Simple schmitt Trigger

Discussion in 'The Projects Forum' started by jj_alukkas, Aug 30, 2009.

  1. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
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    I needed a simple single pulse one-shot ckt without a 555, so I designed this. But I need to get it properly biased? Also the slope of the charging cap tends to be a problem, how to easily correct it?

    [​IMG]
     
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  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Hmm, where you can see "schmitt Trigger"
    I can see none.
    And for me the BJT will be "death" in no time.
     
  3. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    I meant a schmitt trigger like.... Just to create a pulse, thats all... And the NPN listed has a VCE of 45V, so hasn't died after so much testing I did so far, however, I have no idea of its biasing, thats why I'm asking for help.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    The defining characteristic of any schmitt trigger is it's hysteresis, which I don't think you can get from a single transistor. The hysteresis is what causes the squaring.

    I don't see the input to this circuit, but I assume it is the base.

    To square this signal you're going to need 2 transistors, or an IC. You can always use a classic transistor monostable, or make a simple 2 transistor op amp. What do you have against the 555 for this application?

    The believe 2.7V zener is useless in this configuration, as the transistor BE drop rules.
     
  5. Ron H

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    Apr 14, 2005
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    Do you want the LED to come on when power is applied, then go off after a period of time?
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The properly design " single pulse for LED"
     
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  7. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    At times I need a simple ckt to give a single pulse, say a beep for power on as an example, nothing precise or complex. I didnt feel like throwing in a 555 to just give a pulse. Squaring is also not necessary. However, I would like to learn if that is possible too like by using a zenner.

    The 2k7 has a role here as I first built it without the zenner, but instead of the pulse, the output goes down from full to off like exponentially. To make a 'cut-off', I added that zenner and now it gives a good flash, though not a perfect one, which would be enough. However, I was not good at biasing it properly.

    Yes, nothing precise, just it should cut of after a pulse.

    And Jony130, that was what i was looking for.. Thanks.. amd I liked your CAD software, looks intersting.

    The only downside I find with this circuit is that the circuit wont operate again unless the cap discharges. will a 1M across the cap solve that?
     
  8. Ron H

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    Apr 14, 2005
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    The circuit with the Sziklai pair has ≈unity voltage gain, so the output current will fade exponentially.
     
  9. Ron H

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    When power is turned off, the cap discharges through the diode and the 1k resistor in parallel with the circuit.
     
  10. rspuzio

    Active Member

    Jan 19, 2009
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    Since you're using the transistor as a switch as opposed to an
    amplifier, biasing isn't really the issue here. Just think of the
    transistor as a current-controlled device here (let's not get
    sidetracked on chickens and eggs yet again, please :( ) : In
    order for a given current to flow in through the collector, you'll
    need at least that current divided by β going through the base.

    As it stands, the B-E junction would pretty much act as a short
    so the capacitor would charge up with a quick inrush of large
    current the moments you'd turn on power to the circuit. You
    need to connect a resistor series with the capacitor in order
    to limit the current and provide the timing function. Also, as
    Bill said, lose the Zener because it's not doing anything. Here's
    the improved schematic:


    [​IMG]

    Now the resistor R2 limits the current flow --- the exact values
    of C and R2 will depend on how long you want to light to stay
    on. This circuit will function as Ron described. The only thing is
    that, to reset it, you will need to discharge the capacitor. If
    you are using this to turn on a light for a brief time after flipping
    on the power switch, one solution would be to put a largish
    bleeder resistor across C. As long as the bleeder resistor is
    large enough so that the current trickling through it isn't enough
    to keep the LED on once the capacitor stops charging, this
    will do the trick.
    [​IMG]
     
  11. Ron H

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    Add a couple of parts and it will recover rapidly.
     
  12. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    I tested it and it solved the capacitor discharge problem, but exponential discharge still exist. I guess that needs a zenner somewhere to cut of the slope's half. Would that work? I tried it with a zenner instead of diode and works better.

    I have tried this circuit first without the zenner. The resistor was only helpful in increasing the delay time for precise adjustment. The exponential output issue still sustained which was limited to an extend using the zenner. anyway I'll try it again with different values. I know exponential discharge cant be eliminated, but I need to cut it half, say like remove the slope after a particular low voltage.

    The bleeder idea required some trial and error to match values again and so I guess Ron's idea of a diode and resistor is more of an ideal approach.

    With LED, this circuit works good right from the beginning but if I connect a buzzer, I can clearly identify the exponential state noting the decrease in pitch. Everything else is now perfect, only need that voltage cut after a certain discharge.

    Now I find that this circuit is ready for use with LED's and other types. I need a slight adjustment for the buzzer.
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    The exponential discharge is a direct result of declaring squaring is not necessary. It is one or the other, it is why the other circuits exist. If you don't want a curve then you will have to go with two transistors, or an IC. One transistor can not do this.
     
  14. Bernard

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    Aug 7, 2008
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    You can flatten the lumen expo. drop by converting ckt. to a constant current one, & the zener reappears, anode to ground this time. Drop the 1k and add 56 ohm between emitter & gnd.[ math sais 100 ohm, but 56 is a little brighter] Z=2.7V. It will still trail off at end as Z drops out.
     
    Last edited: Sep 1, 2009
  15. Ron H

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    Can you post a schematic? I can't envision what you mean.
     
  16. Bernard

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    Aug 7, 2008
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    Without some positive feedback I see no way to square -up the output, but by converting to constant current can keep a constant output for a whole second before trailing off to infinity[ leaky cap]. Added 10k to shorten trail off.
     
  17. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    This circuit looks exactly as I posted first except, i didnt use that 10K and 68R resistors.

    Thanks for that.
     
  18. Papabravo

    Expert

    Feb 24, 2006
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    And you had the zener pointed the wrong way. You seem to be talking a great deal without listening, or is that just my imagination.
     
  19. jj_alukkas

    Thread Starter Well-Known Member

    Jan 8, 2009
    751
    5
    You simply seem to talk without experimenting, or is that my imagination.

    If I have posted something here, I must have tried all the ways possible before asking for help. I tried the zenner both ways. The way I have drawn seemed to show a better drop at the exponential even with a low valued capacitor rather than the other way which just goes a long exponential and finally gives the drop. Both give results suited for different situations, so I guess that answers your question. I just stated that my circuit was having that zenner though others said it wasn't necessary. Having little or no experience, doesn't make me dumb.
     
  20. Papabravo

    Expert

    Feb 24, 2006
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    That statement is manifestly untrue since the zener was in fact the other way. Experimenting without understanding the underlying principles is a complete and utter waste of time. If that is all you got, then what the heck, build your circuits in all possible combinations and see what falls out. Just don't come moaning to us when things don't work out the way you think they should.

    BTW - I don't have to do much experimenting any more because my stuff works the first time, every time with few exceptions.
     
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