Thermoelectric Phenomena Introduction

The Peltier effect was first observed in 1834 by the Frenchman Jean C.A. Peltier. It describes the heat

current that arises as a result of an electric current through the interface of two different conductors.

The purpose of this project is to measure the maximum cooling of a commercial Peltier element, in

terms of the lowest achievable temperature inside a small refrigerator built using the element. The

project also includes investigating the limiting factors of Peltier cooling and presenting the relevant

theory behind it.

Theory

The heat current, ABQ , at an interface between two conductors (A and B) can be expressed as a function

of the electrical current, I, through the interface, and the Peltier coefficients of the two materials, ΠA

and ΠB. The heat current at the interface is:

IQ BAAB )( (1)

Heat is generated if ΠA> ΠB and the electrical current flows from A to B. If the direction of the

electrical current is reversed, so is the direction of the heat current.

By combining two interfaces, A to B and B to A, a hot and a cold junction can be created. In this case,

heat is generated at one of the interfaces and the same amount is absorbed at the other.

In most commercial Peltier elements, three different materials are used (a P-doped semiconductor (P),

a metal (M) and an N-doped semiconductor (N)). The Peltier coefficients of the three materials are

different, ΠN> ΠM> ΠP. If the materials are arranged according to figure 1, a hot and a cold side are

created, using several interfaces between the materials (and thus maximizing the cooling) with simple

serial electrical connections. The arrangement functions due to the fact that (for the materials given in

this example) heat is generated at junctions of the type M to P and N to M and absorbed at junctions M

to N and P to M.

Figure 1: Sketch of a commercial Peltier element.

The cooling ability of a Peltier element is limited by several factors. Even though the Peltier cooling

increases linearly with the electric current through the element, an increased current does not always

lead to increased cooling. As the electric current increases, so does the resistive heating of the entire

element. At some point, the resistive heating caused by an increasing electric current will be larger

than the cooling caused by the Peltier effect.

At a constant current the temperature difference between the hot and the cold side, ∆T, will be

constant. This gives rise to another limiting factor, namely the heat dissipation on the hot side. If the

heat generated on the hot side is not dissipated into the environment, the temperature of the hot side

will be increased. Since the temperature difference between the sides is constant, this leads to an equal

increase in the temperature of the cold side. The heat conduction through the element depends linearly

on the temperature difference between the sides. Since the temperature difference increases with the

electric current through the element, so does the conduction of heat through the element.

For each pair of hot/cold interfaces (eg. metal-semiconductor-metal), the cooling, P, (eg. the flow of

thermal energy from the cold to the hot side) is related to the Peltier coefficients of the two materials,

ΠA and ΠB (where it is assumed that ΠA>ΠB), the electric resistance of the element, R, the electric

current, I, the thermal conductivity of the element, K, and the temperature difference between the sides

of the element, ΔT, according to equation 2.

TKIR

IP BA

2

)(2

(2)

In this case, since cooling is the parameter of interest, the heat flow is described in this slightly unusual

way (strictly speaking as flow of “cold”, or negative flow of heat).

As seen from equation 2, the Peltier cooling (the first term) depends linearly on the electric current,

whereas the resistive heating (the second term) is related to the square of the electrical current. The

third term describes the conduction of heat through the Peltier element and relates to the temperature

difference between the hot and cold sides, ΔT. The relationship between ΔT and the electrical current is

unknown. At low electric currents, the Peltier cooling is the dominating term, but as the electric current

increases, the resistive heating will increase faster than the Peltier cooling. At a certain value of the

electric current, the resistive heating will be larger than the Peltier cooling, and any further increase in

electric current will actually decrease the cooling.

By setting the derivative of the cooling P in equation 2 with respect to the current I to zero the current

at maximum cooling can be obtained.

0

I

TKIR

I

PBA (3)

From equation 3, the current at maximum cooling, I0, can be determined.

R

I

TK

IBA )(

0

(4)

In equations 3 and 4, the unknown dependence of ΔT on the current, I, is included. This must be

measured experimentally or estimated in order to calculate an optimum current from equation 4.

It should be noted that the above stated equations are valid for a single pair of material interfaces. For a

commercial Peltier element, the contribution from all interfaces must be added in order to obtain total

values of cooling power, current at optimum cooling etc. This requires knowledge of the number of

interfaces in the element, as well as the Peltier coefficients of all participating materials.

For an element consisting of nMP pairs of interfaces between metal and p-doped semiconductor, and

nMN pairs of interfaces between metal and n-doped semiconductor, the total cooling power would be

described by equation 5 (obtained by modification of equation 2). For a symmetrically built Peltier

element, it is of course true that nMP=nMN.

TKIR

InInP MNMNPMMPtotal

2

2

(5)

In equation 5, the Peltier element is considered as a single unit regarding the heat conduction and

resistive heating. The resistance, R, in the equation describes the total electrical resistance of the Peltier

element. It is worth noting that the current is the same through each of the interfaces as through the

entire element (since they are connected in series). For practical purposes, it would also be possible to

assign a total Peltier coefficient to the entire element, thus incorrectly regarding it as a single pair of

material interfaces. In this case, equation 2 can be used, with the total Peltier coefficient replacing the

term (ΠA-ΠB):

TKIR

IP totaltotal

2

2

(6)

Adapted from `Testing a Peltier Element’

home.student.uu.se/hany4287/Peltier%20element.doc

Several experiments can be performed with a peltier element or thermoelectric coolers (TEC)

1. A TEC with its ’hot end’ fixed to a large Aluminium block is provided. Temperature sensors

can be attached to the block and to the cold end of the TEC. The cold end could be surrounded

by thermal insulation to produce a small refrigerator. The cold end sensor may also be kept jsut

above the cold end (to measure the cold air temperatue)

2. Determine the maximum current and voltage the TEC can tolerate. Obvioulsy you should

operate well below these limits

3. Measure the voltage across the TEC, ,the temperature inside the cavity and the temperature of

the Al block as a function of the current through the TEC. Plot the cavity temperature and the

temperature difference as a function of the current and analyze the plot.

The thermal conductivity of the entire refrigerator and the total Peltier coefficient of the element can be

roughly estimated using equation 6 and the polynomial fitted to the curve in figure 6, together with the

assumptions that the thermal flow through the refrigerator walls are linearly dependent on the

temperature difference between the cavity and the surroundings, ΔT. It is also assumed the thermal

conductivity of the Peltier element and the refrigerator walls do not vary with temperature.

For a stationary temperature, the cooling power of the Peltier element (from equation 6) is equal to the

thermal flow through the refrigerator walls (including both conducted and radiated heat). If this flow is

assumed to depend linearly on ΔT through some constant C, equation 6 can be written as:

TKRI

ITCP totaltotal 2

2

(7)

This implies that ΔT can be written as a second order function on the current, I, through the element.

2

2)(I

KC

RI

KCT total

(8)

Determine the Peltier coefficient Πtotal. What information can you get from this number?

What are the factors that affect the measurement? What can you do to make this a better refrigerator?

4. A second TEC can be fixed to the Al block with screws, temperature sensors and a heating

element as shown below:

(you could use thermistors or thermocouples)

figure courtesy: Experiments and Demonstrations in Physics, Yaakov Kraftmakher, Bar-Ilan

Physics Laboratory, World Scientific

5. Apply a signal from a function gerenator with the hot-end on top first and then with the cold-

end on top. Measure the temperature at both ends and plot as a fucntion of time. At the end of

2-3 minutes you could increase the amplitude of the signal and measure for the next 2-3

minutes. Repeat for a few amplitudes. Plot T (temp. difference) as a function of time. Explain

your observations.

6. Repeat the above process now with DC current.

7. Determine the Seebeck coefficient of the TEC by measuring the voltage across the TEC and

T by heating the top side of the TEC using the heater (dont connect a power supply to the

TEC). Repeat this measurement by placing the bottom-side on top.

8. How can you determine the thermal conductivity of the TEC?

9. Review the following writeup (atleast upto fig. 8!) from: Experiments and

Demonstrations in Physics, Yaakov Kraftmakher, Bar-Ilan Physics Laboratory, World

Scientific

10. Attempt to reproduce the experimental tasks and analysis outlined. INFORM THE

INSTRUCTOR IF YOU HAVE SPECIFIC DIFFICULTY IN DOING THIS.

Further Information

PE-127-10-13

Thermoelectric Cooler

1. Demonstration of the Peltier effect and calibration of the thermocouple

2. Insulate thermocouple, keep thermocouple in slab and on top of TEC. Measure,

V, T1, T2, by changing the current through the TEC. Give time for stabilization

after each current. Plot cold-end temperature and temperature difference vs.

current.

3. Repeat the measurement with the thermocouple above the cold-end of the TEC

(in the insulated air-gap).

4. Use both data sets to determine the Peltier coefficient. Can you find the thermal

conductivity of the element? What information can you get from this number?

What information can you get from this number? What are the factors that affect

the measurement? What can you do to make this a better refrigerator?

5. Place the TEC mounted on the circular slab on an ice bath. Apply some heat to

the top end of the TEC with a heater. Measure V, T1, T2 as a function of the heat

supplied. Plot V vs. temp-difference to determine the Seebeck coefficient. Repeat

the experiment with the bigger Aluminium slab and compare your results.

6. Place the TEC mounted (cooling side up) on the circular slab on the large Al block.

Vary the current through the TEC. For each current, apply power to the heater

such that the temperature difference between the sides is nearly zero.

Determine the Peltier coefficient. What is the difference between this method

and that described above?

7. Think of a way to determine the coefficient of performance of the TEC as a

function of current.

8. Place TEC mounted on the circular slab on an ice bath with the cooling side on

top. Apply power to the heater such that the top side is nearly at room

temperature. Monitor the voltage across the module. Use the data to determine

the figure of merit of TEC. Read the manual carefully before doing this. DO NOT

EXCEED A TEMPERATURE OF 60 dec C