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Why won't my TI CD4098B multivibrator give me a pulse

Discussion in 'General Electronics Discussion' started by cliquot22, Mar 17, 2016.

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  1. cliquot22

    cliquot22

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    Mar 17, 2016
    I'm trying to create a ~0.25 second pulse from a leading edge trigger. I found the TI CD4098B chip (datasheet: http://www.ti.com/lit/ds/symlink/cd4098b.pdf) that shows a circuit in the datasheet that will do just what I need (leading edge non-retriggerable pulse). However, I am not able to get the chip to work.

    When I trigger the +TR input with 5V (VDD) or with ground (VSS), the output Q-bar goes to 5V. It seems as long as the trigger input is not floating, the output is 5V. But more importantly, there is no output pulse. I can't get the output to give me the 0.25s pulse I need.

    With the input floating, the output is around 0.6V.

    Here is my circuit which, as far as I can tell, is exactly what it shows in the datasheet.
    Pin 1: NC
    Pin 2: NC
    Pin 3: Vss (gnd)
    Pin 4: Vss (gnd)
    Pin 5: Vdd (5V)
    Pin 6: NC
    Pin 7: NC
    Pin 8: GND
    Pin 9: to pin 11, and output Q-bar
    Pin 10: NC
    Pin 11: to pin 9
    Pin 12: input
    Pin 13: Vdd (5V)
    Pin 14: cap to pin 15 and resistor to Vdd
    Pin 15: cap to pin 14
    Pin 16: 5V

    R=47k, C=10uF -> time ~0.25s
    DSC_0355.JPG
     
  2. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    Draw the circuit.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,513
    2,651
    Nov 17, 2011
    Welcome to electronicspoint.

    Don't let the input float, a CMOS circuit will nottake this well. Add a pull-down resistor to ensure defined low when not triggered, connect the trigger by a pushbutton to Vcc (5V)

    And show us a schematic drawing of your circuit diagram. Reading a circuit from a breadboard is tedious at best.
     
  4. dorke

    dorke

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    665
    Jun 20, 2015
    Apart from not leaving inputs floating,
    which you shouldn't do,as was said above by @Harald Kapp .

    Why do you connect /Q to -Tr (pins 9 and 11) ?
    This will cause a "locking state",
    that is why your circuit doesn't work.:(

    i.e once /Q will be "low" you have a "state lock".
    From that point onward the +Tr has no effect.
    Only a Reset will clear that lock...til the next one occurs.

    Remove that connection and tie the -Tr(pin 11) to VDD,
    that should fix that problem.:)

    BTW,
    Choosing that CD4098B is an odd thing,
    unless you had it in stock;)
     
    Last edited: Mar 17, 2016
  5. cliquot22

    cliquot22

    4
    0
    Mar 17, 2016
    Attached is the schematic (before and after suggested modifications). I'll try this out later today.

    The datasheet said to connect 9-11 to prevent retriggering on another input pulse. But I guess I have to send a reset to get even the first trigger. Since this is for a light sensor, I can just connect the reset to the light level.

    -> Dorke: I chose the CD4098B because that was the results of my internet and Digikey searches. Is there something better for creating a pulse from a rising edge trigger?
     

    Attached Files:

  6. Alec_t

    Alec_t

    2,955
    803
    Jul 7, 2015
    In your modified circuit the photo-sensor will have to produce about 250uA in order to bring the chip out of the reset state. Can it manage that much current?
     
  7. cliquot22

    cliquot22

    4
    0
    Mar 17, 2016
    That might be a challenge. It looks like I would need about 1000lux (http://media.digikey.com/pdf/Data Sheets/Everlight PDFs/ALS-PT243-3C,L177.pdf) which is fairly bright.

    So I should put the output of the phototransistor into the base of a 2N2222 transistor to bump up the current?
     

    Attached Files:

  8. Alec_t

    Alec_t

    2,955
    803
    Jul 7, 2015
    That looks promising. Alternatively, just increase the value of the emitter resistor from 10k to, say, 100k.
     
  9. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Yes,
    555 timer, CD4047, 74HC123, Schmidt trigger gates, etc.

    In your "per-modified original circuit" the mono-pulse is created by a manual switch without any relation to the photo-tr.
    What is it your are trying to do?
    Where do the "outputs" connect-to/activate?

    Is this pulse for generating a "reset signal" for a uC/uP?
    There are simpler solutions for that.
     
    Last edited by a moderator: Mar 22, 2016
  10. cliquot22

    cliquot22

    4
    0
    Mar 17, 2016
    Ok, I think I have it working now. Thanks for all your help.

    I have replaced the switch with a connection to the light sensor. You can see the final layout.
    This circuit creates a reset signal for my ESP8266 board when a light turns on and it tells the board that the light is on. I'm sure there are many easier ways to do this but this seems to work for me. I'll check out the other options when I have time.

    Clipboard01.jpg
     
    Last edited by a moderator: Mar 22, 2016
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