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I've been having a problem trying to build a breadboard circuit.

The problem being, that some IC high outputs doesn't reach a high enough voltage, staying around ~2V. That is not sufficient for a clear high signal, and indeed the logic gates that receive such a signal end up not working correctly.

Here is a pic of my wiring (in the pic there are some disconnected input pins, but the problem persists even when they are all connected):

Annotated picture of wiring

The IC in question is the 74LS173, a 4-bit tri-state register. Link for its datasheet.

I feel like I must be missing something pretty simple, but I can't for the life of me find out what is it.

In any case, thanks in advance for the patience and for taking the time to help!

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    \$\begingroup\$ You need to add a schematic and some explanation about what the circuit is supposed to do. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Your schematic will probably reveal that you have no current limiting resistors on your LEDs. \$\endgroup\$
    – Transistor
    Commented Feb 1, 2019 at 15:18
  • \$\begingroup\$ electronicsandcommunications.com/2018/09/… \$\endgroup\$
    – G36
    Commented Feb 1, 2019 at 15:25
  • \$\begingroup\$ Thanks for taking the time to answer! This particular circuit is just the smallest circuit I could come up with to reproduce the exact problem I've been having, the actual circuit I'm trying to build is more complex than that. But it's just as it says on the tin, it should act as a register, outputting, when necessary, a high signal of at least 2.4V (according to the datasheet). I will try and create a schematic, and edit my post with it. In the mean time, though: I assumed that the IC acted as a current limiter already, is that incorrect? \$\endgroup\$
    – rasputinscousin
    Commented Feb 1, 2019 at 15:27
  • \$\begingroup\$ If you have an LED, and only an LED, between a device output and ground, the output voltage will be clamped to the forward voltage drop of the LED. Per @Transistor \$\endgroup\$
    – mike65535
    Commented Feb 1, 2019 at 15:51

3 Answers 3

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You appear to have some LEDs connected to your outputs. These will limit the voltage to 2(ish) volts if driven direct from the outputs. TTL outputs are not designed to drive LEDs to ground and have a valid logic high level. Their weak drive current is however convenient for driving LEDs like this.

TTL is much better at sinking current than sourcing. If you want to drive a LED, you should connect it to the +ve rail, in series with a current-limiting resistor. Unfortunately, that means it will be 'on' for a low output. This is a small price to pay for driving the LED properly, and delivering valid logic states as well.

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  • \$\begingroup\$ The part can sink up to 24mA to turn an LED on with a low signal - go that route for sure, with a 220, 250, 270 ohm series resistor for current limiting. \$\endgroup\$
    – CrossRoads
    Commented Feb 1, 2019 at 15:31
  • \$\begingroup\$ Thank you so much for the reply! Just tested the output while not driving a LED, and got ~3.7V high output. I knew it had to be something simple, it did not cross my mind that the LED would limit the voltage. I will test the inverse LED wiring you suggested to drive it. It would be inconvenient for my purposes having the LED on for a low logic state, but hey, guess that'll have to do. Once again, thank you for your help! \$\endgroup\$
    – rasputinscousin
    Commented Feb 1, 2019 at 15:41
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    \$\begingroup\$ Somewhere on the web there must be a rant that can be linked to. Here's the short version (which applies to bare LEDs, not modules of some sort): an LED is not a light bulb. You do not drive an LED with a constant voltage. Ideally, you drive an LED with a constant-current source; in the real world you usually drive an LED with a voltage source and a resistor that's carefully selected for the LED you're using and for the power supply voltage so that you get enough light and don't burn out the LED. \$\endgroup\$
    – TimWescott
    Commented Feb 1, 2019 at 16:17
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You can't connect an LED to a typical TTL logic output and still get a valid logic signal on that output. Typical LEDs draw more current than a TTL gate is designed to provide. In particular, TTL gates cannot source much current at all for a logic '1' level.

You can add current limiting resistors in series with the LEDs but you would be much better off using a dedicated driver for those LEDs.

The datasheet for you LS173 states that you can draw no more than 5.2mA from a high logic output, and the output voltage will fall as low as 2.4V if you do. If you have low current LEDs and add current limiting resistors you might make it work.

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  • \$\begingroup\$ Note that if you're driving LEDs from a TTL gate then you're reducing the output's fan-out. To really work to spec, a TTL gate doesn't just need to pull its output up to 2.4V -- it needs to do so within the rise time of the stuff it's connected to. This won't matter much if you're not pushing the circuit to the speed limits of the logic, but if you are, pay attention. \$\endgroup\$
    – TimWescott
    Commented Feb 1, 2019 at 16:20
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Without an schematic it’s hard to be certain but I see two basic issues:

  • The TTL logic families (like the 74LSxx you are using) are not designed to generate a “strong” 1. Its output driving current can be very small when driving a logic 1, as opposed to its sinking current when driving a logic 0.
  • More importantly, you seem to have directly connected red LEDs from the outputs of your IC to ground. Take a look at the I/V curve of a typical red LED and you will see how much output current are you demanding from your IC. No logic family would handle that.
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  • \$\begingroup\$ I see, as far as I can tell both your assessments are correct. The issue is indeed the LED consuming too much current and limiting the voltage of the output. Without the LED, I get around 3.7V as a high output, which is more than sufficient. In any case, thanks for the reply! \$\endgroup\$
    – rasputinscousin
    Commented Feb 1, 2019 at 15:47

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