memristors

=3D dq/

Yeah back in the news again. The wiki article does OK at explaining it.

The resistance is a function of how much charge has gone through the thing.

George H.

On a sunny day (Tue, 13 Apr 2010 11:24:33 -0700 (PDT)) it happened Wanderer wrote in :

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.

er

I =3D dq/

ld in,

Yes, I can see it as something exotic in an IC. But I just don't see it as a fundamental circuit component like resistors, capacitors and inductors. If I had a circuit with one in it I wouldn't know how to solve it. Heck I can't even figure out what value to give it.

On a sunny day (Tue, 13 Apr 2010 12:17:05 -0700 (PDT)) it happened Wanderer wrote in :

If it is at the atomic level, perhaps some QED effects may come into play. That is way beyond what I have learned, but looking at it from the mechanical explanation I think you will see R change after some I is applied. Probably some minimum I is needed to cause the effect, and a lower I to read it? Nothing 'fundamentally' new, just marketing hype. But it is a new type of memory for sure. If they can make that density that they mention then we have a nice part to play with.

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.

..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?

erer

, I =3D dq/

uild in,

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

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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 =3D Vo(1-e^t/MC) or I =3D 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.

I was half-expecting that you come up with e=mc^2 or something like that ;-)

yg

--
http://ygdes.com / http://yasep.org

Well, I overcomplicated things

It's much easier to see that

V - R(Q)*I - Q/C = 0

and I = Q'

So we have

V - R(Q)*Q' - Q/C = 0

which is the nonlinear DE if R(Q) is nonlinear. Knowing the expression for R lets us solve for Q then we can solve for I.

Here is a matlab script that simulates an M-C circuit:

Script - memristor

options = odeset('RelTol',1e-5,'AbsTol', 1e-5, 'MaxStep', 0.001); [T,Q] = ode113(@memcap, [0 5], 0, options);

hold on; plot(T, Q); plot(T, [diff(Q)./diff(T); 0], 'Color', 'red'); hold off;

------------- function - memcap

function dQdt = memcap(t, Q)

V = 1; C = 1;

dQdt = (V - Q/C)./Res(Q);

function r = Res(Q) r = 1 + 0.25*cos(10*Q);

---------------

Res(Q) = M(Q). If M(Q) is constant you get the standard RC circuit.

Showing a non-linear resistor for various parameter adjustments(similar to the code I posted).

We you can see they are all approximately "resistor like". Remember that it comes from an RC circuit with R depending on charge.

Because the charge will eventually reach a maximum value(C*V for this circuit) if the resistance never becomes negative, the resistance must eventually become constant. i.e., The cap will eventuallly max out and the current will decrease causing a steady steady state in the resistance.

Again, the max charge that can pass through the resistor is C*V which is the max charge the cap can hold. Because there is a maximum charge R(Q)->R(C*V) and C*V is a constant, hence R(Q)->constant and which is why the curves seem to become more and more like and R-C circuit as time goes on.

If we replaced the cap with a linear resistor then the charge Q will approach infinity and R(Q) could be drift off. There are a few possibilities that could happen.

What we sorta need to know is if R(Q) is generally increasing, constant, or decreasing as Q->oo. Each one implies a different steady state for different circuits.

I imagine that for true memristors there is no singularities since this would create some impossibile physical situations.

in,

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?

BTW, a capacitor is a special type of memristor(at least in an RC circuit):

V = V0*(1 - exp(-t/RC)) I = V0/R*exp(-t/RC) Q = C*V

==> I = V0/R - Q/C/R

==> V/I = Q/C/(V0/R - Q/C/R)

Hence M(Q) = V/I = Q/C/I.

This is to be expected since the "resistance" of a cap depends on how charged it is(relatively to it's capacity).

M(Q) = QR/(CV - Q)

In fact if we extend resistance into a generalized resistance we can cover just about any device(since they all have V/I characteristics even if dependent on time).

If, say we have R(Q) instead of R,

M(Q) = Q*R(Q)/(CV - Q)

Lets suppose M(Q) = R(Q) then Q = CV - Q ==> Q = CV/2. Which is exactly what happens whence you have two capacitors in series. So it is constistent.

We get a similar but more complex result for an RL circuit if we fudge a bit.

It seems like the tech that will drive all before it. And HP has shown they work down to 3nm Also perfect for building neural nets. Goodbye flash and phase change memory...

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
http://www.blogtalkradio.com/onetribe - Occult Talk Show

nderer

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dt, I =3D dq/

build in,

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

MMM! Send me a pound of fudge!

Phase change memory was a joke the first time the idiot thought of it seriously.