Low Temperature Transport - Department of Physics at UF · grid points inside a sphere|called the Fermi sphere|of some radius kF ... Low Temperature Transport LTT 3 ... Fermi surface

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 (1996), ISBN 0-471-11181-3 QC176.K5.

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 3-7 of Ali to gain amore solid foundation on the topics discussedhere.

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 function ofk) for free electrons. The Fermi energy and Fermiwavevectors 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 volume4πk3

F /3 is

N = 2 · 4πk3F /3

(2π/L)3=

V

3π2k3

F , (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)

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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 =h2k2

F

2m. (8)

Exercise 1 Gold has one free electron peratom. The density and atomic weight of goldare 19.3 g/cm3 and 197 g/mole, respectively.Avogadro’s number is 6.02×1023 atoms/mole,h = 1.05× 10−34 Js, m = 9.1× 10−31 kg, and1 eV = 1.6 × 10−19 J. (a) Use this informa-tion to show that n = 5.9×1022 electrons/cm3,that kF = 1.2 × 108 cm−1, that the deBrogliewavelength λ = 5.2 A, and that EF = 5.5 eV.One may also define a Fermi velocity viavF = hkF /m. Show that gold has a Fermivelocity of 1.4 × 108 cm/s, about 0.5% of thespeed of light. (b) Suppose an electron on thisFermi surface is moving in the x-direction, i.e.kx = 1.2× 108 cm−1. Calculate the differencebetween its energy level and the next higherenergy level (one grid point over in the kx-direction) in eV and as a fraction of the Fermienergy. Take the sample to be 1 cm in size andbe careful that the subtraction of two nearlyequal energies that only differ in a high orderdigit is performed with sufficient 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 goldat room temperature ρ = 2.2 µΩ-cm. Use thisvalue to calculate the room temperature colli-sion time τ for gold. (b) The power P ab-sorbed 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

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kx

ky

kx

ky

kδ

I

Figure 2: Fermi surface of a free-electron metal inthe absence (left) and presence (right) of a trans-port current.

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 gold. Approximately how many gold atomsdoes an electron pass between collisions?

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 the

small 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

ne2

1

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 labsor in different ways. Scattering by lattice vi-brations (phonons) is temperature dependentand the rate 1/τph depends on the intrinsic

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properties 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 is 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 Matthiesen’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 5 T < ΘD(17)

where ΘD ≈ 165 K is the Debye temperaturefor gold.

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 filledand there is an energy gap between this andthe lowest unoccupied band—the conduction

band. 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 thep-n junction diode. The diode energy banddiagram is shown in Fig. 3 (located at theend of this writeup) for the unbiased, forward-biased, and reverse-biased cases. (If you wantto play with the parameters in a p-n junctiontake a look at www.acsu.buffalo.edu/~wie/applet/pnformation/pnformation.html

One side of the crystal is doped p type, theother n type. A thin (∼ few µm wide) 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 havea small number of thermally generated “mi-nority” carriers of the other type (holes in the

January 30, 2008

www.acsu.buffalo.edu/~wie/applet/pnformation/pnformation.html

www.acsu.buffalo.edu/~wie/applet/pnformation/pnformation.html


n region and electrons in the p region).Both holes and electrons would like to dif-

fuse to fill the crystal uniformly but this wouldleave the crystal electrically charged. Instead,a small charge transfer by diffusion leaves theregion around the junction depleted of carri-ers. Consequently, a charged double layer ex-ists at the junction, producing an electric fieldwhich inhibits the diffusion of additional car-riers (of either sign) 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 acancelation 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 onlya small current to a limit set by the thermalgeneration of minority carriers.

The electric current I as a function of ap-plied bias voltage V is consequently given by:

I = I0

(eeV/kT − 1

)(18)

where k is Boltzmann’s constant, T the tem-perature, and I0 is an increasing function oftemperature.

Apparatus

You will make three measurements at temper-atures between about 10 and 325 K:

1. the resistance of a thin gold film;

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

3. IV characteristics of the diode at severaltemperatures.

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 tower on top of the lab table isthe vacuum shroud. At the base of the shroudand inside it, is the expander consisting ofvalves and a sealed column. The samples aremounted on a copper block attached to thetop of the column. High-pressure helium ex-pands inside the column thereby cooling it andthe samples. The helium pressure drops in theexpander and returns to the compressor (bluebox under the lab table) where it heats up asit is compressed back to high pressure. Thehot, high-pressure helium is cooled in a heatexchanger by water from the Cool-Pak (whitebox under the lab table) before going back tothe expander.

There are no user adjustments for the cool-ing accomplished by the helium expansion.When the compressor switch is turned on,all components of the cooling machine (ex-pander, compressor, and Cool-Pak) begin op-erating. A heater (current carrying wirewrapped around the outside of the expandercolumn) is used to adjust the sample temper-ature. With the heater turned off, it takes thecooling machine about 60 minutes to reducethe temperature from room temperature to

January 30, 2008


the minimum achievable temperature (around7 K). You should plan on making measure-ments of the gold film during one day of ex-perimenting and measurements of the diodeon a second day.

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.

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 changes

Figure 4: Temperature changes with coolingmachine on and heater on at maximum power(warming) or off (cooling).

and 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 asthe 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-

January 30, 2008


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. 4 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.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. 4.

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 name

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

for 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, wait

January 30, 2008


for 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.

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. 5. The film is patterned with potential“sidearms” and the resistance measured is theresistance of the part of the sample betweenthese 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 sample re-sistance is 10 Ω. What is the minimum inputimpedance of the voltage amplifier for the errorin the resistance to be no more than 1 mΩ?

The temperature-dependent resistance ismeasured with the Temperature changing - 4wire ohms vi. A picture of the front panel isshown in Fig. 6.

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.

~

pU pL pU pL

pU pLpU pL

TL

Figure 5: 4-wire method of measuring resistance.In the diagram, the “ohms-source” connection onthe digital multimeter is represented as a currentsource while the “ohms-sense” connection is rep-resented as a voltage amplifier. The multimetertakes the ratio of voltage to current to computethe resistance. RL represents the resistance of theelectrical leads, including the cables to the vac-uum feedthrough (small) and the wires leadingto the cold tip (not so small). Rc represents thecontact resistance between the wires and the spec-imen (sometimes small, sometimes not so small).

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 previ-ously, the voltage required for a given currentvaries significantly with temperature and here,you will study this behavior for a commonsilicon diode (rather those used by the tem-perature controller which are specially manu-factured for use as a thermometer). At typ-ical 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. 7 will be used.The Keithley 224 current source delivers the

January 30, 2008


Figure 6: The Temperature changing - 4 wire ohmsvi.

current 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.

The temperature-dependent diode voltageis measured at fixed currents using the Tem-perature changing - V versus T at several I vi(see Fig. 8). 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.

.HLWKOH\

.HLWKOH\

Figure 7: Circuit diagram for the diode measure-ments.

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 shown

Figure 8: The Temperature changing - V versus Tat several I vi.

January 30, 2008


Figure 9: The Temperature steady - I versus V atseveral T vi.

in Fig. 9. 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 youare 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 and gold film, the energy is releasedas heat and may lead to self heating and a de-vice temperature not in equilibrium with the

sample diode used to measure the tempera-ture. While this is probably not a problem formost measurements, it may have an effect atthe lowest temperatures and highest currentfor the diode where the power dissipation getsto 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-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. Set one of the boxes to R1 =10 Ω and the other to R2 = 1000 Ω. Use theKeithley digital multimeter in Ω2 mode to de-termine the actual resistance of the boxes.

Connect the two boxes in series, and mea-sure the series resistance in Ω2 mode. Con-vince yourself that the resistance is the sum ofthe two by setting R1 to 11 Ω and 9 Ω.

Connect the ohms sense terminals acrossR1 and measure the resistance in Ω4 mode.

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12

11

10

9

8

7

6

5

4

3

2

1

12

11

10

9

8

7

6

5

4

3

2

1

G(-)

H(+)

J(-)

K(+)

C(-)

E(-)

D(+)

F(+)

A(-)

B(+)

Al foil

Goldfilm

1N914

Figure 10: Connections to the samples.

Demonstrate that the resistance is that of R1

by setting it to 11 Ω and 9 Ω.Demonstrate that the resistance reading is

insensitive to R2 by setting it to 1100 Ω and900 Ω.

What is the maximum value of R2 for whichthe resistance of R1 = 10 Ω is measured cor-rectly? The minimum?

Measurement of dc resistance

The circuit diagram of the wires to the samplearea is shown in Fig. 10.

1. Remove the outer vacuum shroud. (Youwill need to vent the vacuum system if itis not already at atmospheric pressure.)Unscrew the radiation shield. 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%.

2. Connect first the current leads (D+ C−)to the digital multimeter and next thevoltage leads (B+ A−). 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. 5.

For a rectangular cross section, the relationbetween resistance R and resistivity ρ is

R =ρL

wt(19)

where L is the sample length, w is its width,and t is its thickness. Thus for a thin film of aknown substance, one may estimate the thick-ness from the resistance and measured lengthand width.

C.Q. 1 Assume the gold film has the resistiv-ity of pure gold (2.2 µΩ-cm). Use your roughmeasurements of the film geometry and the 4-wire resistance to estimate the film thickness.

3. 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?

4. Screw on the radiation shield. Check theO-rings on the expander for dirt. (Re-move any you see. You may want to add asmall amount of vacuum grease.) Gentlyslide the vacuum shroud over the O-rings.

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

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6. 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 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.

7. Connect the current leads (D+ C−) andthe voltage leads (B+ A−) to the digi-tal multimeter as required for 4-wire re-sistance.

8. 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.

9. 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.

10. The vi will ask for a file in which to storethe data. Rename the file, save it in theMy Documents folder, and write the filename in your lab notebook. Be carefulnot to select an existing file you will needlater.

11. 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-

3 The value of 3.5 K/min is near the fastest ratethat can be maintained over the full range of temper-atures for both increasing or decreasing ramps. Therate is limited by the smallest slopes of the curves inFig. 4 which are both around 4 K/min.

dicator above the graph should start toramp down and a resistance measurementshould be made and added to the graphfor each degree it falls.

12. 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.

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

14. 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.

15. 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 will

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know 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 19), the resistance is pre-dicted to be

Rpred(T ) = Ri + CT α (20)

where

Ri = ρiL

wt(21)

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/wt becomes C in Eq. 20 and will bedifferent in the two regions. Although Eq. 17suggests that α should change from one to fiveright at ΘD = 165 K, these limiting power lawbehaviors should only be expected well aboveand well below ΘD.

Fit your data to Eq. 20 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 2.2 µΩ-cm—given in Exer-cise 2 and used again in C.Q. 1—is actually the

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

near one.

value for pure gold. That is, it is the intrin-sic resistivity at room temperature and doesnot include any contribution from the resid-ual resistivity ρi. Let ρph(297 K) (= 2.2 µΩ-cm) represent this known room temperatureintrinsic resistivity and let R(297 K) representyour measured film resistance at room tem-perature. Show that the scale factor betweenresistance and resistivity κ = L/wt can thenbe considered experimentally determined as

κ =R(297 K)−Ri

ρph(297 K)(22)

Determine κ and use it with your estimates ofw and L to get an improved thickness deter-mination. How much does it change from thevalue determined in C.Q. 1?

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 gold film can exceed 200. Use your mea-surements to determine the ratio for our film.

According to Eq. 20, 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 another line throughthe low temperature data points with a slopeof five. Change the value of Ri in very smallincrements watching how the results changein the low temperature region. What doesthis step together with the low temperaturefits suggest about this experiment’s ability tocheck the validity of the T 5 behavior?

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Diode measurements

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

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

18. 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.

19. 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.

20. 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.

21. Open the Changing temperature - 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 the

5 Turn 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.

Temperature change at 1 K. Start it bypressing the Run button in the toolbar.

22. The vi will ask for a file in which to storethe data. Rename the file, save it in theMy Documents folder, and write the filename in your lab notebook. Be carefulnot to select an existing file you will needlater.

23. 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 indicator abovethe graph should start to ramp downand measurements the diode voltage ateach of the currents should be made andadded to the graph for each degree it falls.Note in your lab notebook the tempera-ture change during a single set of mea-surements at “one” temperature.

24. 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.

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

26. 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.

January 30, 2008


27. 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 tothe 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.

28. 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.

29. 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

Plot V vs. T at each current on a single graph.Comment on the usefulness of the 1N914 as athermometer. How does it compare with a

DT-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?

The voltage-current relation is predicted tobe

I = I0(eeV/kT − 1). (23)

where I0 changes with temperature. Showthat the -1 is negligible compared to eeV/kT

for all forward biased data points so that aplot of ln I vs. V at each temperature shouldbe a straight line where the measurements andtheory are in agreement. Make a series of suchplots on a single graph from the IV measure-ments at constant T and discuss the agreementwith the prediction. Fit each plot over the ap-propriate region and make a table or graph toshow how the fitted slope and the fitted valueof I0 change with temperature. Discuss whyI0 depends on temperature. (Hint: the answeris not in the writeup.) Does the slope behaveas expected? Explain.

January 30, 2008


Figure 3: Diode band diagram. The three panels show the unbiased, the forward-biased, and thereverse-biased cases.

January 30, 2008