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Rain Sensor Status

Discussion in 'Sensors and Actuators' started by chopnhack, Sep 9, 2014.

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  1. chopnhack

    chopnhack

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    Apr 28, 2014
    LOL @ me

    Apparently I wasn't awake when I wrote this - there is no RMS with DC :oops::rolleyes:

    If I am getting 37 VDC across the coils I am exceeding the relay's capability. The rating is for 30vDC. The resistance in the coil checks out against the spec sheet, so the relay's coil is undamaged. I take the possibilities to be limited to damaged from the start or damaged during soldering. I limited my contact time to each pin to no more than 5 seconds, but the manufacturer's sheet says 3! I also did one pin and worked on another part before coming back.

    Do you think I should try to get a replacement from Digikey? or order a different relay, one with a higher coil voltage or drop the potential some more before hitting the relay?
    Thanks in advance.
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Hmm. I guess that's an "oops" for me. Sorry about that.

    The DC voltage should drop somewhat from 37V when the relay coil is passing current. Say it's 35V and you want 24V across the relay. The simplest answer is to put a resistor in series with the coil. The resistor will drop 11V, so the ratio between resistor voltage and coil voltage is 11:24. The relay coil's resistance is about 3k so the resistor should be 1.375k. Say we use a 1k2 resistor. Dissipation in the resistor will be about 0.1W so that's not a problem.

    270202.002.GIF

    The excessive voltage has probably damaged the relay coil. Sorry about that :-(

    Edit: Did you say the relay coil still measures about 3k? In that case it isn't damaged. What happens if you apply 24V DC directly across the coil? Does it click?

    Edit2: What exact part number did you use for the relay? Some of them have diodes built-in, and for them, the coil must be powered with the right polarity.
     
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  3. chopnhack

    chopnhack

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    No worries mate, even simple projects fall apart around me ;-)

    I don't think it has a diode built in - part number 399-11037-5-ND. I don't have access to 24VDC to check, but I have a 1/4w resistor. 1k5 value ok?

    Yes the coil measures 3k, which is close enough to the spec of 2.88k +/- 10% If it doesn't click over after adding the resistor I will call it a loss due to perhaps too much heat from soldering. I wonder if I can take it apart and check the internal connections w/o damaging it?
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    LOL but that's not true. You've been very careful with your PIC-based control board, and you've caught a lot of problems before they became problems. Just imagine how many problems there must be in there, that you haven't caught! Just kidding. It's practically impossible to do any non-trivial design without mistakes of one kind or another, so don't put yourself down.

    Right. That relay doesn't have anything special inside it. If the coil measures ~3k then it's not damaged, and you didn't damage it when you soldered it. It should click when you apply 24V to it. If it doesn't click, perhaps there's some mechanical damage that's preventing the armature from moving. I don't know why else it wouldn't click.

    Yes, 1k5 is fine for the resistor.
     
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  5. chopnhack

    chopnhack

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    With any luck, if I get some time tomorrow, perhaps I can do a little surgery to that board and add in the resistor. I will keep you posted.
    RE: the PIC - shivers - I can only imagine what gremlins will crawl out of that project, LOL - for sure, I will turn it on the first time from the panel! :eek::p
     
    KrisBlueNZ likes this.
  6. chopnhack

    chopnhack

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    Surgery complete, patient dead... LOL
    I took out one lead of the rainswitch at the Y trace below. I put the resistor in series and connected the r.switch. leg that was removed to the right on the other side. Potential across the coil is 26VDC - no change - the relay is still in its NC position. Relay coil ok, but relay itself not functioning.

    upload_2014-10-14_16-49-13.png
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Hmm. Faulty relay?
     
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  8. chopnhack

    chopnhack

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    It's gotta be - It just poured - I checked the rain switch and it is currently wet status - open circuit across the rain switch and no potential across the coil. I will email Digi-Key now :)

    Thanks for the help troubleshooting.

    Digi-Key was fantastic. Explained about the part, no questions asked, replacement on the way. They offered to overnight it even. I declined and elected to have it shipped the same speed I had originally paid for. Great service.

    After the relay comes in I will check it first before mounting it to the board with some jumpers from the board. Next, I will swap the LED's positive traces.

    All in all, its going to be a functional but ugly as sin board ;-)
     
    Last edited: Oct 15, 2014
    KrisBlueNZ likes this.
  9. chopnhack

    chopnhack

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    Whatta newb... :oops::rolleyes:

    So there was nothing wrong with the relay - sorry digikey!

    New relay came today - talk about fast!! I tested it and heard it click. De-soldered the old one and it too clicked... Inspected the board and realized that it was mirrored!!!! Damn... I first had it wrong in the package because I was reading the schematic as if it was a top down view (they should all be top down!) and then I think I got confused when I was creating mirrored images for the toner transfer method. :mad: To make matters worse, that wasn't the only problem. I set the pin arrangement incorrectly in Eagle so my positive trace to the LED's were not balanced between switched poles but rather one LED was on the pole and the other was on the NC gate... ughhh...

    Here is the new layout:

    upload_2014-10-16_22-26-41.png
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Oops! Well, at least you've solved the mystery!
     
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  11. chopnhack

    chopnhack

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    Learning curve ;-)
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    ski jump cartoon.jpg

    Learning curves... Take them slowly...

    skier buried in snow cartoon.jpg

    ... or you may find you're in over your head!
     
    Harald Kapp, Arouse1973 and chopnhack like this.
  13. Arouse1973

    Arouse1973 Adam

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    Lol been there done that. I even forgot the ground plane on a PCB. You should have seen the amount of wires. I told my boss it was for EMC purposes phew!
    Adam
     
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  14. chopnhack

    chopnhack

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    LOL, good cover ;-)
     
  15. chopnhack

    chopnhack

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    Should I create the common ground plane after the diode? It seems to me that the common of the 28VAC input (see lower left corner) can also go positive - potentially affecting the coil of the relay and draining the capacitor's negative terminal. I don't think it would affect the LED's since they are diodes as well unless it exceeds their reverse breakdown potential, correct? Or am I worrying about nothing? Should I be adding another diode in series with the other 28VAC common to the plane? Let me know what your thoughts are on the matter.
    Thanks all!

    Edit: - updated board layout to correctly reflect NC pin to correct LED/ RS output
    Edit: - @KrisBlueNZ - this might be easier to follow, I wasn't clear above. Thanks!

    upload_2014-10-19_22-57-21.png
     
    Last edited: Oct 20, 2014
  16. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No, that looks good as it stands.

    The 0V rail (the groundplane) is all connected to one side of the 28V AC input. That just means that the other side, that's connected to the anode of D1, ranges from about +36V to -36V relative to that 0V rail.

    D1 couples this voltage onto C1 and conducts during the short periods when its anode is positive relative to its cathode (strictly, when its anode is at least about 0.6V more positive than its cathode). C1 is the reservoir capacitor; it holds its voltage relatively constant, apart from some ripple voltage.

    So, once on every mains cycle, when the voltage from the transformer (at D1's anode) approaches its maximum positive value, D1 finds that its anode is 0.6V more positive than its cathode, and current flows, topping up the voltage in C1. During the rest of the mains voltage cycle, C1 discharges somewhat, through the load current path(s), but only by a few volts. Then it gets topped up by D1 again on the next mains cycle.

    This variation in voltage across C1 due to the rapid charging via D1 then the gradual discharging through the load resistance is the ripple that you get across C1.

    This is a "half wave" rectifier because only one half of the AC voltage (the positive half) is used to charge the reservoir capacitor. The negative half-cycles are not used at all. You can use both halves of the cycle if you use a bridge rectifier instead of a single diode, and this results in less ripple voltage since the reservoir capacitor is topped up twice as often, but that's not worthwhile for this circuit because the load current is pretty low.

    The reverse voltage across D1 (when its cathode is positive relative to its anode) can be as high as 72V, at the negative peak of the mains voltage cycle, when the voltage at its anode is -36V and its cathode is at around +36V.

    From the point of view of D1 and the rest of the circuit, the fact that D1's anode goes as far as 36V negative relative to the 0V rail (the groundplane) is not significant.
     
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  17. chopnhack

    chopnhack

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    Ah, sorry - I forgot that one side is the phase and one side is the neutral and I would do well to determine which is which when I solder them onto the board! If I soldered the neutral to D1's anode and the phase to the ground plane I assume I could blow out the cap?

    I hope tomorrow brings me some time to reprint the graphics and produce a new board!
    Thanks :)
     
  18. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No, it doesn't matter which side is phase and which is neutral. AC voltage doesn't have a polarity! C1 will always be charged the right way because D1 is connected with its cathode to C1's positive terminal, D1 will only conduct on the half-cycles where the voltage coming in at its anode is more positive than the voltage presently on C1, so it will only feed current "into" C1's positive terminal.
     
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  19. chopnhack

    chopnhack

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    Thanks mate for the explanation... I am thick. I keep looking at the ground plane and thinking that it become more positive for 1/2 the cycles...
    I need to leave this be for tonight and perhaps it will sink in tomorrow :confused::rolleyes::oops:;)
     
  20. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No you're not thick! It's just some new concepts to get your head around. Yeah, come back to it a few times.
     
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