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Is it possible to drive a Peltier from PWM?
CONCLUSION1: Current/Voltage load curve is reasonably linear, both driving from current or voltage source will do fine as long as the device is operated within its spec. (Olin Lathrop, Russell McMahon) CONCLUSION2: Don’t drive a Peltier from PWM, power loss due to current increase, grows more rapidly than cooling power.
What kind of power supply do I need for 4 Peltier plates?
Short answer is you need around a 25A 12V (300W) supply. A PC supply would work for that. However, the long answer is those things are really finicky and die easily. You really need to use current and temperature sensing for each plate, and getting rid of 60W of heat per plate is no simple task.
How are Peltier devices used in Electrical Engineering?
\\$\\begingroup\\$. Peltier devices work on current, but usually have significant enough resistance so that voltage control is possible. Peltier devices are one of the few things you do not want to run with pulses, particularly in cooling applications.
What happens when you put a Peltier on a computer?
So you stick a Peltier on your processor, you run it from your computer’s Power Supply Unit (PSU), and suddenly the chip’s colder than ambient. Possibly much colder than ambient – below freezing, in some circumstances.
How to drive a peltier element in Electrical Engineering?
Driving it at the steady current to produce the same overall cooling is more efficient. Of course the microcontroller regulating the temperature will still produce PWM pulses. These pulses need to be filtered so that the Peltier device sees relatively smooth current.
Is the load curve of a Peltier device linear?
Peltier devices have a reasonably linear load curve. They are voltage driven devices but you could safely enough drive them from a voltage or a current source as long as you do not exceed their maximum voltage or current ratings.
Why are Peltier devices not run with pulses?
Peltier devices are one of the few things you do not want to run with pulses, particularly in cooling applications. The cooling effect is proportional to current, but the internal heating due to \\$I^2R\\$ losses is proportional to the square of the current.