Driver Circuit for 5,000 LEDs

Discussion in 'General Electronics Chat' started by Benzodiazepine, Aug 18, 2014.

  1. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
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    Curious about an driver circuit for 5,000 LEDs. Each LED to be supplied with 5mA of current, and the supply voltage at 12VDC.

    So, 5,000 x 5mA equals 25 AMP.

    What would be your solution?
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
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    hi,
    What is the forward working voltage of the LED type or colour.?
    E
     
  3. ScottWang

    Moderator

    Aug 23, 2012
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    5mA current for a 2V/10mA/3mm LED?
     
  4. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
    17
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    Say yellow which is like ~2.2 volt drop right?

    But keep in mind that LEDs are current controlled devices. Not voltage.
     
  5. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
    17
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    Ok make it 10mA then. This then makes it 50 AMP total.

    I don't think modern LEDs need to be driven quite thus hard though.
     
  6. ScottWang

    Moderator

    Aug 23, 2012
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    The spec of 10mA, considering the life of led and brightness , normally I will use 80% of 10mA, that's equal to 8mA, so the leds need 40 Amp total.

    But considering the Vcc for leds is 12V, it can be supply 4 leds and one limiting resistor, so 5000/4 = 1250, I_total = 8mA * 1250 = 10A.
     
  7. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
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    O.K but lets' talk about an circuit to limit the current to 40 Amps.

    What do you suggest?
     
  8. ScottWang

    Moderator

    Aug 23, 2012
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    What's the suggestions:
    Using NE555 and mosfet to do the pwm control.
     
  9. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
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    I have had this thought too. And as an bonus a pot could be used to vary the brightness.
     
  10. ScottWang

    Moderator

    Aug 23, 2012
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    Please review the current of LEDs that it's shows on post #6.
    10A is more easier to 40A, and choosing the Id of mosfet is 4 or 5 times of real values(10A), that is 40A or 50A, and choosing the Vds at least 2.5 times of 12V, that is 30V or bigger, and Rds less then 10Ω.

    Here are some types of mosfet, choosing the good c/p values for what you need.
    FDP5800_Nch_80V60A_4.6mΩ_Vgs10V.
    FDP8440_Nch_40V80A_2.4mΩ_Vgs4.5V.
    IPP042N03L_Nch_30V70A-79W4.2mΩ.
    IRF3205_Nch_55V110A_8mΩ.
    IRFP2907_Nch_75V209A470W_4.5mΩ.
     
  11. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
    17
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    I'm more of an BJT man personally. And I would decompose / bust the 5K LEDs into sections (many PCBs)
     
  12. ericgibbs

    Senior Member

    Jan 29, 2010
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    I appreciate LED's are current devices, for 2.2Vfwd, you could at a maximum have 5 LED's in a series string, leaving a 1V overhead for the series resistor.
    At 10mA per string would require a 100R series resistor for each string.

    This would mean 1000 series strings, so 1000 * 10mA = 10Amps.

    E
     
  13. Benzodiazepine

    Thread Starter New Member

    Aug 17, 2014
    17
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    Yeah would work. The PWM is going to be more effecient though.
     
  14. takao21203

    Distinguished Member

    Apr 28, 2012
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    I would not use resistors for so many LEDs, if you drive them with 2.2 volts, that's 1/5 of 12v, so you only need 10 Amps.

    Could be done with 6 of these chinese dc/dc modules.

    You can drive them directly, just the gradient rises very much over just a few 100 millivolts. You have to adjust with an Amp Meter, and incorporate some 20% reserve for heating LEDs.

    Have done that with all kinds of LEDs, works with LEDs from same batch, just nor different colors/different manufacturer.
     
  15. ericgibbs

    Senior Member

    Jan 29, 2010
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    It will still work with a PWM drive, if you need to vary the intensity of the LED's.
     
  16. wayneh

    Expert

    Sep 9, 2010
    12,085
    3,024
    You need to get over that. A MOSFET switch driven by PWM is ideal for this application (if you don't just use a pre-built DC-DC driver). Using a BJT would be much more difficult because you need to supply the base current.
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    PWM will allow variations in brightness, but you will still need current limiting resistors.

    Personally I don't like 1V overhead, LEDs are fickle on their voltage drop, and a statistically improbable mix could leave one chain very dim.

    I have come up with a library over the years for both PWM and linear LED drivers. For example...

    [​IMG] [​IMG] [​IMG] [​IMG]

    I like using the Darlingtons because they are one device and won't load the voltage reverence.
     
  18. ScottWang

    Moderator

    Aug 23, 2012
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    My calculation had a little mistake on the post #6, for a 12V power, a 2.2V LED and in series 5 pcs then it will be 11V, the rest 1V for the limiting resistor, using 10mA for pwm or using 8mA for a fixed current, for a pwm method the total current will be like as 5000/5=1000, I=10mA*1000=10A, for a fixed current method, the total current will be like as 5000/5=1000, I=8mA*1000=8A, using pwm and mosfet will be a choice, if you using bjt then you will meet the heat problem, you will also get a lots of ideas from others.
     
  19. alfacliff

    Well-Known Member

    Dec 13, 2013
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    bjt's would need base current? a pulsed batch of mosfets would have a large amount of gate capacitance for that many fets, there would be a large drive current there.
     
  20. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    To use a BJT (bipolar junction transistor) as a saturated switch requires 1/10th of the desired collector current flowing through the base; so if you had 10A continuous collector current, you'd need 1A continuous base current.

    With MOSFETs, you simply need to charge and discharge the gate(s); once charged or discharged there is no current required to maintain the state.

    There are MOSFETs available that have very low gate charge such as an http://www.irf.com/product-info/datasheets/data/irlr7821.pdf IRLR/IRLU7821; 12.5m Ohm max when drain current = 12A and Vgs = 4.5v; Qg=10nC (gate charge). That is a VERY small gate charge! Now, the peak gate charge/discharge current may be high, but that's for a small fraction of a second, and the average is considerably lower than would be required to saturate BJT's.
     
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