Experiment 7Bench 7MZU-PE / HAL-7Ready

The Hall Effect: Identifying Carriers

Hall Probe & Electromagnet

Pass a constant current through a Hall bar between the magnet poles and read the transverse voltage.

Duration: 3 hours
Kittel: Chapter 8
Topics: Lorentz forceHall coefficientCarrier concentration

Hall probe between the magnet poles

n-type germanium · t = 1.00 mm

FieldI

Running

Bench instruments

Gaussmeter · constant-current source · nanovoltmeter

Gaussmeter

Sample current

Readings

Voltmeter (raw)

mVNOISE

Flux density B

Sample current

Corrected V_H

Hall coefficient

m³/C

Thickness t

Field reversal

The contacts are never exactly opposite, so the voltmeter reads an offset even at zero field. Measuring at +B and −B and halving the difference removes it.

Derived from your reading

Carrier density n = IB/(e t V_H)

1.000e+21m⁻³

True n

1.000e+21m⁻³

Carrier type

n (electrons)

Drift velocity

6.242e+0m/s

Resistivity

1.600e-2Ω·m

Doped with a donor, so the majority carriers are electrons. The Hall voltage should come out negative for the field direction marked on the magnet.

Hall bench control

MZU-PE / HAL-7

Ready

Strong magnetic field. Keep watches, cards and tools away from the pole gap while the magnet supply is on.

Specimen

Electromagnet

Magnet current3.00 A

The iron yoke saturates, so B stops rising in proportion above about 4 A.

Polarity

⊗ IN⊙ OUT

Field direction

→←

Current direction

Constant-current source

Sample current10.00 mA

SignalBelow noise

Key Equations

V_H = \frac{IB}{nqt} = \frac{R_H IB}{t}

with the Hall coefficient $R_H = 1/nq$ . Its sign identifies the carrier: negative for electrons, positive for holes. Its magnitude gives the carrier density,

n = \frac{IB}{e\,t\,V_H}

Why reverse the field?

The Hall contacts can never be placed exactly opposite one another, so a fraction of the ordinary voltage drop along the bar appears across them. That offset does not care which way the field points, but the Hall voltage does — so half the difference of the readings at +B and −B is the Hall voltage alone.

Procedure

0/4

AWatch the carriers deflectLoad n-Ge, set 10 mA and wind the magnet up to 3 A. Note which edge the carriers pile up on. Switch to p-Ge and check whether the carriers move to the same edge — and whether the voltage keeps the same sign.
BV_H against BHold the sample current at 20 mA and take the magnet from 0 to 8 A in steps of 1 A. At each step press Measure at ±B and record the corrected Hall voltage.
CV_H against IHold the magnet at 4 A and step the sample current from 5 mA to 50 mA. Record the corrected Hall voltage at each point.
DMetals and thin filmsLoad the copper foil at full current and full field. Record what the voltmeter shows and why. Then load the InSb film and compare — both are far thinner than the germanium.

Table 7 — Hall measurements

0 entries

Press Measure at ±B before recording, so the offset is out of your number.

#	Specimen	Type	t/ mm	I/ mA	B/ T	Raw reading/ mV	Offset/ mV	V_H/ mV	n from V_H/ m⁻³	R_H/ m³/C
No readings yet — set the controls, then press “Record reading”.

Analysis & Reflection

Analysis questions

Plot your Task B data as $V_H$ against $B$ . Is it a straight line all the way? If it bends over at high magnet current, is that the sample or the magnet?
From the gradient of your Task C plot, $V_H$ against $I$ , extract $R_H$ and then $n$ . Compare with the value the panel reports.
In Task A the carriers of both signs were pushed to the same edge, yet the Hall voltage changed sign. Explain carefully why.
Explain why the Hall voltage is so much larger in a semiconductor than in a metal, using the carrier densities on the panel.
What would happen to V_H if you reversed both the magnetic field and the sample current at the same time? Try it and explain the result.

Physics problems

A germanium sample 1.0 mm thick carries 20 mA in a field of 0.35 T and gives $V_H = -13.1$ mV. Calculate $R_H$ and $n$ , and state the carrier type.
Show that halving the sample thickness doubles the Hall voltage for the same current and field. Why is a good Hall probe always a thin film?
Copper has $n = 8.5\times10^{28}$ m⁻³. Calculate the Hall voltage for a 50 µm foil carrying 50 mA in 0.6 T, and compare it with the 0.3 µV resolution of the voltmeter.
Combine $R_H$ with the resistivity shown on the panel to obtain the carrier mobility from $\mu = |R_H|/\rho$ , and compare germanium with indium antimonide.