low temperature transport - department of physics

Low Temperature Transport

Experiment LTT

University of Florida — Department of PhysicsPHY4803L — Advanced Physics Laboratory

Objective

Temperature has dramatic effects on thetransport properties of metals and semicon-ductors. The electrical resistance of pure met-als can decrease by several orders of magnitudeas temperature is reduced from room temper-ature to cryogenic temperatures. The junctionvoltage of a p-n junction increases in a charac-teristic way as temperature is lowered, allow-ing the Si diode to be used as a thermometer.In this lab you will study these properties attemperatures between about 10 and 325 K.

References

Charles Kittel, Introduction to Solid StatePhysics, Wiley (2004), ISBN 978-0-471-41526-8

M. Omar Ali, Elementary Solid StatePhysics, Addison-Wesley (1975) ISBN 0-201-60733-6.

Theory

The following is a very brief introduction tothe subject matter involved in this experi-ment. You are expected to read a solid statetextbook such as Chapters 6–9 of Kittel orChapters 3–7 of Ali to gain a good foundationon the topics discussed here.

Metals physics

A metal is a solid in which one or more elec-trons per atom are free to move throughoutthe solid. In the language of band theory,the highest occupied band is partially filled,so that the Fermi surface lies in the middle ofthe band, and there is an infinitesimal energybetween the highest occupied orbital and thelowest unoccupied one.

The simplest model of a metal is the free-electron model, where the atomic potential istaken to be a constant, U0, and the elec-trons are treated as non-interacting. Thenthe Schrodinger equation for each electron be-comes

− h2

2m∇2ψ(r) + U0ψ(r) = Eψ(r), (1)

where h is Plank’s constant, m the electronicmass, and E the energy eigenvalue. Withoutloss of generality, we can take U0 = 0 andfind that there are plane wave solutions forthe eigenfunctions ψ(r):

ψ(r) =

√1

Veik·r, (2)

where V is the sample volume and the fac-

tor√

1/V satisfies the normalization condi-tion

∫ψ∗ψdV = 1. The quantity k is the

wavevector of the plane wave. Its magnitudek is related to the deBroglie wavelength λ via

LTT 1

LTT 2 Advanced Physics Laboratory

Figure 1: Dispersion relation (E as a functionof k) for free electrons. The Fermi energy andFermi wavevectors are indicated

k = 2π/λ and k is related to the electron mo-mentum p via p = hk.

The eigenenergy E appearing in Eq. 1 (forU0 = 0) is the electron kinetic energy p2/2mand thus depends only on the wavevector mag-nitude:

E(k) =h2k2

2m, (3)

The allowed values of k are determined bythe boundary conditions. It is conventionalto use periodic boundary conditions, wherethe sample is taken to be a cube of edge L,V = L3, and the wavefunction is required tobe periodic:

ψ(x+ L, y, z) = ψ(x, y, z), (4)

and similarly in the y and z directions. Thiscondition1 leads to quantized values for each

1One can come to the same conclusions by assum-ing fixed boundary conditions, with ψ(x, y, z) = 0 forx ≤ 0 and x ≥ L, and similarly for y and z.

component of k:

ki = 0,±2π

L,±4π

L,±6π

L, · · · i = x, y, z.

(5)In other words, quantization leads to allowedwavevectors k = (kx, ky, kz) lying on a Carte-sian grid in a three dimensional “k-space” withthe spacing between grid points in all threedirections given by 2π/L. Because the energyE(k) depends only on the magnitude of k, it isconstant on the surface of a sphere in k-space.

Electrons are fermions and obey the Pauliexclusion principle. Hence each of the allowedwavevectors in the grid can be occupied byat most two electrons, one spin up and onespin down. The ground state for all electrons(lowest energy state occupied at zero temper-ature) is then found by filling up the allowedk-space grid, two electrons per point, start-ing from the lowest energies (smallest k) andworking up to higher energies (larger k) untilall the free electrons are assigned. Thus theground state has two electrons at all k-spacegrid points inside a sphere—called the Fermisphere—of some radius kF—called the Fermiradius or Fermi wavevector.

From Eq. 5, the volume of each allowed statein k-space is (2π/L)3. Thus the total num-ber of states N in a Fermi sphere of volume4πk3F/3 is

N = 2 · 4πk3F/3

(2π/L)3=

V

3π2k3F , (6)

where the factor of 2 comes from the spin de-generacy. Note that N should be the sameas the total number of electrons in the solid,which is related to the number density n byN = nV . Using this in Eq. 6 allows the Fermiradius kF to be determined in terms of a fun-damental property, the free electron density.

kF = (3π2n)13 . (7)

August 31, 2021

Low Temperature Transport LTT 3

The surface of the Fermi sphere is called theFermi surface and is where the most energeticelectrons in the ground state lie. Their kineticenergy EF is called the Fermi energy and isgiven by

EF =h2k2F2m

. (8)

Exercise 1 Copper has one free electron peratom. The density and atomic weight ofcopper are 8.96 g/cm3 and 63.55 g/mole,respectively. Avogadro’s number is 6.02 ×1023 atoms/mole, h = 1.05 × 10−34 J s, m =9.1× 10−31 kg, and 1 eV = 1.6× 10−19 J. (a)Use this information to show that n = 8.5 ×1022 electrons/cm3, that kF = 1.4× 108 cm−1,that the deBroglie wavelength λ = 4.5 A, andthat EF = 7.0 eV. One may also define aFermi velocity via vF = hkF/m. Show thatcopper has a Fermi velocity of 1.6× 108 cm/s,about 0.5% of the speed of light. (b) Supposean electron on this Fermi surface is moving inthe x-direction, i.e. kx = 1.4×108 cm−1. Cal-culate the difference between its energy leveland the next higher energy level (one grid pointover in the kx-direction) in eV and as a frac-tion of the Fermi energy. Take the sample tobe 1 cm in size and be careful that the sub-traction of two nearly equal energies that onlydiffer in a high order digit is performed withsufficient precision.

As part (b) of this exercise demonstrates forenergy, the very fine spacing of the quantumstates relative to the highest occupied statesimplies that the wavevectors, momenta, andenergies can often be regarded as continuous.

The free electrons in the metal have chargeand will be accelerated by an external electricfield, E. Applying Newton’s second law, wecan write

F =dp

dt= −eE. (9)

Consequently, in the absence of collisions, theelectron velocities change at a uniform rate

dv/dt = (1/m)dp/dt = −(e/m)E so that af-ter a time t they are all shifted by the amount

δv = tdv

dt= − e

mEt. (10)

Collisions with impurities, surfaces, and lat-tice vibrations (phonons) will stop the accel-eration, relaxing the electrons to equilibrium.If the mean collision time is τ , the average ve-locity of each electron—the drift velocity vd—becomes

vd = − e

mEτ (11)

The electric current density j is the chargedensity times the average velocity

j = n(−e)vd =ne2τ

mE. (12)

This gives Ohm’s law, j = σE, with the con-ductivity σ given by

σ =ne2τ

m. (13)

Exercise 2 (a) The resistivity is the inverseof the conductivity: ρ = 1/σ and for pure cop-per at room temperature ρ = 1.68 µΩ-cm. Usethis value to calculate the room temperaturecollision time τ for copper. (b) The power Pabsorbed by each accelerating electron is F · vand thus has an average value P = |eEvd|which must continuously be dissipated as Jouleheat. Show that this leads to a power dissipa-tion per unit volume (for all electrons) givenby p = E2/ρ. Show that the magnitude of theelectric field that would require (the fairly sig-nificant) power dissipation of p = 1 kW/cm3

is around 5 V/m. (c) At this electric field,what would the drift velocity be? Also ex-press vd as a fraction the Fermi velocity. (d)The Pauli exclusion principle assures that onlyelectrons near the Fermi surface participate inscattering events (and affect the conductivity)because electrons inside the Fermi surface have

August 31, 2021


Figure 2: Fermi surface of a free-electronmetal in the absence (left) and presence (right)of an electric field.

no nearby unoccupied states to scatter into.Part (c) demonstrates that the electrons ac-quire a very small average drift velocity in thedirection of the electric field. However, duringone collision time τ electrons near the Fermisurface travel through the metal without a ve-locity changing collision and thus the averagedistance they move between collisions—calledthe mean free path `—is ` = vF τ . Calculate` for copper. Approximately how many cop-per atoms does an electron pass between colli-sions?

In the absence of an electric field, there is nocurrent and electrons fill the Fermi sphere. Forevery electron with one wavevector k there isanother with the opposite value −k. The twomove in opposite directions and the net cur-rent due to each pair is zero. This is illustratedon the left in Fig. 2 where the Fermi sphere iscentered at the origin. In an electric field E inthe +x-direction, the electrons move with anaverage drift velocity vd in the −x-directionand thus the Fermi sphere is maintained at afinite displacement δk = mvd/h. This is illus-trated on the right in Fig. 2 where the Fermisphere (solid circle) is centered off to the left.The magnitude of δk is greatly exaggerated inthe figure for clarity.

For the shifted sphere, one might argue thatevery electron wavevector is shifted by thesmall amount δk and each contributes to thecurrent producing an overall current densityj = −nevd. Keeping in mind the magnitudeof the shift δk << kF , one might also arguethat the large majority of electrons may stillbe considered to occur in pairs with oppositewavevectors. As for the unshifted sphere, eachsuch pair would contribute nothing to the cur-rent. As the figure shows, only electrons inthe shaded crescent (really just a sliver verynear the original, centered Fermi surface) donot have an oppositely moving counterpart.Thus, one might then argue that only thesefaster moving electrons near the Fermi surfacecontribute to the net current. It turns out thislatter view is more accurate. The number den-sity of such electrons is much lower (≈ nvd/vF )but because they move much faster (≈ vF ),the current density is the same.

Scattering rates

The resistivity ρ and conductivity σ are in-verses ρ = 1/σ and thus according to Eq. 13we can express the resistivity as

ρ =m

ne21

τ(14)

with the interpretation that τ is some meancollision time. This same interpretation means1/τ is an average collision or scattering rate.If τ = 10−14 s, then the electron undergoes1/τ = 1014 collisions per second, i.e., that thescattering rate is 1014/s.

There are basically two scattering mecha-nisms. Scattering by impurities, lattice faults,the surface, or other defects is expected to betemperature-independent with a rate 1/τi thatdepends on how the sample was made. Thisscattering rate can vary greatly between sam-ples of the same material made in different labs

August 31, 2021


or in different ways. Scattering by lattice vi-brations (phonons) is temperature dependentand the rate 1/τph depends on the intrinsicproperties of the metal and does not vary sig-nificantly among different samples. The twoscattering mechanisms are independent andtheir rates are thus additive

1

τ=

1

τi+

1

τph(15)

Substituting this in Eq. 14 implies the resis-tivity can be decomposed into two terms: atemperature independent part ρi due to scat-tering by impurities, etc., and a temperaturedependent part ρph(T ) due to scattering byphonons:

ρ(T ) = ρi + ρph(T ). (16)

This decomposition is known as Matthiessen’srule.

At high temperatures (above the Debyetemperature) the number of phonons per unitvolume is proportional to the temperature andρph is proportional to T . At low tempera-tures, two effects come into play. The num-ber of phonons falls as T 3; moreover, the en-ergy and momentum of these phonons becomesmall, so that they are ineffective in scatteringelectrons. Consequently the resistivity fallsmore quickly than the phonon density. Blochshowed that

ρph(T ) ∝T T ≤ ΘD

T n T ΘD(17)

where ΘD ≈ 343 K is the Debye temperaturefor copper and n is theoretically n = 5 butemperically is in the range 3–5.

At the lowest temperatures ρph goes to zeroand the overall resistivity reduces to ρi; con-sequently ρi is called the residual resistivity.

Semiconductors

In pure semiconductors the highest occupiedband—the valence band—is completely filled

and there is an energy gap between this andthe lowest unoccupied band—the conductionband. Thus at low temperature there are al-most no mobile carriers and the resistivity ap-proaches infinite values.

Semiconductor devices are made possiblebecause semiconductors can be doped withdonors or acceptors to control the electricalproperties. Donors are typically atoms withone more valence electron than the semicon-ductor whereas acceptors have one less. Ex-amples of donors are pentavalent impurities inGe or Si, such as P, As, and Sb; examples ofacceptors are trivalent atoms, such as B, Ga,and In. In general, donor levels lie close tothe edge of the conduction band and accep-tor levels lie close to the edge of the valenceband; the carrier density is equal to the donoror acceptor density.

Donor-doped materials conduct via elec-trons in the conduction band. The mate-rials are called n-type because the electronshave negative charge. Acceptor-doped mate-rials conduct via “holes” (an empty electronstate) in the valence band. The materials arecalled p-type because the holes have positivecharge.

The simplest semiconductor device is the p-n junction diode. The diode energy band dia-gram is shown in Fig. 10 (located at the end ofthis writeup) for the unbiased, forward-biased,and reverse-biased cases. (If you want to playwith the parameters in a p-n junction takea look at www.acsu.buffalo.edu/∼wie/app-let/pnformation/pnformation.html.)

One side of the crystal is doped p type, theother n type. A thin (∼ few µm thick) junc-tion separates these two sides. Away from thejunction region on the p side are negatively-charged acceptor ions and an equal numberof free holes. On the n side are positively-charged donor ions and an equal number ofmobile electrons. In addition both sides have

August 31, 2021


a small number of thermally generated “mi-nority” carriers of the other type (holes in then region and electrons in the p region).

Both holes and electrons tend to diffusethrough the crystal. Electrons from the n re-gion diffuse across the junction and recom-bine with holes in the p region (and viceversa). These recombinations leave the n-region depleted of electron carriers and pos-itively charged and they leave the p-region de-pleted of hole carriers and negatively charged.The recombinations occur over a thin deple-tion layer around the junction. This charged,double layer grows until the electric field itproduces is strong enough to inhibit any fur-ther diffusion of electrons across the junction.

The inhibition is not complete (at least atfinite temperatures) and the most energeticelectrons and holes (those in the tails of theBoltzmann distribution function) can crossthe barrier. At zero applied voltage, the cur-rent due to electrons diffusing into the p regionis canceled by the minority electrons diffusingin the opposite direction. Similarly, there is acancellation between the currents due to holesand minority holes.

A voltage applied across the junction can ei-ther increase or decrease the height of the po-tential barrier. If the applied voltage is suchthat the p region is positive (the p region isconnected to the positive electrode of a bat-tery and the n region to the negative elec-trode), the junction is said to be biased in theforward direction. If the voltage is applied inthe opposite direction, the junction is said tobe biased in the reverse direction. When thejunction is biased in the forward direction, thebarrier height is reduced and the current in-creases rapidly. In contrast, a reverse bias in-creases the barrier height and produces only asmall reverse current up to a saturation limit−I0 set by the thermal generation of minoritycarriers. To create a minority carrier (hole) in

the n-region, an electron must overcome theband gap energy Eg (in silicon about 1.2 eV)via its thermal energy and governing Boltz-mann distribution. Consequently I0 is verytemperature dependent and given by

I0 = I00e−Eg/kT (18)

where k is Boltzmann’s constant and T is thetemperature.

The current I as a function of applied biasvoltage V is predicted by the Shockley diodeequation.

I = I0(eeV/kT − 1

)(19)

where I0 is called the reverse saturation cur-rent because I = −I0 when the diode is reversebiased (V < 0) sufficiently that eeV/kT << 1.

This model assumes all of the current arisesfrom recombination in the depletion region. Asomewhat more detailed analysis of recombi-nation and other conduction processes leads toa modification of the formulas for both I0 andI placing an ideality factor η in the denomi-nators of both exponentials

I0 = I00e−Eg/ηkT (20)

I = I0(eeV/ηkT − 1

)(21)

The ideality factor depends on the details ofthe diode fabrication and typically varies be-tween 1 and 2. It may also vary with volt-age. Even with the ideality factor, there canbe deviations from the predictions of Eqs. 20and 21. For example, real diodes have a se-ries resistance Rs, typically on the order of anohm or so, which would imply that the V inEq. 21 should be replaced by V − IRs. Realdiodes also have a large parallel resistance,typically over 10 kΩ, which adds a small ad-ditional ohmic current to the measurements.In addition, other details of the diode current-voltage-temperature dependences can becomeimportant, particularly at very low tempera-tures.

August 31, 2021


Apparatus

You will make three measurements at temper-atures between about 10 and 325 K. There isan option of making a fourth measurement.

1. the resistance of a thin copper wire about1 m long;

2. forward voltage of a 1N914 silicon diodeat several fixed currents;

3. IV characteristics of the diode at severaltemperatures.

4. (Optional) the resistance of a thin goldfilm;

Refrigerator and temperature controller

The cold temperatures are achieved using aclosed cycle helium refrigerator called “thecooling machine” in this writeup. The prin-ciples of operation are briefly discussed next.

The metal cylinder at the top half of thehelium refrigerator is the vacuum shroud. Atthe base of the shroud and inside it, is the ex-pander consisting of valves and a sealed col-umn. The samples are mounted on a cop-per block attached to the top of the column.High-pressure helium expands inside the col-umn thereby cooling the column and the sam-ples. The helium never enters the volume be-tween the shroud and the column. It staysinaccessible inside the column before return-ing to the compressor (blue box under the labtable) where it heats up as it is compressedback to high pressure. The hot, high-pressurehelium is cooled in a heat exchanger by waterfrom the Cool-Pak (white box under the labtable) before going back to the expander.

The vacuum shroud is exposed on its outsidesurface to the room and stays near room tem-perature while the very cold column is only afew centimeters away. Any gas in the volume

between the shroud and the column would cre-ate a heat conduction path between them andprevent the column from reaching the low tem-peratures needed. That is why the air in thisvolume must be pumped away with a vacuumpump. In addition, there is a separate radi-ation shield between the vacuum shroud andthe column to reduce radiative heating of thecolumn by infrared rays from the room tem-perature vacuum shroud. There are ports inthe shroud and cutouts in the shield near thesample area to allow laser beams, for example,to illuminate the samples. We do not use thiscapability. In fact, the cutouts on the radia-tion shield should be wrapped with aluminumfoil to get a small reduction in the radiativeheating and thus a small reduction in the low-est achievable sample temperature.

There are no user adjustments for the cool-ing accomplished by the helium expansion.When the compressor switch is turned on, allcomponents of the cooling machine (expander,compressor, and Cool-Pak) begin operating.A heater (a current-carrying wire wrappedaround the outside of the expander column) isused to adjust the sample temperature. Withthe heater turned off, it takes the cooling ma-chine about 60 minutes to reduce the tem-perature from room temperature to the min-imum achievable temperature (around 7 K).You should plan on making measurements ofthe copper wire during one day of experiment-ing and measurements of the diode on a secondday. The optional measurement of the goldfilm could be done on the next-to-last day orlast day if time permits.

Temperature Control

A LakeShore Model 330 Temperature Con-troller is used to measure the temperature andcontrol it using the heater. It has two silicondiode thermometers.

August 31, 2021


Diode A is on the sample holder and mea-sures the “sample” temperature.

Diode B is at the top of the expander columnjust below the sample holder and mea-sures the “control” temperature.

The temperature controller can be con-trolled remotely through a GPIB interfaceor locally using the keypad. The controllerwill automatically switch from local to remotemode any time the computer sends a com-mand over the GPIB interface, but you mustplace it in local mode using the top right Lo-cal keypad button to use the keypad. An in-dicator in the upper LED display shows REM(remote) or LOC (local) to indicate the status.

Because it is easy to make mistakes us-ing the keypad, it is recommended that youuse the Control setpoint and ramp program tochange the setpoint. (More on this in a mo-ment.) If, using the keypad, you get into anysettings you do not understand or do not wantto change, hit the Escape key. If you think youmay have changed some parameters and needto get back to the “power on settings” (the set-tings before you changed them), simply turnoff the LakeShore 330 controller (not the cool-ing machine), wait a few seconds and turn itback on.

You ask for a new temperature by changingthe setpoint. The heater power then changesand then the controller waits to see how the(control) temperature changes. The tempera-ture can overshoot its mark at which point theheater power changes in the other direction,etc. The PID parameters (for proportional, in-tegral, and derivative) help determine how thefeedback loop between the thermometer andthe heater operates. In addition to the PIDparameters, there are three Heater Range ormaximum power settings: (Low, Med, High)corresponding to (50, 5, 0.5) W of power.

Because the thermal properties change as

Figure 3: Temperature changes with coolingmachine on and heater on at maximum power(warming) or off (cooling).

the temperature changes, the PID parametersand heater range need to be set differently indifferent temperature zones. These settingshave been determined empirically and can bechanged if you notice that any setpoint tem-perature causes large oscillates or takes toolong to reach. Consult with the instructor onthe proper procedure for doing this.

Changing the Temperature

The lowest reachable (base) temperature isaround 7 K while the highest reachable tem-perature is around 325 K. With the heateroff, it takes about one hour to get from thehighest to the lowest temperature. With theheater on maximum power, it takes about 35minutes to get from the lowest to the high-est. Fig. 3 shows graphs of temperature vs.time for these two cases and demonstrates therapid temperature changes that occur below100 K.2

Some of the measurements will be made ver-sus temperature at 1 K increments while thetemperature is made to fall smoothly in timeusing the ramping feature of the controller.

2The change is largely due to a decrease in the heatcapacity of the components at the lower temperatures.

August 31, 2021


The ramp causes any change in the setpointto be performed automatically at a uniformrate that can be set from 0.1 to 99.9 K/min.Setting the ramp rate to 0.0 make the setpointchange immediately.

Keep in mind that ramping the setpoint isnot necessarily the same as ramping the tem-perature. The controller still simply comparesthe setpoint and control temperature and ad-justs the heater as it tries to get them equal.Thus, the temperature can still overshoot andoscillate due to the controller’s feedback sys-tem. More importantly, the maximum rate oftemperature change is still limited by the heat-ing and cooling powers available, i.e., Fig. 3.

Measurements and LabVIEW programs

A LabVIEW program is called a vi (pronouncevee-eye for virtual instrument). If you makechanges to any of the vis used in this labora-tory, you would use the Save As item in the Filemenu of that vi, and save a copy to your ownarea in the My Documents folder. All Saveitems in the File menu save the vi; THEYWILL NOT SAVE DATA.

Every vi that collects data in this exper-iment will ask you to provide a file namefor storing any resulting data. Work yourway to the My Documents folder (you mightwant to create a subfolder therein) click intoit, and change the default file name (typi-cally DATA.TXT) to something more descrip-tive. Record the filename in your lab note-book. If this step fails, the vi terminates with-out taking any data. If it succeeds, the vi willsave the data to this file when data collection iscomplete and you click on the red-letter STOPbutton near the graph to stop the vi. Mostother ways of stopping the vi will prevent thedata saving step from executing and the col-lected data will be lost.

A vi is run by clicking on the run button

(right-pointing thick white arrow) in the toolbar. This arrow turns black while the vi runsand back to white when the vi stops running.

Instructions for changing the tempera-ture

1. Open the Control setpoint and ramp ratevi. Click on the Run button in the toolbar. This vi reads and displays the sam-ple temperature, the control temperature,the setpoint, and the ramp rate when youclick on the Read button. It sets the set-point and ramp rate when you click onthe Write button.

2. Changing the setpoint immediately:If you just want to get to the desired tem-perature as fast as possible, set the Set-point control to the desired value, set theRamp rate control to 0.0, and click on theWrite button.

3. Ramping the setpoint: First Read thecurrent setpoint and if it is not where youwant the temperature ramp to start (typ-ically the current control temperature)change the setpoint immediately as perthe prior instruction. If necessary, waitfor the temperature to reach the startingsetpoint. Then change the Setpoint con-trol to where you want it to end, set theRamp rate control to the desired rate, andthen click on the Write button.

4. You can leave this vi running as it usesvery little computer resources, or STOPit until you need it again.

Shutting down

Returning the apparatus to room temperaturewith all power turned off starts by shutting offthe cooling machine. With no heating powerapplied, the column can take several hours to

August 31, 2021


return to room temperature. The time canbe reduced by using the heater. The heaterwill work even with the cooling machine offby simply adjusting the setpoint to room tem-perature or slightly above. Starting from thelowest temperatures, with the heater on andthe cooling machine off, the sample area andboth thermistors will rise to room tempera-ture even more quickly than the forty minutesshown in Fig. 3.

However, thermistor readings above roomtemperature do not mean the entire columnhas warmed to room temperature. Most ofthe column can still be very cold at this point.If exposed to room air while still cold, the col-umn will frost over as the water in the humidroom air freezes on the surface. Do not exposethe column to room air until its whole lengthis at room temperature and it is not easy todetermine when this happens.

If the column is still cold, and the heaterpower is then turned off, the thermistor tem-perature would start falling again. To turn offthe heater, simply adjust the setpoint well be-low room temperature. If it does not fall andstays above 300 K for at least 10 minutes aftershutting off the heater, the column is at roomtemperature and can be opened to the roomair.

Measurement of resistance

The resistance is measured by a Keithley 2700digital multimeter in Ohms-4 wire mode. This“4-wire” method allows the resistance to bemeasured without the parasitic influence ofthe resistance of the wires between the me-ter and the sample and of contact resistancebetween the wires and the sample.

The 4-wire method works as shown inFig. 4. The film is patterned with potential“sidearms” and the resistance measured is theresistance of the part of the sample between

Figure 4: 4-wire method of measuring re-sistance. In the diagram, the “ohms-source”connection on the digital multimeter is repre-sented as a current source while the “ohms-sense” connection is represented as a voltageamplifier. The multimeter takes the ratio ofvoltage to current to compute the resistance.RL represents the resistance of the electri-cal leads, including the cables to the vacuumfeedthrough (small) and the wires leading tothe cold tip (not so small). Rc represents thecontact resistance between the wires and thespecimen (sometimes small, sometimes not sosmall).

these sidearms.

Exercise 3 Suppose for all 4 wires going tothe sample, the sum of lead and contact resis-tance is 500 Ω. Suppose further the sampleresistance is 10 Ω. What is the minimum in-put impedance of the voltage amplifier for themeasured resistance to be no more than 1 mΩfrom the true resistance? Hint: The measuredresistance is the voltage across the amplifierinputs divided by the current I supplied by thesource.

The temperature-dependent resistance is

August 31, 2021


Figure 5: The Temperature changing - 4 wireohms vi.

measured with the Temperature changing - 4wire ohms vi. A picture of the front panel isshown in Fig. 5.

This vi uses the Keithley 2700 to mea-sure a resistance in (Ohms-4 wire mode) everytime the temperature decreases by the amountshown in Temperature change, set to 1.0 K bydefault.

Temperature dependence of a diode’sforward junction voltage

When used as a thermometer, a constant cur-rent is passed through the silicon diodes andthe voltage is measured. As described pre-viously, the voltage required for a given cur-rent varies significantly with temperature andhere, you will study this behavior for a com-mon silicon diode (rather than those used bythe temperature controller which are speciallymanufactured for use as a thermometer). Attypical currents of 1 to 10 µA, the diode has ahigh resistance so that a 4-wire arrangementis not necessary. Instead the “2-wire” mea-surement circuit shown in Fig. 6 will be used.

The Keithley 224 current source delivers thecurrent and the Keithley 2700 digital multi-meter measures the voltage. The Keithley 224output is from a triax connector on the backpanel, not the the banana jacks. A triax toBNC adapter cable is used to facilitate theconnections.

Figure 6: Circuit diagram for the diode mea-surements.

The temperature-dependent diode voltageis measured at fixed currents using the Tem-

Figure 7: The Temperature changing - V versusT at several I vi.

August 31, 2021


Figure 8: The Temperature steady - I versus Vat several T vi.

perature changing - V versus T at several I vi(see Fig. 7). This vi also makes measurementsevery time the temperature decreases by theamount shown in Temperature change, set to1.0 K by default. It first uses the Keithley 224to set the current at one of the values displayedin the Current array control on the front paneland then uses the Keithley 2700 to measurethe diode voltage. This is done sequentiallyfor all currents in the array. Because it takesa few seconds to cycle through the measure-ments, the temperature will vary a few tenthsof a Kelvin during the measurements, but thisshould not be a problem.

Diode current-voltage characteristic

More detailed current-voltage characteristicsfor the diode will be measured while the tem-peratures is held steady. The front panel ofthe Steady temperature - I versus V at several Tvi used to make these measurements is shownin Fig. 8. The sample temperatures are setinteractively so you can’t run this vi and thenleave and come back later. A reasonable setof IV curves might be taken at temperaturesaround 10, 50, 100, 150, 200, and 300 K. If you

are at the base temperature when you start,work up from 8 K. If you start from roomtemperature, work down from 300 K.

As is traditional for IV curves, they are plot-ted as I vs. V even though the data is takenwith I as the independent variable. At eachtemperature, V is measured by the Keithley2700 after each of the Current array control val-ues on the front panel has been applied usingthe Keithley 224. The default values from 0.1to 1000 µA in a 1, 2, 5 sequence (plus onereverse bias measurement at -0.1 µA) workswell.

Self heating

A power IV is delivered to any device carry-ing a current I through a voltage drop V . Forthe diode, copper wire (and gold film), the en-ergy is released as heat and may lead to selfheating and a device temperature not in equi-librium with the sample diode used to mea-sure the temperature. While this is probablynot a problem for most measurements, it mayhave an effect at the lowest temperatures andhighest current for the diode where the powerdissipation gets to about 1.5 mW or so.

A sub-vi is used to set the currents and mea-sure the corresponding voltage for each cur-rent in an array of values. This sub-vi can bemade to turn off the current for a user-selectedtime delay between each array element. (It al-ways reads the voltage immediately after set-ting the current.) This parameter is called theIV delay on the front panels of the two IV dataacquisition programs.

If the device temperature does not rise tooquickly in response to an excitation current,this feature may successfully allow the deviceto re-equilibrate with the sample diode ther-mometer before each IV measurement. Set-ting the IV delay to zero turns it off and causeseach new current in the array to be set imme-

August 31, 2021


diately after the prior array value without firstsetting it to zero.

Whether or not the IV delay is used, the cur-rent is always returned to zero after processingall currents in the array.

Procedure

4-wire versus 2-wire resistance

Here you will use two of the “Ohms-ranger”boxes to simulate the effects of lead and con-tact resistance.

1. Set one of the boxes to R1 = 10 Ω and theother to R2 = 1000 Ω. Use the Keithleydigital multimeter in Ω2 mode to deter-mine the actual resistance of the boxes.

2. Connect the two boxes in series, and mea-sure the series resistance in Ω2 mode.Convince yourself that the resistance isthe sum of the two by setting R1 to 11 Ωand 9 Ω.

3. Connect the ohms sense terminals acrossR1 and measure the resistance in Ω4mode. Demonstrate that the resistanceis that of R1 by setting it to 11 Ω and9 Ω.

4. Demonstrate that the resistance readingis insensitive to R2 by setting it to 1100 Ωand 900 Ω.

C.Q. 1 What is the maximum value of R2 forwhich the resistance of R1 = 10 Ω is measuredcorrectly? The minimum? Explain. For anideal circuit, what do you expect for the maxi-mum R2? How would it depend on the actualresistance of R1?

Measurement of dc resistance

The circuit diagram of the wires to the samplearea is shown in Fig. 9.

Figure 9: Connections to the samples.

5. Remove the outer vacuum shroud. (Youwill need to vent the vacuum system if itis not already at atmospheric pressure.)Unscrew the radiation shield. (Weargloves to prevent fingerprints!) Note thecopper wire wound around a small brassscrew on the upper left. (In case you havetime to measure the gold film, use a smallruler to estimate the width of the thingold film, its overall length and the sepa-ration between its voltage sidearms. Becareful not to touch the wires fixedto the film!! It is only necessary to getan estimate of these geometric factors towithin about 25%.)

6. Connect first the current leads (H+ G−)to the digital multimeter and next thevoltage leads (K+ J−). Measure theresistance (2-wire) for each connection.Now connect both sets for 4-wire resis-tance and measure this quantity. Thesemeasurements will allow you to determinethe room-temperature values of the (lead+ contact) resistances shown in Fig. 4.

August 31, 2021


For a circular cross section, the relation be-tween resistance R and resistivity ρ is

R =ρL

πr2(22)

where L is the sample length and r is its ra-dius. Thus, for a wire of a known substance,one may estimate the thickness from the resis-tance and measured diameter.

C.Q. 2 Assume the copper wire has the re-sistivity of pure copper (1.68 µΩ-cm) and is5.1×10−3 cm in diameter. Use your measure-ments of the 4-wire resistance to estimate thelength of the wire.

7. Connect the digital multimeter to the1N914 diode leads (F+ E−) and measurethe resistance. Now reverse the leads andremeasure the resistance. What does thismeasurement tell you about the diode?About the multimeter’s circuit for mea-suring resistance?

8. Screw on the radiation shield. (Usegloves.)

9. Check the O-rings on the expander fordirt. (Remove any you see. You maywant to add a small amount of vacuumgrease.) Gently slide the vacuum shroudover the O-rings.

10. Check that the vent valve is closed.

11. Start the vacuum pump and then openthe vacuum valve. The thermocouplegauge should start to indicate vacuum af-ter a few minutes.

12. Turn on the temperature controller, thedigital multimeter, and the computer.After the pressure goes below 50 mTorr,turn on the cooling machine and do an im-mediate change of the setpoint to 324.9 K

by following the appropriate procedurein Changing the Temperature starting onpage 8. The LED bar on the temperaturecontroller should show that the heatercomes on. If not, ask for help.

13. Connect the current leads (H+ G−) andthe voltage leads (K+ J−) to the digi-tal multimeter as required for 4-wire re-sistance.

14. Do a Read from the Control setpoint andramp vi, check that the setpoint is at 324.9and, if necessary, wait for the control tem-perature to reach 324.9 K.

15. Bring up the Temperature changing - 4wire ohms vi. Leave the switch above thegraph set for Falling and the TemperatureChange at 1 K. Start the vi by pressingthe Run button in the toolbar.

16. The vi will ask for a file in which to storethe data. Create a directory in MyDoc-uments, or better yet, on your personalthumb drive and save it there. Be sure todocument the file name and its contentsin your lab notebook.

17. Make sure the cooling machine has beenrunning for at least a five minutes. Thenramp the temperature controller’s set-point to 4 K at 3.5 K/min,3 again follow-ing the procedure on Changing the Tem-perature. The sample temperature in-dicator above the graph should start toramp down and a resistance measurementshould be made and added to the graphfor each degree it falls.

3The value of 3.5 K/min is near the fastest rate thatcan be maintained over the full range of temperaturesfor both increasing or decreasing ramps. The rate islimited by the smallest slopes of the curves in Fig. 3which are both around 4 K/min.

August 31, 2021


18. If, at any point during the cool down, theshroud gets very cold (shows frost) thereis either a leak or a mechanical contactbetween the cold section and the shroud.Stop the cooling machine and check withan instructor.

19. When T ≈ 220 K, close the vacuum valveand turn off the vacuum pump.

20. The vi should continue to a temperaturebelow 10 K. When the temperature stopschanging, press the red-letter STOP but-ton near the graph and the program willsave the data to the file specified whenyou first started the program. The firstcolumn contains the sample temperaturesand the second column contains the resis-tances.

21. Rather than getting one data point everydegree or so as in the Temperature chang-ing - 4 wire ohms, you will sometimes wantto use the Continuous - 4 wire ohms pro-gram, which takes temperature and resis-tance readings as fast as it can until youtell it to stop by hitting the Stop buttonon the front panel. Remember to stop itthis way to properly save your data to aspreadsheet file. The extra data is partic-ular important near the lowest temper-atures, say from the minimum tempera-ture achievable to around 30 K so you willget a large enough statistical sampling oftemperatures and resistances for analysis.Start the program while the temperatureis at its minimum and then do a slow up-ward ramp in temperature with the Con-trol setpoint and ramp program. You willknow you have good data when the graphof resistance vs. temperature shows lotsof points throughout the desired temper-ature range and shows the scatter of thedata as well as the general upward trend.

Analysis of resistance

Plot R(T )—your measured resistance as afunction of temperature. According to Eq. 16(with Eqs. 17 and 22), the resistance is pre-dicted to be

Rpred(T ) = Ri + CTα (23)

where

Ri = ρiL

πr2(24)

and α = 1 at high temperatures and α = 5 atlow temperatures. The proportionality con-stants in Eq. 17 together with the scaling fac-tor L/πr2 becomes C in Eq. 23 and will bedifferent in the two regions. Although Eq. 17suggests that α should change from one to fiveright at ΘD = 343 K, these limiting power lawbehaviors should only be expected well aboveand well below ΘD/5.

Fit your data to Eq. 23 over the appropriatetemperature range with Ri, C and α as thefitting parameters.4 For the low temperatureregion start by fitting only those points belowa limiting temperature of 30 K. For the hightemperature region start by fitting only thosepoints above a limiting temperature of 275 K.Also plot the residuals R(T ) − Rpred(T ) vs.T and report how the quality of the fit andthe parameters change when you increase ordecrease the limiting temperature.

The resistivity of 1.68 µΩ-cm—given in Ex-ercise 2 and used again in C.Q. 2—is actu-ally the value for pure copper. That is, it isthe intrinsic resistivity at room temperatureand does not include any contribution fromthe residual resistivity ρi. Let ρph(297 K) (=1.68 µΩ-cm) represent this known room tem-perature intrinsic resistivity and let R(297 K)represent your measured wire resistance at

4You may need to include a fixed multiplier withC = βC ′ so that the fit is working with a value of C ′

near one.

August 31, 2021


room temperature. Show that the scale factorbetween resistance and resistivity κ = L/wtcan then be considered experimentally deter-mined as

κ =R(297 K)−Ri

ρph(297 K)(25)

Determine κ and use it with the wire diameterd to get an improved determination of the wirelength. How much does it change from thevalue determined in C.Q. 2?

The ratio of the resistivity at the ice point(273 K) to the residual resistivity ρi is calledthe residual resistivity ratio. Keeping in mindthat ρi arises from impurities and other crystaldefects, higher residual resistivity ratios implyhigher film quality. This ratio for a high qual-ity copper single crystal can exceed 1000. Useyour measurements to determine the ratio foryour wire.

According to Eq. 23, a plot of log(R(T )−Ri)versus log(T ) should be a straight line with aslope α. Make such a plot using the Ri deter-mined by the low temperature fit. Due to ex-perimental error, you may find some negativevalues for R(T ) − Ri at the lowest tempera-tures for which the log function will return anerror. Just ignore these points. Can you seethe two power law regions in your graph? Adda line through the high temperature data hav-ing a slope of one and two other lines throughthe low temperature data points with a slopeof three and a slope of five. Change the valueof Ri in very small increments watching howthe results change in the low temperature re-gion. What does this step together with thelow temperature fits suggest about this exper-iment’s ability to check the validity of the hy-pothetical T 5 behavior?

Diode measurements

22. Start the vacuum pump and then openthe vacuum valve. The thermocouple

gauge should start to indicate vacuum af-ter a few minutes.

23. Turn on the temperature controller, thedigital multimeter, the current source,and the computer.

24. After the pressure goes below 50 mTorr,turn on the cooling machine and do an im-mediate change of the setpoint to 324.9 Kby following the appropriate procedurein Changing the Temperature starting onpage 8. The LED bar on the temperaturecontroller should show that the heatercomes on. If not, ask for help.

25. Connect the diode leads (F+ E−) to thecurrent source and the digital multimeter.Manually adjust the current to 1 mA5 andcheck that the voltage is around 0.6 V. Ifit is not, check your connections; you mayneed to reverse the leads to the diode.

26. Do a Read from the Control setpoint andramp program, check that the setpoint isat 324.9 and, if necessary, wait for thecontrol temperature to reach 324.9 K.

27. Open the Temperature Changing- V versusT at several I program. If you want tochange the currents at which the voltagesare measured, you need to do this beforestarting the program. Leave the switchabove the graph set for Falling and theTemperature change at 1 K. Start it bypressing the Run button in the toolbar.

28. The vi will ask for a file in which to storethe data. Rename the file and save it

5Turn on the Keithley 224, push the SOURCE key,enter the desired current with the keypad, and pushthe ENTER key. Push the OPERATE key if its indica-tor LED is not lit. If the V-LIMIT led flashes, there is aproblem—the 224 could not reach the desired currentwithin the set voltage limit; check the circuit and/orask for help.

August 31, 2021


to MyDocuments or a USB drive direc-tory created earlier. Again, record the filename and its contents in your lab note-book.

29. Make sure the cooling machine has beenrunning for at least a five minutes. Rampthe temperature controller’s setpoint to4 K at 3.5 K/min, again following theprocedure on Changing the Temperature.The sample temperature (indicated abovethe graph) should start to ramp downand for each degree it falls, the programtakes measurements of the diode voltageat each of the currents and adds them tothe graph. Note in your lab notebook thetemperature change during a single set ofmeasurements at “one” temperature.

30. If, at any point during the cool down, theshroud gets very cold (shows frost) thereis either a leak or a mechanical contactbetween the cold section and the shroud.Stop the cooling machine and check withan instructor.

31. When T ≈ 220 K, close the vacuum valveand turn off the vacuum pump.

32. The vi should continue to a temperaturebelow 10 K. When the temperature stopschanging, press the red-letter STOP but-ton near the graph and the program willsave the data to the file specified whenyou first started the program. The firstcolumn contains the sample temperaturesand each subsequent column contains themeasured voltages for one of the chosencurrents.

33. Start the Temperature steady - V versusI at several T program. The programwill wait for you to set the temperature.Do an immediate change of the setpointto the desired temperature according to

the Changing the temperature procedureand then wait for the temperature to sta-bilize before proceeding by pressing theREADY button. The program then mea-sures the IV characteristics as describedpreviously. Due to a bug in the Lab-VIEW program, which makes the firstcurrent/voltage measurement unreliable,this first measurement is performed twice.Use the second one.

34. Repeat for all desired temperatures (10,50, 100, 150, 200, and 300 K are recom-mended). Write down the control andsample temperatures in your lab note-book.

35. After the last set of IV measurements,stop the program using the red-letterQUIT button on top of the graph and theprogram will save your data to the filespecified when you first started the pro-gram. The first column contains the setof currents used. Each additional columncontains the voltages measured for thesecurrents at one particular sample temper-ature which temperature will be at thehead of the column.

CHECKPOINT: Complete all measure-ments of resistance and diode IV char-acteristics.

Analysis of diode measurements

The voltage-current relation is predicted to be

I = I0(eeV/ηkT − 1). (26)

where I0 is temperature dependent and theideality factor η is of order unity. Show thatthe -1 is negligible compared to eeV/kT for allforward biased data points so that a plot ofln I vs. V at each temperature should be a

August 31, 2021


straight line over the region where the theoryis applicable.

From your IV measurements at constant T ,make a series of such plots on a single graphand discuss the agreement with the prediction.Fit a straight line to each plot over the appro-priate region (where the plot is linear). Make atable and graph to show how the fitted slopeand the fitted value of the intercept (ln(I0))change with temperature. Discuss how (andwhy) the slope and ln(I0) depend on temper-ature. Plot the slope vs. inverse temperature,show that it is linear and use the slope of thisgraph to determine e/k and compare with pre-dictions. Should your fit be forced through theorigin? Why or why not? Plot the interceptvs. 1/T to determine the band gap energy Eg.

From the V vs. T data at several I, makeplots V vs. T for each current on a singlegraph. Fit each plot over the appropriate re-gion to find the silicon band gap energy Egand compare with predictions from the litera-ture. Make a plot of the slope of the V vs. Tgraph vs. ln(I) and use its slope to again de-termine the ideality factor. Why do you haveto exclude data at the lowest temperatures?Comment on the usefulness of the 1N914 asa thermometer. How does it compare with aDT-470 commercial diode thermometer usedin our cryostat? (Data from the LakeShoremanual is included with the auxiliary mate-rial for this experiment.) Is there measurablevoltage noise at any current? How accuratelyshould the current be controlled to achieve atemperature accuracy of ±1% at 100 K? at10 K?

August 31, 2021