bi-color LEDs with mixed common lead

If the transistors were sot-23 then they could be incorporated to the led board itself. But if the pic can handle the led directly, its just 3 simple resistors. That would make 48 resistor for the divider plus 24 led's and maybe an equalizing resistor in series with the led.

Question? does this Pic have 24 I/O lines for the led output. Also if space is a premium , then surface mount should not take too much room for the whole thing. JTT

"James Thompson" wrote in news:aa260$45138047$438c85c9$ snipped-for-privacy@ALLTEL.NET:

No matter how you arrange 24 transistors you have to work at it, spend time, money, effort, thought... Transistors won't be needed, the PIC can source and sink 20 mA per pin for each LED.

The op-amp is needed though, it's critical. No way is a 2.5V rail made only from resistors going to be stiff enough given the currents involved, each LED current will push it around to the point where the small headroom between LED Vf and 2.5V is badly afflicted. If you have ten LED in high state they'll maybe pull up the divider voltage to the point where a single low-state LED will draw so much current that it burns out the PIC output feeding it. To keep that rail stable at half/way between the supply, you can use a voltage follower with the non-inverting input on a divider made from 100K resistors. This is very stable, and very efficient, and very easy to build. Just bear in mind that the other 2.5V worth at 20 mA per LED has to be dissipated by that op-amp, which is why I mentioned dual or quad IC's being used as ganged stages to do this. Even with 24 LED's you only need add one small IC, two resistors for a single divider, and one resistor per bipolar two-lead LED. That's the lowest part count, the easiest layout, the cheapest cost, and the best performance you're going to get, I'm sure of it. The only weakness is that LED Vf might be too close to 2.5V, so it's worth choosing LED's with less than the standard 2.2V if you can find some.

Lostgallifreyan wrote in news:Xns9846850B328EDzoodlewurdle@140.99.99.130:

More... I looked at PIC specs, the output voltages don't go to supply and ground rails, but 0.6V short of reach. And PIC's can't be supplied with 6V as I'd hoped, max is 5.5V. With 5.5-1.2=4.3 that means standard 2.2V Vf isn't an option, but there are lots of LED's at 1.8 Vf. The trick is finding bipolar LED's at that Vf. There are lots of little two-LED PCB mound thingers, but I couldn't find an actual single LED-shaped blob with two chips in it.

I see what you mean about using the opamp for the divider. With my resistor divider, I was not meaning only 1 divider for the whole 24 led array. Also they could consider a divider using 2 zener diode in series to stiffen the reference voltage. Still like you say, it would be best if they could find high efficiency led's.

"James Thompson" wrote in news:94e5e$4513fad9$438c85c9$ snipped-for-privacy@ALLTEL.NET:

Zeners are neat, but a tad temperature sensitive. LED currents would change their temperatures enough to risk the headroom available. The PIC should already be on a well-regulated supply, it just needs accurately to be halved, hence the voltage divider and the op-amp.

I take your point about the divider per LED, but I'd rather solder two resistors and one op-amp than 46 extra resistors.

The op-amp would be a neat idea even if there was only a couple of bicolour LED's to think about. It might be worth rethinking the LED format to adapt to this, but I seriously hope that manufactures start making bicolour 1.8 V LED's bearing in mind this kind of use on a 5V/2.5V rail system. 1.8 V LED's are usually high brightness types, so it would allow running on as little as 5mA. For reasons totally inexplicable to me, there don't seem to be standard indicators using this type, and there should be, makers would have no trouble shifting them in thousands.