report on hall effect

8/7/2019 Report on Hall Effect

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REPORT ON HALL EFFECT

AIM

To find out the type of the excess carriers (p-type or n-type) present in a semiconductor bar using HallEffect measurements.

APPARATUS

Hall Probe, Hall effect set-up, Electromagnet, Constant current power supply, Digital Gaussmeter.

THEORY

When the charges flow, a magnetic field directed perpendicular to the direction of flow produces amutually perpendicular force on the charges. When this happens, electron and holes will be separated byopposite forces. They will in turn produce an electric field ( h )which depends on the cross product of themagnetic intensity, H, and the current density, J.

h = R J x H

Where R is called the Hall coefficient.Now, let us consider a bar of semiconductor in which the electric field is applied in y direction andmagnetic field is applied in z direction. The magnetic force experienced by an moving electron will begiven by

Fm = e (V x H)

Therefore the Hall potential will be generated in y direction

Then we could write

R=V/yJH=V.zIH (1)

V -> Hall voltage

I= J yz

(a)One type of carrier

now , E h = v x H J = q n v

Putting in eq. 1

R=EhJH=1nq

=R (b)Two type of carriers

R=h2p-e2n2(hp+en)2Procedure1. Connect the widthwise contacts of the hall probe to the terminals marked lenghtwise contacts

to terminals marked current.2. Switch ON the hall effect set up and adjustment current(say few mA).3. Switch over the display to voltage side. There may be some voltage reading the magnetic

field . this is due to imperfect alignment of the four contacts, and is generally known as theZero Field Potential. In case its value is comparable to Hall voltage it should be adjusted to aminimum possible (for hall probe(general cases, this error should be subtracted from the HallVoltage reading.

4. Now place the probe in the magnetic field and Switch ON electromagnet power supply and

adjust the current to any desired value. Rotate the probe till it become perpendicular tomagnetic field. Hall voltage will be maximum in this adjustment.


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5. Measure Hall voltage for both the directions of the current and magnetic field observations fora particular value of current and magnetic field.

6. Measure the Hall voltage as a function of current keeping the magnetic field constant . plotgraph .

7. Measure the Hall votage as a function of magnetic field keeping a suitable value of current asconstant. Plot graph.Measure the magnetic field by the Gaussmeter

Observations:

TABLE-1: Current Control: 0.18 A ; Magnetic Field: 302 Gauss

Current

Voltage1

Voltage2

Voltage Avg

0.51 0.2 -0.5 0.35

1.05 0.8 -1 0.92.03 1.7 -1.9 1.82.98 2.7 -2.5 2.63.98 3.6 -3.3 3.45

5 4.6 -4.1 4.356.01 5.6 -4.9 5.256.98 6.5 -5.5 68.01 7.6 -6.1 6.85

9 8.4 -6.9 7.6510 9.3 -7.6 8.45


Current

Voltage1

Voltage2

Voltage Avg

0 0 01 1.6 -1.5 1.55

1.97 3.1 -3 3.053.01 4.9 -4.9 4.94.03 6.6 -5.9 6.25

5 8.2 -7.3 7.756.02 9.8 -8.7 9.25

7 11.4 -10.1 10.758 12.9 -11.8 12.35

9.01 14.4 -12.5 13.4510 19.9 -19.3 19.6

Y axis: Voltage(V) ; X axis: Current(mA);


Curre Voltag Voltag Voltag


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nt e1 e2 e Avg1.08 2.5 -2.5 2.52.04 4.7 -4.5 4.63.05 7 -6.7 6.854.04 9.9 -8.7 9.34.99 11.5 -10.7 11.16.01 13.9 -12.8 13.35

7 16.1 -14.9 15.58 18.2 -16.9 17.559 19.8 -19.6 19.7

10 21.9 -21.6 21.75



Current

Voltage1

Voltage2

Voltage Avg

1 2.7 -2.8 2.751.98 5.7 -5.7 5.72.95 8.9 -8.5 8.7

4 11.9 -11.3 11.64.99 14.9 -14.1 14.55.99 17.1 -17 17.057.03 20 -19.9 19.95

8 22.6 -22.1 22.359.01 25.6 -25 25.39.99 28.1 -27 27.55



Curre Voltag Voltag Voltag


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nt e1 e2 e Avg1.07 3.8 -3.8 3.82.03 7.3 -7.4 7.35

3 10.8 -10.8 10.84.02 14.4 -14.4 14.4

5 18 -17.9 17.956.01 21.6 -21.4 21.57.05 25.1 -25.1 25.18.02 28.4 -28.5 28.459.01 31.7 -31.8 31.75

10 34.9 -35.3 35.1


Now drawing plot of hall voltage vs magnetic field keeping current constant at 6 mA.

appliedI

H

hallvolatage

0.18 302 5.13730.39 529 9.1553

0.6 768 13.21950.8 977 16.9215

1 1212 21.2737

slope=0.0177

CALCULATION:

1)R=(V.z)/(I.H)=(V/H)*(z/I) ; where z = 5x10 -2 cm = 5x10 -4 m & I=6 mA

=>R=0.0177x5x10 -4+3 /6=1.475x10 -3 m 3/C

2) Now, we know that: R = 1/(n.q) ; =>n = 1/(R.q);

n = 1/(1.475x10 -3x1.6x10 -19) = 4.273x10 21 m -3 = 4.273x10 15 cm -3

3) p = R/resistivity = 1.475x10 -3+6 /10 = 1.475x10 +2 cm 2/V.s = 147.5 cm 2/V.s

OBSERVATIONS:

1) Hall coefficient , R =1.475x10 -3 m 3/C.

2) R is positive for p type.

3) Carrier concentration,p = 4.273x10 15 cm -3

4) Mobility of holes, p = 147.5 cm 2/V.s


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ANALYSIS:

1. With the increase of constant current the magnetic field is increasing.2. The Hall voltage is generated in mutually perpendicular direction of magnetic field & current

(motion of carrier) ,as we are getting the Hall voltage by measuring along y axis which isperpendicular to both H& I.

3. The V vs I graph is linear keeping magnetic field constant, which interns gives us the value of

resistance (slope of graph).4. The resistance is increasing i.e. slope of V Vs I graph increasing as magnetic field is increasing. This effect is called magneto- resistance is due to the fact that the drift velocity of all the carriersare not the same with the magnetic field. This results in effective decrease of mean free path &increase in the resistance.

5. As it is a p-type of semiconductor the Hall coefficient is positive.

CONCLUSION:

The Hall coefficient, carrier concentration are calculated using Hall effect which helped us to calculatemobility of the carrier in the P-type semiconductor.

report on hall effect

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