NEED HELP! -NE555 issue

Discussion in 'General Electronics Chat' started by Andrew216, Jun 24, 2015.

  1. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    So I have a 3v coin battery going into a voltage doubler that uses a 555. The output is right around 5-6v. I have another circuit made using a 555 timer as well that is an astable multivibrator. Im using this to blink 4 leds with an 80% duty cycle and at 80hz. They both work independent of each other. When I remove the grounds and power from the mulitvibrator circuit on my breadboad and connect the power from the output of the doubler to that circuit along with the grounds i get NOTHING. I need the voltage to be at least 4v for correct illumination of the led. 3v is to dim hence the doubler. And I HAVE to use a coin battery. Any help would be appreciated. Im a Noobie. I have attached my doubler circuit and the astable multivibrator.

    Doubler Components

    • C1 100nF ceramic capacitor
    • C2 1nF ceramic capacitor
    • C3/4 4.7uF electrolytic capacitors
    • D1/2 1N4004 diodes
    • R1 39Kohm resistor
    • 555 timer IC
     
  2. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Better to post a complete schematic so we can see all of the circuit interactions rather than just guess.

    A Cockroft-Walton voltage doubler has a relatively high output impedance; its output voltage sags under load. Calculate the peak load current for the flasher circuit, calculate an equivalent resistance for the output voltage of the doubler, use the resistor to load the doubler, and see what the output sags to. My guess is that it is below the operating voltage for a 555.

    ak
     
  3. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    I am going to draw up my exact schematic now.
     
  4. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    Sorry work turned into a mad house. Did not get a chance to draw my circuit on multisim. Will do that tomorrow. In the mean time, I may have enough room to run 2 coin batteries in parallel. Will this help in anyway?
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Probably nope. If the problem is that you are drawing too much current for one battery, then maybe. Without a schematic, who can tell?

    ak
     
  6. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    NewProject1.png


    Output of the 555 on the right is connected to a 330ohm resistor and a red led that goes to ground.
     
  7. AnalogKid

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    Are the 555's CMOS or bipolar?

    ak
     
    Andrew216 likes this.
  8. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    NE555 so im ASSUMING they are bi polar
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Disconnect the load from the 2nd 555, connect the 2nd 555 to the CW output, and see if it still loads down the first one too much. If so, then the static current through the 2nd 555 is too much for the CW, and gives you an estimate of what is too much. If not, then it's the 2nd 555 plus its load that is too much.

    A bipolar 555 output stage is good for something like 200-300 mA, so you have more than enough drive. Double/triple the two CW capacitors to see if that gets you enough charge.

    ak
     
  10. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    Awesome. I will test this out tomorrow morning and reply with my results. Thank you for helping me I greatly appreciate it!
     
  11. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    Okay, I am still getting loaded down. as soon as the 2nd 555 in any way shape or form is connected to the 1st 555 it drags the voltage back down to the input voltage which is only allowing 1v to reach my diode. I tried increasing the caps with a plethora of values and played with other values like the resistors and diodes. Basically, I need to run the astable multivibrator for about 2 years or so off of a single or 2 coin batteries. Any other suggestions on how I could go about this? Maybe I have been going about this wrong from the start.
     
  12. k7elp60

    Senior Member

    Nov 4, 2008
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    If I were to try is application I would use a cmos 555 and then reduce the value of 330 ohm resistor in series with the red LED, and I would connect the LED and series resistor from Vcc to pin 3. This would save some current as the LED iin your circuit is being driven when the waveform is the greatest time. Changing the drive from Vcc to pin 3 drives the LED when the waveform is at the minimum time.
     
  13. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    There are many many astable circuits that draw way less current than a 555, let alone two of them. A quick change would be to shift to the CMOS LMC555 with 30 times less static current, but there are other circuit options.

    If the 2nd 555 is running at approx. 80 Hz, why? That's too fast to see the LEDs blink, so why is it running at this frequency?
    The 2nd 555 timing resistors are very low, consuming unnecessary current. You can increase both of them by 1000 times, and decrease the capacitor to increase efficiency.
    Same for the 1st 555. Increase the R by 10x and decrease the cap. Also, there is an error in the schematic.
    D1 and D2 are too slow for a CW generator, and have a high Vf. If you can, change these to 1N5817 (or 18 or 19) Shottkey diodes and you'll get more energy out of the CW.

    What is the application for this circuit? Does it have to be visible outdoors? The LMC555 is tested at 50 mA output sink current. Is this enough for your LEDs? What LED current do you need?

    Your overall approach is fine, but there are other parts to do what you want with less work and better performance. For example, the Maxim MAX1682 is a much more efficient, self-contained voltage doubler, but it can deliver only 30 mA.

    Better data = better answers.

    ak
     
  14. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    I want the LEDs to appear on but was thinking that them blinking at a rate that cant be detected would also prolong the life of the battery since it is not always on. (Im sorry Im a noobie) Basically, when a car door is open I want to illuminate the LEDs. Battery life must be about 2 years and can run it off 1 or 2 3v coin batteries. Or, I can make a monostable circuit and have the light shut off after 30 seconds. Either way, I need a battery to last 2 years while illuminating LEDs for the entire time the car door is open or shut off after 30 seconds. I have a few meetings to go to and then I will be able to give more information on current needs and all the other technical stuff I keep leaving out. I apologize this is just my first Design job and I have struggling a bit.
     
  15. ian field

    Distinguished Member

    Oct 27, 2012
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    The bipolar 555 has the minimum guaranteed Vcc of 4.5V, although most samples will work maybe as low as 3V - you haven't a huge margin on a lithium cell that's anything but fresh.
     
  16. ian field

    Distinguished Member

    Oct 27, 2012
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    I'd go for a flyback design and use it in a way that the 555 boosts its own Vcc.

    Feed the battery via an inductor, then a diode to the 555 Vcc pin so the diode would conduct to the Vcc pin if nothing else was going on. Connect a logic level MOSFET; gate to pin 3, source to GND and drain to the junction between inductor and diode.

    When I did this on a 4.8V supply, the Vcc pin shot up to about 30V - I was dead lucky not to blow the 555 since the max Vcc is only 18V.

    If Vcc gets too high, you can convert it into a burst regulator - first you need to connect the reset pin to Vcc via a 4k7 resistor instead of just a wire link, use a NPN transistor to shunt the reset pin to GND when its base is biased on. Next; decide what maximum Vcc you want to regulate at and select an appropriate zener. put about 100R current limiting resistor in series with the zener and put it from base to Vcc - when the voltage gets high enough for the zener to conduct, it switches on the transistor which pulls down reset and stops the 555.

    The inductor isn't too critical - the one I used was liberated from a scrap PC motherboard. Not one with few turns of thick wire - if there's too many to count, that's probably enough. Anywhere between 47 - 220uH is probably about the right ball-park if you have the means to measure it. Whatever inductance you end up with - you can tweak the frequency for maximum output.
     
  17. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    I think the inefficiency of a 2nd oscillator outweighs the power savings of a pulsed load. If the only requirement is that the LED's light up, then an inductive boost circuit will be the most efficient. Thanks to the relatively large Vf of white LED's there are a zillion examples on the innergoogle.

    Transistor switch > inductor > LED and series resistor. No output rectifier, no regulator. Like this:
    http://www.ecircuitslab.com/2012/01/15-v-white-led.html

    Blocking oscillator:
    http://www.homemade-circuits.com/2012/01/how-to-make-simplest-15v-bluewhite-led.html

    Oscillator driving a CW voltage doubler with an active rectifier (3906) and the LED as the output rectifier. The constant current load can be replaced by a resistor once the LED is known.
    http://www.discovercircuits.com/DJ-Circuits/3VwhiteLeddriver.htm

    And finally, the master speaks:
    http://www.cappels.org/dproj/ledpage/leddrv.htm

    ak
     
  18. Andrew216

    Thread Starter New Member

    Jun 24, 2015
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    SO....my circuit now works I have attached a schematic.
     
  19. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Good. What is the right side IC?

    ak
     
  20. Andrew216

    Thread Starter New Member

    Jun 24, 2015
    29
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    Here is a cleaned up version. The other IC is the max1682 to convert my 3v coin battery to 6v and Im using the tlc555 for the monostable multivibrator. Now my next obstacle to is to maybe install a Hall effect so when the door opens it will activate the timer as well as interrupting the timer if the door closes. Any other ideas?
     
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