Getting back into electronics as a hobby, and I'm giving myself some breadboard tasks and drills to do to refamiliarize myself with both simple analog and digital circuitry.

Here's what I want to do: 5VDC source, with a pot (say, 100k). I want to take that analog signal and drive 8 leds which light up between 1 and 8 of them at a time depending on the relative output of the analog signal.

My first thought is 8 bit analog to digital. Got ahold of the ADC0804 chip (https://circuits-diy.com/adc0804lcn-8-bit-a-d-converter-datasheet/)

Simple hookup with a pot driving the V+ relative input, and my output was hooked to 8 leds through some current limiting resistors. I turn the POT and the output is 8 bit binary value displayed on the 8 leds.


But now my thought is, rather than displaying an 8 bit binary value, let's drive 8 leds which turn on successively more lights as the 8 bit output goes up. At first glance it might SEEMS like the same thing, but it's not.

For example, let's say the 8 bit output is 0b10000000. When displaying binary, of course, bit 7 is on, and bits 0-6 are off.

But I don't want to do that. That value is 128, which is half of 255, so the lowest 4 lights out of 8 should be on instead.

I'm racking my brain as to how to convert this without a simple microprocessor, which I can easily do. Probably could do it with TON of logic gates, but is there a 1 or 2 chip solution for this? I'm going through a bunch of data sheets, but since I don't know what this might be called it's difficult to find a prepackaged solution. Maybe I didn't need to convert to 8 bit digital in the first place, and there's a simple driver to accomplish this already?

This truth table might help (I've added some logic tables below the truth table). The logic table is how I could accomplish this with AND and OR gates. Would still love to know if there's a chip designed for something like this.

enter image description here

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    \$\begingroup\$ Decode the three high bits to seven of the LEDs and the last LED to the fourth most significant bit? \$\endgroup\$ Jan 27, 2022 at 2:20
  • \$\begingroup\$ @GrapefruitIsAwesome sounds like a winner, make that an answer. \$\endgroup\$
    – hobbs
    Jan 27, 2022 at 2:28
  • \$\begingroup\$ decode it using what? Understood the high 3 bits essentially count from 0 to 7, how do I convert that to 8 separate leds? This pointed me closer to the right direction. the 74238 gets me CLOSE, as it will light up between 0 and 7 lines, but only 1 at a time... \$\endgroup\$
    – LarryBud
    Jan 27, 2022 at 2:31
  • \$\begingroup\$ I had written an answer only to realize I that decodes logarithmically. Rather than that maybe using a 74xx series bar driver, I'll write that up instead. \$\endgroup\$ Jan 27, 2022 at 2:41
  • \$\begingroup\$ I'm being dumb (long day, I was linear to begin with...) \$\endgroup\$ Jan 27, 2022 at 2:47

1 Answer 1


As you want to (I assume linearly) may the 8 bit level to having one of the 8 LEDs on I would suggest the following mapping: decode the 3 highest bits to one each of 7 of the LEDs and then map the fourth most significant bit to the 8th LED. You won't need to ever consider the lowest 4 bits. See the truth table below.

111xxxxx 10000000
110xxxxx 01000000
101xxxxx 00100000
100xxxxx 00010000
011xxxxx 00001000
010xxxxx 00000100
001xxxxx 00000010
0001xxxx 00000001
0000xxxx 00000000

You can decode the top 3 bits with a 3 to 8 decoder IC like the 74LS138 anding the lowest order output with the 4th most significant bit for the final output.

  • \$\begingroup\$ Close, see my edit above with the truth table. I also want to turn on all leds "below" the top most lit up. \$\endgroup\$
    – LarryBud
    Jan 27, 2022 at 2:45
  • \$\begingroup\$ And if you have something logarithmic like sound, take the leftmost 1 in your value and set that light and all the others beneath it. \$\endgroup\$
    – Dan
    Jan 27, 2022 at 2:46
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    \$\begingroup\$ @LarryBud I see, if someone doesn't write up a more complete answer before me I'll update tomorrow (or sooner) when I get a chance. \$\endgroup\$ Jan 27, 2022 at 2:49
  • \$\begingroup\$ Appreciate it! I will continue to look to see if something is premade for this or if I have to get a little more basic with logic gates. \$\endgroup\$
    – LarryBud
    Jan 27, 2022 at 2:49

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