OBJECT:
To determine the Hall Coefficient
of Copper metal.
To deduce that the transfer of charge is by mobile negative charge carriers (electrons).
To show that the Hall Voltage in a metal does not depend on temperature.
To demonstrate the temperature
dependence of electrical conductivity and the Hall Effect in semi-conductors.
To demonstrate two different conduction mechanisms in doped semi-conductors (extrinsic
and intrinsic).
APPARATUS:
Large Cenco Electromagnet with
coils to handle >15 Amperes D.C.
Plane Pole pieces have circular cross section of 10 cm diameter.
With gap width of 2 cm, uniform fields up to 10 Kilo-Gauss can be maintained.
Solid State filtered 15 Ampere power supply and current regulator.
F.W.Bell Gauss/Tesla Meter (Model 4048) for measuring the magnetic field.
Hall Effect Apparatus, 18 µm
Copper strip, supporting plate, PHYWE11803.00.
Heater voltage power supply, 8 V AC or DC/ 5A.
A regulated control current supply for steady currents up to 20A DC max.
Voltmeter fo
Amplifier for Hall Voltage when AC current passes copper strip.
Hall Effect Apparatus, p-germanium
crystal, supporting plate, PHYWE11802.00.
Heater voltage power supply, 8 V AC or DC/ 5A.
A power supply 12 to 18 V DC for crystal currents up to 50 mA DC max.
Voltmeter for measuring potential drop across the crystal , 300 mV and 1.0 V scales.
Voltmeter for measuring thermoelectric voltage up to 10 mV DC from copper-constantan
thermocouple.
Voltmeter for measuring the Hall Voltage 30 to 100 mV.
Ammeter for measuring the crystal current up to 30 mA.
BACKGROUND INFORMATION:
The basic theory for these operations is found in The Hall Effect, a lab procedure write-up provided by Northeastern State University,OK. The basic equation for the Hall Effect predicts that the Hall Voltage UH is directly proportional to either the applied field B or the control current I (provided that the other parameter is held constant). The Hall coefficient RH and the sample thickness d are constants.
UH = RH ( I B/ d )
Set up the magnet with only the transverse probe of the Gauss/Tesla meter placed at the center between the magnet pole pieces oriented with the magnetic flux perpendicular to the flat side of the probe. Collect data for a graph of the Magnetic Field Strength B versus the Coil Current through the magnet windings.
The Copper supporting plate 11803.00 is displayed in Figure 1. The plate is fitted with an electrical heater and a thermocouple in order to show that the Hall voltage in a metal, unlike that in a semiconductor, does not depend on temperature.
The current conductor proper 1 is a section of a thin copper laminate in strip form; copper thickness is 18 µm, copper width is 25 mm.
The control current (max. 30 A )is passed through the conducting path via sockets 2.
The Hall voltage is taken from sockets 3.
An interfering voltage due to the control current and superimposed on the Hall voltage can be compensated by an adjusting knob 4 on a potentiometer.
The pair of plugs 5 serve as leads for the heating current.
The copper-constantin thermocouple 6 supplies a thermoelectric voltage of approximately 40 µV/K which is taken off the plugs at 7.
The apparatus connections, exclusive of the heater, are now illustrated by the following circuit diagram:
Exercise 1
Without the heater connected, and without current in the magnet coils, slowly increase the copper current to 2 Amps.
Use knob 5 to cancel any offset voltage.
Select a magnetic field strength along the linear part of the graph of Experiment 1. (say near 4 KGauss)
Vary the control current in 1 A steps up to 10A and determine the Hall voltage for each current.
Remember that for each current, the offset voltage must be rechecked.
Plot a graph of Hall Voltage versus Control Current.
Determine the Hall Coefficient for Copper.
Repeat the plotting on the same graph for other magnetic field values up to 8 KGauss.
Are the results consistent with the basic equation for the Hall Effect?
Exercise 2
Select the maximum magnetic field strength from Exercise 1. (say near 8 KGauss)
Keep the maximum magnetic current of 10A from Exercise 1.
Connect a variac through a 6-8 V AC transformer to the heater (sockets 5).
Increase the heater voltage in small increments. Wait a minute to see if the temperature changes. If it does, allow the temperature to stabilize and record the Hall and thermocouple voltages.
Caution: As soon as the thermoelectric voltage reaches 5 to 6 mV (heating time approximately 2 minutes), the heating current should be turned off in order to avoid overheating the support plate.
Increase the heater voltage and repeat the measurements. Try to take about ten measurements before you reach the maximum setting of the variac.
Plot a graph of Hall Voltage versus Temperature.
Do your results confirm that the Hall Voltage in copper does not depend on Temperature?
The Germanium supporting plate 11805.00 is displayed in Figure 1. (Actually the n-Ge plate is shown, but a p-Ge plate will be used in the experiment.) The plate is fitted with an electrical heater and a thermocouple in order to show how the Hall voltage in a semiconductor, varies with temperature.
The semiconductor crystal 1 is a p-doped germanium crystal with dimensions 20 x 10 x 1 mm.
A DC voltage of 12 to 30 V is applied to the crystal via sockets 2.1 and 2.3. A current stabilizer is interposed between sockets 2.2 and 2.3. With the stabilizer in the circuit, the control current can be held constant in spite of any dependence of the crystal resistance on temperature.
The voltage across the crystal is measured between sockets 2.1 and 2.2 in order to determine its resistance.
The Hall voltage is taken from sockets 3.
An interfering voltage due to the control current and superimposed on the Hall voltage can be compensated by an adjusting knob 5 on a potentiometer.
The pair of plugs 4 serve as leads for the heating current.
The copper-constantan thermocouple 6 supplies a thermoelectric voltage of approximately 40 µV/K which is taken off the plugs at 7.
The apparatus connections, exclusive of the heater, are now illustrated by the following circuit diagram:
Exercise 1
Without the heater connected, and without current in the magnet coils, slowly increase the crystal current to 10 mA.
Use knob 5 to cancel any offset voltage.Select magnetic field strengths along the linear part of the graph of Experiment 1. Measure and record the the Hall voltage for Magnet coil currents of 4, 5, 6, 7, and 8 Amperes.
Vary the control current from 10 to 35 mA in steps of 5 mV and repeat the measurements.
Remember that for each current, the offset voltage must be reset for zero with magnet off.
Plot a graph of Hall Voltage versus Control Current for each selection of magnetic field strength.
All the plotting should be performed on the same graph.
Are the results consistent with the basic equation for the Hall Effect?
Determine the value of the Germanium Hall Coefficient from each line plotted.
Exercise 2
Select the maximum magnetic field strength from Exercise 1. (say near 8 KGauss). Keep the maximum magnetic current of 35 mA from Exercise 1.
Connect a variac through a 6-8 V AC transformer to the heater (sockets 4). Turn on the variac and increase the output to well over 50% maximum. As the temperature gradually changes, record simultaneous values of the Hall voltage, the thermocouple voltage and the crystal voltage.
Caution: As soon as the thermoelectric voltage reaches 5 to 6 mV (heating time about 2 minutes for the six volts AC), the heating current should be turned off in order to avoid overheating the support plate.
Plot a graph of Hall Voltage versus Temperature. Determine from your graph the temperature at which the transition from extrinsic to intrinsic conduction occurs.
Plot a graph of Crystal Resistance versus Temperature.