Is a pull-down resistor needed?

Discussion in 'The Projects Forum' started by szabikka, Jan 7, 2016.

  1. szabikka

    Thread Starter Member

    Sep 3, 2014
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    Hi everyone!

    I have a question regarding the circuit in the picture. I am working on an 555 astable circuit that can be turned on and off with the same push-to-make button. As you can see I have connected a bilateral switch to the reset pin of the 555. The bilateral's control is connected to the Q output of a D-flip-flop. I have the following questions since I have never used the 4066 before. I have added a pull-down resistor to the reset line of the 555 that keeps it low when the bilateral is turned off, but I'm not sure if it's needed at all. My other question is concerned about the control pin of the 4066, should I add a pull-down there or connecting it to the output of the flip-flop is enough? Please take into consideration that the switching must be noise-free in this case.

    Thank you for your answers in advance! 555 4066 reset.JPG
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    You need the pull-down resistor. R10 is typically 10k in most circuits but check the datasheet. 100k should work.

    If you are not going to use the control pin, I would hard wire it to the appropriate rail, not to the flip-flop output.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I predict that you will need to de-bounce the input switch. As drawn, you cannot predict how many times the 4013 will change state every time you push the input switch. This part must work properly before you worry about the 555.
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Based on your schematic there is no reason for the 4066 switch. When the 4013's Q output goes high, it closes the switch and pulls the 555 Reset input high. If this really is the desired logical effect, you can eliminate the 4066 and connect IC1a Q to IC2 pin 4, and eliminate R10.

    Follow Mike's advice and add a 100 nF to 330 nF capacitor across R2.

    ak
     
  5. szabikka

    Thread Starter Member

    Sep 3, 2014
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    Thanks for the help everyone! I will redesign the circuit based on your advice.

    Regards,
    szabikka
     
  6. szabikka

    Thread Starter Member

    Sep 3, 2014
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    Dear AnalogKid,

    I have omitted the 4066 as you adviced and ran a simulation of the circuit, but it doesn't work. I think that the output of the 4013 doesn't provide enough current to operate the base of the PNP transistor that is internally connected to the 555's reset pin.
     
  7. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    I think that's unlikely, if 100k (R10) provides enough pull-down in the circuit as shown?
     
  8. hp1729

    Well-Known Member

    Nov 23, 2015
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    Yes, debounce the switch. Maybe use CMOS version of the 555 (7555) and omit the 4066.
     
  9. szabikka

    Thread Starter Member

    Sep 3, 2014
    77
    1
    I usually use 100k resistors as pull-downs or pull ups, simply because I have hundreds of them at home, while I don't have too many 10ks. I have only built this circuit in a simulator so far, and you are right the problem is not with the current that flows into the reset pin. I don't know, maybe the problem is that I'm using a 12V battery as the power source, and that means that the transition between high and low logic is around 6 V, but the output of the 4013 is only 5V when it's high.

    About the debouncing. I have searched the internet for debouncer ICs, and they appear to be hard to come by in my country and the available ones are very expensive. AnalogKid adviced to use a capacitor in series with R2, but it pulls high the clock input of the 4013 for a brief moment when power is applied. Maybe I should connect the capacitor in parallel with R2, but I don't know if it debounces the switch that way.
     
  10. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Can you set the 4013 Vdd as 12V in the sim?
    As for debouncing, you could try a cap across R2 in your post #1 circuit (plus a very small resistor in series with the switch to limit the current through the switch contacts when the cap charges yet still allow a fast enough rising edge for the clock pulse). Or you could use the other half of the 4013 IC configured as a monostable to do the debouncing.
     
    Last edited: Jan 8, 2016
  11. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    Still, to pull down a node, you need just one. :)
     
    GopherT likes this.
  12. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Check the voltage on Q pin of cd4013 flip flop. My simulator does not allow me to change the output voltage and keeps it at 5v. Yours may do the same. If so, then it is not the current, it is that 12 volt is needed for R pin of 555 timer if supply is 12v

    It should work right on a breadboard.
     
  13. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    @szabikka

    I think what you are trying to build is much simpler than the way you are approaching it. This 555 circuit is all over the web, including here.

    It is intrinsically de-bounced...


    tog.gif
     
    Last edited: Jan 8, 2016
  14. dannyf

    Well-Known Member

    Sep 13, 2015
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    If you don't have a pull-down there, when the switch is open, the reset pin is floating. That ***may be*** a problem if the reset pin is of high input resistance -> a quick check on the datasheet would answer that.
     
  15. szabikka

    Thread Starter Member

    Sep 3, 2014
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    You were right I forgot to reset the default voltage of the simulator to 12 V. :) Thanks for reminding me! Now it works like it should.
     
  16. szabikka

    Thread Starter Member

    Sep 3, 2014
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    Thanks for the replies everyone. It works and I have debounced the switch with a low-pass filter. ;)
     
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