Mate Lab Report 2
pdf
keyboard_arrow_up
School
San Jose State University *
*We aren’t endorsed by this school
Course
153
Subject
Electrical Engineering
Date
Apr 3, 2024
Type
Pages
10
Uploaded by ProfWaterBuffaloMaster3217
From: Van Luong Nguyen, Electrical Engineer
Date: March 9, 2022
To: Mr. Mark Benjamin, Manager
Subject: Testing of doped germanium sample for dopant concentration amount
Dear Mark,
The germanium semiconductor sample from the supplier requires testing when
considering such a large quantity purchase of germanium wafers. Meaning the quality could be a
faulty point, but we would need to know whether the doped germanium product should have a
dopant concentration of
to
carriers per
to be determine of good quality.
5?10
14
5?10
15
𝑐𝑚
3
To conduct testing of the germanium product we must detect the dopant concentration
which is where we utilize the Hall Effect. By applying a voltage to a sample of doped
germanium wafers, making sure to observe the resulting magnetic field. With the Hall voltage
and current across the sample being measured using a Hall console across three magnetic field
strengths in forward bias and reverse bias.
The measured dopant concentration is about
. This value is
7. 62?10
14
𝑐???𝑖𝑒??/𝑐𝑚
3
between the expected range of
and
. Based on these sample
5?10
14
5?10
15
𝑐???𝑖𝑒??/𝑐𝑚
3
results, the doped germanium wafers from the prospective supplier is of a high enough quality to
proceed with a bulk order and SemiTech we should pursue this deal.
Van Luong Nguyen
Introduction
The purpose for this experiment was to test the quality of the germanium sample to
determine if it meets the dopant concentration range of between
to
carriers per
5?10
14
5?10
15
. The test were done in a laboratory as to find if the given sample provided by the supplier
𝑐𝑚
3
will meet the standard dopant values.
Testing of the quality of the semiconductor would be defined by the conductivity and
draft mobility of the sample. These two factors are directly related to the carriers concentration.
Thus the Hall Effect would need to be used in this experiment by placing the germanium wafers
in an electric field and a magnetic field as displayed in Figure 1.
Figure 1: Hall-voltage Circuit
With the sample of germanium being placed in the magnetic field perpendicular to the
current, thus developing the electric field. This phenomenon is used to measure the amount and
type of carriers within a semiconductor. With the fact that both types of carriers (holes and
electrons) produce conductivity, though the carrier type can’t be determined just from electrical
measurements. As that would be subjected to the influence of a magnetic field on a moving
charge to determine. This effect occurs due to the charged particle being subjected to the Lorentz
force as the sample creates a voltage potential. Using this equation to calculate the draft velocity:
(1)
𝑉
?
=
?
?
(𝑒*𝑝*?*?)
The n represents the number of carriers or dopant concentration of the sample,
is the current in
?
𝑋
the x-direction,
represents the magnetic field in the z-direction, q represents the charge of the
?
?
electron which is
, t represents the thickness of the sample which is 0.01 cm, and
1. 6?10
−19
𝑉
?
would be the voltage potential created by the carrier.
According to the right hand rule, the force is on the -y direction. Given that the carrier is
in equilibrium, the forces should be in the inversed direction and would have the same
magnitude. Which means
(2)
𝑒 ? (𝑉
?
/?)
=
𝑒 ? 𝑉
?
? ?
?
From equation 1 and 2:
(3)
𝑉
?
=
?
?
? ?
?
𝑒 ? 𝑝 ? ?
We would also need to utilize Ohm’s Law to get resistance and resistivity as well as
conductivity. In these following equations:
(4)
𝑅 = 𝑉/?
(5)
𝑝
=
1/σ
=
(𝑅 ? ?)/𝐿
Where p refers to resistivity,
is the conductivity, A is the cross section area, and L is the length
σ
of the sample. After we would need to calculate the hole mobility μ from the two value of
conductivity and hole concentration with the equation:
(6)
σ
=
µ ? 𝑒 ? 𝑝
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Experimental Procedure
The sample that the supplier sent was .2 cm long, .1 cm wide, and had a thickness of 0.01
cm. It was held by the Hall console, which was used to measure the current passing through the
sample. There was a probe that we would have in the near the middle of the sample to get the
most direct, but to also have it not touch the sample. It was also used to control the level of the
magnetic field and to have it applied perpendicular to the direction of the current. The current
through the sample and the Hall voltage was measured in forward bias of 300, 600, 900 Gauss as
well as in reverse bias of -300, -600, -900 Gauss equivalently. The probe would be zeroed out
and calibrated before the beginning of each gauss measurement. The experimental equipment is
shown below in Figure 2:
Figure 2: Testing Equipment used to control and measure Hall voltage and current
Results
The given data in terms of the germaniun wafer sample measurements has a length of .2
cm, width of .1 cm, and has a thickness of .01 cm. The gathered data is what was measured
throughout the testing of the sample of germanium.
Magnetic Field
sample current (mA)
hall voltage (mV)
3.00E-06
5
14
3.00E-06
3
9
3.00E-06
1
4
6.00E-06
5
26
6.00E-06
3
16
6.00E-06
1
6
9.00E-06
5
38
9.00E-06
3
24
9.00E-06
1
8
-3.00E-06
5
-10
-3.00E-06
3
-6
-3.00E-06
1
-2
-6.00E-06
5
-23
-6.00E-06
3
-13
-6.00E-06
1
-4
-9.00E-06
5
-35
-9.00E-06
3
-22
-9.00E-06
1
-7
Table 1: Collected Measurements utilizing equipments. Bz for Magnetic Field, Ix for Sample
Current, Vy is for Hall Voltage
With this table we calculate the resistance using Ohm’s Law, with it being
in which instance
𝑉?
??
9.00E-06 at sample current 5 would result in a value of
.
38
5
= 7. 6Ω
Bz (V*s/cm^2)
Ix*Bz
3.00E-06
1.50E-05
3.00E-06
9.00E-06
3.00E-06
3.00E-06
6.00E-06
3.00E-05
6.00E-06
1.80E-05
6.00E-06
6.00E-06
9.00E-06
4.50E-05
9.00E-06
2.70E-05
9.00E-06
9.00E-06
-3.00E-06
-1.50E-05
-3.00E-06
-9.00E-06
-3.00E-06
-3.00E-06
-6.00E-06
-3.00E-05
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
-6.00E-06
-1.80E-05
-6.00E-06
-6.00E-06
-9.00E-06
-4.50E-05
-9.00E-06
-2.70E-05
-9.00E-06
-9.00E-06
Table 2: Magnetic Field in (V*s/cm^2) and Magnetic field(Bz) times Current (Ix)
This value of Ix*Bz would be used with the Hall voltage to get the value of p through the Hall
Effect. By plotting the Magnetic Field and Sample Current vs. Hall Voltage to confirm the carrier
concentration of the sample.
Graph 1: Hall Effect in Semiconductor
This is done through equation 3, with the slope for the fitted line means
. Then to
1
(𝑒 ? 𝑝 ? ?)
= 819388 => 𝑝 =
1
(819388)(1.6*10
−19
)(.01)
= 7. 62 * 10
14
𝑐???𝑖𝑒??/𝑐𝑚
3
calculate the mobility we would need equation 6 with
.
µ
=
σ/(𝑒 ? 𝑝)
=
.0263
(1.6*10
−19
)(7.62*10
14
)
=
215. 71 𝑐𝑚
2
/𝑉 * ?
Discussion of Results
The germanium semiconductor condutivity is a determining factor when considering if
the sample is of great quality. This being that the conductivity is needed to get the dopant
concentration amount to see if the sample value fits between
and
carriers per
5?10
14
5?10
15
. From the test we have as sample carrier concentration value of
.
𝑐𝑚
3
7. 62?10
14
𝑐???𝑖𝑒??/𝑐𝑚
3
Thus deeming the sample to be in range of the reference dopant concentration which would
make these germanium sample to be of good quality. Based on this, the doped germanium
sample is fit for use and the bulk order from the prospective supplier should be pursued.
According to the Electrical properties, the hole mobility of germanium is around
. This means there was a significant difference between the theoretical and
1900 𝑐𝑚
2
/𝑉 * ?
experimental value for mobility. A possible source of error present in this experiment was the
couple beginning test in which the direction and calibration of the equipment was not understood
fully at the time which may have changed future readings. With the measurements being a mess
at the beginning we had to retest again under different pretense setting to prevent the same error.
The values were more accurate the second time around, but the constant use after 2 whole sets of
testing could have result in it being off a bit.
Conclusion
The purpose of this experiment was to determine the quality of the doped germanium
wafers to consider a possible bulk order by SemiTech from the supplier through the utiliziation
of the Hall Effect to compare the dopant concentration of the sample to an accepted value for a
good quality sample. With the accepted values being between
and
carriers per
5?10
14
5?10
15
with our value for carrier concentration being
would confirm
𝑐𝑚
3
7. 62 * 10
14
𝑐???𝑖𝑒??/𝑐𝑚
3
that our sample of doped germanium is of great quality. This would mean that pursuing a bulk
purchase of germanium wafers from the supplier would be worth it as it meets quality standards
for similar quality germanium wafers.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
References
Laboratory Notes Materials Engineering 153 Electronic, Optical and Magnetic
Properties,
Department of Biomedical, Chemical and Materials Engineering, San Jose State
University, San Jose CA, 2014, Chapter 4, pp. 1-8
Kasap, S., “Electrical and Thermal Conduction in Solids: Mainly Classic Concepts,” in
Principles of Electronic Materials and Devices
, 3
rd
ed., McGraw-Hill, 2006, Boston, Chapter 2.5
and 2.7
B.G. Streetman, Solid State Electronic Devices (1995), Chapter 3.4.5.
Velasquez Luna, J. (2018). MatE 153 Protocol Hall Effects. Retrieved March 9, 2022
MatE 153 Hall Effect in Semiconductors-1. Retrieved March 9, 2022
Related Documents
Related Questions
The purpose of the current limiting resistor in a Zener diode circuit is to _________.
a. Maintain a constant current through the load resistor
b. Maintain a constant voltage across the Zener diode
c. Maintain a constant current through the Zener diode
d. Drop the additional voltage from the power supply in order to maintain a constant voltage across the Zener diode
Please answer it clearly.
arrow_forward
Two silicon diodes are connected in parallel and in series with resistor 0.33kohms and a battery of 10V. the voltage across the resistor 0.33k ohms is ___. The current flowing through resistor 0.33K ohms is approximately equal to___. The current of one diode is___.
arrow_forward
Can you give me a proper explanation in a structured way?? Please use text to make it easy for me.
arrow_forward
The following are a number of statements about opto-electronic devices and materials. Some are TRUE and some are FALSE.
i. You can make light emitting diodes from any direct band-gap semiconductor.
ii. You can make a laser from silicon.
iii. You can make a light emitting diode from GaAs.
iv. Light emitting diodes work in reverse bias.
v. Photodiodes often work in reverse bias.
vi. A solar cell does not require an external bias voltage to work.
Which of the following statements is TRUE:
(v) and (vi) are TRUE
(iv) and (v) are TRUE
(iii) and (iv) are TRUE
arrow_forward
11
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you


Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Related Questions
- The purpose of the current limiting resistor in a Zener diode circuit is to _________. a. Maintain a constant current through the load resistor b. Maintain a constant voltage across the Zener diode c. Maintain a constant current through the Zener diode d. Drop the additional voltage from the power supply in order to maintain a constant voltage across the Zener diode Please answer it clearly.arrow_forwardTwo silicon diodes are connected in parallel and in series with resistor 0.33kohms and a battery of 10V. the voltage across the resistor 0.33k ohms is ___. The current flowing through resistor 0.33K ohms is approximately equal to___. The current of one diode is___.arrow_forwardCan you give me a proper explanation in a structured way?? Please use text to make it easy for me.arrow_forward
- The following are a number of statements about opto-electronic devices and materials. Some are TRUE and some are FALSE. i. You can make light emitting diodes from any direct band-gap semiconductor. ii. You can make a laser from silicon. iii. You can make a light emitting diode from GaAs. iv. Light emitting diodes work in reverse bias. v. Photodiodes often work in reverse bias. vi. A solar cell does not require an external bias voltage to work. Which of the following statements is TRUE: (v) and (vi) are TRUE (iv) and (v) are TRUE (iii) and (iv) are TRUEarrow_forward11arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning


Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning