Question about L.E.D's.

Discussion in 'General Electronics Chat' started by NM2008, Sep 5, 2011.

  1. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    Hi,

    Could someone tell me is it possible to run 1.5v led from a 20v power supply if current consumption was restricted, or kept under the max allowable current for the led?

    Regards NM
     
  2. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    Assuming the 20V power supply is DC, sure, just calculate the resistor value to limit your LED current. If the 1.5V you speak of is the voltage drop across the LED, a 1kΩ will limit the current to 20 mA; the resistor will need to be 1/2 watt or larger.
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    The answer is yes. You will need a 1K-ohm resistor in series to limit the current to 20mA. A 2K-ohm resistor would bring the current down to 10mA. You may start out with 5K-ohm and see if 4mA will give you sufficient brightness.
     
  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    A 5kΩ resistor will only need to be 1/8 watt or larger.
     
  5. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    Yes assuming DC.
    I ask this as a few weeks back I came across a demonstration of an ignition circuit that gave a 10mm arc, although when an led was placed across the arc gap directly onto each of the electrodes, it did not burn out, but lit the way it should in normal circumstances, with 20v across it.

    Why did this occur?
     
  6. MrChips

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    Oct 2, 2009
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    Hah! That's a different story. An ignition circuit will produce an energy pulse of very short duration. In this case, it is the total amount of energy that mattered. How do you know that it was a 20V pulse?
     
  7. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    Was it that the led was able to pull the supply right down?
     
  8. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    MrChips
    A multi-meter was placed across it.
     
  9. MrChips

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    Ok. We have a few problems here. To generate a 10mm spark you would have to create a voltage in the order of thousands of volts, maybe 5KV. The multi-meter is not capable of reading such high voltage, short duration pulses. Yes, the multi-meter would load down the signal, and so would the LED. Because the ignition coil is a high-impedance source, both the multi-meter and the LED would constitute shorts to ground.
     
  10. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    In other words at the time the coil was not capable of suppling power to any type of load no matter what size? An led would have appeared to be a very light load.
    I imagined that the led would have at least lit bright yellow and then burnt out.

    You said previously that this was to do with energy, therefore wouldnt that outcome mean the circuit was of low energy, or am I missing something here?

    Regards NM
     
  11. MrChips

    Moderator

    Oct 2, 2009
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    Energy is measured as power multiplied by time.

    1 Joule = 1 Watt x 1 second
    1 Joule = 1 Volt x 1 Amp x 1 second

    As an example, if we had a 12V pulse at 1A lasting for 1ms that would be equivalent to 0.012 Joules. If this were transformed into a high-voltage pulse in an ignition coil, the energy would still be the same if the coil was 100% efficient. So we can only guess that the reason the LED did not burn out is because the duration of the pulse was very short.
     
    Last edited: Sep 5, 2011
  12. Wendy

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    Mar 24, 2008
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  13. Kerim

    Member

    Mar 3, 2011
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    Your high voltage output becomes a current source when shorted by an LED. I guess this current is likely small since it is inversely proportional to the transformer coil turns.

    But what is still looking as a puzzle to me is that you measured 20V on the LED terminals. :confused:
    Added:
    Isn't it negative, relative to the forward voltage direction of the LED?

    Kerim
     
  14. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    MrChips
    The coil was switched by a mosfet at 40Hz.
    I didnt get the duration of the pulse which was suppling it.
    What I found interesting was that when an led was placed directly across the coils output, it read roughly 20v across the led terminals. (standard green 5mm led)

    Kerim
    Yes 20V (+/-2V) was measured across the led terminals.

    So if the led was capabable of pulling the HV supply down to this level, then wouldnt this mean that HV circuit at the time would have been harmless if someone came into contact with it? (except for the physical shock).

    Regardas NM
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    It isn't the voltage, it is the current that light LEDs. You can control the current of any voltage, it is a simple application of Ohm's Law and a resistor. Thing is, if a resistor drops a lot of voltage it will dissipate a lot of heat, it will get very hot. This is very wasteful of power.

    One solution is use a lot of LEDs, each LED drops some voltage. If there are enough LEDs the waste goes way down, and the resistor doesn't get really hot.

    One other thing, if you start using too much voltage it will violate our Terms of Service (ToS). At the moment I don't know how much voltage you are talking about. We regularly get requests to use house mains to light LEDs, these threads are closed quickly.

    In some ways LEDs are robust, in others delicate. If you put more than 5V backwards across an LED the odds are the LED is toast.

    You may need to build a diode bridge to convert the voltage to DC, add a large capacitor to filter it, then load it with 20ma to see what voltage you have to work with. A constant current source would do this without calculating resistance, but it will get hot (unless you have some LEDs to absorb the heat).

    [​IMG]
     
    Last edited: Sep 6, 2011
  16. Kerim

    Member

    Mar 3, 2011
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    Without seeing the voltage waveform on the oscilloscope, one has to guess a lot of possibilities. We can start to check what the meter really measures since the signal is not DC or sinusoidal. Then we need to estimate at which voltage and for how long the LED conducts in the reverse direction since it is an AC pulse coming from a transformer (not digital, between Vcc and ground for example).

    In fact, before I bought my first oscilloscope, I had no way but to 'imagine' the shape of the signals on my designed circuits. I may say; 80% of my guesses were wrong even for the circuits that worked fine. :)

    Wish you good guessings :D

    Kerim
     
  17. NM2008

    Thread Starter Senior Member

    Feb 9, 2008
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    Bill_Marsden

    This was completed a month ago, not by me. I only observed and I am going from memory of what I saw, just trying to get a better understanding, this is not a project of mine, purely out of interest only.

    The fact an led rated at say 1.5V 15mA would pull down a kV supply to 20V and appear to operate normally.
    Doesnt this show that the kV supply would be harmless if someone with a skin resistance of say 1000 ohm came into contact with it, wouldnt the supply pull right down similiar to that similiar or less of the led, producing very little current through the skin?

    Thanks for the input,
    Regards NM
     
    Last edited: Sep 6, 2011
  18. debjit625

    Well-Known Member

    Apr 17, 2010
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    You are talking about a system where current is not constant i.e.. not DC.
    It's your multimeter which measured 20V now you have to know this multimeters can't measure every kind of signal,even most can't give you a proper RMS AC (pure sine wave)reading only True RMS meter will give you a True RMS value,most meters are not True RMS type and just use a reference point to do that which is not accurate.
    And about your system where you are running your coil with 40Hz I think it is not even pure sine wave which makes the measurement more complex for a simple multimeter.

    Anyway you have to measure AC not a DC value with your meter.
    The dangerous thing in an electric system is power not voltage,for example a 1000v with 20 micro Amperes (power is 1000 * 0.00002 = 0.02 watts) would not give me a shock ,but a 100 volt with 50 Amperes (power is 100 * 50 = 5000 watts) would give me a shock.Anyway electric system may be unsafe, so just don't assume by the level of voltage only.

    Good Luck
     
  19. Kerim

    Member

    Mar 3, 2011
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    If I remember well, a current, as low as 4mA, may kill a person sometimes. So, in case of 220Vac and 100K skin resistance (about 2mA only), the electric chock can be sensed very well. I lived it for a fraction of a second a couple of times :)

    So, to know if the ignition coil is fatal or not, one may need to put an equivalent load on it (about 100K) then measures the resulting voltage volt hence the current passing in it.

    For instance, the power supplied (V*I) by a source takes the form of a parabola (in function of the load) that starts and ends at zero power (open P=V*0 and short P=0*I). The maximum power that the load can get is when its value is equal to the internal resistance of the supply (assuming no reactive elements).

    In the case of an LED, we may say that we are working near the short circuit condition at which the power delivered is close to zero. But a resistance of 100K to 200K, as of the skin, would likely get a much higher power than of the LED. I hope you got the point.

    Kerim
     
  20. Wendy

    Moderator

    Mar 24, 2008
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    I don't think I have mentioned on this thread (they tend to run together after a while, same questions) but LEDs are ESD sensitive. They will burn out if exposed to much more than 5V backwards. You can fight this two ways, one is the diode bridge I mentioned earlier, to rectify the voltage and make it DC, then the LED never sees a back voltage. The other way is to put two diodes back to back, so the other diode never sees more than the dropping voltage of the first diode (LED).

    What you are referring to on the high voltage supply is loading. This is not a very regulated way to do business. It may work, but it is not necessarily the best way do to it.
     
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