Current control over a peltier using MOSFET

Thread Starter

Luiz Henrique Marques

Joined Sep 26, 2017
6
Hi . guys . i've been working on a project in order to heat a thermal box using 2 peltier cells in parallel.

He's the circuit proposed. when i simulate, as long as my input( 0 to 10V ) maintain 3.8v up to 5v , i current control the cell. Vth=4V
There's a current measure through the 0.1ohm resistance in which voltage drop goes into the controller for current monitoring
ok mano.PNG
problem: i wish i'd have up to 4Amp and 11V drop over the cell when the mosfet is fully "on". However, i'm facing a proporcional 1:1 increase in current and voltage . For example: 5V input generates 5,5Amps and 6V over the cell. that's too much current the resistor could burn! and voltage drop across resistor is in agreement 0,54V

when connect the pair of cells on a separated power supply varying voltage until 10V creates 4Amps. why?!..

thanks in advance :D
 

crutschow

Joined Mar 14, 2008
27,010
Sorry but I really don't completely understand your problem.

But you have no feedback from the current to the op amp, so your current is a non-linear function of the input voltage, since the MOSFET drain current versus Vgs is very non-linear.

For linear control of current, move the 0.1Ω resistor R1 to between the MOSFET source and ground, and connect the (-) input of the op amp to the junction of the MOSFET source and R1.
The current will then be a linear function of the op amp input voltage, Ild =Vin/R1.

Note that the power dissipated in R1 is I²R so at 4A it will be dissipating 1.6W.
R1 should thus be rated for at least 3W.
 

Thread Starter

Luiz Henrique Marques

Joined Sep 26, 2017
6
Sorry but I really don't completely understand your problem.

But you have no feedback from the current to the op amp, so your current is a non-linear function of the input voltage, since the MOSFET drain current versus Vgs is very non-linear.

For linear control of current, move the 0.1Ω resistor R1 to between the MOSFET source and ground, and connect the (-) input of the op amp to the junction of the MOSFET source and R1.
The current will then be a linear function of the op amp input voltage, Ild =Vin/R1.

Note that the power dissipated in R1 is I²R so at 4A it will be dissipating 1.6W.
R1 should thus be rated for at least 3W.
I got it. And What a outra
Sorry but I really don't completely understand your problem.

But you have no feedback from the current to the op amp, so your current is a non-linear function of the input voltage, since the MOSFET drain current versus Vgs is very non-linear.

For linear control of current, move the 0.1Ω resistor R1 to between the MOSFET source and ground, and connect the (-) input of the op amp to the junction of the MOSFET source and R1.
The current will then be a linear function of the op amp input voltage, Ild =Vin/R1.

Note that the power dissipated in R1 is I²R so at 4A it will be dissipating 1.6W.
R1 should thus be rated for at least 3W.

Perfect! however i must switch the 0.1ohm resistor for an 1Ohm otherwise i'd have for 1V input 10Amp. LOL . Just one more question (voltage drop over the peltier = 12v?)
 

crutschow

Joined Mar 14, 2008
27,010
Here's the LTspice simulation of danadak's circuit with some modification.
I used a single supply op amp so you don't need a negative supply.
I added a attenuator on the input so that one volt in equals one amp out.
I had to increase the value of C3 to 1nF to minimize peaking and overshoot in the response.

I reduced the voltage to 9V to minimize the dissipation in the MOSFET, but even then it can approach 10W (red trace) so it needs to be on a good heatsink.


upload_2018-3-21_20-46-40.png
 

danadak

Joined Mar 10, 2018
4,057
The 1K series into the MOSFET gate is to uncouple the effective Cload
as seen by the OpAmp output from the MOSFET gate. That in turn, as
crutschow mentions, helps to maintain phase margin in feedback loop
so part does not oscillate. Issue is OpAmp has non zero output Z, and if
output is loaded with C causes phase shift.

The 100 pF, not sure. Would have to do an s plane analysis to see what it
is doing. Seems like it potentially adds a zero into loop to aid in phase
margin, but not 100% sure w/o doing s plane analysis.

Regards, Dana.
 

dendad

Joined Feb 20, 2016
3,811
I have a couple of questions...
If you are just heating, why use a Pelier device and not just a resistor?
The Peltier will require a fan on the cold side and cost much more that a resistor.

And why use analog control with and not use PWM?

If you want analog, just connect the pass element to the heated box and use it as the heating element.
 

wayneh

Joined Sep 9, 2010
17,152
The Peltier will require a fan on the cold side...
Probably not. The TEC can be damaged by the hot side getting too hot, and that's where good dissipation is essential, but there's nothing wrong with the cold side getting cold.

I think one reason they're popular for heating is simply because of their form factor - a nice flat plate with 2D heat dissipation, not the 3D form of a resistor or a light bulb.
 

Janis59

Joined Aug 21, 2017
1,330
For Peltiers there are two options: use a LDO with drastic power, what anyway is realizable, for example, with means of doubled 2N3055 (each 60V 15A 115W. But the cooler for 230W will not be nice, believe me. Standard PC cooler is just about 100-150W.
For example, (24Vcc-15Vloa)*15A=150W therm

Other option is to create the simple smps circuit, let be on the same 2N3055 but enough with a single one. As the current will dissipate only over completely on and completely off then average thermal flux will be 1/2*2V*15A=15W or apply the most smallest radiator without of any fan. Therefore the holy grail of IR21XX is the absolutely best idea ever. For example 2110 in start-stop regime or UC3845 in dedicated CC regime. Last one may check the circuit into Danyk.cz full of wonders homepage. Switch there English language and choose sth between CC PS. Your basic problem there will be need for rather large current what mostly are near to 20 Amp. And such igbt already cost sth, in distant from bjt what one buy for dollar full lorry. But may find a plenty of igbts actually beyound 5 bucks, or construct a proper way to feed the base instead of gate.
 
Last edited:

Janis59

Joined Aug 21, 2017
1,330
<<Being a heat pump, a Peltier is more efficient than a resistor in heating a system>>
Only in very generalized terms. As actually Peltiers are devices with ultimatively low effeciency, arond 2% to 20%. For Home heating You use the systems with thermal effectivity 800% 1200% or if russian the 400%. But such is impossible if the Peltier distance between hot end and cold end is ca milimeter or two.
 

Janis59

Joined Aug 21, 2017
1,330
<<I think one reason they're popular for heating is simply because of their form factor>>
Sure not. The first reason is abovementioned harshly low efficiency factor. But much more important is the laughfully simple fact, so stupidly simple that only thing we can do is to tear after tear, the heatflux intensity by area in 99,9999999% of all Peltiers are so dense that is hardly chances to use a airflow. I mean the W/cm2 are too high packed, but if not, then effectivity will diminish further. Therefore all the Peltier coolers are built with water cooled cold surface. I was probably first (at least I found no articles before mine) what realized successful air-cooling of it (for NASA needs, and they for sure didnt knew their machinery actually uses something coming from Eastern Europe), and when their sub-company was willing to copiate my success without my permission, they asked Peltier producer company what is the best suited air cooler for this aim. The transplanetary-like megafactory consultants answered - WE dont know ANY method to cool down our Peltiers just by air. That was been one of the biggest proffessional compliments in whole my life, but the system is not a simple nor cheap.
 
Top