Reaney- Ferroelectric Non-Volatile Random Access Memories Flashcards
Abbreviation for ferroelectric non-volatile random access memories
NvFRAM’s
The two fundamental properties of ferroelectrics which are useful in memory applications
The charge storage capacity associated with the high dielectric constant.
The switching characteristics (high field)
How are NvFRAMs layer out and how do they generally work?
Composed of an array of ferroelectric bits which read 0 or 1 depending on whether they are poled positively or negatively. The memory is read by pulsing each bit to interrogate the sign of the polarisation. Thus digital information is stored and read
How does a ferroelectric layer for a NvFRAM look?
Could be a PZT ceramic as the film which has columnar grains going perpendicular to the plain of the film. Attached to the bottom is a Pt bottom electrode and the top would have a top electrode. The layer is about 200nm thick
What happens to ferroelectrics on cooling through a phase transition?
They exhibit a spontaneous polarisation which is reversible in an electric field
What is fatigue in NvFRAMs?
How much of the remnant polarisation remains as a function of the number of times you read and overwrite the bit. Expressed as the percentage decrease in remnant polarisation as a function of switching cycles.
Fatigue free criteria for NvFRAMs
If the individual bits are to be accessed at GHz speeds then the read/write cycles for the lifetime of a computer chip can be as high as 10^15. The polarisation must remain largely constant over this lifetime so as to minimise mistakes during bit interrogation
High signal to noise ratio criteria
The actual readable signal from the bits must be sufficiently large to be read by the circuitry. Mean the remnant polarisation must be large enough to give a signal to noise ratio greater than 0.08C/m^2
Low operating voltage criteria
Handheld devices and laptops are operating at lower power to maximise battery power and decrease heating of components. Means switching of the ferroelectric must be obtained at low voltages maybe less than 1V. Remember field is in V/m
Long retention and low leakage criteria
Once written, the bit must potentially remain polarised for the lifetime of the component if required. Therefore the retention time must be long and the leakage of charge negligible to avoid memory loss.
How can memory loss occur?
In metal oxide ceramics there are always defects. In a bit there is an electric field due to its polarisation. Ions in the ferroelectric material can diffuse in the presence of a field. The main metals are unlikely to move as this would require a very high energy. However oxygen vacancies can move under field to compensate for the remnant polarisation and cause memory loss. Therefore the number of oxygen vacancies needs to minimised when making the ferroelectric material
What is memory?
How long the polarisation remains in place when it’s not being constantly overwritten
Structure of PZT
Perovskite structure with Pb on A site (corners), Zr/Ti on B site (centre) and O on the O site (face centres). This is for cubic unit cell
SBT
Strontium Bismuth Tantalate (SrBi2Ta2O9). Orthorhombic unit cell (or tetragonal at high T) with c axis greater than a. Has alternating perovskite layers (ferroelectric) and Bi2O2 (fluorite) layers. Both layers are charged to cancel each other out. Polarisation lies in the a direction
More about PZT structure and polarisation
With respect to O octahedra, Zr and Ti ions randomly occupy the centre of them. These centres can form a cube between them with the Pb in the centre. Compound exists in either a tetragonal or rhombohedral polymorph depending on Zr:Ti ratio. Polarisation vectors of rhombohedral and tetragonal structure are in <111> or <001> pseudocubic directions respectively
Hysteresis loops for PZT and SBT
Normal shapes. Changes for PZT with Zr:Ti ratio (no simple relationship). SBT is a single compound and loop only changes for different field strengths (stronger field increases all features of loop)
Why might PZT be better than SBT for NvFRAMs?
Remnant polarisation lower and coercivity higher in SBT than PZT. Means PZT considered to have superior switching behaviour and potentially capable of higher signal to noise ratios in a NvFRAM. Also the lower coercive field more suited to low volt switching. So from a purely ferroelectric perspective PZT is better
Fatigue behaviours in PZT and SBT using Pt electrodes
Polarisation (C/m2) vs log(n cycles).
For PZT starts high a 0.35 and constant until about 6-7 on x, then steep decline down to 0.1 at about 11 on x.
For SBT starts lower at 0.15 but remains constant until about 10 and starts to curve down a bit towards 0.1 at 16 on x.
Therefore from fatigue perspective SBT is better
How to solve fatigue problem of PZT
Use different electrode material like RuO2 and then fatigue is negligible.
Mechanism of fatigue in NvFRAMs
Migration of O vacancies which develop dendritic microshorts. Damage caused by switching is in the form of O vacancies building up over time into clusters near the electrodes. Gradually these grow further into the ferroelectric from both electrodes and unaffected material in middle decreases. When the clusters from both sides meet you get a catastrophic decline in remnant polarisation resulting in memory loss
