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Transcript

PHYS-102 Honor's Lab The Hall Effect – Finding the Hall Coefficient ( RH ) 1.) Objectives The objectives of this experiment are to demonstrate the the effects of a magnetic field (B) on a current carrying conductor (semiconductor/metal). If a current carrying conductor is placed in a magnetic field oriented perpendicular to the direction of the current, a voltage is developed across the conductor in a direction perpendicular to both the magnetic field and the direction of the current. This effect is known as the Hall Effect. This effect is useful in determining the nature of charge carriers ( i.e., whether a semiconductor is of n-type or p-type), the number density of charge carriers, and the mobility of charge carriers. Hall Mobility is an expression of the extent to which the Hall effect takes place in a semiconductor material. For a given magnetic field intensity and current value, the voltage generated by the Hall effect is greater when the Hall Mobility is higher. The Hall Mobility is given by the product of the Hall constant and the conductivity for a given material. In general, the greater the carrier mobility in a semiconductor, the greater the Hall mobility. In this lab you will: 1. Study the Hall Effect and determine: ● Hall Voltage V H . ● Hall Coefficient R H . 1. Determine the type of majority charge carriers ( i.e. determine wether the semiconductor sample is of the n-type or p-type ). 2. Determine the charge carrier density ( or carrier concentration per unit volume ) in the semiconductor crystal. 3. Determine the Hall Angle H . The Hall effect has many interesting applications in science and engineering. In material science the Hall Effect can be used to establish plots of majority carrier concentration and mobility vs. temperature. This is an important tool for characterizing semiconductor devices. As will be seen in this lab, the Hall Effect is also the basis for integrated circuit devices which measure magnetic fields. The Gauss Meter used in this lab uses a Hall Effect Transducer. There are a large number of integrated Hall circuits in use today. Many of these circuits are used as contactless switches and mechanical proximity sensors. 2.) Theory A charged particle in a magnetic field experiences a force that is proportional to the magnitude of the charge of the particle, to the velocity of the particle, and to the magnitude of the magnetic field surrounding the charged particle. The direction of the force is perpendicular to both the direction of the magnetic field surrounding the particle and to the direction of the velocity of the charged particle, and follows the “Right Hand Rule” as shown in Figure One. FB =q v X B 1 [ 1] Figure One: The force on a charged particle in a magnetic field. As has been shown in lecture, that a current carrying wire located in a magnetic field will experience a force due to the charge carriers moving through the wire at a drift velocity, vd . This force is is transferred to the wire by the forces that bind the charge carriers to the wire. The force on current carrying wire is proportional to the magnitude of the current I , the length of the wire L , and the magnitude if the magnitude of the magnetic field B . This force can be used to provide a torque on a current carrying loop of wire, which can then be used to create an electric motor. FB =I L X B =N I AX B wire [ 2] [3] This force can also be used to investigate properties of materials. Consider a current carrying conductor in an external magnetic field, as shown in Figure Two. Since the charge carriers experience a force due to the external magnetic field, the charge carriers will be accelerated towards one side of the current carrying conductor, but will be trapped inside the conductor. Figure Two: The force on a positively charged particle in a magnetic field. In Figure Two, a sample of a conductor with a width, w , carries a current I and is placed in an external magnetic field. A positive charge carrier, a proton, travels in the same direction as the current with a drift velocity, vd . Following the right hand rule, the positive charge carrier will experience a 2 force towards the top of the sample, but is trapped in the sample. The force experienced by the charge due to the magnetic field is equal to: = e vd X FB=q vXB B o ∣F B∣=e∣v d∣∣ B∣ sin 90 =e∣v d∣∣ B∣ [ 4] [ 5] Figure Three: The force on a negatively charged particle in a magnetic field. In Figure Three, a sample of a conductor carries a current I and is placed in an external magnetic field. A negative charge carrier, an electron, travels in the opposite direction as the current with a drift velocity, vd . Following the right hand rule, the negative charge carrier will experience a force towards the top of the sample, but is trapped in the sample. The force experienced by the charge due to the magnetic field is equal to: = −e vd X B FB=q v X B ∣F B∣=e∣v d∣∣B∣ [6 ] [ 7] As the charges build up on the top and the bottom of the sample, and electric field evolves. Considering the motion of a positive charge carriers, there is now a force due to the electric field. The net force, shown in Figure Four, know as the Lorentz Force, will be: q v X Fnet =q E B=−e E e v X B 3 [ 8] Figure Four: The forces on a positively charged particle in a magnetic field and in an induced electrical field. The net force is know as the Lorentz Force. Initially, the electric field is non-existent. As positive charges move to the top of the sample and negative charges move to the bottom of the sample, the electric field grows as the electric force become greater. Eventually a steady state evolves and the force due to the electric field and the force due to the magnetic field become equal. The two forces are equal and opposite and the net force approaches zero. q v X Fnet =q E B=−e E e vd X B o ∣F net∣=−e E e v d B sin 90 =0 e E=e v d B E =B v d [ 9] At this point the electric field will equal E=B v d as shown in equation 9. An electric potential difference across the sample will result from the separation of the positively charged top surface and the negatively charged bottom surface. This electric potential difference if known as the Hall Effect Voltage an is equal to: V H =V = E w=v d B w [ 10 ] Where V H in equation 10 is the Hall Effect Voltage, v d is the drift velocity and w is the width of the sample. Using this information, the carrier concentration and the Hall Coefficient, R H , can be calculated. Consider a conducting or semiconducting sample, with dimensions w , l , t , where the magnetic field is equal to B =∣B∣ j . This measured voltage across this sample will be positive if the majority of charge carriers are positive ( see Figure Two ) and negative if the majority of charge carriers are negative ( see Figure Three ). 4 Figure Five: Orientation of semiconducting sample.. I I e n vd A = = =e n v d , where n is the number of A wt wt charge carriers per unit volume. from Equation 10 the Hall Effect Voltage is given by: The current density, J , is equal to J= V H = E w=v d B w= BI ent [ 11 ] If the magnetic field, B , the Hall Effect Voltage, V H , the current, I , the width of the sample, w , and the charge of the charge carrier, e , are known or can be measured, then the charge concentration can be calculated. n= IB etVH [ 12 ] Also the Hall Coefficient , R H , can be calculated. The Hall Coefficient is defined as: R H≡ 1 VHt = ne B I [ 13 ] Therefore, once the Hall Coefficient is known, the concentration of charge carriers n can be found. The Hall Coefficient can also be used to find the mobility of the sample m and the Hall angle ( The angle of the Electric Field which is a combination of the Electric Field produced produced by the Hall Effect and the Electric Field producing the current through the sample). H =tan 5 −1 m B 3.) Equipment and Set-up The Equipment consists of: 1. Power Supply for the electromagnet ( 0 – 16 Volts , 5 Amps). 2. Constant Current Power Supply ( 0 – 50 mA , Ideally 0 – 20 mA ). 3. Gauss meter with Hall Probe. 4. Semiconductor sample ( Ge single crystal ) mounted on PCB. p-type Ge crystal Thickness t : 0.5 mm Width w : 4 mm Length l : 6 mm 5. Keithly multimeter with mV scale for measuring Hall Effect Voltage. 6. Hand-held multimeter with mA scale for measuring current through sample. 7. Hall effect apparatus ( electromagnet, pole pieces, and pillars.) Figure Six: Hall Effect Apparatus – Two coils of wire surrounding pole pieces and two pillars for mounting probes. Figure Seven: Hall Probe for Gaussmeter. 6 Figure Eight: Semiconducting sample mounted on PCB board with connections for Constant Current Power Supply and Voltage Measurement. Figure Nine: Block diagram of Experimental Set-up. 4.) Set-up and Procedure 1. Mount the PCB ( with mounted crystal ) on one of the pillars of the Hall Effect Apparatus and the Hall probe for the Gauss Meter on the other pillar. Figure Ten: Hall Effect Apparatus with probe and PCB mounted on pillars. 2. Connect the Hall Probe to the Gauss Meter. 3. Connect the two coils of the electromagnet in series to the DC Power Supply. 7 4. Connect the Keithly Multimeter to the Millivolt Terminals on the PCB. Multimeter (millivolt scale) Figure Eleven: Hall Effect PCB. Connected to mA Meter and in series with Current Controlled Power Supply. 5. Connect the Constant Current Power Supply to the PCB with the mA meter in series. YOU WILL BE USING THE GREEN TERMINAL FOR POSITIVE AND THE BLACK TERMINAL FOR NEGATIVE UNDER THE GREEN LETTERING CONSTANT CURRENT POWER SUPPLY. MAKE SURE THE SWITCH IS IN THE 0.05 AMP POSITION. Figure Twelve: Constant Current Power Supply. 6. The Gauss Meter is very sensitive. It must be calibrated to zero before it is used. To do this the probe must be moved away from any magnets or sources of magnetic fields, such as transformers and current carrying wires. Switch On the Gauss Meter and place the hall probe away from the electromagnet and from all power supplies. Select the X1 range of the Gauss Meter and using the zero adjustment knob of the Gauss Meter, adjust the reading of the Gauss Meter as zero. Do not switch on the electromagnet at this stage. 7. Now check to see that the voltage across the sample is zero when the magnetic field is zero. Switch ON the constant current and set the current to 5 mA in the constant current source. 8 Keep the magnetic field at zero as recorded by the Gauss Meter. Do not switch on the electromagnet at this stage. 8. Set the voltage range of the multimeter at 0-200mV. If needed set the voltage as recorded by the multimeter to be zero by adjusting the zero set pot provided on the PCB by using a screwdriver. When the current of 5 mA is passed the crystal without application of the magnetic field the Hall Voltage as recorded by the multimeter should be zero. Adjust the zero set pot only with the permission of the lab instructor. Do not switch on the electromagnet at this stage. 9. Bring the current reading of the constant current source to Zero by adjusting the knob of the constant current source. 10. Switch ON the electromagnet ( 12 VDC, 3 A Max ). 11. Select the range of the Gauss Meter to X10 and measure and record the magnet flux density at the center between the pole pieces. The tip of the Hall Probe and the crystal should be placed between the center of the pole pieces. The pole pieces should be as close to the crystal and to the tip of the Hall Probe. POLE PIECES SHOULD NEVER TOUCH THE CRYSTAL OR THE TIP OF THE HALL PROBE. The magnetic flux density should be more than 1500 Gauss for conducting this experiment. 12. Do not change the current in the electromagnet. Keep the magnetic field constant throughout the entire experiment. 13. Vary the current through the constant current source in very small intervals between 0 mA and 20 mA. It is suggested that you start with a reading near 20 mA and then decrease the current by intervals of about 1.0 mA and record the data. Do not exceed 20 mA or you will destroy the crystal. Complete 15 trials at roughly 1.0 mA intervals. Record the Hall Effect Voltage and Current for each interval in the table provided. 14. Plot V H versus I . 15. Compute the the ratio of Hall Effect Voltage divided by the Current in the column provided. V H mV I mA 16. Compute the mean of the ratio of Hall Effect Voltage divided by the Current. 17. Use this mean to compute the Hall Coefficient. VH t RH= I B 18. If the Hall Coefficient is positive, the material is a p-type semiconductor material. If the Hall Coefficient is negative, the material is a n-type semiconductor material. 19. Compute the Concentration of Charge Carriers. 1 n= e=1.602 X 10−19 C e RH 20. Measure the distance between contact points used to measure the Voltage (L). Compute the Resistivity of the sample. V wt r= L LI 21. Compute the mobility. R m= H r 22. Compute the Hall Angle. −1 H =tan m B 9 Name:_______________________________ Sec./Group ______________ Date: _______________ 5.) Observations Width of Specimen w : ________________________ m Length of Specimen l : _______________________ m Thickness of Specimen t : _______________________ m Distance between Contact Points for Voltage Readings L : _______________________ m Magnetic Flux Density B : _______________________ Gauss = ______________ X 10-4 Tesla Trial Number Current I mA Hall Voltage V H mV Ratio of Hall Volatge to Current, VH I 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Mean Value of Ratio:_______________________W VH t :_______________________mC-1 I B Type of Crystal ( n-type or p-type ) : _________________________________ 1 Carrier Concentration n= : __________________________________ carriers per m3 e RH Hall Coefficient 10 RH= Name:_______________________________ Sec./Group ______________ Date: _______________ Width of Specimen w : ________________________ m Length of Specimen l : _______________________ m Thickness of Specimen t : _______________________ m Distance between Contact Points for Voltage Readings L : _______________________ m Magnetic Flux Density B : _______________________ Gauss = ______________ X 10-4 Tesla Trial Number Current I mA Distance between Two Points between which V is measured. l m Hall Voltage V L mV Resistivity V wt r= L IL m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 R Mobility m= H r Mean Value of Ratio:_______________________W :_______________________ m2 V −1 s −1 −1 Hall Angle H =tan m B : __________________________________ 11 6.) Analysis 1.What are the experimental errors that you encountered while conducting this experiment? 2.What are your uncertainties in the measurement of the Hall Effect Voltage, the current through the sample, the magnetic field and the distance between the contact points? 3.Estimate the maximum uncertainty in computing the Hall Coefficient. 12 Name:_______________________________ Sec./Group ______________ Date: _______________ 7.) Not Just for Fun ( 10 points Extra ): Prior to the development of Hall effect transducers, there were various other ways to measure a magnetic field. Consider a coil of wire carrying and alternating current. A magnetic field with a relatively large magnitude will develop in the center of the coil. This magnetic field will alternate at the same frequency as the alternating current running through the coil. In order to measure this field we can use a small “search” coil and watch the voltage induced in this small coil. The magnetic field will be proportional to the voltage induced. Recall that Faraday left us a model of how electricity can be induced from magnetic flux that changes with time. =−N d B d d =−N B⋅ A =−N ∣ B∣∣ A∣ cos dt dt dt Where N is the number of turns, A is the cross sectional area of the coil, B is the magnitude of the magnetic field, and Q is the angle between the field and the area vector. In our case, we will keep the everything constant except for the alternating magnetic field. In this just for fun, you are given an oscilloscope, a function generator, a coil to provide a magnetic field, various cables, and a search coil. Your job is to get a rough idea of how the magnitude of the magnetic field around a current carrying coil varies as the distance from the coil varies. Use the equipment provided. Good luck! Take ten readings about a centimeter apart and measure the magnetic field along the center of the coil, on either side of the coil. Make a quick plot of your data using a spreadsheet. Did you get what you expected? Explain, in detail, how you measured the magnetic field. 13 14 Name:_______________________________ Sec./Group ______________ Date: _______________ Pre-Lab Hall Effect Honor's Lab 1.A conducting rod is pulled through a magnetic field B at a constant speed v . The magnetic field is into the page and the rod is pulled in the positive x-direction. Draw the forces felt by the electron and the proton, due to electric fields and magnetic fields, as the conducting rod is pulled through the magnetic field. At first, there is only a magnetic field. at a constant speed v . The magnetic 2.A conducting rod is pulled through a magnetic field B field is into the page and the rod is pulled in the positive x-direction. As the positive and negative charges move to the top and bottom of the rod, an electric field evolves. Draw the forces felt by the electron and the proton, due to electric fields and magnetic fields, as the conducting rod is pulled through the magnetic field. ++++ E ---- 15 3.A conducting rod is pulled through a magnetic field B at a constant speed v . The magnetic field is into the page and the rod is pulled in the positive x-direction. As the positive and negative charges move to the top and bottom of the rod, an electric field evolves. Prove that the voltage induced across the rod, when a steady state condition is reached, is equal to ∣∣=∣ B∣∣v∣ l , where l is the length of the rod. 4.If the magnitude of the magnetic field is ∣ B∣=2T , and the speed of the rod is ∣v∣=10 m/ s , and the length of the rod is 1.0 meters, what is the potential difference induced across the rod? 16