Connect with us

Controlling 10V 7 Segment LEDs

Discussion in 'Electronic Basics' started by Robert Monsen, Oct 22, 2003.

Scroll to continue with content
  1. Ok, I've got a simple scheme for controlling these using an 'unnamed 5V
    device' :)

    The problem is that I need to be able to switch the 10V common anode, and
    ground the a-g inputs, all using 5V outputs from a shift register. An open
    collector buffer like a 7407 works for the latter.

    Page down to the second section. It shows a schematic for powering the
    common anode using two transistors and three resistors.

    However, this is a pretty common thing to want to do. I'm guessing there is
    an IC that can do it for me without the added transistors/resistors. I spent
    a few moments in the philips device tree looking for something, but didn't
    find anything that would work.

    I looked at analog switches, but their 'on' voltage is about 7V for a Vcc of

    I looked at CMOS buffers, but they have the same problem.

    Is there a part that is cheaper than the 6 times (two transistors/3
    resistors/extra solder and wire runs)?

    (Of course, the other way is simply to drive each 7 segment from its own
    shift register. However, that means 6 shift registers, and 8 open collector
    buffers, and their wire runs)

    Bob Monsen
  2. Lord Garth

    Lord Garth Guest

    You might use a 7406 inverting OC buffer on the digit lines. When the
    shift register output goes low, the buffer output will be open collector.
    From there you can pull it up to the digit supply voltage with a single
    resistor per digit.
  3. Yes, that would work. Lets see:

    Assume a 12V supply, I need something like 50mA through the LEDs, so I drop
    the voltage to ~10V using a 47 ohm resistor. Then, that 40 ohms goes across
    the entire 12V, so it dissipates 300mW, so a 1/2W resistor will work. Also,
    I can use the same part for both the anode and cathode. I think this will

    Its even nicer that I can use the same part for both the anode and cathode.
    I'll remember this trick.

    Thanks for the help...

    Bob Monsen
  4. Not. Posted too quickly.

    Its actually 300mA through the resistor, which means the dissipation is .3 *
    12= 3.6W. Ouch

    P = 50mA * V^2/(V - 10)... so there is a minima for P at 20V, and thus the
    absolute minimum for this technique, given the preconditions, is 2W.

    It still might be ok, but there also might be another way. Also, the 7406
    would have to sink 100mA. Only one pin is on at a time, though, so this
    should be OK.

    Bob Monsen
  5. Use the 7406 outputs to drive PNP transistors. You will need a
    resistor from base to +10 to turn the transistor off when the 7406 is
    OC, and a resistor from base to 7406 output to limit base current when
    the 7406 is on.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day