{"id":98,"date":"2018-11-16T05:58:34","date_gmt":"2018-11-16T05:58:34","guid":{"rendered":"http:\/\/phy12.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=98"},"modified":"2018-11-16T09:38:32","modified_gmt":"2018-11-16T09:38:32","slug":"calculation-of-density-of-states-dos","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/phy12\/chapter\/calculation-of-density-of-states-dos\/","title":{"rendered":"Calculation of Density of States (DOS)"},"content":{"raw":"
\r\n\r\n \r\n\r\n3.3.1\u00a0\u00a0\u00a0\u00a0 DOS in 3D\r\n\r\n3.3.2\u00a0\u00a0\u00a0\u00a0 DOS in 2D\r\n\r\n3.3.3\u00a0\u00a0\u00a0\u00a0 DOS in 1D\r\n\r\n3.3.4\u00a0\u00a0\u00a0 DOS in 0D\r\n\r\n<\/div>\r\n
\r\n\r\n \r\n\r\nCALCULATION OF DENSITY OF STATES (DOS): Quantum Wells, Wires and Dots<\/strong>\r\n\r\n \r\n\r\n3.3.1 DOS for 3 dimensions (Bulk)<\/strong>\r\n\r\n \r\n\r\nConsider the volume in \"k\" space<\/span>\r\n\r\n\"\"\r\n\r\nwhere for a particle in this space\r\n\r\n<\/div>\r\n
\"\"<\/div>\r\n
\r\n\r\n\"\"\r\n

Figure . Electron state is defined by a point in k-space.<\/p>\r\n \r\n\r\n \r\n

Note that the 2 \u00a0arises from the constraints of periodic boundary conditions as proposed to the more general where n<\/em>=0, 1, 2, 3... The volume of a given mode is then= . The number of modes (N<\/em>) in the sphere is,<\/p>\r\n\"\"\r\n\r\nSay the particle in an electron and we consider spin (up and down), then we multiply N<\/em> by 2.\r\n\r\n\"\"\r\n\r\nis the total number of states in sphere. Now consider the density\r\n\r\n\"\"\r\n\r\n \r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n

Figure . Density of states in 3 dimension (Eq.3.48)<\/p>\r\n \r\n\r\n \r\n\r\n3.3.2 DOS in Two Dimensions (well)<\/strong>\r\n\r\n \r\n

Here we have 1D that is quantized. Let's us assume it is the z-direction. The total energy of this system is a sum of the energy along the quantized direction plus the energy along the other 2 free directions. It is expressed as<\/p>\r\n\"\"\r\n\r\n \r\n\r\nwhere for the particle\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\nThe area of a given mode is then ???? with the total number of modes (N) in the area being\r\n\r\n\"\"\r\n\r\nAgain if the particle is an electron and we consider spin, multiply by 2 to get\r\n\r\n\"\"\r\n

This is the energy density of the sub-band for a given or (<\/span>E<\/em>n<\/em>). For each successive there will be an additional \u045b2 and hence another subband. Therefore the density of the states is written<\/span><\/p>\r\n\"\"\r\n\r\nWhere \u0472 is the heavy side function.<\/span>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n

Figure Density of states in 2 dimension. Shaded area presents occupied states.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n3.3.3 DOS in One Dimensions (Wire)<\/strong>\r\n\r\n \r\n\r\nConsider now the situation where there are two dimensions confined and only 1 degree of freedom (say the x-direction). The total energy of the system can be written as\r\n\r\n\"\"\"\"\"\"\"\"\r\n\r\n<\/div>\r\n
\r\n

Figure Density of states in 1 dimension. Shaded area presents occupied states.<\/p>\r\n \r\n\r\n \r\n\r\n3.3.4 Zero dimensions (Quantum Dot)<\/strong>\r\n\r\n \r\n

Here since all three dimensions are confined. The density of states is basically a series of delta functions. The total energy of the system is<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\"\"\r\n\r\n \r\n\r\n3.3.5 More density of states<\/strong>\r\n\r\n \r\n\r\nDensity of states in the conduction band<\/strong>\r\n\r\n \r\n

For this we need to know the probability that an electron will occupy a given stats of energy E. The Probability, P(E), is referred as the Fermi Dirac distribution. In addition we need to know the density of states ( \u2032). The density of states has units of number of unit volume per unit energy. Therefore \u2032 is the number of states per unit volume. The number of occupied states at a given energy per unit volume is therefore<\/p>\r\n\"\"\r\n\r\nHere\u00a0 \u00a0is the Fermi energy.\r\n\r\n\"\"\r\n\r\nthe total concentration of electrons in the conduction band is therefore the integral over all available energies\r\n\r\n\"\"\r\n\r\nwhere?c<\/sub> is the energy where conduction band starts. For the case of three dimensional material\r\n\r\n\"\"\r\n\r\nTaking account into conduction band begins, the density of states can be written as\r\n\r\n \r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\nThe integral is called the Fermi integral or Fermi Dirac integral.\r\n\r\n \r\n\r\nConsider the case where, E\u00a0 \u2212 Er\u00a0<\/sub>\u226b KTand the Fermi Dirac distribution function becomes\r\n\r\n<\/div>\r\n
\"\"<\/div>\r\n
\r\n\r\n \r\n\r\nThis is the expression for the effective density of states of the conduction band.<\/em><\/strong>\r\n\r\n \r\n\r\nDensity of states in the valance band<\/strong>\r\n\r\n \r\n\r\nThe number of holes at a given energy per unit volume is given as\r\n\r\n<\/div>\r\n
<\/div>\r\n
\r\n\r\n\"\"\r\n

Where Ev is the energy where valance band starts. The total concentration of holes in the valance band is the integral over all energies.<\/p>\r\n\"\"\r\n\r\n \r\n

This is the effective density of the states in the valance band.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nSummary<\/strong>\r\n\r\n \r\n

Fermi level of an intrinsic semiconductor\u00a0 If the bulk semiconductor is intrinsic, there has been no doping of the material and hence no extra electrons or holes anywhere. in this situation<\/p>\r\n\"\"\r\n

One can therefore see that at T=0 the Fermi energy of an intrinsic semiconductor is at the halfway point between the top of the valance band and the bottom of the conduction band.<\/em><\/strong><\/p>\r\n \r\n\r\nDensity of states in the conduction band<\/strong>\r\n\r\n \r\n\r\nWe start with the Fermi Dirac distribution for electrons and also the density of states\r\n\r\n\"\"Consider only one of the subband. In this case the density of states simplifies to\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\nNow recall from the previous section that the number of states at a given energy per unit volume<\/span>\"\"\r\n

the total concentration of electrons in this first subband is the integral over all available energies. Rather than use ntot as before let's just stick to n<\/em>c<\/em> from the start<\/p>\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\nSince the band really begins at en as opposed to Ec<\/em> like in the bulk the integral change from\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\n \r\n\r\nDensity of states in the valance band<\/strong>\r\n\r\n \r\n\r\nAs with the conduction band case we need the probability of occupying a given state in the valance band. This denoted \u210e(\u00a0 ) and is evaluated from<\/span>\r\n\r\n<\/div>\r\n
<\/div>\r\n
\r\n\r\n\"\"\"\"\r\n\r\nFermi level position :2D<\/strong>\r\n\r\n \r\n\r\nThe procedure for finding the Fermi level position is the same as in the 3D Consider a spherical volume of\r\n\r\n \r\n\r\n\"\"\r\n\r\nNow the density is\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\nis the number of states per unit volume and the energy density is given as\r\n\r\n\"\"\r\n\r\nDivide by 2 to go back to only 1 spin orientation since in an optical transition spin slips are generally forbidden<\/span>\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\nThe expression applies to either conduction band or valance band. Applying the following equivalence (\u00a0 )\r\n\r\n\"\"\r\n

where is the desired joint density of the states. Now from the conservation of momentum, transition in k are vertical such that the initial k<\/em> value in the valance band is the same k value as in the conduction band (ka=kb=k) where ka is the k value in the valence band and kb is the value in the conduction band. The energy of the initial state in the valance band is<\/p>\r\n\"\"\r\n\r\n \r\n\r\nLikewise the energy of the final state in the conduction band is\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\nWhere for notational simplicity we have used the reduced mass\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\"\"\r\n\r\n2D Well<\/strong>\r\n\r\n \r\n\r\nArea in k-space\r\n\r\n= ??<\/sub> = 4??\r\n\r\n \r\n\r\nWhere the area occupied by a given mode or state is .Here we assume that represents the confined direction\r\n\r\n\"\"\r\n\r\nTogether, the number of modes in the area is\r\n\r\n\"\"\r\n\r\n<\/div>\r\n
Multiply by 2 to account for spin<\/span><\/div>\r\n
\r\n\r\n\"\"\r\n\r\nNow consider the density\r\n\r\n\"\"With the energy density given by<\/span>\"\"\r\n\r\nStarting with the energy density\r\n\r\n\"\"\r\n\r\n \r\n\r\nDivide by 2 to get rid of the spin since formally speaking, spin flip optical transitions are forbidden\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n

wheree??(?) is the desired joint density of states. As before in the 3D case, the conservation of momentum means that transition in k-space are vertical. That is the initial k value in the valance band is the same as the final k value in the conduction band ( = = ) where (\u00a0 ) is the valance (conduction) band values.<\/p>\r\n \r\n\r\nThe energy of the initial state in the valance band is\r\n\r\n\"\"\r\n\r\nLikewise the energy of the final state in the conduction band is\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\nSuch that when replaced into our main expression the desired expression for the joint density of states is\r\n\r\n\"\"\r\n\r\n1D wire<\/strong>\r\n\r\nConsider the length in k-space\r\n\r\n \r\n\r\nLk<\/sub>=2k<\/sub>\r\n\r\n \r\n\r\nThe length occupied by a given mode or state is where\r\n\r\n\"\"\r\n\r\nThe number of states in the given length is<\/span>\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\nMultiply this by 2 to account for spin, we get total number of states as\r\n\r\n\"\"\r\n\r\nConsider the density ie number of states per unit length\r\n\r\n\"\"\r\n\r\nAnd the energy density is given by\r\n\r\n\"\"\r\n\r\nOr alternately<\/span>\r\n\r\n\"\"\r\n\r\nStarting with the energy density\r\n\r\n\"\"\r\n\r\nDivide by 2 to consider only one spin orientation since spin flip transition are generally forbidden<\/span>\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\nNow apply the following equivalence\r\n\r\n\"\"\r\n

wherepj<\/sub> (E) is the desired joint density of states. As before in the 3D and 2D case, the conservation of momentum means that transition in k-space are vertical so that Ka<\/sub>=Kb<\/sub>=K ) where ka<\/sub>(kb<\/sub>) is the valance (conduction) band values.<\/p>\r\n\"\"\r\n\r\nThe energy of the initial state in the valance band is\r\n\r\n\"\"\r\n\r\nLikewise the energy of the final state in the conduction band is<\/span>\r\n\r\n\"\"\r\n\r\nSuch that when replaced into our main expression the desired expression for the joint density of states is<\/span>\r\n\r\n<\/div>\r\n\"\"\r\n\r\nNow to continue towards our final expression we express k fully. Since\r\n\r\n\"\"\r\n\r\nThis leads to the final expression for the joint density of states\r\n\r\n\"\"","rendered":"

\n

 <\/p>\n

3.3.1\u00a0\u00a0\u00a0\u00a0 DOS in 3D<\/p>\n

3.3.2\u00a0\u00a0\u00a0\u00a0 DOS in 2D<\/p>\n

3.3.3\u00a0\u00a0\u00a0\u00a0 DOS in 1D<\/p>\n

3.3.4\u00a0\u00a0\u00a0 DOS in 0D<\/p>\n<\/div>\n

\n

 <\/p>\n

CALCULATION OF DENSITY OF STATES (DOS): Quantum Wells, Wires and Dots<\/strong><\/p>\n

 <\/p>\n

3.3.1 DOS for 3 dimensions (Bulk)<\/strong><\/p>\n

 <\/p>\n

Consider the volume in “k” space<\/span><\/p>\n

\"\"<\/p>\n

where for a particle in this space<\/p>\n<\/div>\n

\"\"<\/div>\n
\n

\"\"<\/p>\n

Figure . Electron state is defined by a point in k-space.<\/p>\n

 <\/p>\n

 <\/p>\n

Note that the 2 \u00a0arises from the constraints of periodic boundary conditions as proposed to the more general where n<\/em>=0, 1, 2, 3… The volume of a given mode is then= . The number of modes (N<\/em>) in the sphere is,<\/p>\n

\"\"<\/p>\n

Say the particle in an electron and we consider spin (up and down), then we multiply N<\/em> by 2.<\/p>\n

\"\"<\/p>\n

is the total number of states in sphere. Now consider the density<\/p>\n

\"\"<\/p>\n

 <\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Figure . Density of states in 3 dimension (Eq.3.48)<\/p>\n

 <\/p>\n

 <\/p>\n

3.3.2 DOS in Two Dimensions (well)<\/strong><\/p>\n

 <\/p>\n

Here we have 1D that is quantized. Let’s us assume it is the z-direction. The total energy of this system is a sum of the energy along the quantized direction plus the energy along the other 2 free directions. It is expressed as<\/p>\n

\"\"<\/p>\n

 <\/p>\n

where for the particle<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

The area of a given mode is then ???? with the total number of modes (N) in the area being<\/p>\n

\"\"<\/p>\n

Again if the particle is an electron and we consider spin, multiply by 2 to get<\/p>\n

\"\"<\/p>\n

This is the energy density of the sub-band for a given or (<\/span>E<\/em>n<\/em>). For each successive there will be an additional \u045b2 and hence another subband. Therefore the density of the states is written<\/span><\/p>\n

\"\"<\/p>\n

Where \u0472 is the heavy side function.<\/span><\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Figure Density of states in 2 dimension. Shaded area presents occupied states.<\/span><\/p>\n<\/div>\n

\n

 <\/p>\n

3.3.3 DOS in One Dimensions (Wire)<\/strong><\/p>\n

 <\/p>\n

Consider now the situation where there are two dimensions confined and only 1 degree of freedom (say the x-direction). The total energy of the system can be written as<\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n<\/div>\n

\n

Figure Density of states in 1 dimension. Shaded area presents occupied states.<\/p>\n

 <\/p>\n

 <\/p>\n

3.3.4 Zero dimensions (Quantum Dot)<\/strong><\/p>\n

 <\/p>\n

Here since all three dimensions are confined. The density of states is basically a series of delta functions. The total energy of the system is<\/p>\n<\/div>\n

\n

\"\"\"\"<\/p>\n

 <\/p>\n

3.3.5 More density of states<\/strong><\/p>\n

 <\/p>\n

Density of states in the conduction band<\/strong><\/p>\n

 <\/p>\n

For this we need to know the probability that an electron will occupy a given stats of energy E. The Probability, P(E), is referred as the Fermi Dirac distribution. In addition we need to know the density of states ( \u2032). The density of states has units of number of unit volume per unit energy. Therefore \u2032 is the number of states per unit volume. The number of occupied states at a given energy per unit volume is therefore<\/p>\n

\"\"<\/p>\n

Here\u00a0 \u00a0is the Fermi energy.<\/p>\n

\"\"<\/p>\n

the total concentration of electrons in the conduction band is therefore the integral over all available energies<\/p>\n

\"\"<\/p>\n

where?c<\/sub> is the energy where conduction band starts. For the case of three dimensional material<\/p>\n

\"\"<\/p>\n

Taking account into conduction band begins, the density of states can be written as<\/p>\n

 <\/p>\n<\/div>\n

\n

\"\"<\/p>\n

The integral is called the Fermi integral or Fermi Dirac integral.<\/p>\n

 <\/p>\n

Consider the case where, E\u00a0 \u2212 Er\u00a0<\/sub>\u226b KTand the Fermi Dirac distribution function becomes<\/p>\n<\/div>\n

\"\"<\/div>\n
\n

 <\/p>\n

This is the expression for the effective density of states of the conduction band.<\/em><\/strong><\/p>\n

 <\/p>\n

Density of states in the valance band<\/strong><\/p>\n

 <\/p>\n

The number of holes at a given energy per unit volume is given as<\/p>\n<\/div>\n

<\/div>\n
\n

\"\"<\/p>\n

Where Ev is the energy where valance band starts. The total concentration of holes in the valance band is the integral over all energies.<\/p>\n

\"\"<\/p>\n

 <\/p>\n

This is the effective density of the states in the valance band.<\/span><\/p>\n<\/div>\n

\n

 <\/p>\n

Summary<\/strong><\/p>\n

 <\/p>\n

Fermi level of an intrinsic semiconductor\u00a0 If the bulk semiconductor is intrinsic, there has been no doping of the material and hence no extra electrons or holes anywhere. in this situation<\/p>\n

\"\"<\/p>\n

One can therefore see that at T=0 the Fermi energy of an intrinsic semiconductor is at the halfway point between the top of the valance band and the bottom of the conduction band.<\/em><\/strong><\/p>\n

 <\/p>\n

Density of states in the conduction band<\/strong><\/p>\n

 <\/p>\n

We start with the Fermi Dirac distribution for electrons and also the density of states<\/p>\n

\"\"Consider only one of the subband. In this case the density of states simplifies to<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

Now recall from the previous section that the number of states at a given energy per unit volume<\/span>\"\"<\/p>\n

the total concentration of electrons in this first subband is the integral over all available energies. Rather than use ntot as before let’s just stick to n<\/em>c<\/em> from the start<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Since the band really begins at en as opposed to Ec<\/em> like in the bulk the integral change from<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

 <\/p>\n

Density of states in the valance band<\/strong><\/p>\n

 <\/p>\n

As with the conduction band case we need the probability of occupying a given state in the valance band. This denoted \u210e(\u00a0 ) and is evaluated from<\/span><\/p>\n<\/div>\n

<\/div>\n
\n

\"\"\"\"<\/p>\n

Fermi level position :2D<\/strong><\/p>\n

 <\/p>\n

The procedure for finding the Fermi level position is the same as in the 3D Consider a spherical volume of<\/p>\n

 <\/p>\n

\"\"<\/p>\n

Now the density is<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

is the number of states per unit volume and the energy density is given as<\/p>\n

\"\"<\/p>\n

Divide by 2 to go back to only 1 spin orientation since in an optical transition spin slips are generally forbidden<\/span><\/p>\n<\/div>\n

\n

\"\"<\/p>\n

The expression applies to either conduction band or valance band. Applying the following equivalence (\u00a0 )<\/p>\n

\"\"<\/p>\n

where is the desired joint density of the states. Now from the conservation of momentum, transition in k are vertical such that the initial k<\/em> value in the valance band is the same k value as in the conduction band (ka=kb=k) where ka is the k value in the valence band and kb is the value in the conduction band. The energy of the initial state in the valance band is<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Likewise the energy of the final state in the conduction band is<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Where for notational simplicity we have used the reduced mass<\/p>\n<\/div>\n

\n

\"\"\"\"<\/p>\n

2D Well<\/strong><\/p>\n

 <\/p>\n

Area in k-space<\/p>\n

= ??<\/sub> = 4??<\/p>\n

 <\/p>\n

Where the area occupied by a given mode or state is .Here we assume that represents the confined direction<\/p>\n

\"\"<\/p>\n

Together, the number of modes in the area is<\/p>\n

\"\"<\/p>\n<\/div>\n

Multiply by 2 to account for spin<\/span><\/div>\n
\n

\"\"<\/p>\n

Now consider the density<\/p>\n

\"\"With the energy density given by<\/span>\"\"<\/p>\n

Starting with the energy density<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Divide by 2 to get rid of the spin since formally speaking, spin flip optical transitions are forbidden<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

wheree??(?) is the desired joint density of states. As before in the 3D case, the conservation of momentum means that transition in k-space are vertical. That is the initial k value in the valance band is the same as the final k value in the conduction band ( = = ) where (\u00a0 ) is the valance (conduction) band values.<\/p>\n

 <\/p>\n

The energy of the initial state in the valance band is<\/p>\n

\"\"<\/p>\n

Likewise the energy of the final state in the conduction band is<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

 <\/p>\n

Such that when replaced into our main expression the desired expression for the joint density of states is<\/p>\n

\"\"<\/p>\n

1D wire<\/strong><\/p>\n

Consider the length in k-space<\/p>\n

 <\/p>\n

Lk<\/sub>=2k<\/sub><\/p>\n

 <\/p>\n

The length occupied by a given mode or state is where<\/p>\n

\"\"<\/p>\n

The number of states in the given length is<\/span><\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Multiply this by 2 to account for spin, we get total number of states as<\/p>\n

\"\"<\/p>\n

Consider the density ie number of states per unit length<\/p>\n

\"\"<\/p>\n

And the energy density is given by<\/p>\n

\"\"<\/p>\n

Or alternately<\/span><\/p>\n

\"\"<\/p>\n

Starting with the energy density<\/p>\n

\"\"<\/p>\n

Divide by 2 to consider only one spin orientation since spin flip transition are generally forbidden<\/span><\/p>\n<\/div>\n

\n

\"\"<\/p>\n

Now apply the following equivalence<\/p>\n

\"\"<\/p>\n

wherepj<\/sub> (E) is the desired joint density of states. As before in the 3D and 2D case, the conservation of momentum means that transition in k-space are vertical so that Ka<\/sub>=Kb<\/sub>=K ) where ka<\/sub>(kb<\/sub>) is the valance (conduction) band values.<\/p>\n

\"\"<\/p>\n

The energy of the initial state in the valance band is<\/p>\n

\"\"<\/p>\n

Likewise the energy of the final state in the conduction band is<\/span><\/p>\n

\"\"<\/p>\n

Such that when replaced into our main expression the desired expression for the joint density of states is<\/span><\/p>\n<\/div>\n

\"\"<\/p>\n

Now to continue towards our final expression we express k fully. Since<\/p>\n

\"\"<\/p>\n

This leads to the final expression for the joint density of states<\/p>\n

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