Input high, pulse at output#1. Input low, pulse at output#2

Discussion in 'The Projects Forum' started by hermhart, Sep 11, 2015.

  1. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    I want to build a circuit that has two outputs. When the input goes high to +12VDC, output#1 sends a 2 second pulse. When the input goes low to 0VDC, output#2 sends a 2 second pulse. The output loads are 500mA solenoids. I'd like to drive them with IRF634 mosfets. So what does the control circuitry look like. It needs to be cheap and not draw more than about 20mA. I know that 555s could create the pulses. The attached is something like what I have in mind.
    1. Is it OK for the control voltage to remain on pin2 of the 555 for extended periods?
    2. Is the use of an inverter chip a good way to approach this requirement?
    3. Is the use of a 555 indeed a good way to produce my 2 second pulse?
     
  2. Roderick Young

    Member

    Feb 22, 2015
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    That's a great circuit in concept. If the switch has any contact bounce, though, you'll end up triggering both solenoids any time the switch transitions either way. So you'll need to add some sort of RC filter to debounce the switch.

    One of the merits of your circuit is that it runs on 12 volts, which I assume is the native power supply. If you try to use a packaged inverter, you will most likely need to have another voltage regulator to supply 5 volts or 3.3 volts. A single NPN transistor could be used for the inverter, and would work at 12 volts.

    If you're going to have a voltage regulator, you might consider simply using a PICAXE 08M2, which costs about $3. It's an 8-pin DIP, no bigger than a 555, but has a microcontroller inside. You can connect one of the inputs to the switch via a voltage divider, and do all the switch debouncing in software. Also in software, you can generate a precision pulse of any length desired on either output.
     
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  3. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    Yes, I hadn't thought about the bounce. RC debounce might work OK. Anyway I want to keep it as simple and cheap as possible. I like the idea of using a transistor as the inverter. I definitely do not want to deal with voltage regulating, microcontroller and software. What's inside my dotted line box is all on the table. None of that has to be like that. Is there some whole different approach to the problem that would be simpler and cheaper? For example I could probably use a relay. I don't think I'd be troubled with bounce. It's simple, but not particularly cheap. I'd prefer all solid state.
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Will there ever be an occurrence where the switch is flipped before one of the 2-second timers has finished its cycle?

    1. Yes, and I want the current cycle to complete anyway (meaning both outputs will be on during the overlap)?
    2. Yes, and I want the current cycle to end immediately when the other timer starts?
    3. No, that won't happen.

    Also, is there any advantage to having only one timer that is directed to the two outputs? In this way, the two output time periods always would be exactly the same, and both could be changed with one adjustment.

    ak
     
  5. eetech00

    Active Member

    Jun 8, 2013
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    Could use a 556 dual timer and save a chip..
     
  6. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    3. No, that won't happen.

    I didn't realize that it would be practical to get two outputs from the same timer. That certainly would be advantageous... one less component for both cost and real estate.
     
  7. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    eetech00... I might eventually thought of the 556... I knew about that but wasn't thinking about it yet. Great idea.
     
  8. AnalogKid

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    Aug 1, 2013
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    Uh...nope. Having a single point of adjustability comes at a cost. The single-timer steering logic adds more parts than eliminating the second timer removes. Life is choice.

    However, since the switch state always is longer than the output pulse, the two timers do not have to be true monostables. They can be boxcar circuits, an R-C timer with no feedback, sometimes called a retriggerable monostable or edge delay. The whole thing, including debouncing the switch, can be done with one hex inverter. Sch to follow.

    ak
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    While the original 555 was a functional little puppy, it did have some personality quirks. The CMOS version is much better, but I still gravitate toward Schmitt triggers for non-critical timing applications. The net difference between this circuit and your original is that yours needs an inverter between the two triggers that is separate from the 555's, and mine has protection diodes after each R-C. After you add input debouncing, everything else is the same body count.

    Note that the parts are what is in my design inventory. You don't need 1% parts, and your FETs and relays are fine. Also, unused inverters are in the signal path because they have to be tied to something.

    ak
    DualPulser-1-c.gif
     
    Last edited: Sep 11, 2015
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  10. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    I like that, AK. That's exciting. I'll give it a try. I like having the Schmitt trigger output on the FET's gate so that the latter does not need pullups or pulldowns. Thanks.
     
  11. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Since the MOSFET gates are direct-driven by 555 or 40106 totem pole outputs, no resistors are needed in either case.

    Playing around with things, it is possible to have one timer element with only 1 chip if you throw in a few steering diodes. There are two edge detectors, OR-ed into one monostable, gated to two outputs.

    ak
    DualPulser-2-c.gif
     
    Last edited: Sep 11, 2015
  12. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    AnalogKid,
    Referring to your schematic of Sept 11, I assume that R2C2 and R3C3 determine the width of the pulse. Is that right? And can I use an electrolytic for C2 and C3? If so, I suppose positive would point to the Resistor. But it seems to me that the polarity would not always be the same across C or is it?
     
  13. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    I'm not seeing what the advantage would be of schematic #11 over #9.
     
  14. AnalogKid

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    Aug 1, 2013
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    1. Yes. those are the two timing component pairs for the two output pulses.
    2. Yes, electrolytics will work. The main timing action is an inverter pulling the left side of each timing capacitor low. Because of this, that should be the negative side of the cap. The positive side goes to the pull up resistor.
    3. The schematic in post #11 has only one timing circuit, and it is used by both outputs. Three advantages: a) no mismatch between the two output widths; b) only one circuit or component to change if you want longer or shorter pulses; 3) design exercise for me, keeping it within one IC.

    ak
     
  15. recklessrog

    Member

    May 23, 2013
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    A 4027 is a great dual JK flip flop and the are many easy ways to use it as a latch/unlatch driver. I designed a power supply that uses one to control output on/off with two push buttons but it is a simple matter to have just one that will give sequential operation.
     
  16. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    Replying to #14... your advantage b) makes sense to me. It's a cleaner design to have to only change one timing circuit. As to a) the pulse width is not critical, and the two of them don't even have to be close to the same. As to c) I'm impressed with the hex package; in both schematics, we just need exactly 6 inverters. That's great. But Schematic #11 has 5 x 1N914, compared to 2 of them in #9. I like fewer components. Actually I don't understand the purpose of the RC,diode between pin 13 to pin 3. How does that work?

    Replying to #15 ... the 4027 looks like a cool device, but I don't think it's quite ideal for my current application. A flip flop isn't quite it. I had considered that actually. What I want is this ... when the input goes high, output 1 sends a pulse. When the input goes low, output 2 sends a pulse. Schematic #9 is pretty much what I want.
     
  17. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    What would you think of using something like a VND5160 for my output. These have the load connected to ground rather than to the supply voltage. That might be a good thing to avoid the possibility of accidentally grounding the load.
     
  18. recklessrog

    Member

    May 23, 2013
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    use the 4027 to trigger two monostables?
     
    Last edited: Sep 24, 2015
  19. hermhart

    Thread Starter New Member

    Feb 16, 2011
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    But then I'd need at least one more chip. I'm trying to keep real estate and overall cost to a bare minimum.
     
  20. hermhart

    Thread Starter New Member

    Feb 16, 2011
    20
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    OK. referring to schematic at #9, I have built the circuit except for the transistors. I wanted to test the switching first. At pin 8 I get what's expected... a 1 second positive pulse. At switching event, pin 9 goes from 0 to about 8 volts so that as the voltage at 9 increases the threshold is reached and pin 8 switches to 0. BUT, I can't get the other branch to work. There the voltage at pin 5 goes from 0 to about 1 volts and no higher. I tried changing capacitors. I also tried changing the IC. I have not yet tried actually swapping the capacitors. Is there something obvious I'm missing?
     
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