555/556 Problem

Discussion in 'The Projects Forum' started by CitrPug, Jun 1, 2011.

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  1. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    I'm trying to make a circuit that contains a 556 dual timer in monostable mode.

    Everything is connected properly and yet I have one timer of the 556 working perfectly, but the other doesn't.

    The output is remaining HIGH. I've tested the timing capacitor and it is reaching 2/3Vcc and then discharging but the output does not go low at this point.

    Any ideas on what could be causing this? I haven't posted a schematic as it is just the standard monostable circuit. I'm using a 100k resistor and a 47uF capacitor in the timing network but I don't think leakage current is a factor here as the capacitor charges and discharges as it should.
     
  2. beenthere

    Retired Moderator

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    You might need to post up the schematic so we can see what might be wrong.
     
  3. SgtWookie

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    If the cap is charging and discharging by itself, then it sounds like you have an astable circuit rather than a monostable circuit.

    Did you accidentally short the reset and output pins together? That could cause an extremely brief low output that would be difficult to see on an analog O'scope or multimeter. On a related note, are you certain that you didn't accidentally connect the output pin to your Vcc supply? (this is almost the same as shorting RESET and OUTPUT together, but you've hard-wired Vcc to RESET instead of leaving it floating; in this case the OUTPUT pin would never go low). Unless your 555 timer is a CMOS version, the output pins won't go above around Vcc-1.2v under a light load due to the Darlington follower output configuration.
     
    Last edited: Jun 1, 2011
  4. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Schematic below...

    The 556 is a CMOS version. The capacitor chargers on connecting the circuit to the power supply and then discharges, it will only charge again on applying the trigger. The capacitor is acting normally, however during all of this OUTPUT is HIGH.

    - RESET and OUTPUT not shorted
    - OUTPUT not tied to Vcc

    Are you saying that RESET should not be tied to Vcc? I thought it was supposed to be when not being used (as I'm not) due to the possibility of transients resetting the timer if RESET is left floating.

    [​IMG][​IMG]
    [​IMG]
     
  5. SgtWookie

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    Yes, the RESET inputs should be connected to +V/Vcc/Vdd/+12v.
    Some people leave them floating, but they should be wired to +V.

    CMOS 555 timers have pretty good current sink capacity (~100mA) but very poor current source capacity (only around 10mA). The bjt (transistorized) version can source or sink 200mA.

    You may have fried the output in your timer if you have been operating a relay coil with it, particularly if you don't have a reverse EMF diode across the coil windings, cathode to +, anode to ground.

    You have not mentioned how much current the relay coil requires to energize the relay.
    You may need to use a transistor or MOSFET to sink current from the relay if it's anywhere near the timers' 10mA limitation.
     
    Last edited: Jun 1, 2011
  6. SgtWookie

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  7. k7elp60

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    Nov 4, 2008
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    I think part of the problem is that both the trigger inputs to the 556 are floating. A 100K to Vcc will hold them high until the 4069 output goes low.
     
  8. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Looking at the datasheet I can only find a value labelled 'Rated Coil Consumption' which at 12v is 33.3mA. Relays are fitted with back EMF diodes. Maybe a BJT would work better though the side that does work seems to work fine with the coil. The relays are also driven separately, never simultaneously.

    Similar project yes, I just need the timer to drive the relays for a set amount of time to power a motor in either direction. Not necessarily to be used in an automotive environment. I did not mention such an environment in the OP of this thread.
    Am I right in thinking you mean they are floating due to the capacitor? Once charged it won't allow current to flow? Is the capacitor necessary?
     
  9. CDRIVE

    Senior Member

    Jul 1, 2008
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    Sarge, somehow I know you'll appreciate me bringing this to your attention for editing.;)
     
  10. SgtWookie

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    Jul 17, 2007
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    Hey, I got rollin' on the capacity, capacitance, calumnious, Copernicus thang.... ;)

    (BTW, edited my prior post; "reverse EMF capacitor" => "reverse EMF diode"
     
  11. CitrPug

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    Feb 20, 2011
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    I hadn't noticed that! However, yes the coils are connected to reverse EMF diodes. 1N4004s.

    Are the triggers floating due to the capacitors? Do I need them there? Is it simply a case of sticking a pull-up resistor on the trigger? 10k or 100k?
     
  12. SgtWookie

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    You need pull-ups on the trigger inputs, and you need the capacitors there.
    Use 100k resistors.
    You should also use clamping diodes to prevent the trigger inputs from going more positive than Vcc when the signal on the other side of the cap transitions low-to-high. Connect the cathode to Vcc, anode to the trigger input. 1N914/1N4148 diodes work very well for this.
     
  13. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    I've put a couple of 100k resistors and 1N4148 diodes in as mentioned. Am I correct in saying they should be in parallel and not in series?

    In parallel I'm still getting one side working, the other not.
     
  14. SgtWookie

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    Each trigger input needs its' own 100k resistor, and its' own diode.

    There is no "parallel" to it.

    Update your schematic and re-post it so that we can see if you understand what we are suggesting that you do.
     
  15. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Ok, sorry. What I meant was the positioning of the diode and resistor in relation to one another and the TRIGGER pin.

    As I read your post, I took it to mean this...

    [​IMG]
     
  16. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    I, personally, would connect pin 2 (trigger) of the 555 to the output of the 4069 via a transistor. You would still need the pull-up resistor, but when the transistor conducts, it will short pin 2 to ground, triggering the timer. I don't know if you are hoping it will trigger when the output of the 4069 is high or low, but you can choose between NPN or PNP, depending on which output state you are using.
    I hope this makes sense, but let me know if you have any questions.
    Der Strom
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    DerStrom8,
    If the output from the 4069 is not a pulse, he needs the cap there - otherwise the output will stay toggled until the input returns low.

    Citrpug,
    Yes, you understood me correctly

    Try reducing the 100k resistors to 10k or 4.7k - see if that helps. I thought you were using CMOS 555 timers, but you might not be.
     
  18. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    I am using CMOS timers.

    You make an interesting comment however...

    The input is normally tied to ground, it pulses high and then returns to ground, would this mean I do not need the capacitor, however, I still need the pull up resistors and diodes?
     
  19. SgtWookie

    Expert

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    The input to the 555/556 timer, or the 4069?

    You're not leaving an input "floating", are you? (not connected to either Vdd or GND).
     
  20. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    No floating inputs as far as I know. The circuit is connected as per the schematic so unless one can be seen there (which I cannot) there are no floating inputs.

    I was referring to the input to the 4069. It is normally low (output high). It is driven high for a second or less (output low - triggering timer) and then returns low (output high once more)
     
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