Hall Effect Lab (Germanium SC) Without Hall Effect Sensor

In summary, the lab method for determining the Hall Effect involves connecting a current source across a germanium wafer, measuring the voltage in the direction of current with a multimeter, introducing a magnetic field, and measuring the voltage again. The difference between the two voltages is assumed to be the Hall voltage. This method is repeated for both p-type and n-type germanium semiconductors. From these measurements, the hall coefficient, magnetic field strength, type of charge carrier, and charge density of the semiconductor can be calculated. However, the main issue is that most methods and equations for calculating these values require a Hall sensor to measure the magnetic field strength, which is not available in this experiment.
  • #1
Jenn.
1
0
1. Lab Method

The method for determining the Hall Effect is as follows:
This is the method that I was instructed to follow:

  • A current source was connected across a germanium wafer

  • The voltage across the wafer (in the direction of current) was measured with a multimeter

  • A magnetic field was introduced

  • The voltage, in the direction of current was measured again

  • The difference in measured voltage 1 and measured voltage 2 was assumed to be the approximate value of the Hall Voltage

  • This method was repeated for both p-type and n-type germanium semiconductors

2. Required Calculations

From these measurement the hall coefficient, magnetic field strength, the type of charge carrier and the charge density of the semiconductor need to be calculated

The Attempt at a Solution



The problem I'm having is that most reports and methods I've found that show equations for the above quantities are for a Hall Effect experiment that uses a Hall sensor to measure the magnetic field strength. This is the main problem I'm having because if I had a value for B I would be able to calculate all of the other values. Is there a way to calculate B with the measurements that I have?

Any help would be appreciated
 
Physics news on Phys.org
  • #2
The hall voltage is perpendicular to the current direction - the magnetic field should not affect the voltage in the current direction - which is usually maintained at a fixed value by a power supply. Check you have understood the instructions properly.

If you know the current and the hall voltage, then you can use the hall effect equations to calculate the magnetic field strength.
 

What is the purpose of a Hall Effect Lab without a Hall Effect Sensor?

The purpose of a Hall Effect Lab without a Hall Effect Sensor is to study the Hall Effect phenomenon in a Germanium semiconductor material. This lab allows scientists to observe and analyze the behavior of electrons in a magnetic field without the use of a Hall Effect Sensor.

What is the Hall Effect phenomenon?

The Hall Effect phenomenon is the production of a voltage difference across an electrical conductor when a magnetic field is applied perpendicular to the flow of current. This effect was first discovered by physicist Edwin Hall in 1879.

Why is Germanium used in this lab?

Germanium is used in this lab because it is a semiconductor material with a high electron mobility, making it ideal for studying the Hall Effect phenomenon. It is also less expensive and more readily available compared to other semiconductor materials.

What are the applications of the Hall Effect phenomenon?

The Hall Effect phenomenon has many practical applications, such as in the measurement of magnetic fields, current sensing in electronic devices, and in the development of Hall Effect sensors for position and motion detection.

Can the Hall Effect be observed without a Hall Effect Sensor?

Yes, the Hall Effect can be observed without a Hall Effect Sensor by using a Germanium semiconductor material and measuring the voltage difference across it in the presence of a magnetic field. This lab provides a hands-on demonstration of the Hall Effect phenomenon without the use of a specialized sensor.

Similar threads

  • Quantum Physics
Replies
0
Views
137
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Electromagnetism
Replies
7
Views
962
  • Electromagnetism
Replies
2
Views
871
Replies
1
Views
701
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top