How are memristors modeled in terms of impedance?

I have been searching the net for memristors, but could not get much of an idea. I'm not interested in how memristors are made (by combining materials), (yet), because I first want to understand what they do, compared to classical RCL components.

I had a look at https://en.wikipedia.org/w/index.php?title=Memristor&oldid=757282109 (intentionally linked to the version I have read), but this left me more than confused. There, I didn't even recognize some formulas for RCL components.

Then I saw Why does electrical impedance have as many parameters as it has? and the answer https://physics.stackexchange.com/a/187301 , which does not explain about how memristors are modeled in terms of impedance.

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    $\begingroup$ Impedance (a complex number describing phase difference and amplitude ratio between AC voltage and current) cannot model such a thing. (Cannot describe diodes either, or things with hysteresis.) Try the electrical engineering stackexchange. $\endgroup$ – Pieter Jan 3 '17 at 16:13
  • $\begingroup$ @Pieter could you explain about that? The wikipedia article lead me to think it was some passive component, some "missing" element to complement R,C,L. $\endgroup$ – Gyro Gearloose Jan 3 '17 at 17:02
  • $\begingroup$ A diode is also a passive component, but it does not have a constant ratio between voltage and current. Or in a ferromagnet: passive, but hysteresis means that there is no fixed proportionality between magnetization and applied magnetic field, so coils where the core has hysteresis are more complicated than a simple lossy inductance. $\endgroup$ – Pieter Jan 3 '17 at 17:39
  • $\begingroup$ @Pieter OK then, I was hoping to grasp "what a memristor is", but it looks it is not so easy. Do you have any proposal how I can change the question/put another one to understand what it is? Wikipedia is not much of a help there. $\endgroup$ – Gyro Gearloose Jan 3 '17 at 17:43
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    $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$ – Qmechanic Jan 3 '17 at 18:08

A memristor is, by definition, a two-terminal element whose constitutive relation is of the type [1]


where $\varphi = \int_{-\infty}^t v(s)\,\mathrm{d}s$ is the flux linkage and $q= \int_{-\infty}^t i(s)\,\mathrm{d}s$ is the charge.

If the relationship $g(\varphi,q)=0$ is linear, the memristor degenerates into a linear resistor and, thus, its impedance coincides with the resistance.

When the relationship $g(\varphi,q)=0$ is nonlinear, the memristor becomes a nonlinear element for which the concept of impedance is valid only in the so-called small signal approximation, that is, when the applied signal around a certain operating point is sufficiently small that the nonlinearity can be neglected. This kind of linearization is well known in circuit theory, and, for instance, it is commonly applied when analyzing diode circuits and transistor amplifiers. Let's see how this can be done in the case of the memristor.

If the relationship $g(\varphi,q)=0$ can be solved for $\varphi$, that is, we can write $\varphi = f_\mathrm{M}(q)$ (at least in a certain interval) the memristor is called charge-controlled and its $iv$ characteristic is given by [1]

$$v = R(q)i,\qquad\qquad(1)$$


$$R(q) = \frac{\mathrm{d}f_\mathrm{M}(q)}{\mathrm{d} q}$$

is a charge-dependent resistance. Since $q= \int_{-\infty}^t i(s)\,\mathrm{d}s$, the charge $q$ at any given instant depends on the past history of the current. But once you have reached a certain operating point $q$, the one of interest, you can think of applying a sine wave with infinitely small amplitude, so that the charge remains virtually constant. Around this operating point, you can see from (1) that the voltage across the memristor is proportional to the current, that is, for small signals the memristor behaves like a resistor with differential resistance $R(q)$. Hence, the small signal impedance will be $R(q)$, independent of frequency.

A similar conclusion can be obtained for the case in which the relationship $g(\varphi,q)=0$ can be solved for $q$.

To sum up, the small-signal (I cannot stress this enough) impedance of a memristor is resistive, independent of frequency, and, because of memristor nonlinearity, dependent on the operating point.

A more detailed analysis of the circuit behaviour of the memristor can be found in Chua's papers [1-3]. In [3], in particular, it is discussed the impedance. For more general information on small-signal analysis you can have a look at [4] (Chua again!).

[1] L. O. Chua, "Memristor - The missing circuit element," IEEE Trans. Circuit Theory, CT-18, 507–519, 1971.

[2] L. O. Chua, "The fourth element", Proc. IEEE, 100, 1920-1927, 2012.

[3] L. O. Chua, "Nonlinear circuit foundations for nano devices, Part I: The Four-Element torus," Proc. IEEE, 91, 1830–1859, 2003.

[4] L. Chua, C. A. Desoer, and E. S. Kuh, Linear and nonlinear circuits, McGraw-Hill, 1987.

  • $\begingroup$ Thank you, now I can see what it is. What got me confused most was that wikipedia mentions magnetic flux linkage. $\endgroup$ – Gyro Gearloose Jan 10 '17 at 11:25
  • $\begingroup$ @GyroGearloose That got many people confused. In the case of an inductor or coupled inductors, the integral of the voltage is indeed the magnetic flux linkage. From the point of view of circuit theory, however, it doesn't really matter the origin of that integral: the integral of a voltage is called flux linkage, and if that integral arises from a non-magnetic phenomenon it's still flux linkage. It appears that when HP developed the device that realizes the memristor, some people refused to acknowledge it as a memristor, because it's not based on a magnetic phenomenon. $\endgroup$ – Massimo Ortolano Jan 10 '17 at 20:35
  • $\begingroup$ This misunderstanding comes from the fact that many people confuses the model with the thing. For circuit theory, if a device behaves at the terminals like, say, a capacitor, it's a capacitor, even if inside there is a house-elf moving charges, instead of two electrodes. $\endgroup$ – Massimo Ortolano Jan 10 '17 at 20:39
  • $\begingroup$ Good point about that house-elf ;-) $\endgroup$ – Gyro Gearloose Jan 11 '17 at 20:12

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