PCS224 Lab 2 Report

Faculty of Science Department of Physics Laboratory Report Cover Page Course Number PCS 224 Course Title Solid State Physics Semester/Year Fall 2022 Instructor Dr. Rebello TA Name Jason Bartlett Lab/Tutorial Report No. 2 Report Title The Hall effect Section No. 07 Group No. 87 Submission Date October 19, 2022 11:15am Due Date October 19, 2022 2:00pm Student Name Student ID Signature* Ibrahim Abdul Aziz ****84022 IA Omid Mazinani ****14393 OM (Note: remove the first 4 digits from your student ID)

Introduction In this lab, which is about the Hall effect, we are going to measure the hall voltage of a Germanium wafer with different currents along its width and its length. This wafer has a current flowing in the L direction, its magnetic field points along the thickness, and finally the hall electric field points along the width. The main goal of this experiment is to calculate the “‘n” (electron concentration) and the mobility using our voltage and current data and also by looking at the sign of the voltmeter and our experimental setup, we could determine if our wafer(chip) is a p-type or n-type material, in addition, we can state what is the majority charge carriers in our material. The measurements that we will be taking in this experiment are the current, voltage along width, voltage along length, and ultimately the magnetic field. Theory We are going to use a variety of different equations and rules to aid us throughout this lab, first of all is how to determine the majority charge carriers in a semiconductor and the sign of the voltmeter. There are two cases for a material, We are using this as an illustration to show n type.

have to use a different equation to express the velocity. We know |Δ??| = ? * |𝑣?| * |?| Rearranging and we get → so now we have an equation that |𝑣?| = |Δ??| / ?|?| represents the velocity. Using our current equation we solve for n → and ? = ?/?|𝑣?|? now that we have the velocity expression we put it in the equation→ / ? = ??|?| where W is the width, t is the thickness, and e is the charge. We cross out the ???|Δ??| width and we have→ / , we can use this further in our report to get ? =− ?|?| ??|Δ??| the n using the plot of I vs . we have the same equation for p concentration as well Δ?? without the negative / . ? = ?|?| ??|Δ??| Equation for mobility → since we know the drift velocity and so µ? = 𝑣? 𝐸 𝐿 𝐸 𝐿 = ∆?𝐿 𝐿 substituting the values and we get an equation for mobility→ µ? = ∆?? ?·? × 𝐿 ∆?𝐿 = ∆??·𝐿 ?·?·∆?𝐿 Procedure 1. Set the Zero-Gauss Chamber to zero and insert the Gauss meter probe inside the hole to “zero” it (wait till you get zero reading). 2. Set the Zero-Gauss Chamber to measure and insert the Gauss meter probe with the lettering facing down in between the magnets. Take measurements of the magnetic field and it’s direction from left, middle and right to get an accurate reading. 3. Connect the cables with banana plug terminations to the germanium wafer board and the power supply in accordance with figure one. 4. Connect the multimeter across the width of the wafer board. Again, refer to figure one. 5. Turn on the current source (power supply) with the current set to 50mA. Do not exceed this and take special care to make sure the wafer is away from the magnet. 6. Adjust the potentiometer until the multimeter outputs a reading of zero. 7. Insert the wafer in between the magnets making sure that the scratch along the plexiglass lines up with the center of the poles. 8. Starting from 50mA, decrement the power supply by 5mA while recording the hall voltage each time. 9. After reaching 1.4mA (the lowest it can go), remove the wafer and connect the multimeter to measure the voltage drop across the length of the wafer 10.Repeat steps five to nine 11. Remove the wafer, disconnect all cables and turn off all devices

Results and Calculations When setting up our experimental setup as the one shown in the lab manual( figure 1) we can see that our current is flowing from the positive lead of the power supply to the negative end of the power supply or in other words it is flowing counter clockwise. Our magnetic field meter was positive so this means that the direction is downwards ,therefore, by knowing the direction of current and magnetic field we can use the right hand rule to determine the force's direction. We put our fingers in the direction of current and curling them in the direction of B( downwards or along the thickness), then we can see that our unknown charge carriers (either positive or negative) are being pushed to the left side of the wafer or in others words, they are being pushed to the side where the positive lead of the voltmeter is connected too. Since, from our experiment, we know the sign of the voltmeter is negative, we can say that negative charges are built on the positive side; the majority charge carriers are electrons, hence it is a n-type material. Magnetic Field: Left Side = 0.241T Right Side = 0.243T Middle Side = 0.204T Average magnetic field=0.229T ±0.0005T I (A) ±0.00005A H (V) ±0.0005V Δ? L (V) ±0.005V Δ? 0.05 -0.197 4.83 0.045 -0.189 4.34 0.040 -0.18 3.85 0.035 -0.171 3.37 0.030 -0.161 2.88 0.025 -0.151 2.39 0.020 -0.041 1.9 0.015 -0.031 1.43 0.010 -0.021 0.95 0.005 -0.011 0.47

Slope of I vs H =|-0.5055|=0.5055 Δ? Slope of H vs L = |-0.0204|=0.0204 Δ? Δ? To find the carrier concentration (n): To see derivation of the formula refer to the Theory section: -( )/( H ) ? = ? * |?| ? * ? *Δ? → ? = ????Δ??/?|?| ? = ???Δ??/|?| Comparing to a line equation y=mx+b→y= =x =slope ? Δ?? ???/|?| n= |?| * ?????/?? n= 0.229 𝑇 ×0.5055 0.001?×1.602×10 −19 ? = 7. 23 × 10 20 ? −3 Unit of n is → we know so we write each as their units→ n= we ? = −?|?| ??|Δ??| − ? ? × ?·? ? 2 ?·?·? cancel and the only things remaining are ? = [ 1 ? 3 ] To find carrier mobility: = (|v d |)/E L (from pre-lab) µ V d = H /(B*W) Δ? E L = L /L Δ? µ? = ∆?? ?·? × 𝐿 ∆?𝐿 = ∆??·𝐿 ?·?·∆?𝐿 y=mx+b → so slope is ∆? ? = µ·?·? 𝐿 × ∆? 𝐿 µ·?·? 𝐿 µ? = 𝐿×????? ?·? = 0.01?×0.0204 0.005?×0.229𝑇 = 0. 17817 ? 2 ?·? ?? 1781 ?? 2 ?·? To find percent error: Theoretical value for n-type germanium = µ ? = 3900 𝑐? ^2/( 𝑉 ·s) % 𝐸𝑟𝑟?𝑟 = (| 𝐸𝑥?𝑒𝑟𝑖?𝑒?𝑡𝑎? − 𝑇ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? |)/ 𝑇ℎ𝑒?𝑟𝑒𝑡𝑖𝑐𝑎? × 100% % 𝐸𝑟𝑟?𝑟 = ( error with n type 1781 − 3900 | |)/3900 × 100% = 54. 33% error with p type %𝐸???? = ( 1781 − 1900 | |)/1900 × 100% = 6. 26% Since our material is n type we have a 54.33% error, we could have said because our wafer is p type because its mobility was close to 1900, however, there are experimental errors and equipment uncertainties which cause an error of 54.33%.

Discussion and Conclusions In this experiment we measured the voltages along the width and length with different currents to see the pattern that they have and also we plotted I and Vh in order to get our n concentration using the useful eation we had in our pre lab. We also plotted Vh and VL to calculate the mobility. One important observation that we made was that, if the magnetic field increased we would have a bigger n concentration or in other words, they are proportional. On the other hand, the mobility is inversely proportional to the magnetic field, if we increase the magnetic field, we have a smaller mobility. Additionally, we came to the conclusion that knowing the sign of the voltmeter and also, how our experiment is set up, are extremely important in order to tell if our material is n type or p type. Wrap-Up Questions 1) It depends on the charge carriers. In the case of n-type semiconductor, the negative and positive charges are pushed to opposite sides of the semiconductor creating an electric field as well as a potential difference. This being a n-type semiconductor, the negative charge carriers outnumber the positive charge carriers making the potential difference reading negative and vice versa for a p-type semiconductor. It also depends on the way the multimeter is configured. The scenario described above, only happens if the positive and negative terminals of the plugs align with the charge carriers. For example, the negative terminal has to be connected to the side collecting electrons to get a negative reading and vice versa to get a positive reading. If it were the other way around, the readings would switch. 2) We expect out conductivity to be greater than the intrinsic conductivity of germanium because our wafer is not intrinsic, since it is n-type it means it’s electron concentration is much greater than the intrinsic concentration→ >ni ? 0 order to calculate the conductivity, we need n,e, and . since we have all the µ values we can find the conductivity of our wafer→ σ = ??µ = (7. 23 × 10 20 1 ? 3 )(1. 602 × 10 −19 ?)(0. 17817 ? 2 ?·? ) = 20. 63 1 ? · ? ?·? = 20. 63 1 ?·Ω 3) Based on the formula for H , where H = (I*B)/(nqt B ), H it can be seen that Δ? Δ? Δ? H is directly proportional to B (the magnitude of the magnetic field) so if all else Δ?