Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Current limiting resistor for led on 7408 gate output

  1. Sep 11, 2008 #1


    User Avatar

    I am building a 2 to 4 decoder with 7408 And gates and need to put LED's on the output as indicators, active high. 2 questions

    1. Would I put a current limiting resistor in series with the LED to ground or put the resistor and LED from the gate output to Vcc?

    2. The LED draws 30mA at 2V, would I use the worst case Vo of the gate to calculate the required R value?

  2. jcsd
  3. Sep 11, 2008 #2


    User Avatar

    Staff: Mentor

    I didn't check, but I don't think you'll get 30mA out of a 7408 gate, especially active high (Ioh at Voh). If you want to keep it active high, put an NPN transistor at each 7408 output, driven by a base resistor, and connect the emitters to GND and the collectors to the LEDs and current-limiting resistors.

    BTW, 30mA is a lot for a standard LED. Are these big LEDs?
  4. Sep 11, 2008 #3
    choose a ttl logic gate (IC)
    so that LED would consume around 5 to 10mA for a mild glow.
    so that you can avoid using a transistor.

    74 series can sink current but cant source, it has an open collector configuration.
    So have your LED anode connected to Vcc and cathode through a current limiting resistor to the gate pin.
  5. Sep 11, 2008 #4


    User Avatar
    Science Advisor

    You could try a AC or ACT family, which is the line of high-speed, high fan-out CMOS logic, say, something like the 74ACT08 from Fairchild Semiconductors: +/- 50mA source/sink on the datasheet (but the output voltage in source mode would be lower than 5V when used in excess of 25 mA).

    This is incorrect. Standard 7400s can both source and sink, UNLESS THE MODEL IS EXPLICITLY OPEN COLLECTOR/DRAIN (like, say, the 74*09):

    However, you still need to check the datasheet in question to see if the chip can, in fact, sink the desired amount of current.
  6. Sep 11, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    The difference is that TTL uses an internal pull up resistor.
    The webpage shows the internal TTL circuitry.
    Unlike CMOS that uses active pull up.
    The internal TTL resistor can provide a sufficient current limit to drive an LED active high with out an external resistor.
  7. Sep 11, 2008 #6


    User Avatar
    Science Advisor

    I believe many of these have < 1 mA sourcing capability, depending on how the high side BJT is designed. In that sense, Chandra214 is correct (in that there is very low sourcing capability, if not an actual open-collector arrangement). Nevertheless, most of these problems can be avoided by using CMOS (AC/ACT) instead of TTL.
  8. Sep 12, 2008 #7

    Here I am, looking the TI ttl databook, what do you know?

    The 7408 has a totem pole output.

    In other words, it has an active pullup and an active pulldown.

    For standard ttl, the pullup doesn't pullup very far, 2.4V at 400uA... not much compared with a CMOS 74hc08...

    So it's either use the pulldown (16mA max) or drive an external transistor.
  9. Sep 12, 2008 #8


    User Avatar

    To confuse matters, it turns out I need to have active high outputs. The 7408 is a nand gate I assume if I put an inverter on the output I will get an active high output. How will the inverter change anything in terms of driving the led's?
  10. Sep 12, 2008 #9


    User Avatar
    Science Advisor

    No, if you want active high, then (assuming output is connected to a resistor, and then resistor to LED anode, and LED cathode to ground) you do not need an inverter (assuming you go with one of the higher-sourcing solutions presented above).

    If, instead, you wish to sink current, but have the LED light up whenever the output is high, and turn off whenever LED is low, you'd need the inverter--which you'd have to select to be able to source enough current!

    So, find yourself some nice high-current (AC or ACT) logic?
  11. Sep 12, 2008 #10


    User Avatar

    I only have available a 7410 Nand (I need 3 inputs) I was going to place a 7404 inverter (all I have available) on the output to get active high output.

    I would like the LED to light up when output is High. Circuits are a 2 to 4 and 3 to 8 decoder.

    How will the inverter change anything in terms of driving the led's?

    How further investigation the LED will draw 10mA at 2Vs.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook