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Help with Water Pump

Discussion in 'General Electronics Chat' started by salty joe, Dec 30, 2011.

  1. salty joe

    salty joe Active Member

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    Pin 12 is fault and 13 is V+.
    With everything turned off and DMM on 200K, I found virtually no resistance between FSM channel 13 fault terminal point and U1 pin 12.
    Ditto for FSM V+ term point and pin 13.

    Should I pull that board?
     
  2. alec_t

    alec_t Well-Known Member Most Helpful Member

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    With everything connected but switched off, and your DMM on 200k, what resistance do you measure between the Fault terminal and ground?
     
  3. salty joe

    salty joe Active Member

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    After about 30 sec., it fluctuated between 31.1 & 31.3K.
     
  4. dave

    Dave New Member

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  5. alec_t

    alec_t Well-Known Member Most Helpful Member

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    My thinking is that something must be conducting enough to pull the Fault signal low. 31k between the Fault line and ground would be low enough to do that. The question is, is the mystery 31k in the PDM or the FSM?
    Try disconnecting the wire between the Fault output of the FSM and the Fault input of the PDM, then with power off measure the resistance to ground (a) at the FSM Fault terminal and (b) at the PDM Fault terminal.
     
  6. salty joe

    salty joe Active Member

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    With the fault wire disconnected, the PDM fault terminal point showed 28.9-29.1K.
    To measure the FSM term point, I had to switch the DMM to 2000K. After 30-60 sec the DMM showed over 1000K and climbing.
    I hooked the wire back up and got the original 31K. I hope that makes more sense to you than it does to me.
    Thanks.
     
  7. alec_t

    alec_t Well-Known Member Most Helpful Member

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    The good news is the FSM measures ok. But if the Fault input (pin 12) of the PDM is measuring ~30k that indicates U1d is leaky, or some gunge is partly shorting pin 12 to ground. If that part of the circuit is gunge-free then a new IC is called for :(.
     
  8. salty joe

    salty joe Active Member

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    Well I scrubbed the back of U1 with a toothbrush and laquer thinner-I was hoping for a film I couldn't see...Looks like it's time for a new IC. :(At least I have plenty of room on the board below PDMs 13 & 14.

    Let's say I got a new IC and gave it a cap, 12V and signal ground. Could I make pin 12 the fault for PDM13, leaving the rest of the new IC disconnected(except for the pins that need grounded)?

    Thank you Alec for tracking down the problem.:)
     
  9. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    There is nothing special about a "channel" of an IC when the parts are identical. Power , ground and the bypass cap have to be included as well as whatever way to put an IC to sleep in the lowest power mode. The suggested "sleep mode" so to speak.

    So, adding one t a different place and just transferring what's needed is OK.

    Another idea is to get yourself a machine pin socket. These http://www.jameco.com/1/1/28049-6100-16-r-6100-socket-ic-16pin-machine-tooled-low-profile.html sockets are EXTREMELY reliable especially if you get them in gold flash. I would not hesitate to use one.

    Here is a 16 pin gold flash: http://www.digikey.com/product-detail/en/110-93-316-41-001000/ED3316-ND/14041 Not sure if you need 14 or 16 pins.

    So, you could cut all of the pins close to the IC and solder a socket to these cut stubs and go from there.
     
  10. salty joe

    salty joe Active Member

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    It sounds like this might be easy. Please tell me if I have this right.

    Install a new CD4093 IC with a new cap, 12V and signal ground. Connect pin 12 on the new IC to channel 13 fault. Ground pin 12 (faulty fault) from the old CD4093 IC. Ground pins 1,2,5,6,8,9 &13 on the new IC. Will that do the trick?
     
  11. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Sounds about right. You will have to cut the 3 pins on the non-working IC. At least that should make it easier.
     
  12. salty joe

    salty joe Active Member

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    As I see it there are 4 pins to not hook up on the new IC-4, 3, 10 & 11. That's the IC you refere to as non-working, right?

    Thanks.
     
  13. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Joe:

    What I was suggesting is that if one gate is bad (Say 1,2, & 3), just move that gate to 1,2,and 3 on another chip. Ground 5,6,8,9 12 and 13 and connect power ground and the bypass cap.

    Maybe not the most elegant fix or even the safest.

    Cutting the pins right at the IC is a removal method, but it also keeps downstairs intact.

    Soldering an IC or an IC socket to those stubs is a way of fixing things. So, is cutting off the IC and moving the connections on the bottom.

    So, I'm just offering repair/rework suggestions.

    Now, if the entire IC was bad, you would have to move/replace it, I got the impression that alec though one section was bad.

    I am in agreement that moving/replacing the whole IC is better.

    Just not sure if you would have thought about the possibility of cutting the pins at the top of the IC. With the right cutters or possibly an x-acto knife you can do it.
     
  14. salty joe

    salty joe Active Member

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    OK, now I have a little better understanding of that IC, it's more or less four chanels in one IC.
    I'll replace the whole IC as soon as my batch of terminal points arrive. I was hoping to see them yesterday.

    Thanks guys.
     
  15. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Replacing the whole IC in PDM 13 avoids any uncertainty about the operation of the three assumed 'non-faulty' gates (U1 a,b,c). But if you wanted to bypass the one suspected faulty gate (U1d) with a band-aid approach here's the band-aid. Only one connection (at D11 cathode) needs to be broken and three new connections made.
    PDM_BandAid.gif
     
  16. salty joe

    salty joe Active Member

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    Thanks , I think I'll replace the whole IC. The difference in time spent would be minimal versus the bandaid. Other than PDM 13 not responding to lost power, the entire system is fabulous.

    Really like your FIRE!
     
  17. salty joe

    salty joe Active Member

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    Put the new IC for PDM 13 on the same board as the FSM. It's all systems go!:) It was well worth the effort. To be alerted that a pump has lost power is a great feature, a super nice touch. Thank you a zillion for tracking that down to the IC, Alec. I try to find my mistakes first, but I don't think I was ever gonna find that one.

    BTW, is there a test for ICs before instalation? The last one that failed surprised me because I don't think I cooked or zapped it.

    I'll run wires for the four additional pumps up to the tank and let you guys know how they do.
    Thanks guys.
     
  18. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    @ joe
    The major killer is a loose connection on the inputs. This can cause the chip to oscillate which causes it to heat up.
    They are affected by static damage as well. It's tough to do static damage in a basement, unless it's your clothing.
    Storing them on aluminum foil covered styrofoam works. Touch the aluminum before touching the IC.

    Yes, you can test them. http://circuits.datasheetdir.com/109/CD4093BC-pinout.jpg

    Get a breadboard, and wire a ground to all of the inputs and attach a source of power. A 9V battery would do.
    At this point the outputs would be high.

    The function is NOT (A AND B); So only when both inputs are low, will the output be high.

    A high is a connection to +V and a low is a connection to ground. You would have to go though the truth table.
     
  19. alec_t

    alec_t Well-Known Member Most Helpful Member

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    That's what we like to hear! Your Christmas should now be merrier :).
    It's possible, but would involve making a test jig so that high/low logic levels could be applied selectively to all the gate inputs (assuming the IC is a logic device such as the 4093). A socket for the IC would be needed, or a breadboard as KISS mentioned.
     
  20. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    An idea for a test jig, would be:
    1. You could use a Complementary BCD rotary switch, and 4 resistors (~10K pull ups) to select the states.
    http://www.mouser.com/ProductDetail...=sGAEpiMZZMsqIr59i2oRcvurPonEjaxB6J0/jAjVQAM=
    2. A 3 pole 4 position to select the gate.
    3. Something to act as a driver. ULN2004, LM334 for instance.

    To use, you would select the gate (1-4)
    and then rotate through the first 4 positions of the BCD switch.

    To make it a bit more versatile, you could use a 3 or 4 pole, 6 position switch and a "programming header", so you could test a hex inverter if you wanted. With some versatility, you could even do a go, no go for the LM324.
     
  21. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Happy Anniversary, Joe!
    It's two years to the day, since you started this thread :)

    Wine.gif
     

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