Tip:
Highlight text to annotate it
X
This is the 14th lecture of this course and the 3rd lecture on Carrier Transport.
In this lecture we will be completing the discussion on this topic. In the last class
we saw that for small electric fields the drift current is proportional to the electric
field and mobility is the constant of proportionality coming in this particular relation between
current and electric field. Similarly we saw that for small concentration
gradients the diffusion current is proportional to the concentration gradient and the constant
of proportionality here which comes in is diffusion coefficient of course with the other
terms also coming in. We also saw that the µ and diffusion coefficient
are related by the Einstein relation. And since the diffusion coefficient is related
to mobility if you know the behavior of mobility as the function of doping, temperature and
so on we can also know the behavior of diffusion coefficient. So the mobility as a function
of doping on temperature is the behavior that one looks for characterizing the semiconductor.
For this parameter µ = q TL/2m we have seen the various scattering phenomenon and we showed
that the mean free time between collisions for ionized impurity scattering decreases
with increasing total doping or decrease in temperature.
Similarly the lattice scattering, the mean free time between collisions decreases when
temperature increases. Then we came to carrier-carrier scattering and we said the carrier-carrier
scattering does not affect the mobility of carriers if the carriers involved are of the
same polarity. So, scattering of electrons by electrons and scattering of holes by holes
will not affect the mobility of electrons or holes. However, the scattering of electrons
by holes and vice versa can affect the mobility. In a semiconductor if electrons are in a majority
then the hole mobility will be affected by scattering with electrons. However the µ
of electrons will not be affected by scattering with holes because holes are in a small number.
The carrier-carrier scattering affects only the minority carrier mobility in a semiconductor.
If the carriers are almost in equal number and their concentration is very high which
can happen if the semiconductor has been injected with large number of extra electron hole pairs
in such a case also the carrier scattering can affect the µ of both electrons and holes.
Otherwise it affects only the µ of minority carriers and that happens only at very high
doping levels. Taking ionized impurity scattering and lattice
scattering as dominant scattering mechanisms we will ignore the carrier-carrier scattering
in this course. We can write taking these two into account, the overall µ formula.
How do we do that? We can write the equation 1/tauc which represents the number of scattering
events per unit time because tauc is the mean free time between collisions. So 1/tauc that
is the total number of scattering events is the sum of scattering events due to ionized
impurity plus all the scattering events due to lattice scattering. Add up these two and you will
get the total number of scattering event. This formula you can easily translate to µ
because you see from here that tauc is proportional to µ. We can therefore write 1/µ = 1/µ
ionized impurity + 1/µ due to lattice scattering, this is the formula. So if you know how µi
and µl individually depend on doping and temperature then you know how the µ depends
on doping and temperature. Now it turns out that µi and µl individually depend on temperature
as shown in this slide.
So µi ~ αT3/2 this mobility increases with temperature because scattering reduces as
a function of temperature. But scattering increases with doping so the coefficient α
is a strong function of total doping and α decreases with total doping. For lattice scattering
the formula is a constant δ × T-3/2 these powers 3/2 and -3/2 are derived from theory
and in practice the powers are somewhat different. In the extrinsic range beyond 300 K the mobility
is mostly governed by lattice scattering. So you can take these two segments of the
mobility and this formula shows that whichever mobility is lower will decide the overall
mobility.
Whenever µI is much lower than µi the mobility is approximately µI and whenever µ l is
much less than µI the mobility is approximately µl. So whatever we have written the behavior
as a function of temperature and doping can be shown on a graph as follows. This is the
line, here you have µ on a log scale and temperature also on a log scale. We are plotting
both on a log scale because mobility versus temperature is a power law behavior as we
saw T power something. So when you plot log you will get a straight line. This is the
straight line corresponding to the lattice scattering so this is µl which is proportional
to T-3/2 and this is the line corresponding to µI which is proportional to T3/2. Now
if you want to sketch the overall mobility for any given doping as a function of temperature
it would be µI for low temperatures and µl for high temperatures so it will tend to go
like that and this is the behavior of mobility. If you want the mobility for a higher doping
then we said the ionized impurity scattering increases with higher doping so you will get
another line here corresponding to a higher doping for ionized impurity scattering so
your mobility for higher doping will tend to go something like this so this is increasing
doping. And the temperature of 300 K would be somewhere here. That is to say that this
is room temperature and beyond room temperature it is almost lattice scattering that is dominated.
This is exactly the behavior as shown here.
This is the practical behavior as a function of doping and temperature. If you are interested
in knowing the mobility at 300 K then at 300 K the mobility as a function of doping is
given by this particular graph its variation with doping mobility decreases doping and
which can be captured in equations as shown here.
The µ = µ max - µ min /1 plus total doping divided by a constant and this ratio is raise
to the power γ + µmin. For electrons and holes in silicon at 300 K the values are given
in this table. So maximum mobility is 1330 for electrons and 495 for holes, the µmin
is 65.5 that is for a very heavy doping levels for electrons and 47.7 for holes. The unit
for mobility is centimeter square per volt second.
Similarly the value of n0 and γ are also given here. Use this formula to find out the
mobility for any given doping conditions at 300 K. And then if you are interested in the
mobility at any other temperature you could use the formulae that we showed earlier i.e.
µ variation with temperature and find out what is the mobility. We will consider a solved
example to illustrate some of these ideas. The mobility is the other important parameter
that we have come across in this particular discussion. Earlier we had seen the parameters
namely: energy gap of a semiconductor that was an important parameter so µ after energy
gap. µ is the next important material parameter that affects the applications of the semiconductor.
Let us look at a picture of mobility in silicon and gallium arsenide and see how the values
of mobility affects the application of these two semiconductors. If you compare the mobility
of electrons in gallium arsenide is much higher than mobility of electrons in silicon.
You can see that at both doping levels 1014 and 1019. This has important consequences
on the application. For e.g. if you want to make devices which can carry high current
or which can work very fast then gallium arsenide is better than silicon provided the devices
are depended on only the movement of electrons. If you have bipolar transistors such as NPN
transistors or you have MOSFETS or field effect transistors, we cannot say MOSFETS because
it is difficult to make MOSFETS in gallium arsenide, so it is field affects transistors
which depend on electrons that is n-type field effect transistors. Then these transistors
in gallium arsenide can carry more current and will be faster than in silicon. That is
the reason why people look for different semiconductors for making devices. So alternatives to silicon
are being considered and gallium arsenide is one such alternative for high speed devices
or high power devices. However, you can see here that the hole mobility
in gallium arsenide is comparable to that of silicon and in fact it is lower at 1014/cm3
doping level. This low hole mobility coupled with a very high electron mobility creates
some problems. In the case of gallium arsenide if you are making circuits which require both
n-type and p-type devices e.g. CMOS. If you want to make CMOS type of complementary devices
in gallium arsenide in a circuit then there will be problem because of size of the p-type
device will be much larger than that of the n-type device in order that the p-type device
carries the same current as n-type device. in complementary structures in circuits depending
on which are based on complementary devices there has to be matching between the performance
of the p-type and n-type devices. To achieve this matching the p-type device geometry in
the case of gallium arsenide semiconductors will have to be much higher than the n-type
geometry but this creates lot of problem in laying out. When you have a large number of
n-type and p-type devices and you want to interconnect all of these then it creates
problems. So, complementary type of circuits is not possible in gallium arsenide because
of this problem. This is how the mobility behavior can affect the applications, the
value of the mobility can affect the applications of the semiconductor.
Let me also just mention that the effective mass in silicon and gallium arsenide of electrons
and holes you have to consider both types of effective mass: the conductivity effective
mass and density of states effective mass. So what we have discussed so far in the context
of carrier transport the conductivity effective mass is what is important whereas a density
of states effective mass is important while estimating the carrier concentration namely
effective density of states and so on. The differences in these two situations i.e.
conductivity situation and density of states situation should be clear by now.
We will take up the topic of resistivity. Resistivity depends on both carrier concentration
and mobility. If you combine mobility and carrier concentration dependence on what kind
of dependence we get for resistivity as a function of doping and temperature? Let us
start with extrinsic silicon first. This is the picture.
Instead of resistivity we are plotting the conductivity which is nothing but the reciprocal
of resistivity. So it is conductivity as a function of temperature where you find that
the conductivity increases from 0 and the region 2 is the extrinsic range.
Here in this range conductivity increases and then starts falling. And finally when
it enters the intrinsic range it increases again. Let us explain this behavior by combining
the behavior of mobility and Carrier Concentration.
Extrinsic silicon: Supposing you take n-type the Carrier Concentration if you were to sketch
as the function of temperature, so electron concentration nn0 then this behavior on a
linear scale is like this, the so called extrinsic range. This is the region 2. This is the partial
ionization range and this is the intrinsic range.
Now if you sketch the mobility on the same graph, this is carrier concentration and if
you sketch the mobility it will be something like this, it will increase and then fall
and somewhere it is in maximum. For the same doping level we are plotting the mobility,
this is mobility µn. Then conductivity σ = qn µn + qp µp. If you are taking n-type
semiconductor then this n is nn0 and we are considering equilibrium conditions or close
to equilibrium conditions and this is pn0. Now pn0 will be much less than nn0 because
these are minority carriers so the σ in n-type semiconductor is approximately equal to qnn0 × µn. We
need not bother about this hole concentration and the hole mobility. So that is why using
µn behavior and nn0 behavior all we need to do is multiply these two curves. It is
evident that you will get a raise and then in this range a fall in conductivity because
of fall in mobility which is happening because of lattice scattering and then because Carrier
Concentration is increasing you again have a raise in mobility and this is something
like this. So that is how one can explain this particular behavior.
So this fall here is because of mobility behavior, lattice scattering and this raise over a small
region in the extrinsic range is because of ionized impurity scattering. In this range
the Carrier Concentration is constant so whatever variation you see is mainly because of mobility.
Now let us look at intrinsic silicon σi.
In this case σ = q(µn + µp)ni because electrons and holes are in equal concentration. Now
the behavior of σi will depend on µn µp and ni. Now ni is increasing continuously
as a function of temperature right from 0 onwards to very high temperatures and this
change in ni with temperature is very rapid as compared to variation in mobility.
If you combine these two: µn and µp have a power law depending on a temperature whereas
ni has an exponential dependence in addition to the power law. So we can write this as
q(µn + µp) T3/2 exp(- Eg/2KT) that is the behavior into some constant A.
If you now also include the power law behavior of mobility then this will be of the form,
here q also you absorbed in a constant so this will turn out to be of the form BTa exp(-
Eg/2kt). The resistivity of intrinsic silicon is not going to be very much depended on a
mobility behavior. It is the carrier concentration behavior because this is the most dominant
variation with temperature as compared to the power law variation Ta where a combines
this 3/2 power as well as the power law because of the mobility. It turns out that 'a' is
something like - 1 if you combine all the power loss and together 'a' it is about - 1.
So if you sketch σi versus T the behavior will be the same as that of the Carrier Concentration
behavior which means σi versus T is same as n i versus T behavior. In other words if
you sketch σi on a log scale versus 1/T on the x-axis on the linear scale then you will
get a behavior like this which is same as that of the intrinsic silicon. This will be
a straight line because the mobility behavior only affects the Ta here, this term and that
is why for intrinsic silicon if you sketch the measured resistivity as the reciprocal
of temperature from the slope of that one can easily get the energy gap because this
behavior is similar to that of the ni behavior intrinsic carrier behavior multiplied by some
constant. If I sketch log σi it will go as proportional to - Eg/2k ×1/T and that is what you get if you take the log.
So slope of this is equal to - Eg/2k. From the slope of the resistivity versus reciprocal
of temperature graph for intrinsic semiconductor one can get the energy gap. Now it is time
to consider a solved example. You listed some of the ideas and to also show how formulae
can be used to make calculations.
The example is, calculate the following for phosphorous doped silicon whose resistivity
is measured as 1 ohm-cm at 300 K. What we are supposed to calculate are the following:
the impurity concentration, the hole diffusion coefficient at 300 K, the resistivity of the
sample at 500 K and the resistivity of sample at 300 K on adding 1 x 1016cm3 atoms of boron.
Basically you have an n-type semiconductor whose resistivity is 1 ohm-cm at 300 K.
Let us start with the first part, what is the impurity concentration?
Data given is 1 ohm-cm at 300 K and it is an n-type phosphorous doped so what is the
impurity concentration? Now, as we have seen the conductivity of this
semiconductor can be written as qnn0 × µn we ignore the hole component of the conductivity
because only majority carrier concentration decides the conductivity so the resistivity
is 1/ρ that can be written as 1/qnno × µn. We know ρ and we need to get nno so nno = 1/q
× ρ × µn. Now difficulty in applying this particular equation in a straight forward
manner is that this µn depends on the nno which we want to determine so µn depends
on the doping. Since we do not know the doping in advance, a straight forward use of this
relation is not possible and some involved calculation is required. The impurity concentration
is nno and we know that Nd will be nno because we assume at 300 K complete ionization and
the majority carrier concentration is almost equal to the doping level ignoring the thermal
generation. One way to solve this problem is to look at a graph of doping versus resistivity
so these kinds of graph are available where resistivity is given as a function of doping
concentration N.
This is the net doping for both p-type and n-type semiconductors. Here you can see the
resistivity of 1 ohm-cm for n-type and read out the concentration. In this particular
graph the finer divisions are not shown so it is difficult to find out. But in graphs
available in books you will find these smaller divisions present and one can determine the
concentration from there. Even then from the graph an accurate determination is very often
difficult so even tables are available for resistivity as a function of doping which
can be used to get the resistivity for a given doping or doping for a given resistivity.
Now, alternately if you want to do a calculation using a computer then it can be done as follows.
What we do is we assume the mobility corresponding to the low doping or maximum value of mobility
here and find out the concentration. Then we use this concentration and do iteration
and find out the mobility corresponding to this concentration using the mobility versus
doping formula. Then use the new value of mobility and determine the concentration again.
When you repeat this calculation two times normally you get a very good result. If you
do that then the calculation proceeds as follows. So µn = µn max = 1330 cm2/v-s. If you do
that then Nd = 1/q ρ × µn max = 1.6 × 10-19 coulombs. You must also put the units simultaneously
on the sides. Ρ is 1-ohm cm and this is 1330 cm2/v-s this is the value of doping and these
are the units. As you can see this gives raise to cm3 and coulomb per second is amperes.
So ampere × ohm = volt so this cancels and you have /cm3 as a unit and this will be equal
to 4.7 ×1015/cm3. Now we use this value of doping and find out the mobility.
You can write using this formula 1330 - 65.5/1+ (4.7 × 1015/8.5 × 1016)0.76+ 65.5 cm2/v-s.
This
formula was shown on the slide earlier. So you use this formula for mobility as the function
of doping at 300 K. Then the result will be a mobility of 1204 cm2/v-s, instead of a 1330
that we assumed earlier. Use this value of mobility instead of µn max and you will get
the new doping. The new doping will be 4.7 × 1015 × (1330/1204) and this is nothing but 5.2 ×1015 cm3 which is
the value of doping or impurity concentration. Next you need to find out the diffusion coefficient
of holes. To find out the diffusion coefficient of holes that is Dp you can use the Einstein
relation µp × Vt (thermal voltage) so you need to know µp. Now µp you can find out
using a formula similar to this because you know the doping. Like you have formula for
µn, there is a formula for µp so µ p = 495 - 47.7/1 + (5.2 ×1015/6.3 ×1016)0.72 +
47.7 cm2/v-s. Please note that whenever you write the magnitude
you must also write the units, dimensions, you should not forget otherwise you go wrong
in calculations. This mobility turns out to be equal to 431 cm2/v-s. Therefore your Dp
is 431(0.026), it is cm2/v-s and this is volts so volts will
cancel and you will get cm2/sec as the unit and this will be 11.2 and that is the diffusion
coefficient.
Now part c is the resistivity of the sample at 500 K. To find out the resistivity at 500
K we need to check two things.
We need to check whether the minority carrier concentration important? The temperature is
high so will the hole concentration be important and will it make a difference to the resistivity.
For this purpose we need to find out the intrinsic concentration first so ni at 500 K can be
found very easily as we had done in this course earlier. We can write this in terms of ni
(300 K)(500/300)3/2 exp[ - 1.12/0.052(300/500 - 1)] which is the energy gap divided by
2(kT) where T is the room temperature that is 0.052 into room temperature which is 300
by the temperature of interest. Make this calculation and you will get the
result as (1.78 × 1014)/cm3. Now the minority carrier concentration would be pn0 = ni2/nn0
which is ni2/the doping so which will be = (1.78 × 1014)2/5.2 × 1015 doping and evidently since this is 1028 and 1015
your order of this particular result will be 1013 so this is of the order 1013 which
is much less than 5.2 × 1015/cm3. You find that pn0 is much less than nn0. So
result is pn0 is much less than nn0. We can still continue to use the relation σ ≈ q
µn nn0. We are neglecting any contribution of pn0
to the resistivity. So this nn0 is nothing but the doping which we can replace by Nd.
However the mobility at 500 K will not be the same as mobility at 300 K. We need to
determine the mobility at 500 K and then we can easily determine the conductivity or resistivity.
Mobility at 500 K/mobility at 300 K can be written as = 500/300-3/2. This is because
we are using the fact that in the extrinsic range above room temperature the mobility
approximately goes as T-3/2 so µ = some constant × T-3/2 so this is the relation we are using.
From here we find that mobility at 500 K is equal to mobility at 300 K for electrons.
And the result at 300 K the mobility was 1204 cm2/v-s so this is equal to into 3/53/2 = 555.6 cm2/v-s so that is
the mobility. You can use this mobility here and then find out the resistivity So ρ ≈ 1/qµ
Nd. And in fact we could directly substitute this value of mobility or alternately what
we find is since the resistivity is depended on the mobility behavior we can write whatever
we have written in terms of mobility directly as an equation in terms of resistivity.
For example, we can write ρ500 K/ρ300 K so ρ is inversely proportional to mobility
so this is equal to the mobility at 300 K/mobility at 500 K of electrons in this case. This is
equal to 5/33/2 and the result therefore is, resistivity at 500 K = 300 K resistivity which
is 1 ohm-cm multiplied by this and the result would be 2.15-ohm cm. So this is the way one
can find out the resistivity at a higher temperature. You can find out the mobility at a higher
temperature and find out the resistivity. But since we already know the resistivity
at a lower temperature it is easier to write it in this simplified form and then get the
resistivity. At a higher temperature actually the resistivity has increased because of the
fall in mobility.
Now we have the final part of the example that we have to show and this is the resistivity
of the sample at 300 K on adding 1 x1016cm3 atoms of boron. This means we have a compensated
semiconductor with donor type impurity of doping 5.2 x 1015 cm3.
And an acceptor doping of 1 x 1016cm3 and you want to know the resistivity if you have this. Clearly, this is a p-type
semiconductor because Na is more than Nd so the Carrier Concentration of holes i.e. the
majority carrier concentration in this case would be pp0 and can be written as Na - Nd.
All impurities are ionized and now we can neglect the electron concentration because
minority carrier concentration will be very small and this is = 4.8 × 1015/cm3. This
is the p-type semiconductor with this particular concentration of holes. Now the resistivity
can be written as 1/q × pp0 × µp where pp0 is given here. Now µp we have to determine.
We cannot use the hole mobility that we determined earlier in this case because now we have to
use total doping of this plus this. The mobility of holes which we have determined earlier
in context of the diffusion coefficient calculation was only because of this particular doping
so we need to recalculate the hole mobility. So we use the formula µp = 495 - 47.7/1
+ 1.52 × 1016/(6.3 × 1016)0.72 + 47.7cm2/v-s is in the denominator ad power is 0.72. In
the numerator we will have to put a sum of this Nd + Na = 1.52 × 1016 and + 47.7 cm2/v-s
this is the hole mobility and the result is 377 cm2/v-s.
Now we can substitute this pp0 that is this 4.8 × 1015 here and 377 cm2/v-s here and
the resistivity would be 1/q × pp0 × µp. This is the resistivity and of course your unit will be ohm-cm because
this is in coulombs cm3 and this is cm2/v-s. We have shown that if you use these units
you will get resistivity in ohm-cm and the result is 3.46 ohm-cm. The resistivity has
actually increased because the sample has become p-type and hole mobility is smaller
than electron mobility and because of combination of these factors you are getting this resistivity.
This is how one can make calculations of resistivity at different temperatures and different doping
conditions. This example should clarify many of the ideas that we have so far discussed.
The next topic we will discuss before closing this particular Carrier Transport discussion
will be on how do you see the mechanism of drift from the energy band point of view?
Even diffusion can be seen from the energy band point of view but we will see how the
drift is seen from the energy band point of view and one can work out for diffusion in
the same manner.
We need to draw the energy band diagram of a semiconductor in which a drift current is
flowing. Suppose this is a semiconductor let us assume an n-type for simplicity, we can
use a p-type it does not matter really. In the semiconductor the electric field is in
this direction.
Now what will be the energy band diagram under this condition? We will assume a uniform semiconductor
which means that the electric field is also going to be uniform. And that means that the
potential will vary linearly as you move from the left end to the right end so this is positive
and this is negative. Now how do we draw the energy band diagram under these conditions?
Recall that the vertical axis in energy band diagram shows electronic energy. We have also
mentioned earlier in the discussion of energy band diagram that Ec represents the potential
energy of the free electron and Ev represents the potential energy of a hole. So if you
want to draw the energy band diagram without an electric field it would look something
like this where this is Ec and this is Ev. This is the potential energy of the free electron
and this is the potential energy of the hole. And this is the first time that the diagram
is drawn with distance as the x-axis. This is energy and this is distance along the sample,
so this is the sample. No electric field is applied under equilibrium conditions.
The Fermi-level here would be somewhere here this is the Fermi level, these dotted lines
shows the intrinsic level. So Ec is the potential energy of the free electron and this is electronic
energy which means that whichever side is negative, whichever position here in the sample
is negative there the electronic potential energy should be shown higher.
Let us look at the sample now, this side is positive so here on this side your Ec will
be lower so when you draw the energy band diagram it will be drawn as something like
this. This is electronic energy, this is Ec potential energy of the free electron, this
is lower and this is higher because this is more negative and as you move up you have
higher and higher electronic energy. The valance band edge Ev will run parallel to the conduction
band edge because when you apply voltage the energy gap does not change in the semiconductor.
The Carrier Concentration is also not changing so the structure of the silicon is not changing.
The only thing that is happening is that you are superimposing an electric field so energy
gap cannot change with distance and that is why this edge Ec is parallel to Ev and this
difference remains equal to energy gap. Now what is the slope of this particular diagram?
Slope of this diagram is nothing but slope of the potential. Ec is energy and if you
divide Ec/q you get the potential. Now your equation is electric filed E = - d φ/dx
where φ is the potential of a positive charge. So when you write electric field equal to
gradient of potential this is potential of a positive charge. So if you want to write
in terms of a negative potential then this negative sign will go away. It is important
to note here that the E here is electric field whereas the E here is the energy. I hope there
will not be any confusion because we have been using this nomenclature throughout the
earlier lectures. Therefore in terms of EcI if I want to write
down replace φ by potential Ec then I can write the same thing as this is equal to d
Ec/q because I want to convert energy into potential so I have to divide by charge by
dx. The negative sign has been removed because this Ec/q is electronic potential. In other
words what we find is at the electric field is equal to 1/ q and d Ec/d x. Clearly see
that the slope of this line is positive 1/q dEc/dx and therefore the electric field is
in the positive direction from left to right. And the slope of this particular line gives
you the electric field provided you take into account the charge q. So that is the important
thing that you see for the energy band diagram when you apply an electric field. Now on this
particular diagram how will you show the motion of an electron, electron will move from right
to left because it will move against the electric field. This motion can be shown as hours.
Let us say electron start from here and after collision it has started moving in this direction
towards the left. So it moves like this and it encounters a collision somewhere here.
At the end of the collision the energy gain in the direction of the electric field is
randomized. In other words, we can show this as the kinetic energy has got dropped to 0
because this difference represents the kinetic energy of the electrons that it has gained
from the electric field. Again it starts moving and encounters a collision and so on so
this is the path.
Now, note here carefully that this is a horizontal line. This line is not going up nor is it
moving down it is perfectly horizontal. This is because the total energy of the electron
between two collisions remains constant. It is losing potential energy as shown by this
line but it is gaining kinetic energy.
Loss in potential energy is exactly equal to the gain in kinetic energy. Total energy
being constant is shown as a horizontal line. So total electronic energy between collisions
is constant but it is gaining kinetic energy seen by the fact that this line here for example
is above this particular line and there is a gap and this gap represents the kinetic
energy that is gain from the electric field. One can similarly draw the movement of the
hole on the energy band diagram which would be something like this the hole moves in this
direction so this is electron and this is hole this is x in the sample. This is how one can look
at the drift from energy band point of view. Note carefully that I have not shown the Fermi-level
here because Fermi-level is strictly an equilibrium concept and under non-equilibrium conditions
one can extend this idea but then one has to introduce what are called Quasi Fermi-levels
so this aspect we will consider later. I am not showing the Fermi-level in this diagram.
It is not necessary to show this level to understand the drift transport. Now let us
summarize our discussion on Carrier Transport.
We covered the following topics in order. First we saw the example of various modes
of transport such as drift, diffusion, thermo electric current and current because of tunneling,
random thermal motion of carriers, the semi classical phenomena of drift, diffusion and
thermo electric current are based on this particular motion. Then scattering mechanisms
which causes the random thermal motion and then we considered the drift and diffusion
transport and how directed motion is superimposed over random motion by application of potential
gradient or concentration gradient. Then we discussed relation between mobility and diffusivity,
this is the Einstein relation. Then we discussed the mobility as a function of doping temperature
and electric field. For small fields you have a linear relation between velocity and electric
field but for large fields the velocity is constant so mobility varies with field. Then
we discussed resistivity as a function of temperature and doping and finally we discussed
the drift from energy band point of view. In the next class we will consider the topic
of Excess Carriers.