Course 1 Laboratory
Second Semester
Experiment:
Measuring Boltzmann
Constant
1
The Boltzmann Constant
The aim of this experiment is to illustrate the effects of thermal excitation of electrons
in a semiconductor. By relating the electric current through a p-n junction to the
thermal energies of the electrons at the junction, a method for measuring the
Boltzmann constant is established.
Theory
The theory behind this experiment is much more complex than the experimental
procedure and analysis. In order to understand how a diode or even a transistor works
first we have to examine how such semiconductor devices are constructed. The basic
building block is silicon (or germanium) which is a group IV element, i.e. it requires 4
electrons to close the next electron shell. Thus silicon makes 4 covalent bonds with its
nearest neighbours (as shown in Figure 1). If, however, silicon is doped with either a
group III (e.g. gallium) or a group V (e.g. arsenic) element then its properties are
modified. Doping involves replacing the existing silicon atoms in the lattice with new
elements. If doping is performed with the group V element then after making the 4
covalent bonds there is an additional electron, and hence the material is referred to an
n-type silicon (n for negative). If, however, a group III element is used then there are
insufficient electrons to make all the required bonds and there is what is called a hole.
Such materials are referred to as p-type silicon as the charge carriers are now not
electrons, but holes. Holes may be used to conduct, as electrons from neighbouring
bonds can fill the vacancies (holes), and the hole appears to jump to the location from
which the electron came. So just like bubbles moving a solution the holes appear to be
real particles, but with a positive charge.
As
As
Figure 1. The formation of n-type (left) and p-type (right) silicon.
Diodes are formed when a junctions is made between n-type and p-type silicon, as
illustrated in Figure 2.
2
n-type
p-type
Electrons (As)
Holes (Ga)
Figure 2. The junction
between n and p-type
silicon in a diode.
Initially, the electrons in the n-type material diffuse into the p-type region filling up
the holes, creating what is called intrinsic silicon (also called the depletion region - as
it is depleted of free charge carriers). However, this process is prevented from
continuing, as although in this region there are no electrons and holes, the influence of
the nuclear charges becomes important. Given that Ga has one less proton than Si and
As one more, the n-type side of the intrinsic region has a nett positive charge and the
p-type a nett negative charge, and hence there in an electric field between the two
sides preventing any further electrons crossing the junction - there is a barrier.
Thus, the flow of charge across a p-n junction is essentially an example of the passing
of charge over an electrical potential barrier (see fig. 3). If V is the potential
difference across the junction, then the energy needed to overcome this is eV, where e
is the electronic charge. The rate at which charges pass over the barrier (i.e. the
current) is directly proportional to the number of charges which have an energy
greater than the barrier height. Free electrons have a range of energies due to thermal
effects, so that the number of electrons with an energy E greater than eV is governed
by the Boltzmann distribution law, N(E)∝exp(-E/kBT), where kB is the Boltzmann
constant and T is the absolute temperature. This is the essential Physics behind the
experiment.
a(i)
n-type
b(i)
junction p-type
E
eVo
n-type
junction p-type
E
e(Vo- V)
iel
i th
ith
position
a(ii)
iel
position
b(ii)
N(E)
N(E)
eVo
E
e(Vo - V)
E
3
Figure 3
Figure a(i) Graph of potential energy of an electron at a p-n junction with zero bias.
The two arrows at the bottom indicate the magnitude of the electron diffusion
current and the thermal current. In this instance low energy electrons are reflected
from the barrier.
Figure a(ii) Graph of the Boltzmann distribution as a function of energy. The
shaded are shows the number of electrons which have enough energy to overcome
the potential barrier; the diffusion current is proportional to this area.
Figure b. Potential energy and Boltzmann distribution graphs for a p-n junction with
forward bias voltage V. The shaded area shows that many more electrons can cross
the junction and so the diffusion current increases. In b(i) the reduction in the barrier
height now allows the low energy electrons to cross.
It is shown in Appendix 1 that for an ideal forward-biased p-n junction, the current I
through the junction when the bias voltage is V is
I = ioexp(eV/kBT)
(1)
where io is the reverse leakage current, and is generally small. However, this is not
the whole story. There are other ways in which charges can cross the barrier in a real
diode, such as surface currents. The effect of these alternatives is to modify equation
(1) to
I = ioexp(eV/kBT) + i1exp(2eV/kBT) + other like terms
(2)
so that in practice the current through a diode has a more complex relation than
indicated by equation (1), and so it is impossible to obtain an accurate value of kB
using a diode. Enter the transistor!
A transistor is made up of two p-n junctions, very close together. The region between
the two junctions is called the base, and an electrical connection can be made to it.
The other regions are called, for reasons we shall see shortly, the emitter and the
collector, and contact can be made to these regions as well. Suppose now that the
diode formed by the emitter-base junction is forward biased. Then the rate at which
charges flow into the base is given by equation (2), with V = Veb. The charges
entering the base have two possible exits, either they can leave by the base contact, or
they can diffuse across the collector-base junction into the collector, and leave via the
collector contact. It so happens that the charges contributing to the extra terms in
equation (2) flow through the base contact, and the charges whose flow is governed by
equation (1) pass into the collector. Thus the relation between the collector current Ic
and the emitter-base voltage Veb is
Ic = ioexp(eVeb/kBT)
(3)
Equation (3) was first derived by Schockley, however it is always called after Ebers
and Moll. (Strictly speaking, it is one of the Ebers-Moll equations).
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Taking logs of both sides of eq. (3) gives an equation which resembles the equation of
a straight line
1n(Ic) = 1n(io) + eVeb/kBT
(4)
[Note. In this theory, it is assumed that there is no contribution to the collector current
from a voltage across the base-collector junction. This assumption is automatically
correct if Vcb is zero. Normally transistors are operated with the collector junction
reversed biased, so that there is a small reverse current. However this is negligible
compared to the emitter current, so equation (3) is still applicable to this case.]
Experimental procedure
The circuit diagram is shown in fig. 4. The transistor is either a pnp or a npn power
transistor. A power transistor is used because the heat sink provides a very efficient
way of maintaining the transistor at a constant temperature. The transistor is
contained in a tube filled with insulating transformer oil, which in turn is immersed in
a water bath. The emitter bias is provided by the 1.5 V battery and 10-turn
potentiometer. The collector current is measured using the op-amp circuit (See
Appendix 2).
The triangular symbol in fig. 4 is an operational amplifier, and it is used to amplify the
current from the transistor and convert it to a measureable voltage. The conversion
gain depends on the value of the feedback resistor, R. Choose a value of R to give you
the most sensitive measurements for small values of Veb initially. At the point where
Vout is close to ~8 Volts, the operational amplifier will saturate. That means even
though the input current may increase the output voltage remains constant (the
amplifier is no longer sensitive to changes in Ic). At this point you should change the
R value to the less sensitive setting.
The readings to be taken and the way to analyse the data to produce a value for kB
should be obvious from the theory. You may find departures from linearity at the
higher currents as the self-heating of the transistor will raise the junction temperature
above that of the surroundings (you should check for this).
The relative accuracy of the data points, particularly for the IC readings which cover
several orders of magnitude, will vary considerably. So make sure you pay careful
attention to errors and the error analysis. Perform a computer fit to the data, you
should use a the Mathcad weighted least-squares fit.
R
ic
c
e
b
+
ic
_
+
Vout= -ic R
Vbe
Figure 4
5
Circuit diagram (schematic). Note that the circuit is drawn for a npn transistor, for a
pnp the 1.5 V battery would be reversed. The feedback resistor, R, is either 1M or
100k , depending on the current needed to be measured.
•
•
•
Extract a value for kB and io and the errors in these values. How do they compare
with your expectations?
It is also possible to change the current by varying the temperature of the
transistor. How would you expect the current to change as the temperature
increases?
Given the apparatus, which method of determining the Boltzmann constant would
be most accurate, varying Veb or T?
References
Barlow R J Statistics (John Wiley and Sons: Chichester, 1989) Ch. 6 (especially
6.2.3. p 102).
Clayton GB Operational Amplifiers - 2nd ed. (Newnes-Butterworths: London, 1979).
Evans D E, Physics Education 21, 296-9 (1986).
Inman F W and Miller C E Am. J. Phys 41 349-51 (1973)
Neudeck G W The PN Junction Diode (Addison-Wesley: Reading. Mass. 1983).
Tipler P A, Physics for Scientists and Engineers (4th Ed) (Freeman, Worth, 1999) p.
1225.
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Appendix 1: The p-n junction
An absolutely pure sample of semiconductor at 0 K has all the valence (outer)
electrons tied up in chemical bonds, and none are free to conduct. At a higher
temperature, thermal energy may break some of the bonds, freeing electrons, and
leaving a bond with a vacancy for an electron, or a hole. The electrons are free to
move in the crystal and electrons form other atoms can move into fill the hole, but
then a hole is produced in the bond from which the electron has come. This latter
process can continue, leading to a motion of the hole through the lattice. Since a
vacancy in a bond has a positive charge (because the fully-bonded atom is electrically
neutral, and there is a missing electron), this is equivalent to a positive charge moving
through the lattice. The semiconductor will now have a non-zero intrinsic
conductivity, with equal numbers of positive (p) and negative (n) charge carriers. The
conductivity will be generally rather small, and strongly temperature dependent,
because the number of charge carriers varies roughly as exp(Eg/2kBT), where Eg is
the energy needed to create an electron-hole pair.
However, if an impurity atom which has an extra electron in the outer shell is added,
this extra electron may be easily detached, and is free for conduction, giving an n-type
extrinsic semiconductor. On the other hand, if an impurity atom which has one less
electron than is need for bonding is added, then there will be a vacancy or hole in the
bonds, and a p-type semiconductor is formed. The conductivity of extrinsic
semiconductors depends on the number of impurity atoms added, and can be much
larger than that of the intrinsic base material. The temperature dependence is much
weaker, because most of the charge carriers come from the impurity atoms, and these
are nearly 100% ionised.
When p-type and n-type semiconductors are brought together, an p-n junction is
formed. This has many interesting and useful properties but the one that concerns this
experiment is the existence of a contact potential difference between the two main
regions. This arises (see fig. 1) because when the junction is formed, there is a flow
of free electrons from the n- to the p-type region by diffusion, and conversely a
diffusion of holes from the p- to the n-type. These flows have two consequences.
Firstly, the electrons and holes will recombine, by producing a region where there are
no free carriers, called the depletion region. Secondly, the n-type will be left with a
net positive charge, and the p-type region with a net negative charge. The charges
attract, and will produce an electric field and hence a potential difference across the
depletion region. This electric field tends to stop the migration of the electrons and
holes, and an equilibrium is established (see below). The equilibrium potential
difference is called the contact potential Vo. (Although discussed in terms of
semiconductors, contact potentials are present whenever any two materials are in
contact). The energy needed to overcome the contact potential is eVo.
In fact, the diffusion currents are not reduced completely to zero. The magnitude of
the potential barrier is determined by the balance that exists between the few electrons
that have enough energy to pass over the barrier from the n-type to the p-type, the
equilibrium diffusion current and a current due to electrons that are freed in the ptype region by the thermal breaking of bonds. The potential gradient in the depletion
region will attract these electrons over into the n-type, giving a thermal current;
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there are thus two electron flows, and the barrier height adjusts automatically until the
two currents balance each other. Exactly the same argument applies to the hole
currents: thermally generated holes from the n-type region will pass through the
depletion region into the p-type region unimpeded, while the barrier prevents all but a
small balancing hole diffusion current, corresponding to the most energetic holes,
passing the other way. The net current is the sum of these four charge flows, and is of
course zero.
Now consider what happens if the potential difference across the depletion region is
altered by connecting a voltage source. If connected in such a way as to increase the
barrier height, then the diffusion currents will be considerably diminished. The
thermal currents will be unaffected, and the observed total current will be a current
almost independent of applied voltage. Such a situation is called reversed bias, and
the current is the reverse or leakage current, io.
When the applied voltage V is applied so as to reduce the barrier to Vo V, the
number of electrons and holes with energies sufficient to overcome the barrier is
increased dramatically. The number of electrons with an energy greater than or equal
to e(Vo-V) is given by the Boltzmann distribution law to be proportional to exp(-e(Vo
-V)/kBT). Since the diffusion current is proportional to the number of electrons with
enough energy to overcome the barrier, it follows that the total current which equals
the diffusion current minus the thermal current is given by
I = Aexp(-e(Vo-V)/kBT) - io
where A is a constant of proportionality. When V = 0, no external bias is applied, so
the current must be zero, as discussed above, hence io = Aexp(-eVo)kBT). The
current with a bias voltage V is therefore
I = io(exp(eV/kBT) -1)
(A1)
For V large and negative (reverse bias), the exponential becomes very small, so I = io
as expected. For V large and positive (forward bias), the exponential is >> 1, so that
I = ioexp(eV/kBT) .
(A2)
which is the working equation, number (1) in the main text.
Note that equation (A1) explains the rectifying action of the junction. For large
negative voltages, the current is io, which is small. However, for an equal magnitude
positive voltage the current is io multiplied by a large exponential factor. Thus
current passes easily for one bias, and there is a high resistance to current in the
reverse direction.
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Appendix 2: The current-measuring circuit
The op-amp circuit used is known as a transconductance circuit, and it functions as a
current-to-voltage converter. Referring to fig. 2, the usual op-amp operation
conditions lead to the following results:
(a)
The voltage between the inverting and non-inverting inputs of the op-amp is
extremely small. This means that the inverting input is a virtual earth, and also
the circuit has essentially zero input impedance, i.e. zero volts across the input
terminals whatever the current.
(b)
The input current must pass through the feedback resistor. Hence there must
be a voltage across the resistor of IcR, and so the op-amp output must be at
-IcR with respect to earth.
Thus, knowing R, the current may be found. The circuit has two interchangeable
feedback resistors, giving two current-to-voltage conversion factors. The resistors are
believed to be accurate to better than 0.3%. The op-amp is powered by two 9 V
batteries, so saturation is to be expected when the output reaches about 7.4 volts or
thereabouts.
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