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Humidistat op amp circuit

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Danen

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I'm extremely new to building circuits. I want to build the exact same circuit below, but first I want to understand how it works. Here's what I think I know:
  • It's a combination of a Voltage Divider and a Comparator
  • As the humidity rises, the resistance lowers and voltage increases
However, I'm lost as to what role the 47k resistor plays in the circuit. I know the 10k pot works at setting a "switching humidity" setpoint. I have also attached the Humidistat's Datasheet as well.

Ideally, I would like to have the humdistat in a chamber with a mister. When the humidity drops below 90%, a signal is sent to the DigiBee. From there, the DigiBee sends a signal to a water solenoid valve connected to the mister. The mister turns on until humidity inside the chamber reaches above 90% and shuts off.


humid%20cct.gif
 

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  • DSA00292754.pdf
    54.4 KB · Views: 253
The data sheet implies that the resistance of the sensor should be measured using AC, not DC like the circuit you show above. I believe that the no-DC across the sensor limitation has to do with electrolysis damaging the sensor.

In the circuit above, the senor, the 47K resistor, and the pot form a Wheatstone Bridge
 
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The 47k combined with the humidity sensor makes a voltage divider, the 47k pulls to ground and the humid sensor pulls to V+.
The 339 is a quad comparator, your just using 2, which ever input is highest voltage wise controls the output, if + is higher than - the output is high or +, if - is higher than + then the output is - or low.
Look like the pair of comparators are being used to tell if the humidity is within a certain range, adjustable by the 2 10k pots.
As mike just said allthough this circuit may work the sensor isnt going to last that long, youd be better generating a frequency with it using something like a 555 chip and reading that with the uC.
 
The data sheet implies that the resistance of the sensor should be measured using AC, not DC like the circuit you show above. I believe that the no-DC across the sensor limitation has to do with electrolysis damaging the sensor.
The 47k combined with the humidity sensor makes a voltage divider, the 47k pulls to ground and the humid sensor pulls to V+.
The 339 is a quad comparator, your just using 2, which ever input is highest voltage wise controls the output, if + is higher than - the output is high or +, if - is higher than + then the output is - or low.
Look like the pair of comparators are being used to tell if the humidity is within a certain range, adjustable by the 2 10k pots.
As mike just said allthough this circuit may work the sensor isnt going to last that long, youd be better generating a frequency with it using something like a 555 chip and reading that with the uC.
Thanks!
The 47k combined with the humidity sensor makes a voltage divider, the 47k pulls to ground and the humid sensor pulls to V+.
The 339 is a quad comparator, your just using 2, which ever input is highest voltage wise controls the output, if + is higher than - the output is high or +, if - is higher than + then the output is - or low.
Look like the pair of comparators are being used to tell if the humidity is within a certain range, adjustable by the 2 10k pots.
As mike just said allthough this circuit may work the sensor isnt going to last that long, youd be better generating a frequency with it using something like a 555 chip and reading that with the uC.
Thanks! That clears a lot up. So I could set the first pot at a lower resistance (higher humidity). The pot (-/low)would be the output until the Humidistat's resistance became lower than the pot. I could tell the DigiBee that while the pot controlled the ouput, the water solenoid turns on.
The valve will remain on until the Humdistat's resistance drops below the resistance of pot #2. In which case I set the DigiBee to turn off the valve.

If my logic is correct, I have another question. How did they come up with a 47k resistor? Was this a random number, or is there a more logical reasoning?

Thanks for the heads up on the ac rating on the Humdistat. Once I get this figured out I'll tackle that issue.
 
WARNING: Apart from the fact that the sensor should have AC, not DC, across it the peak voltage must be less than 1.4V. The 47k resistor is WRONG and results in about 2V across the sensor!
 
WARNING: Apart from the fact that the sensor should have AC, not DC, across it the peak voltage must be less than 1.4V. The 47k resistor is WRONG and results in about 2V across the sensor!
So the bottom resistor rating is simply anything that keeps the volts less than that of the Humidistat's peak?
 
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So the bottom resistor rating is simply anything that keeps the volts less than that of the Humidistat's peak?
Rated voltage(ac) =1.4Vpk right on the data sheet. This comes right from the electromotive series table, which specifies the voltage at which electrolysis occurs.

If you put 2.5V dc across that sensor, it will last about three days. It is obvious that whoever posted the original circuit on the web didn't have a clue.
 
Rated voltage(ac) =1.4Vpk right on the data sheet. This comes right from the electromotive series table, which specifies the voltage at which electrolysis occurs.

If you put 2.5V dc across that sensor, it will last about three days. It is obvious that whoever posted the original circuit on the web didn't have a clue.
Great. At least I feel a little better that the circuit was wrong to begin with. So I need to find a resistor that limits the volts to the Humidistat to at least 1.4V? I apologize for the simplistic questions. What would be the calculation for that?
 
The approach is totally wrong. To do it right, you need to drive the Wheatstone bridge with an AC voltage, not DC. The "detector" which determines if the bridge is in balance or not must be suitable for detecting an AC voltage.
 
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The approach is totally wrong. To do it right, you need to drive the Wheatstone bridge with an AC voltage, not DC. The "detector" which determines if the bridge is in balance or not must be suitable for detecting an AC voltage.
Okay, I think I follow. So I would need something like this?
imagesCADP5GNZ.jpg
 
looks like you have all the advice you need Daren.
 
looks like you have all the advice you need Daren.
I'm getting a whole lot closer than I was before. After doing some reading, I think I understand the basics of a Wheatstone Bridge. R3 and R4 have to be equal? I guess I'm not entirely sure how the value of the two resistors are found. I know R2 would be the Potentiometer.
 
WARNING: Apart from the fact that the sensor should have AC, not DC, across it the peak voltage must be less than 1.4V. The 47k resistor is WRONG and results in about 2V across the sensor!
I calculate about 1.64V across the sensor, but that's at 60%RH... at 30%RH you have 920K vs 47K, or about 20 to 1, or 4.757V across the sensor... I would suggest the driver voltage be limited to the peak sensor voltage...
 
Here's how you could drive the sensor with an AC signal well below its 1.4V limit.
U1a forms an oscillator of ~600Hz. The sensor is fed via C2. The voltage across R5 never goes below -0.4V, so latch-up of the comparators U1b,U1c is avoided. The positive-going peaks of the R5 voltage are compared in U1b,U1c with thresholds set respectively by trimpots P1,P2. Output pulses from U1b,U1c are integrated on C3,C4 to give Hi/Lo logic levels for feeding to the micro.
If all you are interested in is RH ~90% you could do away with one comparator and use a single comparator with hysteresis.
HumiditySensor.gif
 

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  • HumiditySensor.asc
    4.3 KB · Views: 184
Awesome! Thanks Alec and Mike for all of the help. I'll get to work and see what happens.

Sorry dr pepper for what ever might have offended you.
 
Nope not offended, I was remarking on how effective the forum is, you ask a question and someone solves the problem for you, cant be bad.
 
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