Best T-Shirt Ever (Need help with basic electronics)

Discussion in 'The Projects Forum' started by Kwarl, Jun 4, 2010.

  1. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    I am trying to design a t shirt riddled with LEDs, such as each LED will periodically light up.

    Doing this with 10 groups of LEDs was simple enough : a 555 timer network provided a clock signal to a CD4017 decade counter, which dispatched signals to the 10 groups.

    I am trying to do so with 20 groups of LEDs, but no counter longer than a decade counter exists. How would you do it?

    I don't have the slightest idea what wonders electronic parts can do, so any tip, suggestion or idea is welcome :)

    I have attached a truth table that would be one possible solution : if a part such as the one described existed, attaching one of these to each of a decade counter's output would effectively make a 20-counter.
     
  2. beenthere

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  3. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    What's the little arrow symbol that's shaped like an s on wookie/marsden's schematics? it's between every diode.

    EDIT : Also, can I skip the 330 ohm resistor if I'm not cascading? if not, how is its value calculated for 1 or 2 4017-2804 pairs?
     
  4. beenthere

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    That denotes light emission, making the diode symbol into one for for a LED.

    No, that is the current limiting resistor for the LED's. It is common to all, as only one LED is lit at a time.
     
  5. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    okay thank you for explaining that to me :) I found online a function determining R4's value, so this part is sorted out.

    Another question : can the CD4017 drive high brightness (<75 mA) LEDs? So far the information I've found seems to indicate it can't, so I'm curious as to what the alternatives might be.
     
  6. beenthere

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    If you look at the first schematic, you will notice that the 4017 drives the LED's through transistors (Q1, Q2, etc.) Something like 2N2222's are good for more than 75 ma.

    The second circuit uses ULN2804 Darlingtons to handle the current. They have the resistors built in.
     
  7. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    So ULN2804s could drive a <75 ma?
     
  8. beenthere

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  9. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    Thank you for your help there. In fact, checking the datasheets is what got me trouble in the first place - CD4017's sheet recommends currents under 10 mA so I'd become certain that they couldn't do what I wanted them to do.

    One last thing - I assume CR1 and CR2 are current rectifiers? If so, which ones do you recommend?

    EDIT : Also, I fail to realize how a single transistor in series with D1 can replace the whole bunch of transistors in the schematics where there are no ULN2804. However, since it seems like it'd be quite a long explanation, I'm willing to take that on the account of "it's magic" if it doesn't otherwise affects the way the circuit works.

    Unless the arrangement of diodes & inverters replace the need for a transistor? But then I'd fail to see how.
     
    Last edited: Jun 4, 2010
  10. beenthere

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    CR1 & 2 are signal diodes. Something like a 1N4148 would be a common choice for that. They form a diode AND gate, so the clock signal only passes when both diodes have HIGH levels on their cathodes.

    I don't follow your statement. I believe both schematics are accurate.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    In the schematic I posted that uses the ULN2804A Darlington drivers, there were simply not enough channels available; one more was needed.

    Instead of adding another ULN2804A and only using one channel, I simply used a single discrete transistor with a base current limiting resistor.

    It would be better if that transistor were also a Darlington, like a 2N6426/6427, MPSA13/MPSA14, or similar. Otherwise, it would be necessary to compensate for the difference in Vce between the ULN2804A output and a single discrete BJT.

    Note that you need to look at the datasheet for the ULN2804 to determine the output saturation voltage for your load current.

    Here is a Vce(sat) vs Ic plot from Allegro's datasheet:

    [​IMG]

    When calculating your current limiting resistor, you will need to subtract the Vce(sat) from your supply voltage.
     
  12. Externet

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  13. Kwarl

    Thread Starter New Member

    Jun 4, 2010
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    @Externet : Lovely !

    @Everyone : thanks for your help, it's really handy ! :)
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    No.
    The max allowed input current is 10mA, not its output current.

    Ordinary Cmos logic ICs limit the output current automatically. The amount of output current is shown on Texas Instruments datasheets. Into a 2V red LED and no current-limiting resistor the output current is typically 3.5mA with a 5V supply and 18.5mA with a 10V supply. The current is a little less with 3.5V blue or white LEDs.

    Each output transistor is allowed to dissipate 100mW so with 2V LEDs and a 10V supply then a current-limiting resistor is needed.
     
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