Reaney- Ferroelectric Non-Volatile Random Access Memories

Question 1

Abbreviation for ferroelectric non-volatile random access memories

Answer 1

NvFRAM’s

Question 2

The two fundamental properties of ferroelectrics which are useful in memory applications

Answer 2

The charge storage capacity associated with the high dielectric constant.
The switching characteristics (high field)

Question 3

How are NvFRAMs layer out and how do they generally work?

Answer 3

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

Question 4

How does a ferroelectric layer for a NvFRAM look?

Answer 4

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

Question 5

What happens to ferroelectrics on cooling through a phase transition?

Answer 5

They exhibit a spontaneous polarisation which is reversible in an electric field

Question 6

What is fatigue in NvFRAMs?

Answer 6

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.

Question 7

Fatigue free criteria for NvFRAMs

Answer 7

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

Question 8

High signal to noise ratio criteria

Answer 8

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

Question 9

Low operating voltage criteria

Answer 9

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

Question 10

Long retention and low leakage criteria

Answer 10

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.

Question 11

How can memory loss occur?

Answer 11

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

Question 12

What is memory?

Answer 12

How long the polarisation remains in place when it’s not being constantly overwritten

Question 13

Structure of PZT

Answer 13

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

Question 14

SBT

Answer 14

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

Question 15

More about PZT structure and polarisation

Answer 15

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

Question 16

Hysteresis loops for PZT and SBT

Answer 16

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)

Question 17

Why might PZT be better than SBT for NvFRAMs?

Answer 17

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

Question 18

Fatigue behaviours in PZT and SBT using Pt electrodes

Answer 18

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

Question 19

How to solve fatigue problem of PZT

Answer 19

Use different electrode material like RuO2 and then fatigue is negligible.

Question 20

Mechanism of fatigue in NvFRAMs

Answer 20

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