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From: Robert Baer on 16 Apr 2010 04:00
Bill Sloman wrote:
> On Apr 15, 7:02 am, Robert Baer <robertb...(a)localnet.com> wrote:
>> Bill Slomanwrote:
>>> On Apr 14, 7:51 am, Robert Baer <robertb...(a)localnet.com> wrote:
>>>> Jan Panteltje wrote:
>>>>> On a sunny day (Tue, 13 Apr 2010 11:24:33 -0700 (PDT)) it happened Wanderer
>>>>> <wande...(a)dialup4less.com> wrote in
>>>>> <b74e200b-933e-4848-a9b4-a9a66a623...(a)f17g2000vbd.googlegroups.com>:
>>>>>> HP found a way to make memristors
>>>>>> http://spectrum.ieee.org/semiconductors/design/the-mysterious-memristor
>>>>>> and I'm trying to make heads or tails of them. If they came in 0805
>>>>>> packages and you could buy them from digi-key, how would you spec
>>>>>> them? They say it has the units of ohms but if its a constant its just
>>>>>> a resistor, so you can't spec a 3 ohm memristor. Can you spec a 3
>>>>>> Weber/coulomb memristor? Doing unit analysis on the equations
>>>>>> R = dv/di , C = dq/dv, L = dphi/di, M = dphi/dq, V = dphi/dt, I = dq/
>>>>>> dt
>>>>>> I get
>>>>>> RC = M/L = idt/di has units of time
>>>>>> L/R = CM = vdt/dv has units of time
>>>>>> LC = (idt)(vdt)/(dvdi) has units of time squared
>>>>>> R/M (idtdv)/(vdtdi) is unit less
>>>>>> Does this thing break traditional circuit analysis?
>>>>> I did a read very simple explanation:
>>>>> The electrical current moves some atoms in a grid.
>>>>> that changes the resistance permanently.
>>>>> Reversing the current moves the atoms back.
>>>>> This can be done very fast (much faster then programming FLASH).
>>>>> I am sure that the chips that will be marketed will have a controller build in,
>>>>> and you will just be able to interface with it in the usual way.
>>>> ..something like a Coulomb meter or one of the definitions of current?
>>>> Like the amount of silver plated per unit of time?
>>>> And this is claimed to be new?
>>> Do pay attention.
>>> The apparent "resistance" of a Coulomb meter doesn't change with the
>>> amount of metal that has been plated out; the memistor has a "memory"
>>> and its "resistance" reflects its history.
>>> --
>>> Bill Sloman, Nijmegen
>> Are you trying to tell me that the silver plating cell originally
>> devised to measure / be a current standard does _not_ change its
>> resistance as the silver gets plated / unplated?
>
> Not so that you'd notice. The resistance of the silver layer is lot
> lower than the "resistance" of the solution from which it is being
> plated out.
>
> In fact the voltage drop between the electrodes involved is dominated
> by other effects, and the ohmic resistance of the solutions - while
> quite a bit higher than the resistance of the layer of silver - would
> be hard to isolate from the other processes involved.
>
> --
> Bill Sloman, Nijmegen
Have you made any measurements?
Would be interested..
From: Bill Sloman on 16 Apr 2010 11:38
On Apr 16, 10:00 am, Robert Baer <robertb...(a)localnet.com> wrote:
> Bill Slomanwrote:
> > On Apr 15, 7:02 am, Robert Baer <robertb...(a)localnet.com> wrote:
> >> Bill Slomanwrote:
> >>> On Apr 14, 7:51 am, Robert Baer <robertb...(a)localnet.com> wrote:
> >>>> Jan Panteltje wrote:
> >>>>> On a sunny day (Tue, 13 Apr 2010 11:24:33 -0700 (PDT)) it happened Wanderer
> >>>>> <wande...(a)dialup4less.com> wrote in
> >>>>> <b74e200b-933e-4848-a9b4-a9a66a623...(a)f17g2000vbd.googlegroups.com>:
> >>>>>> HP found a way to make memristors
> >>>>>>http://spectrum.ieee.org/semiconductors/design/the-mysterious-memristor
> >>>>>> and I'm trying to make heads or tails of them. If they came in 0805
> >>>>>> packages and you could buy them from digi-key, how would you spec
> >>>>>> them? They say it has the units of ohms but if its a constant its just
> >>>>>> a resistor, so you can't spec a 3 ohm memristor. Can you spec a 3
> >>>>>> Weber/coulomb memristor? Doing unit analysis on the equations
> >>>>>> R = dv/di , C = dq/dv, L = dphi/di, M = dphi/dq, V = dphi/dt, I = dq/
> >>>>>> dt
> >>>>>> I get
> >>>>>> RC = M/L = idt/di has units of time
> >>>>>> L/R = CM = vdt/dv has units of time
> >>>>>> LC = (idt)(vdt)/(dvdi) has units of time squared
> >>>>>> R/M (idtdv)/(vdtdi) is unit less
> >>>>>> Does this thing break traditional circuit analysis?
> >>>>> I did a read very simple explanation:
> >>>>> The electrical current moves some atoms in a grid.
> >>>>> that changes the resistance permanently.
> >>>>> Reversing the current moves the atoms back.
> >>>>> This can be done very fast (much faster then programming FLASH).
> >>>>> I am sure that the chips that will be marketed will have a controller build in,
> >>>>> and you will just be able to interface with it in the usual way.
> >>>> ..something like a Coulomb meter or one of the definitions of current?
> >>>>    Like the amount of silver plated per unit of time?
> >>>>    And this is claimed to be new?
> >>> Do pay attention.
> >>> The apparent "resistance"  of a Coulomb meter doesn't change with the
> >>> amount of metal that has been plated out; the memistor has a "memory"
> >>> and its "resistance" reflects its history.
> >>> --
> >>>Bill Sloman, Nijmegen
> >>    Are you trying to tell me that the silver plating cell originally
> >> devised to measure / be a current standard does _not_ change its
> >> resistance as the silver gets plated / unplated?
>
> > Not so that you'd notice. The resistance of the silver layer is lot
> > lower than the "resistance" of the solution from which it is being
> > plated out.
>
> > In fact the voltage drop between the electrodes involved is dominated
> > by other effects, and the ohmic resistance of the solutions - while
> > quite a bit higher than the resistance of the layer of silver - would
> > be hard to isolate from the other processes involved.
>
> > --
> >Bill Sloman, Nijmegen
>
>    Have you made any measurements?
>    Would be interested..

Never on an actual plating cell, but cells like it were part of my
undergraduate chemistry course, and I had to get back into the subject
when I was measuring the conductivity of aqueous solutions for
Haffmans BV a few years ago.

--
Bill Sloman, Nijmegen
From: Dirk Bruere at NeoPax on 16 Apr 2010 22:13
On 14/04/2010 22:47, whygee wrote:
> George Jefferson wrote:
>> I'*M(int(I)) + I/C = 0
>> I'*[Q/C - V]/I + I/C = 0
>> I*I'*[Q/C - V] + I^2/C = 0
>> Q - V*C + I/I' = 0
>> diff,
>> I + [I'^2 - I*I'']/I'^2 = 0
>> I*I'^2 + I'^2 - I*I'' = 0
>> [I + 1]*I'^2 = I*I''
>> > If you plot this de out you should get the solution if I didn't make
>> any mistakes(We lost C but it will come back).
>
> I was half-expecting that you come up with e=mc^2
> or something like that ;-)
>
> yg

I'm afraid Einstein got it wrong.
It's really:
E = mc^2 + 1

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
http://www.blogtalkradio.com/onetribe - Occult Talk Show
From: Phil Hobbs on 17 Apr 2010 10:42
On 4/16/2010 10:13 PM, Dirk Bruere at NeoPax wrote:
> On 14/04/2010 22:47, whygee wrote:
>> George Jefferson wrote:
>>> I'*M(int(I)) + I/C = 0
>>> I'*[Q/C - V]/I + I/C = 0
>>> I*I'*[Q/C - V] + I^2/C = 0
>>> Q - V*C + I/I' = 0
>>> diff,
>>> I + [I'^2 - I*I'']/I'^2 = 0
>>> I*I'^2 + I'^2 - I*I'' = 0
>>> [I + 1]*I'^2 = I*I''
>>> > If you plot this de out you should get the solution if I didn't make
>>> any mistakes(We lost C but it will come back).
>>
>> I was half-expecting that you come up with e=mc^2
>> or something like that ;-)
>>
>> yg
>
> I'm afraid Einstein got it wrong.
> It's really:
> E = mc^2 + 1
>

erg. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
From: JosephKK on 19 Apr 2010 23:48
On Wed, 14 Apr 2010 11:25:28 -0500, "George Jefferson"
<phreon111(a)gmail.com> wrote:

>> Maybe 'break' is too harsh, but inductors, resistors and capacitors,
>> represent the properties of inductance, resistance, and capacitance. I
>> can draw circuits with ideal inductors, capacitors, and resistors,
>> knowing that the real world ones have all kinds of parasitic
>> properties and non-linearities. There even is an ideal diode I can
>> stick in a circuit. What's an ideal memristor model look like? Does V
>> = Vo(1-e^t/MC) or I = Io(1-e^tM/L) work? The information they're
>> giving doesn't describe a fundamental component to me.
>
>Those components you mentioned are simplifications of a more complex
>process. One has to use maxwells equations to truly describe a real circuit.
>Just so happens that there are many fundamental things involved that happen
>to have nice simple mathematical approximations.
>
>The equations you are giving are for a simple circuit which is found by
>solving a differential equation. It's not always that simple. Although
>because of superposition and because they are linear we can usually
>decompose a circuit into those mathematical idealizations.
>
>On wiki it says
>
>"Each memristor is characterized by its memristance function describing the
>charge-dependent rate of change of flux with charge M(q) = dPhi_m/dq"
>
>For a capacitors and inductors one has similar laws(they just seem more
>natural).
>
>It seems like you want to use memristors in your circuits and are not sure
>about how to model them?
>
>As wiki says, a memristor is simply a charge dependent resistance, just like
>a normal resistor is also a temperature dependent resistance. We also have
>pressure and temperature dependent resistors and charge, pressure, time,
>temperature, and length dependent resistors.
>
>So to use a memristor in a circuit you have to use a resistor with a
>resistance that depends on the charge. What charge?
>
>Lets take a simple circuit
>
>V -- R --- GND
>
>V - I*R = 0
>
>But suppose R is charge dependent,
>
>V - I*R(q) = 0
>
>What is q? q' = I so
>
>V - I*R(int(I)) = 0
>
>(that makes it much more complex right htere.
>
>V is constant, R is unknown but depends on the type of memristor. Suppose
>that R(q) = q^2 just so we can get somewhere.
>
>Now we have the equation
>
>V = I*(int(I))^2
>
>We must solve for I. Can we? The above is known as an integral equation.
>This is the next logical step after differential equations. In fact, every
>ordinary integral equation can be written as a differential equation by
>differentiating enough times.
>
>0 = I'*int(I)^2 + 2*I^2*int(I)
>0 = I'*int(I) + 2*I^2
>int(I) = sqrt(V/I)
>0 = I'*sqrt(V/I) + 2*I^2
>2*I^5 = I'^2*V
>
>which is a nonlinear differential equation. (assuming I did everything
>right).
>
>
>In any case you can see the complexity involved simply because R is not
>constant.
>
>One can hypothesis current dependent resistors, capacitance dependent
>resistors, etc... Of course we must realize that the memristor is "resistor
>like" exactly because it has a V/I type of charactoristic. V/I is constant
>for resistors, charge dedependent for memristors, Diodes are exponential,
>and other devices have other types of dependents.
>
>Most basic electrical components that are simple do not have an explicit
>time dependence in V/I.
>
>Lets try to find the DE for an MC circuit
>
>V --- M --- C --- GND
>
>V - I*M(int(I)) - Q/C = 0
>
>diff,
>
>I'*M(int(I)) + I/C = 0
>
>I'*[Q/C - V]/I + I/C = 0
>
>I*I'*[Q/C - V] + I^2/C = 0
>
>Q - V*C + I/I' = 0
>
>diff,
>
>I + [I'^2 - I*I'']/I'^2 = 0
>
>I*I'^2 + I'^2 - I*I'' = 0
>
>[I + 1]*I'^2 = I*I''
>
>If you plot this de out you should get the solution if I didn't make any
>mistakes(We lost C but it will come back).
>
>
>
>Note capacitance C = dq/dv and we say it is a voltage dependent charge. M =
>dphi/dq is a flux
>dependent charge.
>
>so M = C*dphi/dv
>
>so in some sense M has a capacitive effect.
>
>In fact, it seems to me, that a memristor is just the next step in defining
>a more general device. All devices can be mathematically modelled and we
>started with the most natural first. A more general device would be charge,
>current, and voltage depedent. This would cover memresistors. After all, we
>are just trying to model phenomena and modeling always progresses from the
>simple to the complex.
>

So in a very rough sense we are looking at something like dielectric
absorption in a semiconductor instead of an insulator.
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