Timeline for 7 segment binary to hex
Current License: CC BY-SA 3.0
11 events
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Jan 25, 2018 at 12:38 | comment | added | Richard the Spacecat | @GoswinvonBrederlow Yep, using NOT gates is basically the traditional way to drive LEDs with TTL, afaik. Join that with Lurker Larry's suggestion, and you're good to go. :) | |
Jan 25, 2018 at 8:59 | comment | added | Goswin von Brederlow | I like this. I would probably connect the outputs of the "ROM" to NOT gates to actually drive the LEDs. Otherwise for 8 you have to drive all 7 segments with a single output pin of an 74138. That would make it 4 ICs + lots of diodes. | |
Jan 24, 2018 at 14:41 | comment | added | emrys57 | 1n4148 diodes cost 40 UK pence per hundred at Farnell. I remember making just this circuit. In 1975. | |
Jan 24, 2018 at 6:07 | comment | added | Lurker Larry | The trick to use here is to program the dark segments, since there are fewer of them. Also use the '154 4-line to 16-line decoder. :) | |
Jan 23, 2018 at 17:30 | comment | added | supercat | If you're even more adventuresome, it might be fun to work out a more "economical" approach if you're willing to put two diodes in series. For example, if one used two diodes for each "corner" to light up a top or bottom segment along with the adjoining edge (eight diodes), that would save two diodes from each of 0, 2, 3, 5, 8, one from 7, and one or two from 6 and 9 (depending upon the font). A simple diode ROM layout is probably easier to build, but working out a cheaper solution might be a nice puzzle. | |
Jan 23, 2018 at 15:45 | comment | added | stark | A 22V10 costs about $2 and can do this in one chip, but I admire your intrepidity. | |
Jan 23, 2018 at 13:44 | history | edited | Richard the Spacecat | CC BY-SA 3.0 |
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Jan 23, 2018 at 13:38 | history | edited | Richard the Spacecat | CC BY-SA 3.0 |
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Jan 23, 2018 at 13:33 | review | Low quality posts | |||
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Jan 23, 2018 at 13:16 | review | First posts | |||
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Jan 23, 2018 at 13:14 | history | answered | Richard the Spacecat | CC BY-SA 3.0 |