, is switched on. In that
case, a transient current does indeed flow through the FET in question,
and yes, if the FET in question is in fully "on" state, as long as the
other FET is still in the in-between state, that one will have a
sufficiently high resistance to bear the majority of power dissipation.

In such a transitory state of /other/ parts of the circuitry, a totally
switched-on FET will indeed not consume any noteworthy power thanks to a
much lower voltage drop /compared to its neighbor/ -- but power
consumption will still happen, namely in that neighbor.

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> Um... no. I was contesting the following statement of yours:
>
> "[...] a transistor used in digital circuits is either in a state where
> the current flow is zero ("off") OR the voltage drop is zero ("on"), so
> heat output is usually zero."
>
> More specifically, I contested the claim that the normally-zero heat
> output in the "on" state was due to the magnitude of the voltage drop.

So if you came up with some transistor where the output resistance was 
not near-zero in the "digital on" state, power consumption shouldn't be 
affected?

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Am 19.05.2016 um 09:11 schrieb scott:
>> Um... no. I was contesting the following statement of yours:
>>
>> "[...] a transistor used in digital circuits is either in a state where
>> the current flow is zero ("off") OR the voltage drop is zero ("on"), so
>> heat output is usually zero."
>>
>> More specifically, I contested the claim that the normally-zero heat
>> output in the "on" state was due to the magnitude of the voltage drop.
> 
> So if you came up with some transistor where the output resistance was
> not near-zero in the "digital on" state, power consumption shouldn't be
> affected?

Not in the stable state of the system, no.

Except maybe for the few transistors at the boundary to the outside
world, which may see pull-up resistors and external AC loads.


The system's behaviour during state switching is another matter of course.

Thanks to P=V^2/R, increasing the FETs resistance in the "digital on"
state would actually _reduce_ power consumption -- provided you kept
your design unchanged otherwise. However, switching delay would also
increase linearly with resistance, since you would also reduce the
current, and hence the amount of charge you could transfer to/from the
gates per unit of time.

To counteract that increase in switching time, you would have to
increase the voltage linearly, which would increase the power
consumption quadratically. Thanks to the aforementioned reduction in
power consumption from the increased R, you'd end up with the power
consumption increasing linearly with FET resistance if the operating
frequency is to remain the same.

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It's not connected to any actual quantum hardware, but I just ran across 
this nifty little quantum circuit simulator that you can run from within 
your browser: http://algorithmicassertions.com/2016/05/22/quirk.html

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> It's not connected to any actual quantum hardware, but I just ran across
> this nifty little quantum circuit simulator that you can run from within
> your browser: http://algorithmicassertions.com/2016/05/22/quirk.html

Neat, that looks way better than the IBM one [reads a bit more] oh wait 
yes he notices that too ;-)

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On 23/05/16 23:42, Kevin Wampler wrote:
> It's not connected to any actual quantum hardware, but I just ran across
> this nifty little quantum circuit simulator that you can run from within
> your browser: http://algorithmicassertions.com/2016/05/22/quirk.html

Thanks for the link, Kevin.

This is in the best traditions of hackerdom (in its old sense):
"I need a tool to do a job. The tools to do this job are either
expensive, inefficient, awkward to use or unavailable; therefore I will
build my own. I will also give it to the rest of the world for free."

Bravo, Craig Gidney

John
-- 
Protect the Earth
It was not given to you by your parents
You hold it in trust for your children

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