Help with simple LED shift register project?

Discussion in 'The Projects Forum' started by 2soar, Nov 8, 2009.

  1. 2soar

    Thread Starter New Member

    Nov 8, 2009
    Working on designing a simple battery operated circuit that would create a randomly generated display of 8-10 LEDs (bi-directional) either red or green. The display would remain until another random sequence started (by push button). See block diagram..Help please! Thanks!
  2. BillB3857

    Senior Member

    Feb 28, 2009
    Your shift register will have at least two inputs. One input is the clock to force the shift and the other will be data. There may or may not be some enable input depending upon the device you use. Your drawing would use the 555 as the clock input but I didn't see anything for data. Are you looking to have only one LED on or random patterns of multiple LEDs on? Random patterns could be generated by running two more 555 timers at much higher frequencies than the one used for the clock. The outputs of the other two would feed an AND gate and the output of the AND would feed the Data line of the shift register. That would be my concept. Others will probably have better ideas.
  3. SgtWookie


    Jul 17, 2007
    Something like the attached schematic?

    Press S1 to turn all of the LEDs off.
    Tap S2 to turn some of the LEDs on.
    Adjust R14 to set the clock speed for the 555 timer.
    The LEDs will flash in the pattern set by S2, shifting around and repeating.

    It'll run on 6v to 9v.
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    For bi-color LEDs, you could connect each LED to the shift register, and the other leg of the LED to a 7414 inverter from that same output of the shift register.

    That way, if the register is zero/low, the other leg of the LED will be high, giving you one color, and if the register is high, the inverted side will be low, giving you the other color.

    7414 or similar inverters drive enough for an LED if you aren't running them super bright.
  5. SgtWookie


    Jul 17, 2007
    A neat idea, but unfortunately the 74xx series is limited to supplies in the range of 4.5v to 5.5v. Some of the 74C series might go to 6v, but not 9v.

    A 4049 is a CMOS hex inverter. A 40106 is also a hex inverter with Schmitt-trigger inputs; the former is probably less expensive.

    The LEDs will not be very bright, as the 3k3 (3.3k) resistors limit the LED current (by necessity of the CMOS IC's specifications themselves) to around 2.1mA when Vdd=9v; that's for a red LED with a Vf of 2v @ 2mA.

    If you wanted to "push things" to make the LEDs brighter, you could reduce the LED resistors to around 1.8k for 4mA current, but that may prevent inverters connected to the outputs from working properly. 2.2k would give about 3.1mA LED current, they would be a bit brighter, and the added inverters should work OK.
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    In the original PDF, I didn't see any voltage or logic type, just the shift register.

    I've done the sloppy but expedient way for temp testing of "doubling up/piggybacking" two hex inverters for double the current output, but I don't recommend that as a standard practice. :D
  7. k7elp60

    Active Member

    Nov 4, 2008
    Here is a schematic of one I created some time ago. It is pretty random as the the three oscillators made from the 74AC14 are all different frequencies. The 74AC138 will sink or source about 20mA and only one LED is on at time so only one resistor is needed.