memristors [Design]

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From: George Jefferson on 13 Apr 2010 17:52

"Wanderer" <wanderer(a)dialup4less.com> wrote in message
news:b74e200b-933e-4848-a9b4-a9a66a6238d4(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?

In what way? It is not a simple linear device and hence won't be like a
normal cap, ind, or res. It is more like a diode in how the analysis would
need to be carried out. No doubt there are simple approximations for it.

"As long as M(q(t)) varies little, such as under alternating current, the
memristor will appear as a resistor. If M(q(t)) increases rapidly, however,
current and power consumption will quickly stop."

"M(q) approaches zero, such that ?m = ?M(q)dq = ?M(q(t))I dt remains
bounded but continues changing at an ever-decreasing rate. Eventually, this
would encounter some kind of quantization and non-ideal behavior."

"# M(q) is cyclic, so that M(q) = M(q ? ?q) for all q and some ?q, e.g.
sin2(q/Q).
# The device enters hysteresis once a certain amount of charge has passed
through, or otherwise ceases to act as a memristor."

Essentially the resistance is "programmed" by by charge. You gotta know how
much charge you have to be able to know what the resistance is. To know what
the resistance is you have to know the charge. A diode uses current. To know
what the current going through the diode is you gotta know voltage across
it. To know the voltage across it you gotta know how much current is going
through it. Nature seems to be able to figure all this stuff about
automagically.

In any case it seems that all resistors are "memristors" but some are better
than others. As far as I can tell it seems that what we call a resistor is
really a memristor that has does not have much memory but all resistors have
some memory effect. This is stated in the wiki page:

"If M(q(t)) is a constant, then we obtain Ohm's Law"

And simply the resistors of today have a fairly constant M(q(t)). Once we
find more materials that exhibt the non-linear effect then we can exploit
it.

It seems the way to "spec" them is to give M. For resistors M is
approximately constant. To figure out what M you need you would have to
model the circuit or have some intuition. It is much more difficult because
of the non-linearities involved(hence why we naturally started with a
constant M).

If these devices do become mainstream then you will see a graph of M in the
datasheet or some quantity such as the slope of M if M is fairly linearly.

From: Robert Baer on 14 Apr 2010 01:51

Jan Panteltje wrote:
> On a sunny day (Tue, 13 Apr 2010 11:24:33 -0700 (PDT)) it happened Wanderer
> <wanderer(a)dialup4less.com> wrote in
> <b74e200b-933e-4848-a9b4-a9a66a6238d4(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?

From: Bill Sloman on 14 Apr 2010 03:56

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

From: Wanderer on 14 Apr 2010 10:22

On Apr 13, 5:52 pm, "George Jefferson" <phreon...(a)gmail.com> wrote:
> "Wanderer" <wande...(a)dialup4less.com> wrote in message
>
> news:b74e200b-933e-4848-a9b4-a9a66a6238d4(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?
>
> In what way? It is not a simple linear device and hence won't be like a
> normal cap, ind, or res. It is more like a diode in how the analysis would
> need to be carried out. No doubt there are simple approximations for it.
>
> "As long as M(q(t)) varies little, such as under alternating current, the
> memristor will appear as a resistor. If M(q(t)) increases rapidly, however,
> current and power consumption will quickly stop."
>
> "M(q) approaches zero, such that ?m = ?M(q)dq = ?M(q(t))I dt remains
> bounded but continues changing at an ever-decreasing rate. Eventually, this
> would encounter some kind of quantization and non-ideal behavior."
>
> "# M(q) is cyclic, so that M(q) = M(q ? ?q) for all q and some ?q, e..g.
> sin2(q/Q).
> # The device enters hysteresis once a certain amount of charge has passed
> through, or otherwise ceases to act as a memristor."
>
> Essentially the resistance is "programmed" by by charge. You gotta know how
> much charge you have to be able to know what the resistance is. To know what
> the resistance is you have to know the charge. A diode uses current. To know
> what the current going through the diode is you gotta know voltage across
> it. To know the voltage across it you gotta know how much current is going
> through it. Nature seems to be able to figure all this stuff about
> automagically.
>
> In any case it seems that all resistors are "memristors" but some are better
> than others. As far as I can tell it seems that what we call a resistor is
> really a memristor that has does not have much memory but all resistors have
> some memory effect. This is stated in the wiki page:
>
> "If M(q(t)) is a constant, then we obtain Ohm's Law"
>
> And simply the resistors of today have a fairly constant M(q(t)). Once we
> find more materials that exhibt the non-linear effect then we can exploit
> it.
>
> It seems the way to "spec" them is to give M. For resistors M is
> approximately constant. To figure out what M you need you would have to
> model the circuit or have some intuition. It is much more difficult because
> of the non-linearities involved(hence why we naturally started with a
> constant M).
>
> If these devices do become mainstream then you will see a graph of M in the
> datasheet or some quantity such as the slope of M if M is fairly linearly..

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.

From: George Jefferson on 14 Apr 2010 12:25

> 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.

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