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I’m supplying 9 V through a 5 V regulator. This regulator powers five IC quad gates. This is all for a 4-bit ripple carry adder.

With that aside, could I just power the LEDs (which will visualize the 4 Sums and the last Carry) with the 5 V, connecting GND to an LED to a Sum output on the IC, for example? I think that's what this guy (Ben Eater) did in this video:

Video

I'm really novice to electronics, so this question may sound silly to the more experienced.

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  • \$\begingroup\$ hobby-hour.com/electronics/ledcalc.php \$\endgroup\$
    – Bruce Abbott
    Commented Apr 8, 2017 at 5:46
  • \$\begingroup\$ Funny how people upvote an good answer to a bad question. Ironic? \$\endgroup\$
    – Roman
    Commented Apr 8, 2017 at 6:26
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    \$\begingroup\$ If a question attracts a good answer it can't be that bad. \$\endgroup\$
    – Bruce Abbott
    Commented Apr 8, 2017 at 7:02
  • \$\begingroup\$ @BruceAbbott Right! And I will elaborate or improve my question if only it was addressed to me less indirectly than a simple down vote. \$\endgroup\$
    – Roman
    Commented Apr 8, 2017 at 7:06
  • \$\begingroup\$ If it's any consolation I upvoted you. But downvotes are addressed directly to you - that's what they are for! electronics.stackexchange.com/help/privileges/vote-down \$\endgroup\$
    – Bruce Abbott
    Commented Apr 8, 2017 at 7:20

2 Answers 2

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Some logic chips have definite current limits built into their output structure, but many do not. Depending on the logic chips being used you may well be able to direct connect the LEDs from output to ground, but I'd suggest you should not.

The forward voltage rating of your Green LEDs depends on the silicon elements and dopants used, but is likely somewhere from 2.2 to 3.3 V when conducting.

This might help you for the various LED colors: enter image description here

What I would suggest is that you should connect your LED with a 1K Ohm resistor directly to +5V and ground. This will allow you to measure the forward voltage (even if it's not very bright) across the LED. From this voltage you can then calculate a series resistor to suit your application and set the LED maximum current.

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    \$\begingroup\$ It might be prudent to mention there are 100k different indicator LEDs of all colours and ranges of power and Vf is not strictly true. e.g. There are different chemistries for green (old 2V and new 3.2V) that affects brightness (Iv in mcd or Cd) and Blue and white are the same but vary according to power and chips size and quality with chip brightness in 5mm types varying from 1mcd and some white up 30,000 mcd @ 20mA \$\endgroup\$
    – D.A.S.
    Commented Apr 8, 2017 at 5:46
  • \$\begingroup\$ Hence why it's important to test the Vf for the LEDs the OP actually has on hand. \$\endgroup\$
    – Jack Creasey
    Commented Apr 8, 2017 at 5:52
  • \$\begingroup\$ and learn to lookup spec in search windows at Digikey and compare \$\endgroup\$
    – D.A.S.
    Commented Apr 8, 2017 at 5:58
  • \$\begingroup\$ Please note that due to technological progress most white and blue LED have a Vf of 3.1V, today. UV leds tend to reach slightly higher voltages. \$\endgroup\$
    – Ariser
    Commented Apr 8, 2017 at 10:18
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    \$\begingroup\$ @LarryBud : ledlight.com/images/lenscolor.jpg \$\endgroup\$
    – kando
    Commented May 2, 2023 at 2:39
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Green Leds typically have a forward voltage of 2.2 to 3.6V depending on their manufacturing process. In no situation would it support 5 volts across it. For 5V, you will need a series ballast resistor of (Voltage Source - Forward Voltage) / forward current

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  • \$\begingroup\$ Except those few with built-in resistors \$\endgroup\$
    – D.A.S.
    Commented Apr 8, 2017 at 5:47
  • \$\begingroup\$ HB (high bright) Green comes in Aqua 515 nmD and true Green 525nmD where D refers to dominant eye corrected wavelength \$\endgroup\$
    – D.A.S.
    Commented Apr 8, 2017 at 6:07

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