Astable 555 circuit LED staying half lit.

Discussion in 'The Projects Forum' started by campeck, Sep 6, 2009.

  1. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3
    I have built this circuit here.

    [​IMG]

    And on the output I have this.

    LED1
    [​IMG]
    LED2

    .68 hz with 10uF cap and R1=10k and R2=100k

    LED 2 turns on and off fine. LED 1 stays dimly lit when supposed to be off.
    This is my first half successful circuit.
    The solution and the reason for it would REALLY help me in understanding more about what's going on.

    small video showing the problem
    http://www.youtube.com/watch?v=ErCcbK8K3Lw


    Thanks Guys!

    PS. any other cool things you can think to show me with the 555 please do!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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  3. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3
    ah
    supply voltage 5v
    LED resistors 470ohm
    LED's = 2.1v 25ma

    (these are the lowest resistors I have besides 10ohm.)
     
  4. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    LED1 never is never completely off because the output pin of the 555 never swings up to the 5V Vcc when high. This leaves a potential difference across the LED. Try inserting a diode or two in series with LED1.
     
  5. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3
    Hey Cdrive.

    That was it! pin 3 shows 3v max on my multimeter. I new that current was going through (why it was lit) but wasn't sure WHY.
    my limited knowledge being a crutch...I did think of diodes as a solution but didn't know how to implement. Then when you mentioned it I remembered that diodes drop .7v across them

    I put one 1N4148 diode in front of the LED expecting It to not work and need another. But it worked. And it works with 2 as well.

    with Just the LED (KEY: Voltage across = LOW - HIGH)
    LEDv=1.79 - 1.96

    One diode
    LEDv = 1.63 - 1.94
    diodev = .37 - .69

    2 Diodes
    LEDv = 1.44 - 1.91
    diode1v = .23 - .67
    diode2v = .23 - .67

    So using this
    [​IMG]

    So without extra diodes it was in the glowing region when LOW
    with one diode it was below the glowing range LOW
    with 2 The same is true.

    Is all my thinking correct?

    But why do the diodes go from dropping .7 to .3?

    Thanks!
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    If you read those articles I pointed out, it will explain in detail. Start with the 555 functional schematic.

    [​IMG]

    The 555 can go to about 1.2 volts of the positive voltage. For 5V this is around 3.2V. The negitive voltage it goes to about .1V, which is good. Being about to switch to the power supply voltages is refered to as rail to rail, which the 555 only does on one side.

    Actually you can do this quite easily, if the LEDs are the 2.1 Vf types, in other words, modern red LEDs. You just have to be aware of the why the 555 does what it does. Even the older red LEDs work OK at 1.5 Vf, you have to compensate for the drop across the Darlington transistor output.

    Vf is the forward dropping voltage of LEDs, the newer ones are different than older versions, as the intensity per ma has gone way up due to manufacturing differences. With older LEDs is it around 1.5V, with newer it is around 2.1 to 2.5V.

    As I explain in LEDs, 555s, Flashers, and Light Chasers this is the functional diagram of a 555 output.

    [​IMG]

    Boosting your supply voltage gives you a bit more to work with, but mostly you have to adjust the resistors according to the supply voltage and what kind of LEDs you are dealing with.
     
  7. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
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    Yeah I am reading through your material right now. at 6 am.
    I am like a sponge!
    Thank you so much for the help.
     
  8. CDRIVE

    Senior Member

    Jul 1, 2008
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    It's difficult, if not impossible, to add anything to what Bill posted. He's very thorough with 555 theory and operation. I also must confess that I've not kept up with the advancements of LEDs. I knew that they produced a hell of a lot more light than they did in the 60's but the details provided by Bill was quite enlightening!;)
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    A diodes' Vf (forward voltage) is dependent upon the current flowing through them.

    Here is a plot that I generated from actual measurements of a 1N4148 diode:
    [​IMG]

    Note that over 20mA, the diode's Vf rapidly increases. This is typical when a diode is reaching it's maximum rated forward current.

    Here is another plot that I generated from measurements of a 1N4002 rectifier diode:
    [​IMG]

    Note that the 1N4002 is rated for a maximum average 1A current, and this plot only goes up to 100mA.
     
  10. CDRIVE

    Senior Member

    Jul 1, 2008
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    Bill, I just took a look at what you did in 'Circuit#1' in your post. Clever! ;)

    I've said this before but I still think we need a Thumbs Up Smilie.
     
  11. CDRIVE

    Senior Member

    Jul 1, 2008
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    Sgt. Wookie answered this very well. Keep in mind that semiconductor junctions have real (integral) resistance in series with the junction. So when threshold is reached, this small resistance will continue to increase the voltage drop across the diode and become more pronounced. Heating will also affect the junction.
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    I'm always running into trouble with that. Most times the effect is so subtle that you barely notice it, and can ignore it as trivial. It's the times it's not that bite ya.

    You want LEDs to impress, check these out.

    http://www.philipslumileds.com/products/line.cfm?lineId=19

    http://www.futureelectronics.com/en...ters/colour/Pages/7191828-LXM2-PL01-0000.aspx

    I think any LED that takes 350 - 700ma (and produces corrispondingly more light) is impressive. Most of what we know about LEDs still apply, everything is just scaled up, yet the LEDs themselves are physically small.
     
    Last edited: Sep 6, 2009
  13. Bob S

    New Member

    Jun 15, 2009
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    1
    Last edited: Sep 6, 2009
  14. Audioguru

    New Member

    Dec 20, 2007
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    the voltages on your simulation are completely wrong:
    1) The output high of a 555 that has a 10V supply is typically +8.8V when the current is 15mA. Not +10V.
    2) The forward voltage of an LED that has a current of 15mA is much more than only 0.6V.
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    If you look at post #6, you'll see why. Funny thing, both voltages were off. Their SPICE model is pretty off, the 555 always get 1.2-1.4 volts less than Vcc on the plus side, and is extremely close to ground (around 0.1V) on the negitive side.

    CMOS versions don't have this problem, but they source a lot less current.
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Looks like Bob_S used Linear Technology's LTSpice with the default colors changed a bit.

    LT's model for the 555 timer is "idealized". I do wish Linear would change the library default to better reflect a "real-world" 555 bjt timer, but one always has the option to add a new model to the library, once one learns how to do so. The idealized 555 timer was created to make the simulation run faster rather than accurately model a true 555.

    The "LED" in Bob_S's simulation is actually just the default diode that one gets when first placing a generic diode on a schematic. In the schematic editor, one can right-click on the diode and select from a list of various library models. The default library supplied with LTSpice is not exhaustive, but certainly usable.
     
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