Wheel of fortune circuit

Discussion in 'The Projects Forum' started by crash563, Sep 16, 2015.

  1. crash563

    Thread Starter Member

    Feb 25, 2013
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    Hi I am of need of a circuit operation for the following circuit. I beleive after building it the not gates act like a nonstabke vibrator but I'm not sure how excalty this circuit works. The counter part I do understand.
     
  2. pwdixon

    Member

    Oct 11, 2012
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    Top left is an oscillator, bottom left is just a buffer that when the button is pressed drives the led's to ground and turns them on. The oscillator obviously clocks the counter, the button press also stops the oscillator pausing the counter.
     
  3. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    Yes , but how does the configuration in the left produce the oscillations
     
  4. pwdixon

    Member

    Oct 11, 2012
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    R1 and r2 charge c2 forming the timing circuit for the input to the osc amp.
     
  5. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    That oscillator isn't what you think it is. IC1D and IC1E form a non-inverting buffer with hysteresis, which can act as a latch. It is IC1A with feedback through D1 and R2 that turns it into an oscillator. The original designer is playing some games with the resistor value ratios. For example, R1 can override the hysteresis feedback coming in through R2, and R4 might figure in there depending on the output state at pin 2.

    ak
     
    Last edited: Sep 16, 2015
  7. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    If possible can you eblaborate on that because I have never seen a schmitt trigger configuration like this before.
     
  8. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    What a neat circuit; with insufficient + bias at pin 1 , nothing moves. Charge up C3 & ckt. oscillates at about 50 Hz. As C3 discharges it takes + bias longer to recover & at about 8 sec. later C3 down to 2.5 V , osc. stops. Vcc = 6V. IC = 4009.
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    As the overall circuit slows down, R3 always discharges C3 through D1 at the same rate. If you tied the three inverters (pins 13, 3, and 5) to pin 10 instead of the switch, then the LEDs would blink off briefly when changing from one LED to the next, and as the circuit slows down the off time would be constant and the on times would get longer and longer until the circuit stops on its final value. VERY nice.

    ak
     
    Last edited: Sep 18, 2015
  10. crash563

    Thread Starter Member

    Feb 25, 2013
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    Thx guys those are excellent explanations
     
  11. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Here's what it looks like:
     
  12. crash563

    Thread Starter Member

    Feb 25, 2013
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    Yes but this circuit is not easy to breadboard as the all the leds do not light up
     
  13. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    They all light up, but one at a time; that's the intent of the circuit.

    ak
     
  14. ISB123

    Well-Known Member

    May 21, 2014
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    I think that he is saying that its hard to breadboard because of many connections he has to do which results in some LEDs not lighting up because of improper connection.
     
  15. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    yes
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    That's part of the "fun" of electronics, and it never goes away completely. About 2 weeks ago my hands thought they knew more than my brain. I love the smell of fried transistors in the morning.

    ak
     
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