555 monostable multivibrator leading edge trailing edge

Discussion in 'General Electronics Chat' started by john74, Feb 1, 2016.

  1. john74

    Thread Starter New Member

    Aug 22, 2015
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    Im studying timer circuits...and it seems to be a bit confusing. Studying the 555 monostable multivibrator in particular the input and output pulses...it states that , with a negative going input pulse, the leading edge of the negative input pulse starts the leading edge of the output pulse which is positive...ok...fine but is this the same as saying - the output pulse starts when the input (trigger) pulse starts?

    Or is it a matter of terminology...or how its worded...it seems like it should be simple but yet its confusing.Because it seems like I read other places where its not clear, sort of contradictory.

    Also what if the input trigger pulse was positive...how would that relate to the output pulse? Would the leading edge of the input still start the leading edge of the output. I appreciate any thoughts, thank you so much
     
  2. hp1729

    Well-Known Member

    Nov 23, 2015
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    Yes, that could be worded better. Your understanding is correct. The trigger input is a low edge clock (1/3 VCC). The output goes high, Discharge* is released and the capacitor starts to charge. When the charge gets up to 2/3 VCC the output goes low. In astable operation Discharge* goes low and the capacitor starts discharging. When the charge on the capacitor gets down to 1/3 VCC the trigger point is reached again, the output goes high, Discharge* is released and a charge cycle starts again.

    Inside the 555 is an S-R latch with voltage comparators on the inputs. The Set side is the Trigger input and a voltage divider set to 1/3 VCC, The Reset side is the Threshold input and a voltage divider set at 2/3 VCC. This 2/3 VCC point is available at the Control input. Discharge* is an NPN transistor coming from the output of the latch.
     
  3. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi John,
    Look at page #4 of this 555 App for info regarding the Trigger.
    E
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,014
    3,234
    The negative going (falling) edge of the trigger input starts the positive going (rising) edge of the output pulse.
    Note that the trigger pulse can't stay low longer then the output pulse period or the output will stay high as long as the input is held low.
     
  5. dl324

    Distinguished Member

    Mar 30, 2015
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    The timer requires a voltage lower than 1/3Vcc to trigger. If you apply a positive pulse, triggering would occur on the falling edge of that pulse.
     
  6. hp1729

    Well-Known Member

    Nov 23, 2015
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    Good ap note. Thanks.
     
  7. john74

    Thread Starter New Member

    Aug 22, 2015
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    Very helpful thankyou
     
  8. john74

    Thread Starter New Member

    Aug 22, 2015
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    Thank you all for your responses, very helpful...thankyou
     
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