LED help !!

Discussion in 'General Electronics Chat' started by sluggs, Dec 7, 2007.

  1. sluggs

    Thread Starter New Member

    Dec 7, 2007
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    Hi, Can you help
    Im using a 24v dc supply connected with a resistance 1.5 kohm to light a L.E.D, but i find once the power is removed there is a delay to turn off, can this be reduced.
    I need it to flash on /off when required, but the flash seems so slow.
    Quite new to this, so please keep it simple.
    many thanks
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
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    Well the problem appears to be the output filter capacitors have no low impedance path to discharge when the AC power is removed. You can spped the discharge by placing a bleeder resistor across the the output. The resistance and the power handling capability will depend on the size of the capacitors. The LED and the 1.5K resistor might draw 14 mA so you need someting that will draw 50-100 mA or 240 - 470 Ohms. Such a resistor will dissipate 2.4 Watts so a 5 watt resistor should do the trick.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Papabravo's idea of using a bleeder resistor will work just fine.

    However, keep in mind that by increasing the load on your power supply, you will reduce it's ability to power a "real" load, and you will also likely increase the presense of "noise" on the output voltage.

    As Papabravo mentioned, if you're using a standard red LED (15mA @ 2.4V), the current through the LED and resistor will be approximately 14.4 mA. If the item you intend to power with this supply draws 144 mA, the LED will extinguish in 1/10 of the time it does now.
     
  4. sluggs

    Thread Starter New Member

    Dec 7, 2007
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    thanks for the replies, there is no capacitor in this circuit, if one was put across the LED would this drain the LED quicker

    24v dc + --------- 1.5 kohm -----{ LED }--- 24v -
     
  5. Papabravo

    Expert

    Feb 24, 2006
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    The capacitor of which I spoke is NOT in your circuit -- it is in the POWER SUPPLY. You may not think it is there, but trust me on this one -- it is there.
     
  6. sluggs

    Thread Starter New Member

    Dec 7, 2007
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    I see thank you
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    If all you want is to turn the LED on and off more quickly, why not place a switch between the LED and the power, rather than turning the power supply on and off?

    But if your goal is to show when there is power available at the output of the power supply, leave things as they are now.

    Keep in mind that a large capacitor charged up to 24 volts can discharge a LOT of energy if the output of the power supply is shorted. You really don't want to be fiddling around with the wires when there is still power stored in the supplys' output capacitor.
     
  8. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
    I think you could put a +20v zener in series with the resistor, recalculate for a +5 volt supply and be happy. You should not see the LED light until you have at least +20+led forward voltage and it should die quickly at power off.

    I've not tried it though I did use this with an SCR to make a 'protect my computer from a nefarious wallwart' crowbar circuit. Worked well and smoked a couple of &#%$ wall warts. (CNN is running a story about the C64....)
     
  9. 3BwEH

    Active Member

    Feb 25, 2007
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    0
    but if u used a push button in series with those remember to put a capacitor in parallel with it to reduce the delay in pushing the button and releasing it
    or try to use a regulator to reduce the voltage source u r using and place a capacitor in parallel with that as well
    i mean use a 5V with a 300 ohm ann u should get the same current as before
     
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