Dielectrics- Capacitor Types Flashcards

1
Q

Class I capacitors

A

Non-ferroelectric materials with no ferroelectric domains. Low to medium permittivity (15-500). Low dissipation factors (tanδ under 0.003). Stable temperature dependence (TCε +100 to -200 MK^-1)

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2
Q

Class II capacitors

A

Ferroelectric materials with ferroelectric domains. High permittivity (2000-20,000). Moderate losses (tanδ between 0.003 and 0.03). Less stable temperature dependence.

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3
Q

Class III capacitors

A

Barrier layer capacitors. εeff over 10,000. Similar to class II but operate only at low voltages. Based on thin-layer effects. E.g surface layer/grain boundary not bulk

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4
Q

Problems with BaTiO3

A

Undergoes several phase changes which cause large peaks in the permittivity. Domains form and their movement causes large losses. Breakdown (current flow) can occur because of oxygen loss at surface leading to Ti3+ ions forming an energy level in the band gap for electrons.

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5
Q

What to aim for when modifying BaTiO3

A

Need to make εr large and temperature independent and tanδ as small as possible at 25°C. Also need to avoid breakdown

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6
Q

Idea of negative-positive oxide (NPO) ceramics

A

Mix two dielectric materials. One has a negative permittivity change with temperature (-TCε) and the other has a positive permittivity change with temperature (+TCε). These cancel each other out and lead to a stable permittivity with temperature

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7
Q

Lichtenekers rule of mixtures for predicting resulting permittivity

A

Vtotal=Vphase1+Vphase2

ln(εmix)=v1ln(ε1)+v2 ln(ε2)

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8
Q

Idea of A site doping

A

Changing the size of the A site cation changes the space for the B site cation. This will change how significantly it goes off-centre and change the ferroelectric stability.

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9
Q

A site doping with Sr2+

A

Smaller ionic radius that Ba. Shrinks the cell uniformly and reduces the space for the Ti to move off-centre. Encourages the material to go cubic so it reduces Tc away from the functional range to make TCC more stable. But this will decrease the εr so need to reach a compromise.

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10
Q

A site doping with Pb2+

A

Similar ionic radius to Sr2+. The ion doesn’t have a spherical charge distribution due to it having full s orbitals (similar for Bi and Sn). Ion naturally moves off-centre and stabilises Ti displacements. Pb also contributes to polarisation and permittivity. Stabilises the tetragonal phase and increases Tc away from functional range

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11
Q

Idea of B site doping

A

Larger 4+ ions can replace Ti4+. Because they are larger they have less free space to move off-centre. Reduces ferroelectric stability by pinching the phase transitions (could go straight from rhombohedral to cubic).

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12
Q

B site doping with Zr

A

Zr-O bond distance greater than Ti-O. The ZrO6 octahedra don’t contain a dipole moment. Confusion broadens the temperature range over which the applied field can shift the active Ti4+ ions. Ba(Ti1-xZrx)O3 forms a complete solid solution. x roughly 0.15 commonly used as εr,max occurs at about 50°C. Permittivity rises then falls either side. Increasing part is due to localised relaxations from Zr defects encourage Ti relaxation (movements) increasing the polarisation

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13
Q

Homogeneous

A

The defects are distributed evenly through the material. Assume all the ions move readily into the material- uniform grains

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14
Q

Heterogeneous

A

The ions have different diffusion/mixing behaviour. Leads to uneven distribution of defects across the grains. Produces a core shell microstructure

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15
Q

Core shell phases

A

Core is unreached BaTiO3 where Tc is 130°C
Shell has variable composition where Tc is lowered. Most commonly achieved with doping via rare earth captions e.g Dy, Y, Ho

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16
Q

How does the core shell structure work?

A

Rare earth captions have limited solubility and mobility in BaTiO3 so won’t perfectly mix without grinding and long sintering. There are many different defect mechanisms with potentially different effects of Tc, conductivity and TCC. Each phase has different permittivity-temperature profile but current/filed must pass through both so their effects combine creating a flatter response