Powering LEDs by stacking chip output

Discussion in 'Analog & Mixed-Signal Design' started by ttshaw1, Jul 18, 2016.

  1. ttshaw1

    Thread Starter New Member

    Jan 24, 2016
    I'm preparing to build an audio spectrum analyzer based off the attached schematic. One change I'm making is that instead of using ten LEDs for each band, I'm going to use more like 30, or some multiple of that, depending on how big I want to make it. I'd like to take the LEDs from a strip like https://www.superbrightleds.com/mor...-with-lc2-connector/1465/#/tab/Specifications
    so that I don't need to solder each individual LED, and because these are much cheaper than buying individual LEDs off Digikey.

    The first of my problems is related to that - each LED on the schematic will be replaced by one strip containing 3 LEDs and a resistor, probably 300R. The LM3914 is current limited. Should I expect the resistor to matter?
    This pictures shows the schematic for a strip that has red, green, and blue, LEDs. The one I want to use should have the same layout with only a red channel, although I can't find a good datasheet for it.

    The CD74HC4017 decade counter has a maximum current of 20mA per the data sheet. That won't be enough to keep the LEDs bright if all 30 (ten sets of three in parallel) in a band need to be lit. The schematic says "THREE STACKED 74HC4017." Does that mean I can duplicate the inputs and outputs on another one (or more) to add to the current? If not, how can I get around the current problem? MOSFETS/transistors draining a dedicated power supply to the LEDs?

    For those curious, the other changes I'm making include using a 555 instead of the binary counter and adding another set of LEDs, decade counter, and LM3914 to display the output of another MSGEQ7 running at a different clock speed for more bands,
  2. BobTPH

    Active Member

    Jun 5, 2013
    You would be better off to use a transistor between each output and the LED strips. If you paralleled the outputs of 3 chips, there is no guarantee that they would be the same, thus leading to a short when one output was high and the other low.

  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
    Agree with Bob. The HC4017 20 mA output current rating is not at full output voltage, so lamp brightness will suffer. For better control of the LED current and way more brightness, add NPN drivers to the 4017 outputs, PNP drivers to the 3915 outputs, and reverse the LEDs. Once you buffer the outputs there are many more options for drivers, LED power supply voltage, current, etc.

    Last edited: Jul 19, 2016
  4. SLK001

    Active Member

    Nov 29, 2011
    I don't see how you are going to get a spectral display the way you have shown your circuit.

    Also, don't use a CMOS device to drive the LEDs directly. Either use a NPN as suggested, or use an N-Channel MOSFET, placed between the last LED and ground, to do the switching.
  5. hp1729

    Well-Known Member

    Nov 23, 2015
    Where on the data sheet are you getting that 20 mA output capability? I see 5 mA, and not at the full output voltage at that.
  6. k7elp60

    Active Member

    Nov 4, 2008
    The LM3914 can be set up to determine the current from Vcc to each of the output pins. If you use a higher voltage for Vcc, like 9V or 12V you can put mutliple LED in series on each output pin without a series resistor to limit the current.
  7. tonyStewart

    New Member

    May 8, 2012
    there are a couple specs you should understand about LEDs,
    Pulse/Avg ratio may not support Muxing more than 4:1 and your design is 7:1 ( most even less)
    ... unless you get really bright LEDs and use 4mA avg for max brightness.
    You will need a high side common anode driver at 7x the average current, so consider driver needs.
    7x average current 10x parallel current = 70x avg current total x4mA = 280mA

    You want to limit the Anode voltage fluctuation on brightness.
    This is called load regulation error and means your Anode driver output impedance must be much lower than load. Rule of thumb is pick a number between 1&10% of load .. (LEDs +'Series R)

    Remember this , that High voltage (CD4017) CMOS is about 300 Ohms output at 12V
    ... and 5mm 65mW LEDs have an ESR of 16 Ohms approx. (more with less current)
    Thus load is 16Ohms/10LEDs=1.6 Ohms then add current limiting R and your common Anode driver must be or drive 200mA with Say 0.2V drop which becomes a 1 Ohm driver resistance. FETs are easy for this job. Darlington's drop too much V but can work if careful.

    It would be best to define your visual expectations and work backwards.
    Brightness in xxx mcd to driver current requirements then choose best configuration.
    Common anode or cathode...etc.