Connect with us

Optoisolated Triac Water Heater Controller

Discussion in 'General Electronics Discussion' started by moniker, Jun 19, 2010.

Scroll to continue with content
  1. moniker

    moniker

    4
    0
    Jun 19, 2010
    Hi All,

    First post here. I hope I don’t bore everyone.

    I have just spent some time designing a small water heater controller and would like some help from those with more experience than I have. Mechanical thermostats have too much dead band and this design should provide very little hysteresis. PID controllers are way too expensive and complicated. So, I gave it a shot. I welcome suggestions, constructive criticism, and sage advice. I am sure I did not get it all right.

    I am concerned about the wattage of the resistor on the AC side of the Triac. Any help is appreciated.

    I calculated the current through an LED that will show the state of the heater. I shot for 5ma but am not sure I got it right.

    I shot for 30ma for the IR LED in the optoisolator. Same question there. Did I get it right?

    The Triac is a 12A job that I think should get ‘er done with ease. Maybe a heat sink is necessary?

    The schematic is attached. I hope you will give it a look and tell me what you think.

    Thanks
    Moniker
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    My first suggestion would be to consider using a MOC3041 as triggers on zero crossing and will generate less RFI

    edit: Oh, you also need to consider the thermal lag between the heater and the sensor. This may introduce several degrees of temperature variation even if there is no hysteresis.
     
  3. moniker

    moniker

    4
    0
    Jun 19, 2010
    reply

    Steve,

    Thanks for the part number.

    There is over and undershoot on the heater. But the deadband on the mechanical thermostat is way wider than what this circuit will provide when it is up and running properly. A couple degree swings will put me in heaven compared to what I have now.

    I hope someone will check my work to see if I got it right. I am not sure I have the logic right re: the opamp inputs. :-o)

    Thanks

    Moniker
     
    Last edited: Jun 19, 2010
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    As it is now, the heater will turn on when the water gets hot, and off when it gets cold.

    You probably want to reverse the position of the thermistor and the pot.

    Do you know the resistance of the thermistor at 94F? is it around 10K. Your pot should be approx double the expected resistance at the trip point so that it is approximately in the middle of its travel.
     
  5. moniker

    moniker

    4
    0
    Jun 19, 2010
    Steve,

    Thanks for the interest and help.

    From looking at some general curves it seems the thermistor resistance will drop to around 6-8K at 100C. So I am figuring it will be a bit higher at 94C. Maybe about 7-9K. Doubling that will result in a pot that is around 15-20K. I used 20K in the schematic simply to use a common number. When I build the real circuit I will measure the resistance in a boiling water bath. I am at sea level so that would be @100C.

    As I said, I am having trouble wrapping my head around whether this circuit will turn on as temp lowers or rises.

    It seems to me that the balance set point is on the + input and the measured result is on the - input. So if temp decreases the - input gets closer to ground and that signal is inverted to a positive out which turns on the transistor and the triac creating heat. If temp increases the - input goes more positive and the signal is inverted to a more negative out which turns off the transistor and the triac therby shutting off the heater. But I surely could be wrong in these assumptions.

    If the above is not the way it works, then I will have to swap the pot and thermistor. Can you help me understand it better?

    Thanks

    Moniker
     
    Last edited: Jun 20, 2010
  6. jackorocko

    jackorocko

    1,284
    1
    Apr 4, 2010
    Now, in the schematic you have the thermistor resistance at 100k ohms at 25C. That means the negative input has 9.5~ volts. The two resistances make a simple voltage dividing network. Vout = (R2/R1+R2)*Vin http://en.wikipedia.org/wiki/Voltage_divider
    In other words, if R2 is Greater then R1 the voltage will be > 50% of Vin. If R2 is Less then R1 the voltage will be < 50% of Vin for easy reference. Of course if R2 and R1 are equal, the voltage will be 50% of Vin

    Also, with a 741 op-amp hooking the neg(-) power rail up to ground will only let the output drop as low as ~2 volts if I remember correctly. It wasn't too long ago I tried to design a circuit that was much like your's. When I was finished I decided to use a lm339 comparator. It was easier for me too use and it looks like it'd be easier for you too. I'll post my schematic below in case you was wondering what I did with the comparator. It looks eerily similar to your design :)

    R4 & R5 set the reference voltage to 6V
    R6 & R7 is the voltage that will be compared to the reference voltage
    R8 is there to control the current of the output, since the lm339 has an open ended collector. If I misspoke, steve will let me know. Most of what I am repeating here is what I learned from steve anyway.


    [​IMG]
     
    Last edited: Jun 20, 2010
  7. moniker

    moniker

    4
    0
    Jun 19, 2010
    Hi All,

    Any suggestions for a rail to rail replacement for the 741? That should remove all concerns about saturating the transistor.

    Thanks

    moniker
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-