Memristor: Perbedaan antara revisi
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|accessdate=2008-04-30}}</ref><!-- when did Chua say that? http://www.eetimes.com/news/latest/showArticle.jhtml?articleID=207403521&pgno=2 In 1971? At the HP talk a few years ago? two weeks ago when informed? now? -->
[[Image:Memristor-Symbol.svg|right|70px|thumb|Memristor symbol.]]
The memristor is formally defined<ref name="chua71"/> as a two-terminal element in which the [[Magnetic Flux#Magnetic flux through an open surface|magnetic flux]] Φ<sub>m</sub> between the terminals is a function of the amount of [[electric charge]] ''q'' that has passed through the device. Each memristor is characterized by its '''memristance''' function describing the charge-dependent rate of change of flux with charge.
: <math>M(q)=\frac{\mathrm d\Phi_m}{\mathrm dq} </math>
Noting from [[Faraday's law of induction]] that magnetic flux is simply the time integral of voltage,<ref>{{cite book|first=Heinz |last=Knoepfel|title=Pulsed high magnetic fields|location=New York|publisher= North-Holland|year=1970|pages=p. 37, Eq. (2.80)}}.</ref> and charge is the time integral of current, we may write the more convenient form
: <math>M(q(t))=\frac{\mathrm d\Phi_m/\mathrm dt}{\mathrm dq/\mathrm dt}=\frac{V(t)}{I(t)}</math>
It can be inferred from this that memristance is simply charge-dependent [[electrical resistance|resistance]]. If ''M''(''q(t)'') is a constant, then we obtain [[Ohm's Law]] ''R(t)'' = ''V(t)''/'' I(t)''. If ''M''(''q(t)'') is nontrivial, however, the equation is ''not'' equivalent because ''q(t)'' and ''M''(''q(t)'') will vary with time. Solving for voltage as a function of time we obtain
: <math>V(t) =\ M(q(t)) I(t)</math>
This equation reveals that memristance defines a linear relationship between current and voltage, as long as charge does not vary. Of course, nonzero current implies instantaneously varying charge. [[Alternating current]], however, may reveal the linear dependence in circuit operation by inducing a measurable voltage without ''net'' charge movement—as long as the maximum change in ''q'' does not cause [[small signal model|much]] change in ''M''.
Furthermore, the memristor is static if no current is applied. If ''I''(''t'') = 0, we find ''V''(''t'') = 0 and ''M''(''t'') is constant. This is the essence of the memory effect.
The [[power consumption]] characteristic recalls that of a resistor, ''I''<sup>2</sup>''R''.
:<math>P(t) =\ I(t)V(t) =\ I^2(t) M(q(t))</math>
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.
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