| |
Example |
Your
Case |
| TEC
Minimum Values |
|
|
| DT max |
65°C |
_______°
C |
| Qmax |
30
W |
_______watts |
| Imax |
5.6
A |
_______amps |
| Vmax |
8.2
V |
_______volts |
b.
Determine the ratio of DT/DTmax:
| |
Example |
Your
Case |
| DT(from
step 2) |
35°C |
______°C |
| DTmax
(from step 5a) |
65°C |
_______°C |
| DT/Tmax
(calculate) |
35/65
= 0.54 |
_______
°C |
c.
On the performance graph (See Figure 3) draw a horizontal line on
the upper graph corresponding to the DT/DTmax (.54 in this
example).
d.
Divide the total heat load (from step 1) by the value of Qmax
for the TEC selected (from step 5a):
| Q/Qmax
= |
9
W/ 30 W |
0.3 |
| Q/Qmax
= |
______W/______
W= |
_______ |
e.
At the intersection of the horizontal line (drawn in step 1c) and
the value of Q/Qmax (just calculated in step 1d), plot
a vertical line on the performance graph.
f.
Record the value of I/Imax at the intersection of the
vertical line just plotted and the bottom I/Imax AXIS.
I/Imax
= .6
I/Imax
= ______
2.
Calculate TEC Current
Using
the value of Imax (obtained in step 1a), and the value
of I/Imax(just obtained in step 1f), calculate the TEC
current (I).
| TEC
Current = |
(Imax)X |
(I/Imax)
= |
Amps |
| = |
5.6A |
0.6
= |
3.4A |
| = |
______
A |
______
= |
______
A |
3.
Calculate the TEC voltage:
a.
Draw a horizontal line through each of the intersections of the vertical
line (drawn in step 1e) and the two curves in the lower half of
the performance curve.
b.
Record the values of V/Vmax at the intersection of the two
horizontal lines drawn and the left vertical axis. These values
determine the range of V/Vmax for the TEC selected. For large
DTs (small values of Q/Qmax), the voltage will
correspond to the high end of the range, and for small DTs (large
values of Q/Qmax), the voltage will correspond to
the low end of the range.
High
(V/Vmax) = .68
Low (V/Vmax) = .54
c.
Multiply Vmax (obtained in step 1a) by each value
of V/Vmax just obtained to get the range of TEC voltage.
| High
TEC Voltage = |
(Vmax
X |
(V/Vmax)
= |
Volts |
| |
8.2V
X |
0.68
= |
5.6
volts |
| = |
______
X |
______
= |
______
volts |
| Low
TEC Voltage = |
(Vmax
X |
(V/Vmax)
= |
Volts |
| |
8.2
X |
0.54
= |
4.4
volts |
| = |
______V
X |
______
= |
______
volts |
4.
Calculate maximum TEC power:
Multiply
TEC current (from step 2) by High TEC voltage (from step 3c).
| TEC
Power = |
3.4
Amps X |
5.6
Volts= |
19
watts |
| = |
______
Amps X |
______
Volts = |
______
watts |
5.
Calculate power dissipated into the heat sink.
The power dissipated
into the heat sink is the sum of the total heat load and the input power
to the TEC.
| Q
h = |
9.0
+ |
19.0
+ |
28.0
watts |
| = |
______W
+ |
______W
= |
______
watts |
At
this point you should have several candidate TECs with their approximate
performances and now be ready to order.
Figure 3
|