### TEC Selection in Thermal Management

```TEC Selection
in Thermal Management
With the push towards reducing component size,
increasing power dissipation and lowering junction
temperature, engineers must utilize different and
more effective means for achieving their goals.
Thermo electric coolers (TECs) are one such
means to achieve their ends. TECs have been
on the market for decades and their application
in component cooling can be attractive, if they
are designed properly or chosen correctly. Even
though their COP is much lower than a Carnot cycle
efficiency, they can solve some problems if the
fundamentals of a TEC are well understood by the
engineering team.
The optimum COP of a TEC can be shown to be:

ΔT 

 M − 1−
T 
T 
c
c
COP =


ΔT  M + 1 




Where,
Tc = cold side temperature (K)
ΔT = Th-Tc temperature difference across TEC (K)
Th = hot side temperature (K)
M=
√ 1 + ZT
Z=
α2
RK
avg
M = Figure of merit
α = Seebek coefficient (μV/K)
R = electrical resistance (Ohm)
K = thermal conductivity(W/m.K)
Tavg = (Th+Tc)/2 (K)
The term outside the bracket is the Carnot
efficiency. Lets assume Carnot efficiency of 10 or,
ΔT
= 0.1
Tc
For the best material in the market today, which
is Bi Te, the value ZTavg ≈1, so COP = 1.30. This is
13% of the Carnot efficiency, if it works under ideal
conditions [1].
Newer materials have ZTavg ≈2, so the COP = 2.28,
which means the TEC works at 22.8% of the Carnot
efficiency under ideal conditions. Even under ideal
conditions, the COP is much lower than the COP of
a vapor compression cycle.
Bell [1] has devised a thermodynamic cycle to
increase the COP of the TEC by isolating the TE
elements on the hot and the cold sides, compared
to traditional ways of assuming constant cold and
hot temperature on the sides of the TEC. Figure 1
shows the traditional concept and the enhanced
concept. In this concept, the temperature of the
two fluids on the hot and cold side is not uniform,
as opposed to the standard way of assuming
that they are constant. The fluid on the hot side
SEPTEMBER
2012|Qpedia
|Qpedia
MARCH 2012
11
Figure 2. COP of a TEC as a Function of Percentage of
Current at Different Temperature Differences [2]
Figure 1. (a) Traditional Application of a TEC, (b)
Enhanced Thermodynamic Cycle of a TEC [1]
gradually heats up from ambient to Th and the
fluid temperature on the cold side decreases from
ambient to Tc. The heat exchangers are separated
so each element of the TEC, as opposed to the
entire TEC, is in contact with a separate heat
exchanger.
Comparing the traditional concept in (a) with
the enhanced cycle (b), it is evident that for the
enhanced cycle at each axial location x,
∆T(x)< ∆T0
Tc(x)>Tc(x1)
Hence
∆T(x)/Tc(x)< ∆T0/Tc(x1)
By looking at the Equation 1 for COP, the smaller
∆T/Tc leads to a higher COP. The authors [1]
elaborate on the same concept with the fluids
coming in two different directions (counter flow).
Their analysis shows that this concept increases the
COP several times.
12
Figure 2 shows the COP of a TEC as a function
of the percentage of maximum rated current for
different temperature differences. The graph clearly
shows that the maximum current causes the COP to
degrade very quickly. In fact, at lower temperature
differences, i.e. less than 20 oC, the effect is
more pronounced. The graph also shows that at
low temperature differences, the percentage of
maximum current is about 10 to 20% for optimum
COP.
Let us apply this TEC to a hypothetical heat sink
and see what the effect on the case temperature is.
Assume we have a component dissipating 120
watts. We want to compare the Tcase before and
after the application of the TEC. Figure 3 shows the
arrangement of the heat sink, TIMs and the TEC
before and after the application of a TEC.
The following assumptions are made [2]:
Tamb = 40 oC
Θfins = 0.18 oC/W (heat sink thermal resistance)
P = 120 watts (Power dissipation)
ATEC = 64 cm2 (TEC surface area)
Apackage = 9 cm2 (package surface area)
ΘTIM2 = ΘTIM3 = ΘTIM4 = 0.2 cm2-oC/W
(thermal interface resistance)
Figure 3. Left (Heat sink Assembly), Right (Heat Sink and TEC Assembly)
Figure 4 shows the resistance paths between the
case temperature and the ambient with just a
simple heat sink (left) or a combination of heat sink
and a TEC (right).
For a simple heat sink assembly, the case
temperature can simply be calculated from:
TCase - Tamb = Px(θTIM2 + θSpread + θFins)
TCase = 73.9 oC
For the heat sink and TEC assembly, we assume
that we can implement a TEC with a COP of 3.
From Figure 2, at a COP of 3 at optimum point, the
TEC has a temperature difference of ∆T = 14.5 oC
across the TEC. With a COP of 3, the power input to
the TEC is 40 W to pump 120 W. It is to be noted
that, if we try to use a TEC with a higher COP, the
temperature difference across the TEC will be much
lower and the use of a TEC will not be justified. The
case temperature can be calculated by solving the
following simultaneous equations:
TCase - TCold = Px(θTIM2 + θSpread + θTIM3)
THot - Tamb = (P+ 40)x(θTIM4 + θFins)
THot = TCold + 14.5
TCase = 67.5 oC
This is a 6.4 oC reduction in temperature. In terms
of thermal resistance:
For a simple heat sink Rca = 0.283 oC/W
Figure 4. Resistance Path for Heat Sink Assembly (Left),
Heat sink and TEC Assembly (Right) [2]
For a heat sink and TEC assembly, Rca = 0.229
o
C/W, which means using a TEC has resulted in
a 23% reduction of the case to ambient thermal
resistance. However, it should be noted that an
extra 40 W is spent to run the TEC.
SEPTEMBER
2012|Qpedia
|Qpedia
MARCH 2012
13
Now the question is: How many TECs do we need to
use? Assume we are using 4 TECs. Each TEC has to
pump 30 W. Assume a typical ∆Tmax = 68 oC which
is typical of commercial TECs:
∆T/∆Tmax = 0.21
Figure 5 plots ∆T/∆Tmax as a function of I/Imax, the
line of optimum performance and the values of Q/
Qmax. From the figure, it can be found that Q/Qmax
= 0.17, which leads to Qmax = 30/0.17 =176 w. If
we had used 3 TECs, then Qmax = 40/0.17 = 235 W.
With fewer TECs, the base of the heat sink might
not be fully covered, which contributes to extra
just use one TEC, then Qmax = 120/0.17 = 705 W.
It could be quite a challenge to find a TEC with this
characteristic; however, the recent advancements in
TECs have made this choice available.
Typically, for the TEC to enhance performance, a
very high performance heat sink is required; so,
the extra work input to the TEC, which causes a
raise in heat sink temperature, will be compensated
for by the temperature difference across the
TEC. Application of a TEC in component cooling
is exciting and also requires detailed analysis and
consideration of parameters such as coefficient
of performance, availability of the required
TEC, increase in power consumption, spreading
resistance, reliability, increase in cost of deployment
and detailed heat sink design.
References:
1. Bell, L., “Use of thermal isolation to improve
thermoelectric system operating efficiency”,
BSST LLC
2. Johnson, D., Bierrschenk, J, “Latest
developments in thermoelectrically
enhanced heat sinks”, Electronics Cooling
Magazine, August 2005
Figure 5. Typical Performance Curve of a TEC [2]
14
JUNE 2012 |Qpedia
MARCH
|Qpedia
October2012
2009|Qpedia
15
```