Experiment 11: Hall Effect & Energy Gap in Germanium
We will see if the charge carrying particles are negative in n-doped germanium, and if they are
positive in p-doped germanium. We will also measure the energy gap between the conduction
and valence bands in germanium.
We will observe the sign of the charge carriers using the Hall effect. The Hall effect is a voltage
across a current carrying conductor in a magnetic field, as opposed to the voltage along the
conductor described by Ohm’s law. The sign of the Hall voltage depends on the sign of the
charge carriers. To measure the energy gap, we will heat the sample. As the temperature goes
up, more electrons will have enough thermal energy to cross the gap into the conduction band.
The rate at which the number of free electrons increases with temperature tells us the material's
energy gap because more electrons can make it across a narrow gap than a wider one. The
number of free electrons will be measured two ways: It is related to both the Hall voltage and
the material’s electrical resistance.
Details:
The Hall effect: The pictures below both show a current toward the right. In one case, it

 
consists of positive particles going right; in the other, negative particles going left. F = q v B
says that either kind of particle is pushed toward the top of the sample by the magnetic force.
So, if positive particles are flowing through it, the top gets charged positive, but if the particles
are negative, the top is negative. We measure whether the top is positive or negative with a
voltmeter.
To measure the energy gap, Eg: The sample is heated. As the temperature goes up, more
electrons have enough thermal energy to cross the gap into the conduction band, so the number
of free electrons per unit volume, n, increases. It was
stated in class that
n
CE 1 / 2 dE
0
e (E
E F ) / kT
1
where C is a collection of constants. This was written for
a metal where E = 0 is normally taken to be at the bottom
of the conduction band. For a semiconductor, it's more convenient to say E = 0 at the top of the
valence band, as shown. This changes the expression to:
- 2 -
n
C E Eg
Eg
e(E
1/ 2
E F ) / kT
dE
1
It can be shown from this that
ln n
- Eg 1 + const.
2k T
If you’d like to see how, it’s in the box:
Substitute u = (E-Eg)/kT, which makes E – Eg = (kT)u. Differentiating, dE = (kT)du. In the
denominator, (E – EF )/kT = [(E – Eg) + (Eg - EF)]/kT = u + (Eg – EF)/kT.
Throughout the range of integration, the exponential in the denominator is much larger than the 1
next to it, so the 1 can be dropped with little effect.
n
C kT u
E Eg
1/ 2
eu e(E
kT du
Eg ) / kT
C kT
e
3/ 2
E g EF
u 1 / 2 du
u 0
eu
How to evaluate the integral is not important; all that matters is that you get some number when
you do. Also substitute in the fact that, in a semiconductor, EF is about ½Eg so Eg - EF = ½Eg.
n = C (kT)3/2(some number)
eEg/2kT
Taking the log of both sides,
ln n = ln C + ln [(kT)3/2] + ln (some number) - Eg/2kT
The ln [(kT)3/2] term varies more slowly than the Eg/2kT, so to a good approximation, it can be
considered constant. Combining it with the other constants,
ln n
- Eg 1 + const.
2k T
As explained in the text, the Hall voltage is VH =IB/(nqt). (I is current, B is the magnetic field, q
is charge, and t is the sample's thickness.) Taking the log of both sides,
ln VH = -ln n + some constants.
Substituting the equation from the box in for ln n,
- 3 -
Notice that if you plot ln VH on the y axis, and 1/T on the x axis, Eg/2k will be the slope of the
graph.
The Hall voltage is measured between the crystal's top and bottom, across the direction of the
current. There is also a voltage difference along the direction of the current, due to Ohm's law.
Let's call this VΩ to distinguish it from VH. Putting the textbook's formula for resistivity into the
formula for resistance, and that into Ohm's law, you get VΩ = I[mel/(nq2τA)]. Everything in that
equation, including the current, is kept constant during our experiment, except for n. The more
charge carriers, the less resistance. So, if we take the log of both sides,
ln VΩ = -ln n + some constants
Making the same substitution for ln n as before,
So once again, Eg/2k is the slope. You will do graphs for both voltages, and average the results.
PROCEDURE:
The experiment will be performed by the instructor with student assistance. The rest of the class
will observe, and then analyze the results on their own.
A. n - germanium.
1. Turn on the power supply, with the knobs at 0. This makes about 30 mA flow through the
crystal, toward the right. Bringing the N pole of a bar magnet near the sample will establish a
field which points away from you. This creates exactly the situation pictured on the first page of
this experiment. Notice that the positive lead from the voltmeter goes to the top of the crystal, so
the sign of the Hall voltage will be the same as the sign of the charge carriers. What is the
result?
2. Place the crystal between the poles of the electromagnet. Turn the right hand knob on the
power supply all the way up, and then adjust the center knob for a Hall voltage of about 50 mV.
(Don't expect a voltage until both knobs are turned.) The crystal is now in a field of about .3
- 4 tesla.
3. The board contains a copper-constantan thermocouple, which is a type of electronic
thermometer. How its voltage translates into temperature has been built into your data table.
4. A heating element is also built into the mounting board. Throwing a switch will put about 5
amps of alternating current through it. A few students are to monitor the thermocouple voltage,
The Ohm's Law voltage, and the Hall voltage, announcing the data to the class. The
thermocouple is not to pass 5 mV since higher temperatures will cause damage.
(Data below 3 mV is not useful because we are not working with pure germanium: it has been
"doped" (contains impurity atoms). Electrons from donor atoms dominate the conduction process
at lower temperatures. As the temperature rises, more and more valence electrons are thermally
excited into the conduction band, until there are enough for them to dominate instead. Only at
this point does the discussion given earlier become valid.)
5. The instructor will replace the n-Ge with another board containing p-Ge. This will take a few
minutes; you may want to start on your graphs while you wait.
B. p-germanium
Repeat just step 1 of Part A.
Analysis:
1. Plot two graphs: ln (VH) vs. 1/T and ln (VΩ) vs. 1/T.
2. Calculate the energy gap from each graph, as described earlier. The results will not be
especially close to each other, due to some complicating factors.
3. Average the two values. Consider the uncertainty in this average to be the difference between
it and either of the individual results.
4. In your conclusion, compare this to the accepted value in the text.
- 5 Report on Experiment 11: Hall Effect & Energy Gap in Ge.
n - type: Were charge carriers positive or negative?
p - type: Were charge carriers positive or negative?
1 ( K-1)
T
2.77 x 10-3
VT
(in mV)
2.7
2.71 x 10-3
3.0
2.67 x 10-3
3.3
2.62 x 10-3
3.6
2.56 x 10-3
4.0
2.52 x 10-3
4.3
2.48 x 10-3
4.6
2.44 x 10-3
4.9
Find Eg and uncertainty:
Textbook Value = __________
VH
(in mV)
VΩ
(in V)
ln(VH)
ln(VΩ)
- 6 -
- 7 -