2011
DOI: 10.1002/adfm.201102208
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A Parallel Circuit Model for Multi‐State Resistive‐Switching Random Access Memory

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Cited by 35 publications
(43 citation statements)
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“…Additionally, TaON layer can be utilized to reduce switching power and build multi-bit RRAM through engineering such a serial resistor and developed methodology. [26][27][28] Regardless of the presence of a TaON layer, the TaO x subphase at the lower interface of the bottom Pt leads to higher carrier tunneling due to its higher localized empty defect density compared to the case at the upper interface. This is because O-rich species at the surface will compensate for oxygen vacancies that drift toward the surface (Fig.…”
mentioning
confidence: 99%
“…Additionally, TaON layer can be utilized to reduce switching power and build multi-bit RRAM through engineering such a serial resistor and developed methodology. [26][27][28] Regardless of the presence of a TaON layer, the TaO x subphase at the lower interface of the bottom Pt leads to higher carrier tunneling due to its higher localized empty defect density compared to the case at the upper interface. This is because O-rich species at the surface will compensate for oxygen vacancies that drift toward the surface (Fig.…”
mentioning
confidence: 99%
“…1 RRAM can also exhibit >2 resistance states that may promise multi-bit storage. [2][3][4][5][6][7][8] In principle, if each single cell can display 2 N distinguished states, then for the same storage capacity the required die area will scale with 1/N. Stated alternatively, for the same die area, each 2 N -state layer of a 2D memory has the same storage capacity of N 2-state layers of a 3D memory.…”
mentioning
confidence: 99%
“…2,7 The fact that these approaches seem to be applicable to all types of RRAM, irrespective of their underlying switching/conduction mechanisms, suggests a common understanding of their existence and control may be possible. However, no such understanding has been provided to-date.…”
mentioning
confidence: 99%
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“…(16) to find R E (I,J). After subtracting R E (I,J) from R(I,J,), we obtain R F (I,J,) for all the devices in this favorable case.…”
mentioning
confidence: 99%