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Commodore64
modified 11 years ago

Alternate Bargraph Demo

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10:48:39
If you don>t have an LM3914 nor a bunch of OPA chips, you can use standard CMOS gates to build your own bargraph circuit. The Trick is that a CMOS gate (40xx series) sees a high between 0.7V and 1V (usually at 0.8V). Use a voltage divider like this and the corrospending LED will light up when the input of the gate receives enough voltage. You can use any 40xx series chip you like. Inverters are good since you get more of them per chip but any chip is good (e.g. the common CD4011) In this simulation, the logic gates inputs work different so I had to use that much voltage. In reality, you need not more than 1V per LED so you can use as many as 12 LEDs if you power the chips with 12V. The problem is that the input voltage may never be higher than the supply voltage! In a 5V system, you can use 3 or 4 LEDs which is enouggh for most purposes (e.g. LOW-MED-HIGH indicator)
published 11 years ago
faceblast
11 years ago
hah neat
xDR1TeK
11 years ago
Inverters have wide noise margins which makes them a terrible choice and require reasonable calculation for error approximation. There is an alternative by using quad comparator pkg then 2 are only needed. Plus using resistance dip array.
Commodore64
11 years ago
The idea is not a perfect bargraph since then you'd use a LM3914. The idea is to use leftovers! It doesn't matter if there is noise st all. Even if the signal is on the edge to light the next LED, all the noise could do is to mske the next LED glow dim creating some kind of "half step"
reinhold
9 years ago
Y don'tcha leave the DC away and change the ladder R to 10 ohm...in R and GND R to 40 ohm and the LED R to 100?

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