help on what display to use

Discussion in 'The Projects Forum' started by amaynew, Oct 19, 2007.

  1. amaynew

    Thread Starter Member

    Jul 9, 2007
    18
    0
    Guys, i wanted to use a 7448 BCD decoder, what kind of seven segment do i use? should it be a common cathode or a common anode?? thanks:)
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The device has active HIGH outputs with internal pull-ups, so a common cathode display is the one to use. My TTL manual is not clear about the output current, so you may need 300 ohm current limiting resistors in each output line.
     
  3. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    The output current of TTL logic family is 2mA when sourcing and 16mA when sinking. Since TTL gives more current when sinking, you should consider using the 7447 open-collector decoder instead (unless you want to use 2mA for each LED, since that is the maximum that the 7448 will source). Besides being open collector, the decoder uses false logic on it's output, proper for a common anode display. Although the decoder can sink up to 16mA per output, I would advice you to limit the current to below 8mA for each, for safety reasons (I often use a safety factor of 2, but that is just my opinion). It will be enought to light the LEDs.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    Checked my TI TTL Databook, vintage '76.
    7448 MAX sink current is 6.4 mA, has an integrated 2k pull-up resistor. LS version is 6mA sink. Recommended sink is 4mA

    The max it could source would be 2.5mA, just using the integrated resistor. Mighty puny. You COULD get up to about 8.8mA source using additional pullup resistors. What value to use for pullups depends upon the LED display rating. If they would display with 8.75mA @ 2.1v, you could use 330 Ohm pullups. Resistor selection is dependent upon your display's rating and the limitation of 6.4mA max sink current; if your pull-up resistors are too low in value, you'll burn out the IC's outputs. Too high, and the displays won't light. Also, I'm talking running at maximum output here; recommended is 4mA. Running at max = short IC life. You'd really need to use a driver IC, like a ULN2003 - that contains 7 NPN Darlington pair transistors. ULN2803's contain 8 Darlingtons. Those will sink a lot of current. You'd need to use resistors to limit the current.
     
  5. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
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    The 7448 should be sourcing current, not sinking. It has active high outputs. TTL chips can only source to 2mA. So it is better to use the 7447, which has active low outputs, but can sink up to 16mA. The datasheet for the 74LS47 chip from Fairchild mentions 24mA (Iol - sinking). The datasheet for the 74LS48 chip mentions only 2mA (Ioh - sourcing). The chips are pin compatible, which means you can easily replace the 7448 by the 7447, and use the a common anode 7-segment display instead of a common cathode one.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    cumesoftware,

    Yes, I actually started to mention 7447's, along with their more modern counterparts (74247s and 74LS447s) - but that would've been outside of what the OP originally asked about; perhaps they stumbled on a large stash of 7448's; I don't know. In any event, it seems likely that if they want to stay with the 7448's, they'll need some sort of driver IC (or heaven forbid, discrete components)
     
  7. amaynew

    Thread Starter Member

    Jul 9, 2007
    18
    0
    im using a 7448 IC coz that is what's available in my case, i've read a spec sheet that says that 7447 is a compliment of 7448. I just got confused on what display to use because i got confused on my schematics putting the 14 segment display directly to VCC. anyways thanks guys
     
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