Hello,

I have a window A/C unit that I need to get the room to 60 degrees but the unit will only allow me to set it to 64. I am trying to control the temp by using an Inkbird temperature controller that turns the A/C unit off and on via the outlet based on the inkbird's sensor. In an effort to get this to work, I have tried relocation the A/C unit's thermistor and insolate it but the results are inconsistent.

Can anyone tell me if I can provide a set temperature reading on the sensor circuit be removing the thermistor and replacing it with a 9K ohm resistor so that it reads a constant 30C? I got that resistance number from an LG service manual but I'm not sure it is correct either. My main concern is if that sort of circuit will work as long as I get the resistance value correct, or do I need other components in the circuit.

The A/C unit is an LG 15k BTU LW1516ER.ASWAHDP.

 

Hello,

I have a window A/C unit that I need to get the room to 60 degrees but the unit will only allow me to set it to 64. I am trying to control the temp by using an Inkbird temperature controller that turns the A/C unit off and on via the outlet based on the inkbird's sensor. In an effort to get this to work, I have tried relocation the A/C unit's thermistor and insolate it but the results are inconsistent.

Can anyone tell me if I can provide a set temperature reading on the sensor circuit be removing the thermistor and replacing it with a 9K ohm resistor so that it reads a constant 30C? I got that resistance number from an LG service manual but I'm not sure it is correct either. My main concern is if that sort of circuit will work as long as I get the resistance value correct, or do I need other components in the circuit.

The A/C unit is an LG 15k BTU LW1516ER.ASWAHDP.

There are as many different thermistors as ants in an anthill. You'll have to remove it, cool it, measure it and then pray it works the way you think it works. Or, you can measure it and then replace it with a thermistor that has a different temperature profile.

 

Why do you need to get the room to 60°F? That's 15.5°C which would officially be too cold for anyone to be expected to work.

 

Why do you need to get the room to 60°F? That's 15.5°C which would officially be too cold for anyone to be expected to work.

Why do you expect he plans to work in that room? He obviously has a special need like an application test. We used to test Asphalt residues at 60°F to insure they don't solidify in transport. Way easier and more accurate in a 60°F room ("field lab" or mobile hole trailer).

 

You could look at the voltage on the thermistor over the range you are interested in. Then add a small op-amp circuit with trim pots for gain and offset. You could then change the gain or the offset until the unit turns on at 60F. You would only need to change the gain over a narrow range say from 0.9 to 1.1 and the offset similarly after you characterize the voltage vs temperature output of the thermistor circuit you have now.

 

Hello,

I have a window A/C unit that I need to get the room to 60 degrees but the unit will only allow me to set it to 64. I am trying to control the temp by using an Inkbird temperature controller that turns the A/C unit off and on via the outlet based on the inkbird's sensor. In an effort to get this to work, I have tried relocation the A/C unit's thermistor and insolate it but the results are inconsistent.

Can anyone tell me if I can provide a set temperature reading on the sensor circuit be removing the thermistor and replacing it with a 9K ohm resistor so that it reads a constant 30C? I got that resistance number from an LG service manual but I'm not sure it is correct either. My main concern is if that sort of circuit will work as long as I get the resistance value correct, or do I need other components in the circuit.

The A/C unit is an LG 15k BTU LW1516ER.ASWAHDP.

Most of the responses appear to be assuming that you want the modified temperature sensor to continue controlling the unit such that it has a target of 60°F. But from what you are saying, it sounds like you just want to defeat the temperature control altogether so that the unit stays on constantly until you decide to turn it off. Is that what you are seeking? If so, one option is to simply put the sensor in a place that is always warmer than the set point -- that back of the unit is possibly an option since, presumably, if the outside air temperature is below 64°F the window AC unit probably isn't being needed. Your idea of replacing the thermistor with a fixed resistor should also work. Just measure the resistance when the thermistor is well above 64°F and replace it with a fixed resistor of similar value.

If, instead, you want the controller to still control the unit but shift its range, then it might be as simple as replacing the thermistor with a different one or possibly putting a fixed resistor either in series or in parallel with the current one. What will work depends on the nature of the thermistor you are currently using.

 

It is also possible that the unit would be incapable of getting to and/or maintaining 60° even if it ran continuously. Some people just want to emulate Sisyphus by continuing to push that boulder up the hill.

 

There are as many different thermistors as ants in an anthill. You'll have to remove it, cool it, measure it and then pray it works the way you think it works. Or, you can measure it and then replace it with a thermistor that has a different temperature profile.

Thank you. That may be the better solution.

Why do you need to get the room to 60°F? That's 15.5°C which would officially be too cold for anyone to be expected to work.

Not relevant. Thank you.

You could look at the voltage on the thermistor over the range you are interested in. Then add a small op-amp circuit with trim pots for gain and offset. You could then change the gain or the offset until the unit turns on at 60F. You would only need to change the gain over a narrow range say from 0.9 to 1.1 and the offset similarly after you characterize the voltage vs temperature output of the thermistor circuit you have now.

I like where you are going. Sounds like that would give me the true control I'm looking for but I fear that is a bit over my head. It would require quite a bit more research on my part to be sure I am even understanding the circuit that needs to be created.

Thank you.

Most of the responses appear to be assuming that you want the modified temperature sensor to continue controlling the unit such that it has a target of 60°F. But from what you are saying, it sounds like you just want to defeat the temperature control altogether so that the unit stays on constantly until you decide to turn it off. Is that what you are seeking? If so, one option is to simply put the sensor in a place that is always warmer than the set point -- that back of the unit is possibly an option since, presumably, if the outside air temperature is below 64°F the window AC unit probably isn't being needed. Your idea of replacing the thermistor with a fixed resistor should also work. Just measure the resistance when the thermistor is well above 64°F and replace it with a fixed resistor of similar value.

If, instead, you want the controller to still control the unit but shift its range, then it might be as simple as replacing the thermistor with a different one or possibly putting a fixed resistor either in series or in parallel with the current one. What will work depends on the nature of the thermistor you are currently using.

I am really open to any solution.

I think the best solution allows the unit to run itself.

I did not explain correctly before. The unit is 15k BTU. The room is less than 100 sqft but there are obstructions between the A/C unit and the area needing to be at temperature. The A/C can be set to 60F but the are in question never gets below 64F. I am currently using an inkbird temperature controller to control the unit based on the temp in the area I need and to allow me to configure the trigger points. I have relocated the thermistor of the A/C unit to outside the window but this only works when the outside temp stays above 60F. It does not always stay above 60F and then the temp in the room goes up.

I think that switching the thermistor may be my best solution if I can also put it on a longer wire.

It is also possible that the unit would be incapable of getting to and/or maintaining 60° even if it ran continuously. Some people just want to emulate Sisyphus by continuing to push that boulder up the hill.

Unit is 15k BTU, Room is less than 100 sqft. That is not an issue.

 

I like where you are going. Sounds like that would give me the true control I'm looking for but I fear that is a bit over my head. It would require quite a bit more research on my part to be sure I am even understanding the circuit that needs to be created.

Thank you.

I like WBahn's idea better: Just place a resistor in parallel or series with the existing thermistor.

 

Unit is 15k BTU, Room is less than 100 sqft. That is not an issue.

Area is not the issue -- it is volume and insulation. 15K BTU might not be enough for some rooms in LA, Vegas, or Miami.

 

Area is not the issue -- it is volume and insulation. 15K BTU might not be enough for some rooms in LA, Vegas, or Miami.

Yup, rooms with high sun load and no insulation. But, you can theorize and theorize and still be wrong - just let the OP do the experiment. We are not getting bonus points if we predict the A/C unit doesn't have enough cooling capacity and we end up being right. So the OP has the motivation and resources to override the controller and test it - so the OP should test it. Right? Or is there some crazy reason for giving up before he starts?

 

I like WBahn's idea better: Just place a resistor in parallel or series with the existing thermistor.

If I'm understanding that correctly, it would establish a sort of maximum resistance creating a minimum reading. Do I have that right?

Area is not the issue -- it is volume and insulation. 15K BTU might not be enough for some rooms in LA, Vegas, or Miami.

If that was the case, the unit would be running non-stop trying to reach the set temperature. This is running for less than a minute at a time every 10 minutes or so. I would love to just add more airflow to the needed area but too much airflow is detrimental to my situation.

Yup, rooms with high sun load and no insulation. But, you can theorize and theorize and still be wrong - just let the OP do the experiment. We are not getting bonus points if we predict the A/C unit doesn't have enough cooling capacity and we end up being right. So the OP has the motivation and resources to override the controller and test it - so the OP should test it. Right? Or is there some crazy reason for giving up before he starts?

Thank you.

 

I also like the idea of just adding a fixed resistor in series or parallel with the thermistor (depending upon whether the thermistor has a positive or negative temperature coefficient).

 

If I'm understanding that correctly, it would establish a sort of maximum resistance creating a minimum reading. Do I have that right?

Depends on some specifics that haven't been laid out yet.

Here's the basic idea:

Measure the resistance of the existing thermistor at several points, say perhaps every 10°F from 50°F to 90°F is probably fine. There are some pretty straight forward ways to do that pretty easily with reasonably accuracy, provided you have a decent thermometer that also covers that range.

The goal is then to combine the thermistor with other resistors such that something like 40°F to 70°F maps roughly to that same range. Doesn't need to be very close at all. You then make a table of the mapping so that you know what to set the temperature controller at to get close to a desired temperature.

But consider some alternatives first that might be better. You mentioned that it is only one part of this room that you are really interested in. Have you considered building an insulated chamber around that area and putting the window unit directly into that space? It wouldn't take much to make a highly insulated chamber that could be disassembled and stored out of the way when not in use.

 

Depends on some specifics that haven't been laid out yet.

Here's the basic idea:

Measure the resistance of the existing thermistor at several points, say perhaps every 10°F from 50°F to 90°F is probably fine. There are some pretty straight forward ways to do that pretty easily with reasonably accuracy, provided you have a decent thermometer that also covers that range.

The goal is then to combine the thermistor with other resistors such that something like 40°F to 70°F maps roughly to that same range. Doesn't need to be very close at all. You then make a table of the mapping so that you know what to set the temperature controller at to get close to a desired temperature.

But consider some alternatives first that might be better. You mentioned that it is only one part of this room that you are really interested in. Have you considered building an insulated chamber around that area and putting the window unit directly into that space? It wouldn't take much to make a highly insulated chamber that could be disassembled and stored out of the way when not in use.

I took a 3 week electrical fundamentals class like 5 years ago, so please forgive my ignorance.

Can I simply use a multimeter to measure the resistance across the thermistor while it is unplugged from the A/C unit and at the various temperatures? Not sure if I have to apply a specific voltage and measure the current draw, because I don't think I have the equipment for that.

Thinking about it and trying to remember, do I have it correct that a resistor added in series with the thermistor will create an offset and a resistor in parallel will change how it scales with temperature.

Funny you should mention it, creating an insulated chamber around the area I need to be cooled is exactly what I have done. Trouble is I can't have that cold, dry, and fast moving air blowing directly in the area. It needs the humidity to be balanced and brought into the area at a slow but steady rate. This also needs to work 24/7/365.

I currently have it working for the most part but it still cuts off at times. The thermistor is on a long wire and so I was able to run it to the outside but that failed the next morning when when the outside temp dropped below 60F for a couple hours. I now have it sandwiched between a couple of pieces of insulation but it is still cutting off too often at night and I would like something more definitive and permanent.

 

Can I simply use a multimeter to measure the resistance across the thermistor while it is unplugged from the A/C unit and at the various temperatures?

Yes, that would be the typical way to do measure its resistance.

a resistor added in series with the thermistor will create an offset and a resistor in parallel will change how it scales with temperature.

Either way, a resistor will affect both offset and scale.
Whether you put it in series or parallel depends upon whether you need a lower resistance or a higher resistance to lower the temperature that the air conditioner turns off.
Post your measurements and we can suggest an appropriate value and connection configuration for the added resistor.

 

If I'm understanding that correctly, it would establish a sort of maximum resistance creating a minimum reading. Do I have that right?

It just 'shifts' the Temperature vs Resistance curve. Example: If the thermistor has a positive temperature co-efficient (PTC) then the resistance increases with temperature and with a negative temperature co-efficient (NTC) the resistance decreases with temperature. By measuring two points: The resistance of the thermistor at 60F and the resistance at 64F we can then shift it. Completely hypothetical but practical: say the resistance at 60F it 10K ohms and the resistance at 64F is 10.5K ohms, this would be a thermistor with PTC. So what you do is take the difference, in this case you want it to look like 10.5K ohm when at 60F, so you would place a 500 ohm resistor in series with the thermistor. Now if it is the other way. That is an NTC, say it is 10K ohms at 60F and 9.5K ohms at 64F, then we want the resistance of the NTC to be 9.5K ohms at 60F. So in that case we need to place it in parallel, the formula for the value in that case is:



This is just the parallel resistance formula for two resistors.
R64 is the resistance of the thermistor at 64F
R60 is the resistance of the thermistor at 60F
R is the required resistor you will need to place in parallel with the thermistor that has NTC.

So we solve for R:

Thus in the example provided that would be 10Kohms*9.5Kohms/(10Kohms-9.5Kohms)=190Koms.

And the proof just plugging it back in for parallel resistance:

R=10Kohms*190Kohms/(10Kohms+190Kohms) = 9.5 Kohms

verified.

Do you have a standard ohm meter? You will need that to measure the resistance at 60F and at 64F. Obviously we will need an accurate way of having the thermistor at 60F and 64F. Be sure you give the thermistor time to reach ambient temperature (soaking time), I would recommend one hour for that.

Because you are only interested in the 60F point, it does not matter that it 'throws' the accuracy off at any other point. Unless you decide to change your desired turn on point to something other than 60F, in which case you will need an entirely different resistance value.

 

It just 'shifts' the Temperature vs Resistance curve. Example: If the thermistor has a positive temperature co-efficient (PTC) then the resistance increases with temperature and with a negative temperature co-efficient (NTC) the resistance decreases with temperature. By measuring two points: The resistance of the thermistor at 60F and the resistance at 64F we can then shift it. Completely hypothetical but practical: say the resistance at 60F it 10K ohms and the resistance at 64F is 10.5K ohms, this would be a thermistor with PTC. So what you do is take the difference, in this case you want it to look like 10.5K ohm when at 60F, so you would place a 500 ohm resistor in series with the thermistor. Now if it is the other way. That is an NTC, say it is 10K ohms at 60F and 9.5K ohms at 64F, then we want the resistance of the NTC to be 9.5K ohms at 60F. So in that case we need to place it in parallel, the formula for the value in that case is:

View attachment 276032

This is just the parallel resistance formula for two resistors.
R64 is the resistance of the thermistor at 64F
R60 is the resistance of the thermistor at 60F
R is the required resistor you will need to place in parallel with the thermistor that has NTC.

So we solve for R:
View attachment 276034
Thus in the example provided that would be 10Kohms*9.5Kohms/(10Kohms-9.5Kohms)=190Koms.

And the proof just plugging it back in for parallel resistance:

R=10Kohms*190Kohms/(10Kohms+190Kohms) = 9.5 Kohms

verified.

Do you have a standard ohm meter? You will need that to measure the resistance at 60F and at 64F. Obviously we will need an accurate way of having the thermistor at 60F and 64F. Be sure you give the thermistor time to reach ambient temperature (soaking time), I would recommend one hour for that.

Because you are only interested in the 60F point, it does not matter that it 'throws' the accuracy off at any other point. Unless you decide to change your desired turn on point to something other than 60F, in which case you will need an entirely different resistance value.

Thank you so much! This is exactly what I was looking for. A clear, detailed, and complete explanation of the factors at hand. Even provided the formulas with explanations for those too.

After the trolls I have been fighting off to find a true expert that doesn't load everything down with sarcastic tangents and heaps of condescension, you sir are a breath of fresh air.

I will perform the tests, see what I come up with, and reply back once I know more.

Thank you, Sir!

 

It just 'shifts' the Temperature vs Resistance curve. Example: If the thermistor has a positive temperature co-efficient (PTC) then the resistance increases with temperature and with a negative temperature co-efficient (NTC) the resistance decreases with temperature. By measuring two points: The resistance of the thermistor at 60F and the resistance at 64F we can then shift it. Completely hypothetical but practical: say the resistance at 60F it 10K ohms and the resistance at 64F is 10.5K ohms, this would be a thermistor with PTC. So what you do is take the difference, in this case you want it to look like 10.5K ohm when at 60F, so you would place a 500 ohm resistor in series with the thermistor. Now if it is the other way. That is an NTC, say it is 10K ohms at 60F and 9.5K ohms at 64F, then we want the resistance of the NTC to be 9.5K ohms at 60F. So in that case we need to place it in parallel, the formula for the value in that case is:

View attachment 276032

This is just the parallel resistance formula for two resistors.
R64 is the resistance of the thermistor at 64F
R60 is the resistance of the thermistor at 60F
R is the required resistor you will need to place in parallel with the thermistor that has NTC.

So we solve for R:
View attachment 276034
Thus in the example provided that would be 10Kohms*9.5Kohms/(10Kohms-9.5Kohms)=190Koms.

And the proof just plugging it back in for parallel resistance:

R=10Kohms*190Kohms/(10Kohms+190Kohms) = 9.5 Kohms

verified.

Do you have a standard ohm meter? You will need that to measure the resistance at 60F and at 64F. Obviously we will need an accurate way of having the thermistor at 60F and 64F. Be sure you give the thermistor time to reach ambient temperature (soaking time), I would recommend one hour for that.

Because you are only interested in the 60F point, it does not matter that it 'throws' the accuracy off at any other point. Unless you decide to change your desired turn on point to something other than 60F, in which case you will need an entirely different resistance value.

Thank you for the complements. I would like to point out that this was WBhan's idea to use a resistor in series or parallel, my idea of using an op-amp was unnecessarily complicated. I just took his idea a step further and showed an example. Much success hope this works for you. Looking forward to hearing back on how well this turns out for you.