Air Conditioner controlled by dew point transmitter 4-20mA loop

[GaryB]

I am trying to control my central air-conditioner by dew point instead of temperature.

I would like the air-conditioner to turn on at a dew point of 55 degrees Fahrenheit and then shut off the air-conditioner at a dew point of 50 degrees Fahrenheit. This means a latch holding the air-conditioner on would occur at dew point 55 and the latch would be broken at 50 degrees Fahrenheit. This pattern would happen repeatedly and it would keep the dew point between 50 and 55 degrees Fahrenheit.

I found a dew point transmitter that outputs the dew point level in the 4...20mA = 0...80°F format. This would mean every mA of current would represent a 5 degree change in dew point. This would mean my latch turning on the air-conditioner would occur at 15mA and the latch would be broken at 14mA.

What would be the easiest way to accomplish the set-points of 14 and 15 mA? I could use a resistor as a load, go through an optic-isolator, and then use a couple compactors, then latching relay and another relay to break the latch?

Is there an easier way?

 

[c_mitra]

I guess your basic idea is simple and effective. Why don't you use the opto-isolator after the comparators? You can directly turn on-off the air-conditional using the output from the opto-isolator.

 

[c_mitra]

You need to set up your logic table as follows:

dew point > 55; on
dew point <55 and dew point >50; off- not so sure. Please check.
dew point <50; off

You also must use sufficient hysteresis so that the compressor is not turned on every other minute. Other technical details will come later.

 

[GaryB]

dew point >55; latch on, power on
dew point <55 and dew point >50; if latch is set, power on; if latch is not set, power off
dew point <50 ; power off

True. I know that it would take my air-conditioner a long time to make a 5 degree Fahrenheit change in temperature, but I do not know how long it would take to make a 5 degree change in Fahrenheit dew point. It would seem logical that the 5 degree swing in dew point would happen faster than the 5 degree swing in temperature. To guard against very short runs on the compressor, I may want to have the latch stay on for a minimum of seven minutes from the time the latch is first initiated.

 

[c_mitra]

I do not see the condition that unsets the latch.

 

[GaryB]

dew point >55; latch on
dew point <50; latch reset
latch on; compressor on
latch reset; compressor off

Not sure how to write the logic table for minimum run time of 10 minutes for compressor.

 

[c_mitra]

Not sure how to write the logic table for minimum run time of 10 minutes for compressor.

That is easy!

Everytime you turn on the compressor, you save the local time (wall clock) in a variable.

Everytime you need to turn off the compressor, you check the current time with the saved time. If the difference is more than 10 mins, then turn off else (do nothing)

 

[BradtheRad]

You also must use sufficient hysteresis so that the compressor is not turned on every other minute.

Agreed. Common advice says that after turning it off, wait a few minutes before you turn it on again. The reason is because liquid refrigerant remains in the compressor at shut-off. The non-compressible liquid bogs down the compressor, causing excessive Ampere draw on startup. Therefore we should give it time for the liquid to turn to gas.

 

[c_mitra]

Agreed. Common advice says that after turning it off, wait a few minutes before you turn it on again. .

Absolutely correct. I modify my comment in #7 as follows:

When you turn off the compressor, save the clock time in a variable (time1)

When you want to turn on the compressor, note the clock time (time2) and if (time2-time1) is greater than 10 mins, only then turn on the compressor.