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[L2.5 Slide 1] Well now that we're comfortable

with some key concepts of quantum mechanics,

we can talk about a really important topic

for semiconductors.

The topic of this lecture is density of states.

[Slide 2] So let me remind you that we have these valance bands

that is mostly filled.

There's a top of the valence band

and a bottom of the valance band and that is so far down

that we don't have to worry about it.

We have a conduction band that consists of a set of states

that are mostly empty.

There's a bottom of the conduction band

there's a top of the conduction band

that we generally don't need to worry about

because it's so high in energy.

It's going to be important for us to know now

how the states are distributed within these bands.

They're not uniformly distributed within energy.

Now, we know that most of the empty states

in the valence bands,

the holes are near the top of the valance band.

Most of the electrons in the conduction band

are near the bottom of the conduction band.

So we only really need to be worried

about how the states are distributed in energy

near the edges of the band.

That's what our focus will be in this lecture.

[Slide 3] The key quantity that we're going to be discussing

is called the density of states

and it's simply a way to count the number of states

between a certain energy and an energy plus dE.

The number of states in that range is the density of states

times the width of that energy range.

So this density of states

is the quantity that we will see in detail,

we will make use of in detain in the next unit.

We want to learn how to appreciate what it is

and how to calculate it for semiconductors in this lecture.

We're going to take a simple description of energy

versus crystal momentum and these simple descriptions

work near the top of the valence band

and near the bottom of the conduction band,

and that's going to be adequate for most semiconductor work.

[Slide 4] So let's begin by thinking about a large chunk

of semiconductor crystal with a volume omega

with the length Lx in the x direction,

Ly in the y direction, Lz in the z direction.

So this is, let's think of this as a large

but finite volume that we have carved out

of an infinite bulk volume.

Well, there's going to be a finite number of states

because the states come from atoms

and there are a finite number of atoms in this volume.

We want to count the states.

There are different ways of doing this.

They end up all giving you the same answer.

I'm going to take one particular way

that makes a lot of physical sense.

Since this is a piece of a volume from an infinite solid

I'm going to assume that at this end of the solid

and at the other end of the solid

the wave function is the same.

We'll assume periodic boundary conditions

so that everything just repeats itself

throughout all of space in the x direction,

in the y direction, and in the z direction.

[Slide 5] So let's just look in the x direction.

This is x equals 0.

This is x equals Lx, which is, we're going to measure things

in units of the lattice spacing

so capital N is the number of atoms

and a is the lattice spacing.

Well, we know what the solutions to the wave equation are

in this region.

They're these bloch functions.

So periodic part times a wave solution.

So at x equals 0 we have to have the same wave function

at x equals L and we know that the periodic part will be the same

because that repeats itself at every lattice spacing.

So this condition imposing the periodic boundary conditions

leads us to conclude that e to the ikx times Lx

must equal 1

in order to satisfy these periodic boundary conditions.

Well, we know that some integer number of 2 pis,

either the i 2 pi or some integer number of 2 pis

will give us 1.

So, we have to require that kx times Lx

equals an integer number of two pis.

So the integer is one, two, three,

it just keeps repeating itself.

Solving for kx we see that kx is two pi over the length

that we're doing all of our counting of states in

times this integer j.

If we plot that we'll get a set of states.

So whenever we apply boundary conditions to a wave function

we get a set of discrete states.

We now have a set of discrete k states.

The difference is that we're thinking about this,

it's not a particle in a box problem.

This is a very large volume.

These are very finely spaced states.

Now, we're going to ask the question.

In some range dk, how many states are there?

Well, we know that each of the states

are spaced by 2 pi over Lx.

So the number of states in that region

is the width of the region dk

divided by the spacing between states,

then we'll multiply by 2.

Because each state can hold an electron with spin up

or with spin down and this quantity we will say

is the density of states in k-space

times the width of the region in k-space.

So we conclude that the density of states in k-space

is the length that we're doing the counting in

divided by pi.

This is in 1 dimension, okay?

And remember, we always have to remind ourselves

to account for the factor of two for spin

for our electrons.

Now, we can also look at this result a little more closely.

The length is some integer number of lattice constants.

That means I can insert N times a here for Lx

and then I have kx is two pi over a.

The integer j divided by the number of atoms

in this region N, capital N.

Now, as I let j get bigger and bigger,

eventually it will hit N.

When it hits N then the phase of this exponential

has gone from zero to 2 pi.

If I continue to let j increase,

I'm just going around the circle again

and getting another set of results

that are completely equivalent.

So, the unique solution to this equation

are between k equals 0 and k equals 2 pi over a.

That means the maximum k that can occur is 2 pi over a.

This is similar to the Brillouin zone

that we talked about in the previous lecture.

There is a region in k-space of which 2 pi over a

in which the unique solutions are found.

We can shift this region.

Sometimes it's usually convenient to shift it

so that we go from minus pi over a to plus pi over a,

but the width of that region,

over which we have a unique set of answers,

is 2 pi over a.

This is the Brillouin zone.

[Slide 6] Okay, alright so we've gone through the counting in 1D

and we've seen that the density of states in k-space

is two for spin, L over 2 pi.

So the result is L over pi.

If we were to do this in 2D, if we had a 2D material,

a quantum well that we've made

with semiconductor heterostructure or whatever,

the density of states would be two for spin

but now they're spaced L over 2 pi in the x direction

and L over 2 pi in the y direction.

So we get L squared, which is the area,

over 2 pi squared, which is 4 pi squared.

So when we reduce this we get A over 2 pi squared.

In 3 dimensions we have the factor of two for spin.

Now we have states spaced in x, y, and z.

Lx times Ly times Lz is the volume of this material

that we're doing the counting of our states in.

This is 2 pi cubed.

So we get omega over 4 pi cubed.

In 1D the width of the region in k-space is dk.

In 2D it's dkx, dky.

In 3D it's dkx, dky, dkz.

It's a volume in k-space.

Alright, now the important thing to point out

is this applies to any type of band structure,

any type of E of k, it's independent of E of k.

The states are, the density of states in k-space

is the same no matter what the material is.

[Slide 7] Now we will more commonly ask ourselves

a different question.

We will ask ourselves the question

how are the states distributed in energy?

Then it gets a little bit different.

So let's assume we have a parabolic band,

energy is h-bar squared k squared over 2m.

This would be the bottom of the conduction band here.

We've seen that the states are spaced uniformly in k.

At any particular k I have a corresponding energy.

If you simply map these states

to their corresponding energies you can see now

that they are not distributed uniformly in energy.

How they are distributed will depend

on the shape of the E versus k curve.

So what we're going to want to do now

is to figure out how these states are distributed in energy

and find a density of states in energy space.

What we will do is to count the states in k-space,

map them into a corresponding region in energy,

and we will relate

the number of states in this region in k space,

the same number of states have to occur in this region

in energy space, and that will allow us

to define a density of states in energy.

I'm going to write this as D prime of E.

dE is the number of states in that region,

and that's in this region right here.

And that's equal to N(k) dk, the density of states in k-space

times the width of the region in k-space.

That's the number of states in this region of k-space

and they are the same.

So by using this relation we will be able to deduce

from the known density of states in k-space,

the density of states energy space.

So let's do that.

[Slide 8] So first example

we will do it for a 1 dimensional nanowire.

These used to be academic exercises years ago.

Today they're real exercises because we can produce

semiconducting nanowires

and we often need to know what the density of states is.

We'll assume just a single subband.

And recall that there are different subbands.

If we assume a single subband with an energy E sub n,

then the kinetic energy

is associated with the motion in the x direction,

it's h-bar squared kx squared

over 2 times the effective mass.

[Slide 9] Alright, so here's our problem.

Here's our k in one dimension, here's our region dk.

We're going to map the states in that region

because there's corresponding regions in energy space.

Well, we know that the number of states in this region

in k-space, we've worked that out.

It's N of k dk,

is L over pi times the width of that region in k-space.

That has to be the same total number of states

in the energy, in the corresponding energy range.

That's D prime of E dE.

Now, I'm going to drop the prime.

Typically when we talk about density of states

we are interested in the density of states per unit length.

We're not interested in the actual number of states

because it's sort of arbitrary.

If I take double the length

I'll get double the number of states.

So, D, 1D of E,

the density of states that we're going to compute,

is the density of states per unit length.

So I have to divide by L.

The units then are going to be the number of states,

not really a unit, per unit energy, per unit meter

for the length, the units of density of states.

[Slide 10] Alright, so now we are prepared

to calculate the 1D density of states.

This is the number of states in energy space.

This is the number of states in the corresponding k-space.

We have our relation

between the number of states in k-space.

We have our dispersion.

We can put it all together.

This comes from the density of states in k-space.

We can now use the dispersion to compute dE

from this relation.

We can also compute k from the dispersion,

and we can then compute dE in terms of energy,

plug it all into this relation,

and we find the density of states in energy

is one over pi h-bar square root of effective mass

over 2 times the energy

with respect to the bottom of the conduction band.

This applies of course only when the energy

is greater than the bottom of the subband

because there are no states below that.

So this is the density of states

and we can pick the first subband, second subband,

third subband, or whatever.

[Slide 11] Now, I should just point out

we've done the counting over here for plus kx

and we found the number of states

that are in that corresponding energy range.

The band structure is symmetric.

There's an equal number of states over here at minus kx

that map into the same energy range.

So I need to multiply by 2.

And this is my final answer.

This is the density of states, the number of states

between E and E plus dE as a function of energy,

and we've derived the result that we were after.

Remember, we've assumed parabolic bands.

So the answer could be different

if we had a much different band structure.

[Slide 12] So in general we'll have multiple subbands,

we'll have a density of states for each one of those.

Note that the density of states goes to infinity

at the bottom of the subband when E is equal to E sub n.

And we'll have a series of those.

If this is the first quantized level

of the particle-in-a-box type solution,

we'll have a density of states associated with electrons

that are in that subband.

If this is the bottom of the second subband quantized state

then we'll have a density of states

associated with that as well, and on and on.

[Slide 13] Alright, that's 1D.

Let's see if we can do it for 2 dimensional electrons.

We can produce 2 dimensional electrons

by artificial semiconductor structures.

There are also 2 dimensional materials these days

that people are quite interested in.

So, consider a thin semiconducting slab of thickness t.

The electrons are confined in the z direction.

So I'll have particle-in-a-box subbands

associated with that.

The electrons are free to move in the x-y plane

and we would like to count the states

within this area A.

We're going to assume again that the bands are parabolic

so the energy is the bottom of the subband

plus h-bar squared times k in the x-y plane squared

divided by 2 times the effective mass.

[Slide 14] So let's have a look at that.

We're going to approach it the same way.

We'll take the density of states

times the width and energy range.

We'll map that to the corresponding number of states

in the k-space range.

And the way we'll do this is we will assume

a 2 dimensional k-space.

So there is a parabolic band.

So energy is h-bar squared k squared over 2 m,

that's this circle, and we'd like to know

how many states are there between k and k plus dk.

The red ring around here is width dk.

The area of that region

that we're asking to count the states in

is the circumference of the circle, 2 pi k

times the thickness of that little ring, which is dk,

and we're asking how many states are there

in that red area.

Well, each state has an area of 4 pi squared over A,

as we've seen

when we apply the periodic boundary conditions.

And then we multiply by two for spin, okay.

So, when we do this calculation

we have A over 2 pi squared.

We're dividing by A because we're going to compute

the density of states per unit area.

And then we have the volume of k-space, 2 pi dk.

We use our dispersion to compute 2 pi k dK

in terms of energy.

We do a little bit of math and we end up

with this simple expression

for the 2 dimensional density of states.

Note that the 2 dimensional density of states

is proportional to the mass divided by pi h-bar squared.

Of course these states only exist

if we're above the bottom of the subband.

[Slide 15] Okay, so the density of states in 2D then

is quite different from the density of states in 1D.

Let's do 3D, and most of the course

we will be concerned with 3 dimensional semiconductors.

So, we will assume a simple dispersion.

Parabolic bands, the constant energy surfaces are spheres.

We will equate the number of states in the energy range

to the number of states in the corresponding k range.

We now have a volume of k-space, d cubed k.

We have computed the density of states in k-space,

so we know that.

We will equate that, the volume, the volume of k-space

that we're going to do the counting in then

is the surface area of the sphere, 4 pi k squared

times the thickness of a ring around that sphere dk.

So that's the volume of k-space

that we are counting the states in.

Well, now we'll make use of the dispersion

and we will express k squared dk in terms of energy.

When we do that algebra we end up finding

that the density of states is proportional

to the square root of the effective mass

times the effective mass

and it varies as the square root of energy.

So we've calculated the density of states now

for 1 dimensional electrons, 2 dimensional electrons,

and 3 dimensional electrons all under the assumption

that the bands are parabolic.

[Slide 16] Now, things can get a little more complicated.

I'll just mention this;

although, we won't go into the calculation.

Recall that the band structure of silicon

is the bands are parabolic

but they have different curvatures in different directions

and there are six bands at the same energy.

So the constant energy surfaces,

the minima of the conduction bands are located

along a [1 0 0] direction, or these equivalent

[0 1 0] direction, [0 0 1] direction.

There are six of these valleys along these six directions

and they have different effective masses

in the x and the y direction.

So that band structure is a little more complicated.

We've been assuming a simple spherical band

centered at k equals zero.

We could compute the density of states for silicon as well.

It would be a little more complicated calculation

but it's something that you go through

in slightly more advanced courses.

I'll simply give the answer.

You can wrap everything together

and you can make it look like our simple result.

If we express, if we define a quantity

that we will call the density of states effective mass,

this is not a real effective mass.

The real effective masses are related to the curvature

of E versus k.

But this is a definition such that if we use it

in our simple spherical band approximation,

we'll get the correct answers

and the correct densities of states.

So, when we are computing the density of states

in a semiconductor, we have to be careful in this formula

to use the appropriate density of states effective mass.

If it's a simple spherical semiconductor

centered at k equals zero, it's just the effective mass

associated with the curvature of the band.

If it's a more complicated material like silicon

it is a more complicated quantity.

So you should remember that.

[Slide 17] Alright, so some things that we should take away.

This is the expression for the density of states.

It's worth remembering that the density of states varies

as the three halves power of the effective mass.

So if I'm dealing with a semiconductor

that has a light effective mass,

I know it's going to have a small density of states.

It's also worth understanding that it varies

as the square root of energy.

So as I increase in energy I will get more and more states.

So, for example if we look at silicon

it has a much higher density of states effective mass

than gallium arsenide.

We would expect then the density of states of silicon

to be significantly higher than gallium arsenide.

[Slide 18] So just to summarize,

we've derived the density of states in 1D, 2D, and 3D

assuming parabolic bands.

If we plot them we can see that the density of states in 1D

goes as one over the square root of energy

when the electrons get down near the bottom of the subband

the density of states goes to infinity.

In 2D there are states above the bottom of the first subband

and the density of states is independent of energy

above the bottom of that subband.

In 3D the density of states is zero

at the bottom of the conduction band

and then it increases as the square root of energy.

So, these are quantities

that we will make significant use of

in the next unit of this course.

[Slide 19] As an exercise, if you'd like to test yourself

and see if you understand the basic concepts here

compute the density of states for graphene.

Graphene does not have a parabolic band structure.

It is a 2D material but it will have a density of states

that is significantly different

than the 2D density of states for parabolic bands

that we just worked out.

The answer will be something like this.

The density of states will increase linearly with energy

as we go above E equals zero

and as we go below E equals zero.

So you might see if you can derive that result.

If you can do that then you understand the process

of deriving a density of states for any band structure.

[Slide 20] Alright, we've covered a lot of ground

but we've introduced this important quantity

that we're going to need to use again and again

throughout the course.

They quantity is the density of states

and it's one that we're going to frequently use.

The important points to remember

are that the density of states

depends on the dimension of the semiconductor.

We have different density of states in 1D, 2D, and 3D.

The density of states in k-space is constant

and does not depend on band structure

but the density of states in energy space

does depend on band structure and can be quite different

for different band structures.

For 3 dimensional bulk semiconductors

with parabolic bands it's worth remembering

that the density of states is proportional

to the three-halves power of effective mass

and to the square root of the energy

above the bottom of the conduction band.

So, these are some important concepts.

The quantum mechanical background that we've had

helps us understand where these come from.

We will make significant use of them,

especially the 3D density of states

in the next unit. Thank you.

For more infomation >> ECE Purdue Semiconductor Fundamentals L2.5: Quantum Mechanics - Density of States - Duration: 24:19.

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Poll: Big Majority Says United States On The Wrong Track Under Donald Trump | The 11th Hour | MSNBC - Duration: 3:48.

For more infomation >> Poll: Big Majority Says United States On The Wrong Track Under Donald Trump | The 11th Hour | MSNBC - Duration: 3:48.

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United States files criminal charges against Chinas Huawei - Duration: 2:45.

United States files criminal charges against Chinas Huawei

The Justice Department on Monday unveiled two sets of charges against Huawei. One set accuses Huawei of trying to steal trade secrets from T Mobile . Another claims the company worked to skirt US sanctions on Iran.

The agency also revealed formal charges against Meng Wanzhou, Huaweis chief financial officer. Meng was arrested in Canada in December, and the United States is looking to extradite her.

"Today we are announcing that we are bringing criminal charges against telecommunications giant Huawei and its associates for nearly two dozen alleged crimes," Acting Attorney General Matthew Whitaker said in a statement. "China must hold its citizens and Chinese companies accountable for complying with the law."

FBI Director Christopher Wray said at a press conference that Huawei "relied on dishonest business practices that contradict the economic principles that have allowed American companies and the United States to thrive."

"The prosperity that drives our economic security is inherently linked to our national security," Wray said. "And the immense influence that the Chinese government holds over Chinese corporations like Huawei represents a threat to both."

A 10 count indictment, which was filed in Washington state, claims that Huawei worked for years to steal T Mobiles proprietary phone testing technology, known as "Tappy."

Huawei also obstructed justice when the Washington based telecom company threatened to sue it, according to court documents.

The government also unsealed a 13 count indictment against Huawei and Meng in Brooklyn federal court. That document details an alleged scheme by Huawei to deceive financial institutions and the US government about its business in Iran.

The company and Meng are said to have claimed that an affiliate, Skycom, was a separate company in order to conduct business in the country, when in fact it was a subsidiary.

The charges from the United States increase pressure on Huawei, one of Chinas tech champions and a crucial part of the countrys efforts to dominate the rollout of superfast 5G wireless networks.

The United States has said for years that Huawei devices pose a national security threat. But government efforts to rein in alleged abuses by the company have escalated in recent months.

Mengs arrest in December set off a geopolitical firestorm and has strained relations between the US, Canada and China. China has repeatedly called the detention of Meng, who was , a political move.

But the United States on Monday reiterated its intention to move forward with the extradition process. Meng is due back in court again in Canada on February 6.

Congress is also considering legislation that would ban the sale of US parts to any Chinese telecom firm that has violated US export control laws or sanctions, a measure that expressly targets Huawei.

"It has been clear for some time that Huawei poses a threat to our national security, and I applaud the Trump administration for taking steps to finally hold the company accountable," Senator Mark Warner, a Democrat, said in a statement Monday.

Importantly, the charges come as the United States and China are racing to cut a deal on trade before March 1, when US tariffs on dollar 200 billion in Chinese goods will otherwise rise to 25 percent from 10 percent . Vice Premier Liu He is scheduled to travel to the United States for two days of talks starting Wednesday.

Its not clear how Mondays announcement will factor into negotiations. President Donald Trump previously suggested that he may intervene in the Meng case if it would help reach a trade deal with China.

For more infomation >> United States files criminal charges against Chinas Huawei - Duration: 2:45.

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Use Custom Review States in the Data Viewer - Duration: 2:13.

In this tutorial you will see how to use custom review states in the Data Viewer.

Data Viewer can be configured with custom data review workflows.

For example a medical monitors can be alerted when critical medical data is entered and ready for review.

Similarly, monitors can be alerted to data that is ready for Source Document Review.

Navigate to the Review page.

The default columns in the Data Viewer are 'Forms with queries', 'Form State', and 'Source Verify'.

During study setup up to 5 additional review workflows can be configured by the Study Designer.

Each review workflow contains 3 review states.

The custom review states appear as additional columns.

In this example, you can see two forms in the AE/CM visit in the Monitor Review column, under All Forms.

Select the cell showing these forms.

And click the Go to RecordListings button.

Note that filters are set to values based on the cell you selected.

In this case, All Sites, AE/CM Visit, and Monitor Review: All Forms.

The forms can be moved to the Reviewed custom review state by selecting the review state

and providing a comment.

Select the Medication Number cell of the first row of the form.

In the Review drop-down list, select Monitor Review: Reviewed.

Type any comment and click Continue.

Click the Review button to see the updated list.

You can see that one form has moved from All Forms to Reviewed.

A review state does not change if the data is edited after a form has been marked as reviewed.

However the filter 'Updated Since review' can be used to highlight forms that have been

edited or entered since a review was last completed.

To learn more, see the documentation on the Oracle Help Center.

For more infomation >> Use Custom Review States in the Data Viewer - Duration: 2:13.

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Four States Dental Care - Lumineers / Fast Braces (012919) - Duration: 2:08.

For more infomation >> Four States Dental Care - Lumineers / Fast Braces (012919) - Duration: 2:08.

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Now Trump praises states for requiring schools to offer 'Bible literacy' classes - Duration: 4:47.

Donald Trump made a bid to re-energize evangelical supporters on Monday by stating his support for 'Bible literacy' classes in U

S.public schools.The move comes at a fraught political moment for the president, as the White House is stinging from a lost battle over his long-promised border wall – and some conservatives are doubting whether he can win the longer border war with Democrats

'Numerous states introducing Bible Literacy classes, giving students the option of studying the Bible

Starting to make a turn back? Great!' the president tweeted.'Fox & Friends' had aired a brief segment about the phenomenon a half-hour earlier, with Christian Broadcasting Network Chief Political Analyst David Brody delivering the Good News

The occasional church-going Trump had an unlikely relationship with Bible-belt Republicans during the 2016 campaign, outperforming Southern Baptists like Sens

Ted Cruz and Lindsey Graham and nondenominational evangelicals Rick Perry and Scott Walker

His nominal relationship with pews and pulpits showed, despite holding his thumbworn childhood Bible aloft during a September 2015 Values Voter Summit speech

'It brings back so many memories,' he said then of his Presbyterian upbringing.Four months later during an appearance at Liberty University, he botched the name of a New Testament book, drawing chuckles by calling it 'Two Corinthians

' 'Two Corinthians, 3:17, that's the whole ballgame, he said."Where the spirit of the Lord, right? Where the spirit of the Lord is, there is liberty

And here there is Liberty College, Liberty University.It's so true.' 'We are going to protect Christianity,' Trump declared that day, two weeks before the Iowa Caucuses

Evangelical leaders have long complained about the disapearance of God from public schools, a phenomenon that has fueled the growth of home-schooling

USA Today reported on Sunday that lawmakers in at least six states are pushing back, with bills that would nudge the study of both the Hebrew Bible and the Christian New Testament back into high schools

'The Bible is an integral part of our society and deserves a place in the classroom,' Republican state Rep

Aaron McWilliams of North Dakota told the paper.The American Civil Liberties Union, which opposes official mentions of religion in public institutions, has identified Bible literacy bills this year in Florida, Indiana, Missouri, North Dakota, Virginia and West Virginia

Legislatures defeated similar measures last year in Alabama, Iowa and West Virginia

Tennessee passed a weaker version, and Kentucky Gov.Matt Bevin signed a 'Bible studies' bill in 2018

That law gives schools guidelines for teaching the Bible as literature, but doesn't require it