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I bought one of these circular bargraphs from Sparkfun. A 16 LED bargraph with two common cathodes.

I'd like to drive it from a single chip to free up I/O on my micro, and not take a bunch of cycles running soft PWM. There are many chips, like the TLC5940 that can do this, but they can only sink current, not source it. A 74HC595 solves the first, but not the second, and is limited to 70mA total.

I've tried every combination of search term I can think of and have only come up with 7 segment drivers, which are overkill and never have individual dimming. So does anyone know of a chip I'm looking for? Or am I better off tossing this and making my own out of individual LEDs?

Thanks.

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Looking at the TLC5940 datasheet, it looks like it could work. This probably isn't the best way to do it, but if you put the current limited resistors ahead of the LED pins to Vcc, and then tie the output pins of TLC5940 in between the resistor and LED, when the output is active, the LED will go out. Those outputs are essentially open collector.

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  • \$\begingroup\$ Wouldn't this invert the output? Would this cause weirdness with functions any functions of the driver? \$\endgroup\$ – mordac Aug 11 '12 at 2:09
  • \$\begingroup\$ Yes, it would invert the output. It shouldn't cause any problems with the driver, at least electrically. Logic wise, he will just have to as if it were an active low device. I've never used this IC, but I don't see anything in the datasheet that would keep it from working. \$\endgroup\$ – Matt Young Aug 11 '12 at 3:26
  • \$\begingroup\$ Alright, I'll give it a try, thanks. It'll certainly make for an interesting debug. Shouting "why won't it turn off?!" instead of "why won't it turn on?!" \$\endgroup\$ – mordac Aug 11 '12 at 6:23
  • \$\begingroup\$ I don't think that would work. @mordac needs a driver that can source current. \$\endgroup\$ – m.Alin Aug 12 '12 at 13:12
  • \$\begingroup\$ That's the purpose of the pull up to Vcc. That keeps the LED on, and when the IC sinks the current, it goes off. \$\endgroup\$ – Matt Young Aug 12 '12 at 15:17

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