Time Delay Relay to Permanently Open

Discussion in 'The Projects Forum' started by markmain, Dec 28, 2008.

  1. markmain

    Thread Starter Member

    Dec 28, 2008
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    I'm trying to make a circuit that will start a timer for 4 minutes (approximately) when it receives a positive input signal of 5VDC to 9VDC; the timer will stop and reset if the input signal drops at any time during that period.
    If the constant 4 minute limit is reached (which means one of my sensors is probably broken) then it will permanently keep a different line (which also uses 5VDC to 9VDC/10mA) in a permanently open state from what was originally a closed state at startup, and it will permanently power a buzzer (5VDC to 9VDC/7mA).
    I need this to be a small simple and dependable circuit that will run for years to come. Any help is appreciated, especially a diagram.
    Thank you for your help.
     
  2. markmain

    Thread Starter Member

    Dec 28, 2008
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    As I re-read my note I realized that I should mention why I need it to handle 5VDC to 9VDC; I plan to build 3 of these for 3 separate projects. The first project has a 5VDC transformer, the second project has a 9VDC transformer, and the last one has a 6VDC transformer.

    And so, the voltage will be constant for each project, and that will certainly make things simpler.

    I've been searching around and can't quite figure out a simple way to do this. Thanks again for help.
     
  3. KMoffett

    AAC Fanatic!

    Dec 19, 2007
    2,574
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    markmain,

    This is a circuit that should work for you. The timing is pretty much independent of supply voltage. R6 is not needed for the 5v or 6V supply. R6 should be equal to 4V/(Ibuzzer+Irelay) for the 9V supply.

    Ken
     
  4. markmain

    Thread Starter Member

    Dec 28, 2008
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    Wow, thank you for the schematic. I've been working a lot on this and so this will really help. Thank you.

    I understand all of what you have in it except for one thing that I want to confirm: I assume that pins 10 & 11 on the CD4001 are unused; if this is not true please let me know (e.g. they are hooked up to the source)
     
  5. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    Pins 10 & 11 are outputs and should be left unconnected to anything
    I was surprised that no one else had jumped in on your request by now.

    Ken
     
  6. markmain

    Thread Starter Member

    Dec 28, 2008
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    One other question about R6.

    My buzzer is a MSR516NPR from Mallory Sonalert Products; it's suppose to handle anything from 5 to 16VDC.

    Do I still need the resistor R6?

    If so, I assume you are saying that I divide 4V by the total ohms of the buzzer plus the relay... correct?

    My relay is a Crydom LC241 (Normally Open relay), input is 15mA@5VDC 300 Ohms; and the buzzer didn't show ohms, but the graph (on their specs) looks like 7mA @ 9VDC, and 2mA @5VDC and slightly more than 3mA @6VDC... and that confused me since I was expecting Ohms Law to work; using V=IR I get 1286 Ohms for the 9VDC, 2000 for the 6VDC and 2500 for the 5VDC. I'm new, learning, and confused on that point.

    If you can clarifiy what I need to do for resistance, that would be a great help, I don't want to fry the part buzzer since it cost about $8.

    Thanks again and Happy New Year!
     
  7. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    Auck! I lost my schematic! Oh, well:

    For the 9vdc version, connect the Sonalert (+) to the left side of R6, and make R6 approximately equal to the the relay's coil resistance...270 is a standard value and close enough. For the 5v/6vdc version, just eliminate (short out) R6.

    Ken
     
    Last edited: Dec 31, 2008
  8. markmain

    Thread Starter Member

    Dec 28, 2008
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    I've been studying your plan with great interest and I'd really like to learn how to adjust the time so that I can use this for future projects.

    I've read 4 different datasheets on the chip and they didn't help me much. I can see by your note that it's done via the resistor R1 & R2, but I need some more help, because I'm doing something wrong in my calculations because I calculated that you'd need 292911 Ohms on 0.1 uF for it to be 34.14 Hz.

    34.14 Hz = 1 / (292911 Ohms * 0.0000001 F)

    If you have a moment to explain how to adjust the time I would be grateful because I'm having a tough time finding info on this chip. I'll keep looking, because I'm trying not to just ask dumb questions.

    Thanks for the schematic, I'm learning a lot from this, it's very exciting.
     
  9. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
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    On Motorola 4060 datasheet, the formula for freq is:

    freq = 1/[ 2.3 * R * C ].

    That works out to be about 128KΩ. It seems you have missed the factor 2.3 in your original calculation.
     
  10. markmain

    Thread Starter Member

    Dec 28, 2008
    13
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    Thanks, I'm learning. I've been up all night on this, now I'm going to bed. I'm really having fun working on this. Happy New Year everyone!
     
  11. markmain

    Thread Starter Member

    Dec 28, 2008
    13
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    I ordered the parts today from Digikey and I just bought a variety resistances for those 25-turn pots that they have to I can play around with the Circuit that Ken showed me. I'm new at this, but working hard to learn, I've been studying on this solidly over the holiday since it was posted.

    I'd appreciate some coaching if my understanding is wrong, but here is how I think that I control the time:

    Q=this will be the divisor of the pin that I want to estimate the pull time (in seconds); so for pin 3 Q will be 16384 (2^14).

    Pulse Time in Seconds for the Pin = Q * 2.3 * C1 (in Farads) * (R1 + R2)

    So, using Ken's example,

    Pin 3 Pulse Time will be 16384 * 2.3 * 0.0000001 * (100000 + 31000) = 493.64992 seconds (8.23 minutes)

    And the CD4001 is being used to permanently lock the power onto the 2N3904 transistor once it receives it's first pulse.

    Thanks again for the help, this has really given me the boost that I needed to improve what I know.
     
  12. markmain

    Thread Starter Member

    Dec 28, 2008
    13
    0
    I really like how you used the CD4001, it's opened up a whole new world to me--I'm not so afraid of these chips now.

    After studying this it left me wondering what the advantage is in tying CD4001 Pin 1 to CD4060 Pin 12 (reset); maybe I don't understand how the 4060 would reset. I can see that if Pin 1 were set to high the CD4001 would reset.

    As a side note, I plan to break the connection that you have from the switch to the CD4001 Pin 14 and supply Pin 14 with it's own power directly, because once power reaches CD4001's Pin 3 (which will be permanent until reset) I will force the switch to open (permanently until reset).

    Thanks again for the schematic, it gave me something to study and learn.
     
  13. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    The momentary high reset on the 4060 clears the internal counter so all the outputs are set low, as power is applied. After the reset returns low, the binary count sequence starts. With 4011 pin-1 and 4060 pin-12 tied to the same RC network (C2/R4), the 4011 and 4060 both get a momentary high.

    Should be OK.

    Ken
     
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