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Thread: Insert led in my case (analog modulation for normal led)

  1. #11
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    hi dave
    would you be willing to share that buffer circuit

  2. #12
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    Quote Originally Posted by badger1666 View Post
    hi dave
    would you be willing to share that buffer circuit
    The attached will turn full on at roughly 3.2V for green and 2.4 V for red and protect the leds and the amplifier/DACs is is attached to.

    See circuit BufforedTechniColor Attached

    Increasing Rg slightly will lower the turnon thresholds at the expense of some linearity in the red side. You could try 150K or 200K For Rg, some place around 470K you may see latchup favoring one color.

    It uses zener diodes as clamps for each direction to avoid damaging the leds if left open circuit.

    Steve
    Attached Thumbnails Attached Thumbnails BufforedTechnicolor.bmp  

    Last edited by mixedgas; 10-25-2009 at 16:55.
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  3. #13
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    Silly thought, but if you take the input directly into the non inverting input of an opamp, Connect one end of the bi colour LED directly to the opamp output and the other end of the LED to a resistor to ground, and then connect the junction of the led and the resistor to the inverting input, then you will have an LED current that is equal to Vin/R with no threshold.

    Make R about 470 ohms and 5V will give you a nads over 10mA into the led, and it will be safe with 10V up it as LEDs are normally specified to 30mA.

    Just a thought, but why would the above current source not work?

    Regards, Dan.

  4. #14
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    Dans idea is attached, with a buffer resistor added. either way works, On my way you need to play with Rgain and the leds may be protected a bit better. I suffer the lack of early turn on in my way, settling for more protection of the leds.

    Either way is a winner.

    See attached.
    Attached Thumbnails Attached Thumbnails DansIsource..bmp  

    Last edited by mixedgas; 10-25-2009 at 17:58.
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  5. #15
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    You can protect the LED on mine too (and it only needs an extra resistor that does not effect the brightness), split the resistor and take the feedback from the split.

    Then the current in normal mode is set by Vin/(resistor to ground) and the fault current is set by (Vs - Vled)/total resistance.

    Assume say 2V for the led and +-15V rails, and that we want say 10mA at 5V input, then if both resistors are 470 ohms, the maximum fault current will be ~13mA.

    This obviously needs an opamp that can swing 12V while sourcing 10mA with 15v supplies.
    100K series on the input will make next to no difference to something like a '084 (and would in fact protect the opamp if the thing saturates due to excessive input as it will limit the current due to differential mode input voltage limits being exceeded.

    The one win with my approach is that current is directly proportional to applied voltage (no dead zone).

    Regards, Dan.

  6. #16
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    Dan's improved I source with LED protect,

    See attached
    Attached Thumbnails Attached Thumbnails DansImproved.bmp  

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  7. #17
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    Your version of my concept doesn't work (The upper resistor is too high for a 12V rail, better make it more like 220R (Which is also OK at 15V rail with most LEDs)).

    A slight tendency to over design is not actually a bad thing!

    I suffer from intermittent bouts of it (You should see the first draft of my KTP oven controller, more parts then the space shuttle, most of them belt, braces and a bit of string).

    Now the trick is to modify that current source so the input bridges across the differential input on the ILDA connector.....

    The downside of the differential version is that the common mode impedance is lower then you would really like, but see drawing.

    Regards, Dan.
    Attached Thumbnails Attached Thumbnails dif current  source.png  


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