How to read Peltiers’ performance chart?
This time we have an article instead of a video. It is quicker and maybe more illustrative with pictures and text.
In this article I will show you how to read and understand the performance chart of Peltier coolers. We will look at two different charts. One is the voltage vs. DT chart, and the other is the cooling power (Qc) vs. DT chart. Before diving into the charts and their interpretations, we should discuss a few definitions and abbreviations.
Voltage [V]: The voltage applied on the Peltier cooler.
Current [A]: The current running through the Peltier at a given applied voltage. The current at a given voltage depends on the temperature of the device.
DT [°C]: Temperature difference between the hot side and the cold side of the Peltier cooler. Example: Th = 50°C, Tc = 2°C, DT = 48°C.
Cooling power (Qc) [W]: The amount of heat that the Peltier cooler can move from the cold side to the hot side at a given current, voltage and DT.
Typical performance charts for TEC12703
Voltage - DT curves at different currents for TEC12703 at 50°C hot side temperature.
Cooling power - DT curves at different currents for TEC12703 at 50°C hot side temperature.
So, above you can see the two most important and informative performance charts for the TEC12703 Peltier cooler, defined at 50°C hot side temperature. Manufacturers usually define these charts at 27°C and 50°C hot side temperature. As you can see, there are two common things on these charts, DT and current. Therefore, we will use these parameters to navigate between these charts.
In this example we will look at the DT = 20°C case. So, if we maintain the hot side at 50°C, the cold side will be at 30°C. We will see how this can be realized at different voltages and currents and what amount of cooling power we can obtain.
1.) First, we need to draw a vertical line on the Voltage-DT chart at the desired DT value. This vertical line will intersect the lines of the different currents.
2.) Then we check the voltages at each current. To obtain this, we draw a horizontal line at each point where DT intersects the current charts.
3.) We calculate the rejected heat based on the current and voltage.
4.) Finally, we move to the cooling power-DT chart, draw a vertical line at the desired DT and note down the different cooling powers at different currents.
First, we start with the voltage-DT chart. We know that we want DT = 20°C, so we first look it up on the X-axis. Then, we draw a vertical line at DT = 20°C.
Once we have the vertical line, we see that the line intersects the lines given for different currents.
Now, we draw horizontal lines towards the Y-axis from each intersection in the same way as it is shown on the second image. This will reveal the voltages needed to maintain the given currents at a given DT.
Now, we know the current and voltage values, we can calculate the wattage which is simply the product of them. P = U * I. The result gives the amount of heat rejected by the device.
Finally, we move to the cooling power-DT chart. Similarly to the voltage-DT chart, we draw a vertical line at DT = 20°C and then we draw horizontal lines towards the Y-axis at each intersection. The obtained values are the cooling power values.
After going through these steps, we can see the provided cooling powers at different voltages, currents and DT values. As an example, we will look at the 2 A curve.
The DT = 20°C intersects the 2 A curve at 7.7 V. This means that if the hot side is maintained at 50°C, the applied voltage is 7.7 V and the current is 2 A, the cold side is at 30°C. At this point, the rejected heat is 15.4 W. Now, we check the cooling power-DT curve at DT = 20°C. We see that the 2 A curve is intersected by the DT = 20°C vertical line at 20 W. Therefore, the cooling power is 20 W. Now we know both therms that appear on the hot side. The total heat on the hot side is 15.4 W + 20 W = 35.4 W. We need a cooler which can dissipate at least this amount of heat.
If you need cooling power, you have to increase the voltage which results in an increased current. For example, you can reach 4 A at DT = 20°C if you apply 14.1 V on the Peltier device. At this point, the device will generate 56.4 W heat. The cooling power is 33 W. The total heat on the hotside therefore 56.4 W + 33 W = 89.4 W. This requires some good cooling, such as a CPU cooler or a similar sized heatsink. From this example, you can see that the heat generated by the current is almost two times larger than the cooling power. This brings us to the COP (Coefficient Of Performance), which is COP = Qc / P = 33 W / 56.4 W = 0.585.
Just a fun thought experiment. If you want to build a Peltier-based air conditioner, you would need roughly 2000 W of cooling power. That is 60 units from the example above. Additionally to that 2000 W which comes from the cooling, you would also have to deal with 56.4 W * 60 = 3384 W heat as a result of Joule heat produced by the current. So in total, you should be able to dissipate 5384 W of heat. Also, you will need 60 power supplies that can provide 14.1 V and 4 A. For 60 units, that is 240 A. This little calculation shows how impractical is to build a Peltier-based air conditioner. All those attempts on YouTube are fake. A few Peltier will not provide enough cooling power and even if you manage to build a system with around 2000 W cooling power, the amount of electricity needed would make the system very uneconomical.
Some things worth to note:
You can always obtain the real DT if you measure the current and voltage. Once you know the real DT, you can get the real hot side temperature by measuring the cold side temperature.
If you need much higher voltage than it is suggested by the performance chart to maintain the desired current, your hot side is warmer than it should be! Check the cooling, check the contacts between the Peltier and the hot side heatsink!
The rejected heat from the P = U * I term is usually much higher than the cooling power. The Peltiers are very inefficient coolers.
