Temperature Dependence of Resistance (last edited June 14, 2013)

Dr. Larry Bortner

 Learning Objectives

Explain quantitatively and qualitatively how the resistance of conductors and semiconductors change with temperature.

Compare and contrast the shell theory of an atom with the band theory of a solid.

 Purpose

To investigate how the resistances of a conducting sample and a semiconducting sample vary with temperature.

 Background

 Resistivity and Resistance

If we apply a voltage V across an object that has a length L and cross-sectional area A, there is a flow of charges from one end to the other measured by a current I. One of the characteristics of the material that makes up this object is the resistivity ρ (“rho” not “pee”). As a measure of an object’s resistance to a charge flow, we define its resistance as

 

(1)                     

 

If ρ does not depend on V, then V and I are directly proportional to each other and the resistance is said to obey Ohm's law (the verification of which we leave to another experiment).  The question we consider today is how R changes with temperature. We already know that the length and cross-sectional area changes as the sample expands and contracts. But these variations are inconsequential compared to how the resistivity changes and we ignore them in our investigation. This means that the resistance dependence on temperature is directly proportional to the resistivity dependence. Values of the resistivity or its reciprocal, the conductivity, have been tabulated for many materials.

 Material Classification

Resistivities can vary wildly, from 1016 Ω·m for diamond to 10-8 Ω·m for copper. One way to classify materials is based on their electron transport, or resistive, properties. Materials with low resistivities (high conductivities) are conductors.  Most metals fall under this category. Those materials that have high resistivities, where it takes a lot of energy to get an electron from one end to the other, are insulators. And the ones that have intermediate values are semiconductors.

 Temperature and Molecular Energy Levels

How do we explain the movement of charge through solid materials? And how do we account for temperature? Beyond the macroscopic notion of temperature relating to hot and cold, on the molecular and atomic level, temperature is a measure of the average kinetic energy per particle. The most probable kinetic energy of an electron at an absolute temperature T is 2kT, where k = 8.61739×10-5 eV/K is Boltzmann’s constant. Any model for resistance we come up with must be based on the principles of quantum mechanics that reflect the interaction of matter and energy at the atomic level.  A quantum mechanical description (a mathematical model) of solids focuses on what happens to the discrete electronic energy levels of an isolated atom as more and more atoms are brought together, bond to each other, and form a lattice.  Calculations show that if N identical atoms are brought together, each electronic level of the isolated atom gets split into a group of N distinct levels.  The energy spacing between these levels is extremely small, decreasing as N increases. (See Fig. 1.)

 

Description: Figure 1.jpg

Figure 1 Allowed energy levels in an atom and in a solid composed of N identical atoms.

 

For a macroscopic sample, N is so large (~1023) that the levels in a particular group are essentially continuous. The group is then called an energy band.  The various bands are separated by energy gaps; that is, differences between the highest allowed energy in one band and the lowest allowed energy in the next highest band. An electron cannot have any energy between these two energies.

 

As with atomic levels, the bands in solids may be filled, partially filled, or empty. 


     The lowest energy band that is not completely filled is called the conduction band.

     The highest energy band that has at least one electron (at absolute zero) is the valence band.

Whether the valence band is filled and, if it is, the size of the energy gap between it and the conduction band determine the charge transport properties of the solid. 

 Band Theory of Electron Transport

The kinetic energy of an electron in the conduction band can easily be raised by an applied electric field, even a fairly small field. Because there are so many empty levels nearby, the electrons are mobile within the solid.  Moreover, metals that have the band structure shown in Fig. 2 have many electrons that can participate in the conduction process; the valence band and the conduction band are the same. The large number of mobile electrons gives rise to very low values of ρ.       

                       

Description: Figure 2.jpg

Figure 2 Typical band structures of the three resistive types of materials.

 

In an insulator, the valence band is filled and the conduction band is separated from it by a relatively large energy gap of value ε. When an electric field is applied across the sample, the electrons in the valence band cannot respond to it because no higher energy levels are readily accessible. The energy needed by an electron to reach the nearest unoccupied energy level (in the conduction band) is larger than the energy provided by the field and no electrons are able to participate in the conduction process.  Consequently, insulators have very high values of resistivity. (Of course we can always crank up the field so that it provides enough energy to reach the next level, but this is called the dielectric strength and is the point where the insulator breaks down.)

 

For semiconductors, the valence band is also filled, but the energy gap to the next higher band is small compared to an insulator, small enough that an appreciable number of electrons cross the gap because their thermal energy is high enough. Electrons that are in the conduction band can now participate in the conduction process because, as in a metal, many unfilled energy levels are available. This gives semiconductors a reasonable conductivity (but weaker than a normal conductor).  Moreover, those electrons thermally promoted out of the valence band leave behind vacant levels, referred to as holes.  An external electric field enables electrons in the next lower band to be promoted into these vacant levels and this in turn leaves behind other holes.  The net result is an induced migration of holes through the lattice which can be treated as a flow of positive charge carriers.  Negative (electron) and positive (hole) currents in semiconductors are fundamentally important to the design and operation of many modern electronic devices. In addition, the additional mode of electrical conduction (the movement of the holes) further enhances the conductivity of semiconductors. 

 

Note that an electron thermally excited into the conduction band does not remain there indefinitely.  Such an electron remains in the upper band for an average lifetime before losing some of its thermal energy and “falling” back to the valence band.  But by that time a different electron from the valence band will have gained enough energy to move up. Given the enormous number of electrons in any real sample, there is a fairly constant number of electrons in the conduction band at any instant.  The number of electrons thermally excited into the conduction band (and an associated number of holes in the valence band) is orders of magnitude less than a same-sized conducting sample, so the electrical resistivity of semiconductors is more than that of metals and less than that of insulators.  For both semiconductors and insulators, as the temperature and hence the available thermal energy to valence band electrons increases, the number of electrons in the conduction band increases. Thus their resistivity decreases as the temperature increases.

 R(T) for Conductors

For conductors, the number of conduction electrons doesn’t really change with temperature. What happens as the temperature increases is that the ions locked in the lattice vibrate at larger amplitudes and the probability that a conduction electron collides with an ion increases. From this behavior we expect the resistivity of a metal to increase with temperature.

 

Over a wide range of temperatures (far from absolute zero and far from the melting point) we find empirically that the resistance of a metal varies linearly with temperature. The functional form of the resistance R at a temperature T in Celsius is

 

(2)                     

 

where Rref is the resistance of the sample at the reference temperature Tref and α is the temperature coefficient of resistance (or temperature coefficient of resistivity). Note that α is constant and positive for a given reference temperature. It is also the fractional change in resistance with temperature:

 

(3)                     

 

(Eq. 3 is a definition and is not the way to calculate α.) Values of α at room temperature for various metals are shown in the following table:

 

metal

α (°C-1) at 20 °C

antimony

0.0036

brass

0.002

copper

0.00393

gold

0.0034

iron

0.005

nichrome

0.0004

nickel

0.006

palladium

0.0033

platinum

0.003

silver

0.0038

tantalum

0.0031

tin

0.0042

tungsten

0.0045

zinc

0.0037

 R(T) for Semiconductors

 

For semiconductors with energy gap ε the expression for the resistance is

 

(4)                     

 

where R0 is a constant and T is the absolute temperature in Kelvins. Applying the calculus part of Eq. 3 we find that the temperature coefficient of resistance is

 

(5)                     

 

This is a negative value, as we reasoned it should be in our earlier discussion. (The resistance decreases as the temperature increases.)

 

The energy gaps for a few semiconductors at room temperature are listed here:

                                                                                                                       

semiconductor

ε (eV) 

Si

1.11

Ge

0.66    

GaAs

1.43

PbTe

0.29

 Procedure

You need the following items:

*        Science Workshop interface


*        temperature sensor (u{T}=0.5°C)

*        Fluke 45 multimeter (u{R}/R= 0.1%)

*        Fluke 8050A multimeter (u{R}/R= 0.2%)

*        coil with leads

*        thermistor with leads

*        Bunsen burner and ring stand

*        ice water

*        beakers

*        banana-banana leads, banana-alligator clip leads

*        (optional, provided by student) cell phone with camera or stand-alone digital camera

 

1.     Click on these menu items on the Windows desktop: Data Studio Experiments> Second Quarter> Thermometer. A DataStudio window opens up with a digital display of the temperature in degrees Celsius from the sensor connected to the interface box. Click on the Start button in DataStudio to start the display.

2.     Connect the coil (your conducting sample, encased in red plastic) to the Fluke 8050A, COMMON and V/kΩ/S. Turn the power on and be sure the kΩ and 200Ω buttons are depressed. The units displayed are Ω .

3.     Connect the thermistor (semiconductor, long aluminum tube) to the Fluke 45, COM and VΩ. Color coding is important here: red to red and black to black. Turn on the power and press the Ω button. Hold the tip of the thermistor probe between your fingers. A noticeable change in resistance should occur. If it doesn’t, try switching the leads.

Checkpoint! Have the TA check your connections before you proceed further.

4.     Start a data table on your data sheet, three columns with the following headings and record the room temperature data (the temperature from DataStudio, the resistance from the Fluke 8050A, and the resistance from the Fluke 45):

                        T (°C)  Rcoil (Ω)      Rthermistor (kΩ)

5.     Light the Bunsen burner and put one of the beakers with about an inch of water on the ring stand. Fill another beaker with ice.


6.     You want the resistances of the conductor (coil) and the semiconductor (thermistor) at particular temperatures. Put the Science Workshop sensor, the coil, and the thermistor in the water. (Be sure that the leads are not near the flame when you do this.) The ends of the sensor and the thermistor have to be in the water and the coil has to be completely submerged.

a.     Record T, Rcoil, and Rthermistor every 10°, starting at 20° or room temperature, then 30°, then 40°, etc. as the water warms up.

i.      You may want to have one partner responsible for the Fluke 8050A reading and the other responsible for the Fluke 45 reading. This is one of the few cases where it is OK to copy your partner’s data after it has been taken. (Generally you want both people recording the data independently.)

ii.     Another way to take the data is to take a picture of the computer screen and the two voltmeter displays, then read the data off that picture.

b.     Your highest temperature point will be when the water is boiling. (the T reading may not be 100°C). Record T, Rcoil, and Rthermistor every 10°, starting at 95°, then 85°, then 75°, etc. as the water cools down.

c.     The water will start cooling down immediately after you turn off the heat. Monitor the values and record any that you missed.

d.     Add a few pieces of ice and stir the water a little with all three elements in it to speed the cooling. Use caution: too much ice will cause the water to cool down quicker than you can take the data.

e.     You may need to drain some of the water into the empty beaker as you add ice.

f.      You’ll have to add more and more ice when the temperature gets below 20°C.

g.     After 5°, you want to record one last point at the lowest temperature you can get to, at or near 0°C.

h.     The total number of temperature-resistance pairs for each element should be at least 20.

 Analysis


1.     Start a generic lab spreadsheet, entering in your names and the name of the experiment. In the blank spreadsheet, enter values of the defined temperature of ice water in kelvins and Boltzmann’s constant in eV/K.

2.     For the coil enter in labels for T (°C) and Rcoil (Ω), then enter in the data values.

3.     For the thermistor, enter in labels of T (°C), T(K), 1/(2kT) (eV-1), Rtherm(kΩ), and Rtherm(Ω). Enter in the data values of T (°C) and Rtherm(kΩ).

4.     Fill out the table for the semiconductor using appropriate formulas.

5.     Plot R vs. T for the coil and include a trendline. Is this a linear relationship? Does the resistance increase or decrease as the temperature increases?

6.     Do a linear least squares fit of the coil data to find αRref and its uncertainty. Refer to Eq. 2.

7.     From this fit find the resistance at 20°C and its error. This is Rref and u{Rref}.

8.     Find α and u{α} for your sample. From the table in the Background, what is your best guess as to what metal makes up the coil? Justify your assertion.

Checkpoint! Have the TA check your calculations before you proceed further.

9.     Plot R in Ω vs. T in K for the thermistor. Include a trendline that best fits the points. Is this a linear relationship? Does the resistance increase or decrease as the temperature increases?

10.  Do a semilog plot of R in Ω vs. 1/(2kT). Include a trendline that best fits the points. Is this a straight line on the graph? (Step 18 in the Analysis of the Capacitors write up gives details for semilog plotting in Excel 2007.)

11.  Do an exponential least squares fit of R in Ω vs. 1/(2kT) to find eε and u{ε}.

Checkpoint! Have the TA check your calculations before you proceed further.

12.  Calculate the energy gap ε. From the table in the Background, what is your best guess as to the semiconductor that makes up the thermistor? Justify your assertion.

 Questions

1.     Given the fact that the resistivity of semiconductors decreases as the temperature increases, is it reasonable to expect there to be some elevated temperature at which the resistance of a semiconducting sample would disappear completely?  Why or why not?

2.     Propagate the uncertainty for Rref from R = mT + b without using Rule 8 in the Lab References directly.