monostable timer

Discussion in 'General Electronics Chat' started by goldendragon, Apr 10, 2013.

  1. goldendragon

    Thread Starter New Member

    Apr 10, 2013
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    Hi,

    Need some help. I'm out of ideas already I still can't figure out what I did wrong.
    My problem is that the LED is always in the ON state. I triggered the timer only once, sending a "low" signal to pin 2. after sometime t=(1.1x470ux11k) the LED should turn off but it continues to turn On. When I simulate it in the MultiSim it works fine.
     
  2. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
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    Welcome to AAC.

    Try removing R3. Also, are you releasing the switch or keeping it switched on (0V to pin 2)? Just checking. :D

    If that doesn't work, what is the voltage between GND and pin 2 when the LED is on beyond the expected time?
     
  3. goldendragon

    Thread Starter New Member

    Apr 10, 2013
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    I tried removing R3 but still it continues to turn on. Also I tried removing the transistor and connect the LED and 220Ω directly to pin3 and it emits a small light. Do you think my timer is busted?I'll try changing it and see what happens.

    Also I can't check the voltage between pin 2 and GND, cause I don't have any meter with me. All I have is the simulation in Multisim, If it works.I try it on the actual board.
     
  4. elec_mech

    Senior Member

    Nov 12, 2008
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    Leave pin 5 unconnected for now or add a 0.01uF capacitor between pin 5 and GND.

    What are you using as a switch? A toggle, rocker, momentary push button, etc.?

    If the switch is left connected (pin 2 left connected to GND) beyond the timing cycle, the output will remain high after the timing cycle is complete.

    If you're using a momentary switch, you press it quickly and let go. If you're using a latching switch, e.g., toggle, rocker, etc., you flip it ON (so pin 2 is connected to GND) then quickly flip it off.
     
  5. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
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    Try this switch input method, also un-connect pin 5 as this alters the trigger threshold
     
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  6. goldendragon

    Thread Starter New Member

    Apr 10, 2013
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    I'm using a momentary push button (NO), what would 0.01uF in pin 5 do to the timer?

    I'll try that switching method and see what happens.

    thanks to the both of you... :)
     
  7. elec_mech

    Senior Member

    Nov 12, 2008
    1,513
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    Dodgydave recommended adding what is called an edge-trigger. It allows a single, brief pulse to pass to the trigger input when the switch goes from high to low. If the switch stayed in the low position beyond the timing cycle, it would not keep the 555 turned on. More on its operation here.

    Note there is the addition of a capacitor AND another resistor. There is a pull-up resistor connected to both ends of the cap.

    Since you're using a momentary switch and assuming you're releasing the switch before the timing cycle is finished, it sounds like there may be another problem. Can you take a picture of the physical circuit you've put together and post it?
     
  8. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    Although I don't understand some of your resistor value choices, and I don't know why you are tying pin 5 to 12V, the circuit does function as you drew it in post 1. When the switch is closed very briefly, pin 3 goes high, which turns the 2N3904 on and lights the LED. After about 10 seconds, pin 3 goes low and the LED is extinguished.

    I believe you have a wiring error, component(s) failure, or both. If you are using one of those 1/4" pushbuttons, you probably have it wired wrong; they are easy to get the pins confused.
     
  9. goldendragon

    Thread Starter New Member

    Apr 10, 2013
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    it says here http://home.cogeco.ca/~rpaisley4/LM555.html#5 (no. 5) that if I connect 1.8k to pin5 and 12v I could double my time.

    In the image 6KΩ and a trimmer 10KΩ is in series.

    What I wanted to achieve is that I could turn on the LED and vary the time from 5s to 15s. After achieving it, I'll substitute 220Ω and LED with a 12v relay so that I could drive a 12V load.

    Btw, I'm making making my own power supply from 220V 60hz source, then a step down transformer to 15V 1amp, then to a full bridge rectifier (4pcs 1N4001), 1000μF for smoothing and 7812 and a 0.1μF for regulation. Will these affect the performance of the timer? and also is there another way to drive a 12V load without using a relay?

    thank you.. :)
     
  10. elec_mech

    Senior Member

    Nov 12, 2008
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    Okay, there are some discrepencies in your physical circuit:
    • The LED is reversed. The lead closest to the flat side should be connected to the transistor.
    • There is no 220Ω between the transistor base and pin 3. This could allow the transistor to get too much current and burn out. For now, I'd suggest removing the transistor altogether and just connect the LED to pin 3 directly and through the 220Ω to GND.
    • Can't quite tell how R4 is connected. You appear to have an 8kΩ resistor in series with a pot. If so, add a jumper between the pot's center terminal and the third unconnected terminal (the one not connected to the resistor).
    Everything else appears okay. I'd also suggest adding a 0.1uF cermanic and a 1uF electrolytic cap as close to the power pins of the IC as possible - this will cut down on any noise from the IC to anything else in your circuit.

    I can't really speak to the design of the power supply, but it sounds okay.

    What is the load and how much current does it require?
     
  11. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    As elec_mech has already pointed out, the polarity on the LED is reversed. In addition, a 220Ω resistor in the LED power circuit will pass about 50mA at 12V, which is too much for most common LEDs, and will burn them out or shorten their life substantially. To limit the current to 20mA, you need more like 620 ohms; a 1k will likely provide all the brightness you need.

    The base resistor to Q1 should probably be 1k.

    There's a wire from ground to column 21 that doesn't seem to do anything.

    And if your PB switch happened to be normally closed, that would cause a problem. It' probably normally open, but might be worth checking.

    I can't see the power connections to the breadboard. I assume that you know that the ground bus on one side of the breadboard is not connected to the ground bus on the other side of the breadboard unless you have a wire between the two busses. Likewise, with the positive power rails.

    The LM7812 should have a .33uF cap on the input to ground and a .1uf on the output to ground. This is in addition to a large filter cap on the power supply. A 10uF on the LM7812 output is also recommended. The 555 puts a lot of noise on the power rails and this can cause erratic operation.

    You can use a MOSFET to switch 12VDC instead of the relay. It depends upon the load's voltage, current, and inductance.
     
    Last edited: Apr 15, 2013
  12. goldendragon

    Thread Starter New Member

    Apr 10, 2013
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    ok, I'll try putting LED to pin3 in series with a 220Ω to GND. What do you mean by adding a jumper between pot's center terminal and the unconnected terminal? Will I just short the two terminal?

    Where should I put 0.1uF cermanic and a 1uF electrolytic cap? in pin 1 and 8?Sorry I'm not good with electronics term.

    It is just a 12v motor from a toy car. I don't know how much current will it require and how to get it. what should I do to get it's current?

    I've checked the PB(N.O.) and it's working fine.Yes, It has a wire between the two busses.

    I'll try adding 0.33uF cap on the input of the LM7812. I've already got .1uF on the output side of the regulator. Also I'll research about MOSFET. Do you know any site where I could learn it's basic?

    http://home.cogeco.ca/~rpaisley4/LM555.html#32 I saw that he uses two transistor 3904 and 3906 to drive a 24v relay. Will it do if I change the 24v relay with a 12v load, assuming that I change 24v to 12v already?

    Thank you.
     
    Last edited: Apr 16, 2013
  13. elec_mech

    Senior Member

    Nov 12, 2008
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    As tracecom pointed out, you probably want to use a larger resistor value for the LED. Looks like you're using a standard red diffused LED. They have a forward voltage of ~2V. 12V - 2V = 10V, 10V/0.02A = 500Ω minimum. I'd suggest 1kΩ resistor for now. That will give you at least 10mA which is plenty.

    Yes.

    Yup. At least as reasonably close to pin 1 and 8 as possible.


    The best way would be to use a current meter (found on most multimeters). While it is straight-forward to do, it is easy to short something if you're not familiar using test equipment and I think you mentioned not having a meter, so let's try another route. Is there a part number on motor? See if you can look up the make and model and find the specs.

    Here is a pretty good resource for MOSFETs. They include a motor example. Pay attention to adding a diode to counteract back EMF. A MOSFET should work just fine for a motor, the key will be making sure you select a big enough MOSFET in terms of current.
     
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