8 buttons - push 1, lock out other 7

Discussion in 'The Projects Forum' started by RainbowVideo, Apr 10, 2012.

  1. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
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    I've been away from this stuff for too long, I'm sure the solution is simple. Show me the way and I can build it. Thanks.

    -8 momentary pushbuttons C/NC/NO
    -12 VCD power supply
    -1 LED light controlled by each button
    -When any button is pressed, need to lock out the other seven until released
    -interested in old school (electro mechanical) and semi-conductor solutions.
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,118
    3,042
    That always makes me chuckle. Why do people say that? Many things are really, really hard or even impossible. Even the over-unity gang thinks they're just one tweak away from success. Maybe humans are basically optimistic, to a fault. :)

    Anyway, this sounds like a game buzzer kind of thing, to indicate who hit the button first? There have been other threads on this topic here.
     
  3. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
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    actually it's for a museum display (push the button, highlight the feature) the lockout is to foil obnoxious kids. It's 'old school' but it's what the client wants. Thanks for the game suggestion. I'll forage around.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    Below is a circuit that should do what you want. It uses 2-input OR gates to act as a latch when a button is pushed. This also turns on one of the CD4066 switches which connects the Inhibit line to ground and blocks all other push buttons until the original button is released. Shown is the circuit for two push buttons. 8 pushbuttons requires two CD4071 and two CD4066. The inhibit line is connected to all OR gates and CD4066 switches.

    The CD4071 can only output about 2mA to drive the LEDs so if you want more than that you will need to add an NPN emitter-follower to drive each LED and also reduce the resistor value in series with the LED.

    Remember that any unused CMOS inputs must be grounded (although with 8 circuits, you won't have any).

    PB Lock.GIF
     
    Last edited: Apr 11, 2012
  5. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
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    Thank You Crutschow! I'll order the parts today and put together a prototype. I'm told the indicator lights (LED spotlights) will draw approximately 1 amp, thus I will need to add the NPN emitter-follower that you suggested. I have to admit that I can build things like this and I (somewhat) understand what is happening, but to specifiy components or figure out the circuits is way above my pay grade and capabilities. I'm reluctant to ask, but would you sketch out how the NPN emitter-follower goes into the circuit and specify the component? Over sizing things, even at an additional cost, if possible, is worthwhile as I my goal is to make this as reliable as possible for them. Again, deepest thanks for your assistance.
     
  6. svdsinner

    Member

    May 17, 2011
    39
    2
    That is a lot of current, and you might want to use the NPN follower to power a relay (don't forget protection diodes) and let the relay switch the spot light.

    You might be able to get away with just a power transistor (assuming you spec it carefully), but it may end up costing more for a big transistor that can handle that than a small transistor plus a relay. Also, a transistor of that capacity might generate enough heat to be a concern.
     
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  7. crutschow

    Expert

    Mar 14, 2008
    13,006
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    1A will require good-sized driver. What is the voltage across the LEDs at 1A? LEDs normally have a few volts drop unless they have a built in circuit to limit the current to the desired value.

    Edit: Hold off on ordering the parts until I come up with the LED driver circuit.
     
    Last edited: Apr 11, 2012
  8. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
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    I'm the front end of the is project (control). My contribution ends at a labeled terminal strip with 16 lugs. I'll email the display person and see if I can get the exact lighting instrument he will be using and the specs. Thanks again for the help. I'll get back to you as soon as I hear from him.
     
  9. crutschow

    Expert

    Mar 14, 2008
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    Here is the schematic with some MOSFET drivers for the LEDs. (They are configured as common-source, not emitter-followers as I previously suggested).

    I selected these transistors but any N-MOSFET with at least a 2A rating and a maximum ON resistor of 0.5 ohms or less should work.

    Note: The switches are shown in the normally closed position (button not pressed).

    Edit: Just noticed that the MOSFET I specified is not readily available. Here's a higher rated device which gives you more margin and it's still less than $1 apiece.

    PB Lock.GIF
     
    Last edited: Apr 11, 2012
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  10. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
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    Once again, thank you. I'm anxious to get the parts and get one built. I'll post after I've built and tested the prototype. Thanks!
     
  11. crutschow

    Expert

    Mar 14, 2008
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    Did you see my edit in my previous post about the different transistor?
     
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  12. t06afre

    AAC Fanatic!

    May 11, 2009
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    If you struggle I think this can be done quite simple with only relays (8 total). But I have not gone to deep into this problem. So I may be wrong ;). Just keep on with your current work. And if you have problems with your current design. I may take a second look at your problem.
     
  13. crutschow

    Expert

    Mar 14, 2008
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    You are correct. I originally thought it would take a many pole relay but, after some thought, it can be done with one diode and one DPDT relay per switch. See attached. It shows the circuit for three switches. It just needs to be expanded to the eight required.

    The capacitor is to provide current to the relay coil and prevent it from chattering during the time the relay contacts are moving and there is momentarily no direct power connection from the supply to the coil.

    This alternate will likely cost more than the solid-state design since relays tend to be relatively expensive and also less reliable.

    Relay Inhibit Ckt.gif

    Edit: If you connect a diode across each relay coil (cathode to positive) that would help sustain the coil current during the relay transition time and you probably wouldn't need the capacitor.
     
    Last edited: Apr 12, 2012
  14. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    The following CD4017 circuit will meet most of your requirement, except that the "focus" of the output can still be changed while the first user has not yet released the push button.


    To lock out the selection mechanism until the last user has released the push button, one need to use more involved IC circuits or to use a type of push button that has two pairs of contacts and make the modification to the original circuit as shown in the second drawing.

    Note: An improved design in next post removes the requirement of the push button with two pair of contacts


    [​IMG]

    [​IMG]
     
    Last edited: Apr 13, 2012
  15. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    The attached design do away with the requirement of push button with two pair of contacts.

    Note the value of the pull up resistor has been increased to 470K, polarity of the diodes 1N4148 reversed and 33K resistor connected in parallel with each diode.

    Using 50Hz or 60Hz as clock source is acceptable but each of the LEDs would indicates a brief flash of 20mS as the counter "sequent" to the desired output.

    If this is not acceptable visually, one can use a simple 555 timer circuit to generate clock pulses of 1KHz to 10KHz. One such clock generator can serve practically unlimited number of push button circuits.

    If the lamp takes 1A, 4017 output cannot drive them directly. You would need to add the following driver circuit to each output of the 4017 in order to drive them properly.

    [​IMG]


    [​IMG]
     
    Last edited: Apr 13, 2012
  16. crutschow

    Expert

    Mar 14, 2008
    13,006
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    Attached is the last solid-state circuit I posted but using logic symbols rather than the IC packages to allow easier understanding of the circuit operation.

    The OR gates provide a latch action when its respective PB is pressed which also turns on its respective LED.

    The transmission gates pull the inhibit line low when any PB is pressed. This generates a logic low on all the other PBs, preventing them from turning on its LED.

    PB Lock Logic Ckt.gif
     
  17. RainbowVideo

    Thread Starter New Member

    Apr 10, 2012
    16
    0
    Thanks to all for your continued interest in this. Let me pose a question. From a reliability standpoint, would you choose the solid state route or the electromechanical route. My goal is to walk away from this installation and never need to visit it again. The indicator lights on the buttons are LED and the lights being illuminated are LED so fixture maintenance should be an issue, just the control circuit. (And for what it's worth, although I can build the solid state circuit, the relays make more sense to me.

    Also, on the original design, how many watt resistor would you recommend for the .5 ohm resistors.

    Again. Many thanks.
     
  18. crutschow

    Expert

    Mar 14, 2008
    13,006
    3,232
    From a reliability standpoint the solid-state design should be better (and cheaper). The MOSFET I show at the bottom of my post #9 has a maximum current of 11A and maximum voltage of 60V so there's plenty of margin there. About the only thing that would be likely to cause a failure is a miswire when connecting it up.

    Typical relays have a mechanical life of over a million operations (check data sheet for exact value) so you can estimate how long it will last in this application. You should use a relay rated for at least 5A.

    In the original design I show 5k (5000 ohm) resistors since that was all the current the circuit could provide. But for your circuit you will need to determine if it requires a resistor or if it's built into the lamp. If a resistor is required you will need to determine its value for the desired current using Ohm's law R = V/I where V is (12V minus the drop across the LED) and I is the desired LED current. The resistor power dissipated would be I^2*R.
     
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