74 series ic output voltage

Discussion in 'General Electronics Chat' started by robby991, Apr 5, 2008.

  1. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Hello everyone, I have some 74 series ic chips (inverter 7404, nand 7408 etc.) and I was wondering if I can control the output voltage of these. I have the outputs connected to LEDs, but the 2V that I am getting out of the logic chips are not enough to power my LEDs. They need around 3V. Is there anyway to do this?
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The problem is the current.

    Take a look at 74S03, 74S05, and 74S22. They have "open collector" outputs that are capable of sinking up to 20mA. They can't source current.

    Note that the "LS" and "L" versions cannot sink as much current as the "S" version.

    You can also use your 7404's and 7408's to sink the current of the LEDs you want to light - but don't try to use more than 5v for the LED's anode supply voltage, as you'll damage the totem-pole outputs. Totem-pole output TTL IC's have internal current limiting resistors for the upper "source" transistor. Open-collector TTL devices don't have the upper "source" transistor nor the current limiter resistor. Have a look at some datasheets for typical outputs of these devices. http://www.ti.com is one source.

    Basically, you use your LED's typical mA @ Vf rating (eg: 20mA @ 2.8v) to calculate the limiting resistor, along with your supply voltage, and the drop of the output transistor of the IC. We'll assume 0.6v for the the transistor's voltage drop.
    Let's say that your supply voltage will be the same 5V you're using for the IC's Vcc.
    Rlimit = (Supply Voltage - (Vf(LED) + Qdrop) ) / I(LED)
    Rlimit = (5V - (2.8V + 0.6V)) / 20mA
    Rlimit = (5 - 3.4) / 0.02
    Rlimit = 1.6 / 0.02
    Rlimit = 80 Ohms
    Use a resistor that is at least 80 Ohms.
    The closest standard 5% tolerance resistor is 82 Ohms.
    A chart of standard resistor values is on this page:
    http://www.logwell.com/tech/components/resistor_values.html
     
  3. Papabravo

    Expert

    Feb 24, 2006
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    I hope you are not trying to drive the LEDs directly with the TTL outputs. A TTL part is more than sufficient to drive a LED from a 5V suppy if you connect it through a current limiting resistor and pull the cathode low.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    A 74xxx TTL IC is spec'd to sink only 16mA but some can sink more.
    A 74LSxxxx low power.high speed TTL IC is spec'd to sink only 8mA but some can sink more.

    74Cxxxx and 74HCxxxx are Cmos, not TTL and have different spec's.
     
  5. mik3

    Senior Member

    Feb 4, 2008
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    If the ic's are not capable of driving the leds amplify the output signal by using transistors. But first check it with the datasheet of each ic.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Following on mik3's suggestion, you could also take a look at ULN2003/ULN2803 driver IC's. These handy devices contain (respectively) seven or eight open-collector Darlington drivers, each output can sink up to 500mA, and some versions can handle upwards of 30V on the outputs. They can be driven directly by TTL or CMOS up to 5v. You can use them for driving LED displays, small motors, lamps, all kinds of things. They are inexpensive (under $1 US most places) and will save a lot of wiring and space.

    If you're using higher voltage CMOS, use ULN2004/ULN2804, or add another 6k-7k resistance between the CMOS output and the Darlington driver input.
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    You are right SgtWookie, i forgot of that nice chips!!
     
  8. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Thanks for the help. Another problem I am having, I have the output of these 74 series ICs as a control line to CMOS analog switches (CD4016BE). Once the output of the 74 chips goes high (2V), it is supposed to open the gate to allow 5V to pass which activates an LED on the output of the analog switch. The problem is, the CD4016BE needs 3V from the the control line to activate the n and pmos transistors in the chip, but I am only getting 2V out of the 74 series chip. Once I give it 3V the LEDs light up at the output. What would be the best way to solve this problem? Amplify the output of the 74 series chips to get it above 3V? Here is the data sheet for the switches.

    http://focus.ti.com/lit/ds/symlink/cd4016b.pdf

    p.s. My supply voltage to all these chips is 5V.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    If your 74 series is not an L-type (low-power) you could add a pull-up resistor between 470 Ohms to 4.7k ohms to the output.
    If it IS an L-type, try using a 1k-10K pull-up resistor.
    (Pull-up = one side of resistor connected to Vcc, the other connected to an output pin of the device)
     
  10. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    I tried that but it did not really work. I added a 50 ohm resistor to pull the output of the 74 to pull it up to about 1.9V, just before the LED activates. Anything lower than this pulls the output to over 2V activating the LED. When the output of the 74 is then activated in the circuit, the pulse is about 2.6V, just shy of 3, not enough to open the gate of the analog switch. Am I doing this right?
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    If you were using open-collector output 74-series ICs it would.

    Apparently, you're using totem-pole output 74-series TTL.

    You could use an NPN transistor to "fake" an open-collector output - but your signal level would then be inverted.

    Wire a 1K resistor from the output of your TTL gate to the base of an NPN transistor, like a 2N2222 or 2N3904, etc. Connect the emitter to ground, and the collector to +5v using a 10K resistor. Your inverted output for the CMOS input is taken from the collector.
     
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