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Chapter 5. Liquid crystal cell alignment

The static LC cell alignment is determined by the boundaryconditions (on the 2 glass surfaces) and the elasticdeformation energy of the LC molecules.

5.1 LC director

LC optics is determined by it director. The LC director is thedirection of the rod shaped LC molecule.

θ

φ

x

y

z

Director

The director angles are θ and φ. They are usually expressedas functions of z, the distance inside the LC cell.

Azimuthal angle: θ = θ(z)

Polar angle: φ = φ(z)

(Note: 90o - θ is called the tilt angle α.)

2

This is called the one-dimensional case (1D). The variation ofθ and φ is only along the cell direction z. There is no x,ydependence. This is good enough for all TN, STN modeling.

For active matrix displays, and for small pixel sizes, it isnecessary to consider θ(x, y, z) and φ(x, y, z). They are called3D cases, and are obviously much more complicated.

Commercial softwares are available for both 1D and 3Dsimulations of LCD.

1D software:DIMOS from Autronics.LCDSoft from HKUST.Shintec

3D software:LIQUID from TMS.3DIMOS from Autronics.Shintec

Once θ(z) and φ(z) is known, the optical properties such astransmission etc of the LCD, can be calculated. So it isnecessary to understand how θ(z) and φ(z) change as afunction of applied voltage.

We shall try to understand the director distribution, i.e. θ(z)and φ(z), under 2 situations:

1. No voltage. The director distribution is determined byboundary conditions.2. Applied voltage. The director distribution will deform, andthe optical properties will change.

With no voltage, the alignment of the LC cell is determined bythe boundary conditions on the 2 cell surfaces.

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5.2 LC director alignment on the boundary surface:

The alignment or director distribution inside the LC cell isdetermined by the boundary conditions of θ(z) and φ(z) at z=0and z=d.

Liquid crystals near the surface of the alignment layer arefixed in direction by the alignment layer.

Different kinds of boundary conditions:

(1) Homogeneous alignment . It is obtained by rubbing thealignment layer. It can also be obtained by UV photo-alignment. The director is nearly flat with a small pretilt angle.

The alignment layer is usually a layer of polyimide. Therubbing direction determines the alignment direction and thedirection of the pretilt angle.

Rubbingdirection

Director

Pretiltangle

The type of PI determines the pretilt angle. It is usually in therange of 1 – 8o. Low pretilt are used in TN and high pretilt areused in STN.

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(2) Homeotropic alignment: The director is nearlyperpendicular to the cell surface. It is obtained by applying asurfactant to the alignment surface. This type of alignmentcondition is not used often.

(3) It is also possible to obtain any pretilt angle, especiallynear 450 by evaporation of SiO2 at oblique angles. Obviously itis not amenable to mass production and is used mostly inresearch.

Different methods of alignment:

1. Rubbing of PI

2. Linear polarized light photopolymerization (LPP)

3. SiO2 evaporation

4. Langmuir-Blodgett film: This is a layer of surfactant. Can beused to control the pretilt angle. Not amenable tomanufacturing.

Given the boundary conditions of θ(z) and φ(z), it is straight-forward to obtain the static alignment condition of the LC cell.

For example:

To obtain left-handed 90o twist TN:

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Rubbingdirection oftop plate

Rubbingdirection ofbottom plate

90o

The director will have an anticlockwise twist as you look downat the cell.

As will be explained later, this particular alignment will give thebest 6 o’clock viewing.

Most alignments are obtained by rubbing the alignment layer.Alignment can also be achieved by other means, such as SiOevaporation, and more recently UV light irradiation of thesurfaces. However, the most reliable method in manufacturingis still rubbing of cured PI.

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5.3 Elastic deformation of the LC director:

The director of a LC cell is like a spring. It has a preferredalignment which will minimize its elastic energy.

There are 3 kinds of elastic deformations: bend, splay andtwist. They affect important LC properties such as thresholdvoltage and response time.

In the continuum theory, the Frank’s free energy density for adeformed LC is given by

F = 2

1 K11(∇⋅n)2 + 2

1 K22(n⋅∇xn)2 + 2

1 K33nx∇xn2

n = directorK11 = splay distortion elastic constantK22 = twist distortion elastic constantK33 = bend distortion elastic constant

Note that this elastic energy is similar to a stretched spring:

F = 2

1 kx2

Usually, K33 > K11 > K22 and they are all ~ 10-12 N. For MBBA,K33/K11 = 1.3 and K33/K22 = 2.9.

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Diagram to illustrate three kinds of director distortions: ( B.Bahadur Vol. 1 pg 166)

If there is a natural twist (chirality) for the LC, then the twistterm is biased by a permanent twist qo, and Frank’s freeenergy becomes

F = 2

1 K11(∇⋅n)2 + 2

1 K22(n⋅∇xn - qo)2 +

2

1 K33nx∇xn2

= 2

1K11(∇⋅n)2 + 2

1 K22(n⋅∇xn)2 + 2

1 K33nx∇xn2

- qoK22 n⋅∇xn + 2

1Κ22qo2

Noting that n = (sin θ cos φ, sin θ sin φ, cos θ),it is easy toshow that F can be written as

F = 2

1k1(θ)2

θ•

+ 2

1k2(θ) 2

φ•

+ 2

1k3(θ) φ•

+ 2

1Κ22qo2

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where k1(θ) = K11 sin2 θ + K33 cos2 θ

k2(θ) = (K22 sin2 θ + K33 cos2 θ) sin2 θ

k3(θ) = 2qo K22 sin2 θ

F depends on θ, φ which are functions of z. Such F is called afunctional . The director distribution is determined by aminimization of F by varying the functions θ(z) and φ(z). Thisis the calculus of variations .

Theorem: For any functional F which can be written as

),,,( φφθθ••

= FF

∫d

Fdz0

can minimized by solving the Euler-Lagrange equations

∂

∂=∂∂

•θθF

dz

dF

∂

∂=∂∂

•φφF

dz

dF

which are equations for θ(z) and φ(z). This is a general resultand appears in many other branches of physics andmechanics.

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In the present case, ∫d

Fdz0

is the total elastic energy stored in

the LC cell. Minimizing the integral means finding a stableequilibrium configuration.

The Euler-Lagrange equations in general have to be solvednumerically. They can be solved for some special cases.

(1) One constant approximation. K11 = K22 = K33 = K

Here k1(θ) = K

k2(θ) = K sin2 θ

k3(θ) = 2qo K22 sin2 θ

(2) No twist. Only splay-bend deformation. φ•

= 0.

(3) Twist only. No splay-bend. θ• = 0.

5.4 Some special examples:

(1) Homogenous cell (H-cell)

Also called the Nn alignment. It is obtained by anti-parallelrubbing of alignment layers.

Here θ(0) = θ(d) = 90o – α

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Assume that there is no doping (qo = 0). From the BC, it isobvious that the solution will have no twist. Therefore abovecase (2) applies.

F = 2

1k1(θ)2

θ•

Assume further that K11 = K33 Then there is only one Euler-Lagrange equation for θ

0)( 11 =•θK

dz

d

The solution is

θ(z) = Az + B

where A, B are constants to be determined by the BC. Usingthe BC given, we find that

θ(z) = 90o - α for all z.

This solution is obvious actually. However, if we do not makethe one constant approximation, or if there is doping, even thissimple case can become very complicated and needsnumerical solution.

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α

z

(2). Homeotropic cell:

It is also called the Np alignment. It is obtained by ahomeotropic treatment of the alignment layers. The biundarycondition is like a homogeneous cell with

θ = 90ο, φ = 0 for all z.

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(3) Splay cell (S-cell)

Obtained by parallel rubbing of the alignment layers.

Here θ(0) = 90o – α; θ(d) = 90o + α;

α

z

The Euler-Lagrange equation is the same as before and thesolution the same

θ(z) = Az + B

for some A.B. Using the BC here, the solution is

θ(z) = 90o + d

dz )2( −α

This is a splay cell as shown. There is only splay deformation.

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4) Bend cell (B-cell)

Also called the π-cell.

The boundary conditions are the same as the S-cell. Thedifference is that the tilt angles are larger, so that a benddeformation will be preferred over the splay deformation. Thebend deformation can also be obtained if we apply a constantvoltage to a S-cell.

The general solution is the same as the S-cell.

θ(z) = Az + B

But now the BC can be written as

Here θ(0) = 90o – α; θ(d) = 90o + α − 180ο;

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The solution is therefore

)2

1)(90()(d

zz o −−= αθ

Difference between the B-cell and the S-cell:

θ

z

90o+α

-90o+α

90o-α

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(5) Twist cell (T-cell)

The boundary conditions are φ(0) = 0, φ(d) = φο

θ(0) = θ(d) = 90o – α

For simplicity, we can assume that θ = constant throughoutthe cell. (This is actually not strictly correct.)

Then above case (3) is valid. Also, we shall use the oneconstant approximation. Therefore

F = 2

1K2(θ) 2

φ•

+ 2

1K3(θ) φ•

+ 2

1Κ22qo2

The Euler-Lagrange equation reduces to

0)( =•φ

dz

d

The solution is given by a linear variation

φ(z) = Az + B

where A,B are to be determined by the BC.

For a 90o twist TN LCD, the director distribution is given by

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θ = θο

φ(z) = d

z

2

π

This is the most common LCD. φ is twisted by 90o. The tiltangle (90o-θ) is constant.

(6). Hybrid aligned cell (HAN cell):

The boundary conditions are

θ(0) = 0; θ(d) = 90o

φ(z) = 0.

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In the one constant approximation, the solution is the same asthe B-cell or S-cell:

θ(z) = Az + B

Substituting in the BC, we obtain:

θ(z) = d

z

2

π

In general, the solution can be quite complicated. It is shownbelow:

θ

z

K11<K33

K11>K33

d

90o

In general, K11<K33. It is better to use numerical solutions:

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General classification:

Twist cell (T-cell):

Picture of T-cell has been shown before. It is the mostcommon kind.

It should be noted that the T-cell can be made without anyinternal stress by adding chiral dopants, so the LC has anatural twist already. This is not the case for other kinds ofdistortions (S-cell and B-cell), where the internal elasticenergy cannot be compensated.

ECB cell

All non-twist cells operate as a birefringent plate. Thereforethey are all electrically-controlled birefringent (ECB) LCDs. Atwist cell operates by polarization rotation.

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The total birefringence of a non-twist cell (S-cell, B-cell, HANcell) is given by:

∫∆

=d

dzzn

0

))((

λθπδ

where we have emphasized the fact that the birefringencedepends on the tilt angle.

5.5 Details on T-cells:

1. Pitch:

Pitch is the distance covered when the director twists by 360o

(one complete cycle).

If thickness = d,twist angle = φ,then pitch p is given by

p = φπd2

e.g. a 5µm thick TN cell has a pitch of 20µm.

If the twist is induced by the boundary conditions, then this isa forced pitch. The LC can be doped with a chiral dopant tohave a natural pitch of 20 µm as well. In that case, there willbe no elastic deformation energy.

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2. Right twist and left twist amibiguity:

A RH 90o twist cell can also have a 270o LH twist under thesame rubbing conditions.

Similarly, a LH 180o twist cell can also be a 180o RH cell.

This leads to anti-twist domains and grain boundaries –defects.

These defects can be eliminated by (1) doping and/or (2)having high pretilt angles.

3. Doping:

For a twist cell, it is possible to add a chiral dopant to give thedirector a natural twist. There are dopants that give RH twistand there are some for LH twist.

Pitch = 1/(dopant concentration x HCP)

HCP = helical twist power

e.g. S811 from Merck.

HCP = -10.1 to -11.6 /µm depending on the LC

∴ a 0.4% doping concentration => p = 90 µm, LH twist

In order to prevent reverse twist in TN displays, the inducedhelical pitch by the dopant should be 10-30 times the cellthickness d.

e.g. d = 6 µm => p = 60 – 180 µm. Therefore 0.4% is OK.

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Commercial dopants:

Merck: S811 (LH): HTP = -10.9 /µmCB15 (RH): HCP = +6.5 /µm

Roche (Merck):LH: CM5715A, CM9111S, CM9209F, CNRH: CM5815C, CM9207E

For sophisticated use, have to mix + and – dopants tocompensate for temperature dependence.

4. d/p ratio:

Here p is the pitch induced by the dopant, not the pitchinduced by the cell boundary conditions. For the TN cell, p is10-30 times d. Therefore d/p = 0.03 – 0.1.

For STN, it is better for the pitch to match the twist angle. Forexample, if the twist angle is 180o, then the pitch should be

p = 180

360d = 2d

Therefore d/p = 0.5

In actual practice, a d/p value of 0.25 – 0.5 is used for STN.This d/p affects the steepness of the electro-optic curve for theSTN.

The actual d/p value used also depends on K11, K22, and K33,and other parameters. It affects not only reverse twistdislocations, it also affects the hysteresis and slope of thevoltage response curve. Actually, it is possible to play with thed/p value to obtain bistable phenomenon – the bistable twistednematic (BTN) display.

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Reference: H S Kwok, Z L Xie, T Z Qian and P Sheng, Study ofSwitching Behavior in Bistable Nematic Liquid Crystal Displays,Liquid Crystals , Proceedings of SPIE, pp 22-30, 1997.

5. 6 o’clock and 12 o’clock viewing:

90o TN cells are made mostly for 6 o’clock viewing. There are4 possible ways to rub the alignment layers to give a LCdirector distribution that is symmetrical about the 6-12 o’clockaxis:

Top glass

Bottomglass

Left handedtwist

Right handedtwist

The top 2 configurations will give better 6 o’clock viewingwhile the bottom 2 will give good 12 o’clock viewing.

There are 4 other possible rubbing combinations, which willgive director distributions that are symmetrical about the 3-9o’clock axis. Check these as an exercise! Theseconfigurations are used generally.

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5.6 Bistability of LC alignment

1. BTN: bistable Twisted Nematic cells:

For a twist cell with boundary conditions that favor a twist of φ,twist angles of φ±2Nπ should also be allowed, for any N. Thisambiguity is usually eliminated in practical cells by doping theLC so that the natural pitch and the LC cell pitch areagreeable.

For a cell with thickness d, the d/p ratio for the of φ±2Nπ twiststate is

d/p = φ/2π ± N

Consider the 2 states φ and φ+2π, the elastic energy as afunction of the natural twist (helicity) qo is given by

F = 2

1k2(θ) 2

φ•

+ 2

1k3(θ) φ•

+ 2

1Κ22qo2

= 2

1K22 2

φ•

+ qoK22 φ•

+ 2

1Κ22qo2

=2

1K22 (qo-d

Φ)2

where we have used the result for a T-cell

Φ=d

zφ

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Also d/p = qod/2π

φ φ+2π

d/p

φ+π

F

Hence if the d/p ratio of the cell is between the values,bistability will occur, with both twist states equally possible.

Known BTN with (-90o, 270o), (0o, 360o), (90o, 450o) havebeen demonstrated by us.

2. Pi-cells: Bistable Bend/splay cells (BBS):

A bend cell can also be bistable with a splay cell. For a largepretilt angle, the splay deformation may have the same elasticenergy as a bend deformation. Hence bistability will occur.

The total elastic energy for the S-cell or B-cell is

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F = dzdz

dzKzK

d 2

0

233

211 ))(cos)(sin(

2

1

+∫

θθθ

If we take the one constant approximation, then the B-cell andthe S-cell have the same elastic energy if α = 45o. This makessense. In general, the integral has to be evaluated numericallyto find the tilt angle for which the bend and splay energies areequal.

In both the BTN and the BBS cells, the addition of a biasvoltage will alter the energy balance and give interestingeffects.

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Chapter 6. Electromechanics of LC cells -Director deformation under an electric field:

6.1 Introduction

When a voltage is applied to a LCD, 2 effects can happen –current effect and field effect. In early days with LCD operatingin the dynamic scattering (DS) mode, there is current thatgenerates turbulence to scatter light. Nowadays all LCDoperates with field effect.

Resistivity of LC material:

ρ ~ 1010 – 1013 Ω−cm

with the low end used in TN displays and high end used inAMLCD. For example, a 6 µm TN segment display with asegment area of 3 mm2 will have a resistance of 200 MΩ. It ispractically an insulator.

The RC decay time constant is independent of the size of thesegment or pixel.

τ = ρεr

For εr = 10, and ρ = 1010, the RC time constant is 8.85 ms. Anyvoltage applied will decay quite rapidly, and constant refreshis needed. For AMLCD, this RC decay is too fast.

6.2 LC dielectric anisotropy

In a LC, the dielectric constant is anisotropic, just like therefractive index:

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=

z

y

x

εε

εε

00

00

00

For LC, εx = εy. It is customary to write ε⊥ = εx = εy, and ε// = εz.Then

= ⊥

⊥

//00

00

00

εε

εε

We can define the dielectric anisotropy as

∆ε = ε// − ε⊥.

For most LC, -6 < ∆ε < 50

Positive anisotropy: ε// > ε⊥

In this case, the LC would like to align parallel to the electricfield at high voltages. (Homeotropic alignment). Most LCD areof this type. This can be seen in the above example for TNcell. At high fields, the midplane tilt is all 90o. At theboundaries, the LC cannot move much due to stronganchoring effects.

Negative anisotropy: ε// < ε⊥

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In this case, the LC would like to align perpendicularly to theelectric field at high voltages. (Homogeneous alignment).

The data sheet from Merck details the dielectric constants ofvarious fluids.

The dielectric anisotropy depends on T, λ, and ω. Same asbirefringence ∆n. But note that their dependence can be quitedifferent as they are at very different frequencies.

Below is an example of the T dependence of ∆ε.

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Note that it is possible to have a negative ∆ε and a positive∆n. There is no conflict since they are defined for differencefrequencies. Usually, large ∆ε means large ∆n as well.

6.3 Dielectric energy

The dielectric energy stored in the LC cell is dependent onε, which is dependent on the director distribution due todielectric anisotropy. Hence the alignment of the LC celldepends on the applied field. The optical properties of the LCcell is determined by its director. Thus the optical properties ofan LC cell can be controlled by the application of an externalelectric field. This is the electro-mechanical-optic effect.

The effect of the electric field on the LC alignment is throughthe dielectric term. Recall from elementary electromagneticsthat the electrostatic energy in an isotropic medium is given by

U = 2

2

1Eε

where ε is the dielectric constant. ε of a material is highlydependent on the polarizability of the molecules. Polarmolecules are more polarizable. For example, ε of H2O is 80.glass (silica) has a dielectric constant of 3.8.

For an anisotropic medium, the electrostatic energy is betterwritten as

u = ED •2

1

30

Since there is cylindrical symmetry, we can assume that theLC lies on the x-z place as shown:

θ

x’

z’

x

z

In the principal axes (x’,z’) of the LC molecule

=

⊥

'

'

0

0

'

'

// z

x

z

x

E

E

D

D

εε

In the laboratory frame (x, z), this relationship is transformedto

=

z

x

zzzx

xzxx

z

x

E

E

D

D

)()(

)()(

θεθεθεθε

The laboratory frame and the molecular frame are related by acoordinate rotation by θ. It is easy to show that

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εxx(θ) = εzz(θ) = ε⊥ sin2 θ + ε// cos2 θ

εxz(θ) = ∆ε sin θ cos θ

εzx(θ) = -εxz(θ)

It is also common to write εzz(θ) as

)(sin)( 2// zzz θεεθε ∆−=

Notice that because 0=×∇ E

Therefore 0=∂

∂=∂

∂z

E

z

E xy

Applying the boundary conditions yields Ex = Ey = 0.

Therefore, only the z-component of E is nonzero.

E = Ez(z) z

However, Dx does not have to be zero because

Dx = εxz(θ) Ez(z)

and Dz = εzz(θ) Ez(z)

Additionally, 0=•∇ D

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implies that 0=∂

∂z

Dz

or Dz(z) = D = constant

Therefore from above

)()(

θεzzz

DzE =

and u = )(2

1 2

θε zz

D

Therefore the total dielectric energy stored in the LC due tothe applied voltage is given by

FE ∫∫∫∫ ∆−==

d

z

dzADudV

02

//

2

)(sin2 θεε

The physics is that if an external voltage is applied, θ(z)should be such that FE is minimized.

Let us consider the simple case that there is no elastic energy.The effect of minimizing FE is easy to see.

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(1) If ∆ε > 0, then FE is minimum if θ(z) = 0o for all z. Thiscorresponds to the homeotropic alignment. Therefore theexternal electric field tends to move the LC moleculesparallel to the field for positive anisotropic LC. For suchcases, it is better to have the LC homogeneously alignedin the V=0 condition.

(2) If ∆ε < 0, then FE is minimum if θ(z) = 90o for all z. Thiscorresponds to the homogeneous alignment. Thereforethe external electric field tends to move the LC moleculesperpendicular to the field for positive anisotropic LC. Forsuch cases, it is better to have the LC homeotropicallyaligned in the V=0 condition

Therefore the dielectric force tends to rotate the LC moleculesdepending on the sign of ∆ε. This force interacts with theelastic mechanical force of LC alignment to produce theelectro-mechanico-optic effect. Notice that the dielectric forceis proportional tot he square of D, and does not depend onthe sign of V.

6.4 Euler-Lagrange equations

The alignment of the LC cell in the presence of an electric fieldis obtained by minimizing the total F.

F = 2

1 K11(∇⋅n)2 + 2

1 K22(n⋅∇xn)2 + 2

1 K33nx∇xn2

+ )(2

1 2

θε zz

D

The calculus of variation will still lead to the Euler-Lagrangeequations:

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∂

∂=∂∂

•θθF

dz

dF

∂

∂=∂∂

•φφF

dz

dF

The E-L equations are:

0)(

12 23..

1 222

1 =

−−−+

•••

θεθφ

θφθθθθ

zzd

dD

d

dk

d

dk

d

dkk

tconskk tan2 32 =+•φ

Commercial softwares are available to solve the Euler-Lagrange equations numerically to give θ(z) and φ(z) at anyvoltage. We shall look at some simple cases here and derivesome general features of the E-L equation.

The applied voltage is related to D and Ez(z) by

∫∫ −=•−=d

zz

d

z

dzDdzzEV

00))((

)(θε

For homogeneous LC cell, EdDd

V ==⊥ε

, since D = ε⊥Ε

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For homeotropic LC cell, EdDd

V ==//ε

, since D = ε//Ε

There are a few constants of integration which are useful insolving the E-L equations:

=−•∂

∂F

F

θθ constant

There are powerful softwares to solve the E-L equations. The1D versions give

θ = θ(z,V) and φ = φ(z,V)

while the 3D versions give

θ = θ(x,y,z,V) and φ = φ(x,y,z,V)

There are several ways to visualize the dependence of thedirector orientation on V: (1) Complete data plot, (2) molecularplot, and (3) a plot of just the midplane tilt angle. They areshown below:

Example: LC alignment of a 90o TN cell at various voltages.(Both θ(z), φ(z) plots)

Alpha = tilt angle beta = twist angle

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Molecular plots:

37

38

6.5 Threshold voltage and Frederick transition:

As the voltage is increased from 0 volts, the LC alignmentdoes not always change immediately. There is a finite voltagewhen the alignment starts to change. This is called thethreshold voltage.

We can qualitiatively derive the threshold voltage for the caseof H-cell with positive anisotropy. First, φ = 0 for a ll z and V.Using the one constant approximation, the E-L equation for θbecomes

∆−=

θεεθθ

2//

2

2

2

sin

1

2 d

dD

dz

dK

This can be simplified for small ∆ε << ε// to

θθεε

θcossin

2//

2

2

2∆= D

dz

dK

Nowθ

θ θθ d

d

dz

d•

•=

2

2

So we have a first order differential equation in θ•

with a solution given by

39

θθεε

θ 22 coscos2//

−∆=

•m

KD

where we have applied the boundary condition thatθ(d/2) = θm. 90o - θm is called the midplane tilt angle.

Hence, the solution for θ(z) is given by the integral:

−∆=∫

− 2coscos //22

dz

K

Dd

m m

εεθθ

θθ

θ

In particular, at z = d, we have the equation for θm

∆=∫

− 2coscos //

2/

22

d

K

Dd

m m

εεθθ

θπ

θ

This equation can be used to solve for θm for any appliedvoltage. If D = 0, θm = 90o which is always true. If D is large,then θm should be near 0. As D decreases, θm shouldapproach 90o. It can be shown that

2coscoslim

2/

2290

π

θθ

θπ

θθ=∫

−→ m mm

d

Hence the behavior of θm is schematically like

40

D

θm

90o

Now for a H-cell, ⊥

=εDd

V

Hence the threshold voltage is given by

εεεπ

∆=

⊥

KVth

//

Above the threshold voltage, the LC director become distortedmore and more and eventually becomes homeotropic. Butthere is always some residual anisotropy.

This is the Frederick transition voltage. In general, for otherdeformations,

εεεπ

∆=

⊥ii

thK

V //

where Kii = K11 for B-cellKii = K22 for T-cellKii = K33 for S-cell

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or chiral nematic LCD, the threshold voltage is given by

επ

∆= 222 K

p

dVth

Interesting fact The HAN cell has no threshold.

e.g. K11 = 12.6x10-12N, ε// =108x10-12F/m, ε⊥=42x10-12F/m,Vth=3.53 volts.

Typical threshold for TN LCD is 1-1.5V. Usually it is listed onthe catalogs, together with other LC parameters.

Example of the mid-plane tilt angle dependence on voltage fora TN cell.

0

30

60

90

0.00 1.00 2.00 3.00 4.00 5.00

Voltage

Mid

plan

e til

t ang

le

42

Notice that the midplane tilt has a lower threshold than thecapacitance dependence on voltage.

Also, the LC cell is just like a capacitor. It is possible tomeasure or calculate its capacitance, as a function of voltage.

At low voltage, d

AC ⊥= ε

At high voltage, d

AC //ε=

Thus one can simply measure the capacitance as a functionof V to get an idea of the tilt angle change, and also measurethe threshold voltage. For example:

0.00E+00

1.00E-05

2.00E-05

3.00E-05

0.00 1.00 2.00 3.00 4.00 5.00

Voltage

Cap

acita

nce

43

6.6 Response times:

Another LC property that is closely related to the elasticconstants is the viscosity ν, η.

η = ρ ν

where η is the dynamic viscosity, ρ is the density and ν is thekinematic viscosity. LC density is typically 0.98 – 1.02 gm/cm3,same as water.

For high speed operation, need small viscosity. General rule

More polar molecules => larger νLonger Y, Z groups => larger νMore benzene rings => larger νMore side chains => larger νLow temperature => large ν

Also notice that large ∆n requires more polar molecules. Solarge birefringence and higher viscosity goes hand in hand.

Typical ν = 5-100 mm2/s at room temp. It can be larger than5000 at low temp.

The dynamics of LC molecule realignment due to externalfields involves solving the Erickson-Leslie equations. Theresults for the H-cell and low twist TN cells are

22

2

πεητ

iion

KV

d

−∆=

and2

2

πητii

offK

d=

44

The total response time is defined as

τ = τon + τoff

For high twist cells and cholesteric cells, the formulas aremodified to:

2

2

2

2

p

K

d

V iion

πεητ

−∆

=

and2

2

πητii

offK

p=

where p is the pitch of the twist.

Notice that the response time is proportional to d2. Thereforevery thin cells are needed for fast response time, e.g. for videoapplications. Also, strictly speaking, ν and η are also tensorsdependent on the flow directions.

Example: ν = 100 mm2/s, ρ = 1 gm/cm3, K11 = 12.6x10-12 N,d = 6 µm. Substitute into formula, get τd = 30 ms. This is atypical number for TN.

Typical response times for LCD:STN 100-200 msTN 10-50 msH-cell 2-5 msB-cell 0.5 msFerroelectric 20 µs

The response time of ferroelectrics are very fast because it isbased on a dipole force rather than a dielectric energy force,i.e. the force is first order in E rather than E2.