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Expphyfinal

The document outlines an experiment to determine the Hall coefficient (RH) of a semiconductor and estimate its carrier density using the Hall Effect. It explains the theoretical background, experimental design, procedure, and observations needed to measure the Hall voltage and calculate RH and carrier density. The results confirm the linear relationship between voltage, current, and magnetic field, while also addressing safety precautions and potential sources of error during the experiment.

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0% found this document useful (0 votes)
26 views8 pages

Expphyfinal

The document outlines an experiment to determine the Hall coefficient (RH) of a semiconductor and estimate its carrier density using the Hall Effect. It explains the theoretical background, experimental design, procedure, and observations needed to measure the Hall voltage and calculate RH and carrier density. The results confirm the linear relationship between voltage, current, and magnetic field, while also addressing safety precautions and potential sources of error during the experiment.

Uploaded by

loremipsum9423
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 8

Experiment No.

Title: Hall Effect

Aim: To find the Hall coefficient, RH of a given semiconductor and hence estimate the carrier
density
Theory:

If a current carrying conductor placed in a perpendicular magnetic field, a potential


difference will generate in the conductor which is perpendicular to both magnetic field and current.
This phenomenon is called Hall Effect. In solid state physics, Hall effect is an important tool to
characterize the materials especially semiconductors. It directly determines both the sign and
density of charge carriers in a given sample.

Consider a rectangular conductor of thickness d kept in XY plane. An electric field is


applied in X-direction using a Current source, so that current I flow through the sample. If w is the
width of the sample and t is the thickness. Therefore, current density is given by
𝐼
𝐽= (1)
𝑤𝑑

Fig.1 Schematic representation of Hall Effect in a conductor.

If the magnetic field is applied along negative z-axis, the Lorentz force moves the charge
carriers (say electrons) toward the y-direction. This results in accumulation of charge carriers at
the top edge of the sample. This set up a transverse electric field EH in the sample. This potential
difference along y-axis is known as Hall voltage VH and this effect is called Hall Effect.

A current is made to flow through the sample material and the voltage difference between
its top and bottom is measured using a volt-meter. When the applied magnetic field B=0, the
voltage difference will be ideally zero.
We know that a current flow in response to an applied electric field with its direction as
conventional and it is either due to the flow of holes in the direction of current or the movement
of electrons backward. In both cases, under the application of magnetic field the magnetic Lorentz
force,
𝐹𝑚 = 𝑞(𝑣 × 𝐵)
causes the carriers to curve upwards. Since the charges cannot escape from the material, a vertical
charge imbalance builds up. This charge imbalance produces an electric field which counteracts
with the magnetic force and a steady state is established. The vertical electric field can be measured
as a transverse voltage difference using a voltmeter.
In steady state condition, the magnetic force is balanced by the electric force.
Mathematically we can express it as,
𝑒𝐸 = 𝑒𝑣𝐵 (2)
where 'e' the electric charge, 'E' the hall electric field developed, 'B' the applied magnetic field and
'v' is the drift velocity of charge carriers.
The total number of charge carriers in the above sample is 𝑛𝐴𝑙, where 'n' is the number
density (i.e. number of charge carriers per unit volume) of the charge carriers in the conductor of
length ‘l’, breadth 'w' and thickness 'd ', the cross-sectional area 𝐴 = 𝑤𝑑 and thus 𝐴𝑙 is the volume.
The total charge is thus 𝑛𝑒𝐴𝑙. If this total charge exits the sample in a time T, then the current is
𝑛𝑒𝐴𝑙. It means each charge carrier travels a distance l in a time T and hence has a velocity 𝑣 = 𝐿.
𝑇 𝑇
Therefore, the current 'I ', defined as charge crossing a cross-section of the conductor in a time T,
can be expressed as,
𝐼 = 𝑛𝑒𝐴𝑣 (3)
Using (1) and (2) the Hall voltage VH can be written as,
𝐼𝐵
𝑉𝐻 = 𝐸𝑤 = 𝑣𝐵𝑤 =
𝑛𝑒𝑑
𝑉 =𝑅 𝐼𝐵 (4)
𝐻 𝐻 𝑑
by rearranging eq. (4) we get
𝑉𝐻𝑑
𝑅𝐻 = 𝐼𝐵
(5)

where RH is called the Hall coefficient. 1 1


𝑅 = ,∴ 𝑛 = (6)
𝐻 𝑛𝑒 𝑅𝐻𝑒

The value of 𝑅𝐻 is calculated from the measured parameters using eq. (5) and that of n from eq.
(6).
Design of Experiment:
An electromagnet which can produce a magnetic field of B=1000Gauss in a gap of 30mm
between the pole pieces is selected. An electromagnet (Fig. 2) consists of two solenoids with solid
cylindrical iron cores such that the flat surfaces of the cylinders facing each other are North and
South poles thus producing uniform field in the gap between the pole pieces. The sample is
mounted on an insulating wooden sample holder. The sample holder is placed between the pole
pieces such that the magnetic field lines (parallel to axis of the pole pieces shown by the red dashed
line in Fig. 2) are perpendicular to the sample. A Hall probe is placed in the gap very near the
sample such that its sensing element is also perpendicular to the magnetic field.
The electrical contacts for current and voltage measurements are made by pressurecontacts.
The pressure contacts are made of stainless steel plates as shown in Fig. 3. The pressure can be
varied by tightening the screws which hold the contacts down. Ideally, it is expected that the tips
of the current contacts should be aligned in a straight line; similarly, voltage contacts should also
be aligned. In practice, there is always a misalignment as shown schematically in Fig.3 for the
voltage contacts. This leads to a voltage V0 even in the absence of a magnetic field whena current
is passed through the sample. This has to be subtracted out from the voltage obtained with field to
obtain Hall voltage.

Fig. 2 Sample and Hall probe are aligned such that they are parallel to each other and perpendicular to the axis of the
pole pieces and hence the magnetic field.

Fig. 3 Stainless steel plates (four blue pieces) are held down on the sample by the respective screws. Misalignment of
the voltage probes can be seen.
Apparatus: Hall probe, Sample mounted on holder, Electromagnet with power supply, current
source.
Procedure:
1. Adjust position of Hall probe to be perpendicular to the pole pieces of the electromagnets.
2. Switch on the electromagnet. Increase the current through the electromagnet and measure the
magnetic field B using the Gaussmeter. Adjust B to the desired value (1000G).
3. Place the sample in pole pieces of the electromagnet such that it is perpendicular to the magnetic
field. Keeping B constant, vary current I through the sample in suitable steps and note
corresponding values of voltage V.
4. Switch off the electromagnet. Keep the sample away from electromagnet. Measure voltage V0
without field for the same current values as in step 3. Hall voltage VH = V-V0.
5. Plot VH versus I and find the slope m.
6. Find Hall coefficient RH= md/ B
7. Also find charge carrier density 'n' using n=1/ RH e.
Observations: 1. Thickness of the probe (Given), d = 0.5 mm = 0.005 m
2. Constant magnetic field, B = 1000 Gauss = 0.1 Tesla
Observation Table:

I VH= V-
(mA) V(mV) V0(mV) V0(mV)
0.5 4.3 1.5 2.8
1 11 5.7 5.3
1.5 17.5 9.5 8
2 23.4 13.5 9.9
2.5 29.8 21.7 8.1
3 35.6 26.2 9.4
3.5 42.2 30.5 11.7
4 48.8 35.1 13.7
4.5 53.7 39.2 14.5
5 60 43.6 16.4
5.5 67.1 47.5 19.6
6 76 50 26

Graph and Calculations:


1) Draw VH versus I and find the slope m.

I (mA)
2) Using the numerical method, outlined using an example below, to find the slope m.

xi yi (xi-x̅)yi (xi-x̅)2
0.5 2.8 -7.7 7.5625
1 5.3 5.3 1
1.5 8 12 2.25
2 9.9 19.8 4
2.5 8.1 20.25 6.25
3 9.4 28.2 9
3.5 11.7 40.95 12.25
4 13.7 54.8 16
4.5 14.5 65.25 20.25
5 16.4 82 25
5.5 19.6 107.8 30.25
6 26 156 36
∑(xi − ∑(xi − x̅)2
x̅ = y̅ = x̅)yi = =
3.25 12.11667 584.65 169.8125

𝑚𝑑
Calculate (i) 𝑅𝐻 =
𝐵
=
3.44291498 × 0.005 = 0.172145749
0.1
1
(ii) 𝑛 =
𝑒𝑅𝐻

= 1 = 3.626111762674551 × 1019
-19
1.602×10 ×0.172145749
Results:

Hall Coefficient of the given material 𝑅𝐻 =0.172145749 𝑚3⁄𝐶


Charge carrier density of the given material 𝑛 =3.626111762674551 × 1019 𝑐𝑎𝑟𝑟𝑖𝑒𝑟𝑠⁄𝑚3
Conclusions:
In conclusion, the Hall effect was verified. As long as the magnetic field and the
current stayed below some threshold, there was a linear relationship between the voltage
measured, and
the current and B field applied
Questions:
1) Explain the working principle of a Magnetic encoder which detects rotational position information.

Ans - Magnetic rotary encoders


Magnetic rotary encoders rely on three main components: a disk, sensors, and a
conditioning circuit. The disk is magnetized, with a number of poles around its
circumference. Sensors detect the change in magnetic field as the disk rotates and convert
this information to a sine wave. The sensors can be Hall effect devices, which sense a
change in voltage, or magnetoresistive devices, which sense a change in magnetic field.
The conditioning circuit multiples, divides, or interpolates the signal to produce the desired
output.
The resolution of a magnetic rotary encoder is determined by the number of magnetic
poles around the disk and by the number of sensors.The primary difference between
incremental and absolute encoders, regardless of sensing technology, is that absolute
versions assign a unique binary code, or word, to each measuring position. This allows
them to track the encoder’s exact position, even if power is discontinued. he simplest magnetic
encoder consists of a permanent magnet and a magnetic sensor. The
permanent magnet is attached to the tip of a rotating body such as a motor shaft, and the
magnetic sensor is fixed in a state where it is mounted on a PCB board at a position where it
receives the magnetic field generated by the permanent magnet. When the permanent magnet
attached to the motor shaft rotates, the direction of the magnetic field detected by the magnetic
sensor changes, as a result the encoder detects the rotational position and speed of the motor
shaft.

2) What precautions must be taken to


(i) minimize the errors in the data in this experiment (Hint: check if all the assumptions in the
derivation of the formula are reproduced in the performance of the experiment)
(ii) safeguard persons, materials and instrument while performing the experiment (Hint: (a) large
magnetic field, (b) how fast the current through a solenoid can be increased or decreased (c) effect
of magnetic field on electronics). What are the WHO guidelines for magnetic field human safety?
3) How can the type of majority charge carrier be found out from the Hall effect experiment?

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