Starting on a new PIC project using bi-color LEDs. First thought was to use a hex inverter to change colors between RED and GREEN. However it occured to me that if there were a bi-color LED (RED/GREEN) that had a mixed common lead it would save me 4 74ALS04 hex inverters, the board real estate and the board size. I can find common anode and common cathode LEDs but, so far, no mixed common. By mixed common I mean that the anode of one LED is internally connected to the cathode of the other LED.
"Carrie" wrote in news:76pPg.8395$ snipped-for-privacy@tornado.texas.rr.com:
I've not heard of them, but if you can make a rail of half your supply voltage (must source and sink, so op-amp driven is best), you could use two-lead bicolour LED's, and a single resistor for each. High output would send current through it one way, low would reverse it. You save a resistor, each LED needs only two connections, and although it might be overkill for a handful of LED's it might be ideal for lots of them. The main problem I can see is the need for a very low value resistor and a very well regulated supply if it's 5V, because half of that doesn't leave much headroom.
Hadn't thought of that. I am trying to save having to use 4 hex inverter chips. Each chip has 6 inverters so I am using a total of 24 bi-color LEDs. If I understand your suggestion correctly I will still need 24 op-amps????
"Carrie" wrote in news:A%pPg.9000$ snipped-for-privacy@tornado.texas.rr.com:
I hope not. :) Assuming 24*20 mA, you'll need maybe a quad op-amp with all four stages in parallel, or better, with 6 LED's per stage, to handle that much current.
I don't know the PIC chips, but if their outputs can source and sink 20 mA, this will work if you choose a quad op-amp that can dissipate 1 watt. If you can get by with 10 mA or less per LED, a dual op-amp should be enough. All the op-amp is doing is providing a ground rail so a single logic output can supply the LED in either state.
They are common, but becoming rarer. Historically, the first 'bi-colour' LEDs, used this design. Latter designs with three leads, are commonly called 'tri-colour' LEDs (though of course the 'bi-colour' units can also generate the third colour by being fed from AC). You can turn a tri-colour design into a bi-colour unit by just joining the outer pins on the package, and hence these are generally getting to be more common. Look at:
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What you have at the moment, is normally called a 'tri-colour' design.
An old style bipolar LED has the two led's arranged so the voltage one way lights up one LED, the opposite polarity lights up the other (both the left and right sides are connected together): --->|--- --|
I am confused as well. I looked at the link provided and it showed a 2 lead LED. Description says it is a RED/GREEN LED. I will assume then that you are guaranteed a color will light up no matter what as long as one is positive and the other is negative. This will give you either a RED lite or a GREEN lite. Nice but once in curcuit there is no way to change the color. Expensive way to avoid having to figure out which is the anode or the cathode of as single color LED.
No, what I need is a LED that will lite up RED OR GREEN, under a PIC control. It will never have to go off. It will either be RED or GREEN.
Either connect them back to back and drive the control lead above and below zero or connect the two non-common leads to two output pins on the chip and get 3 colors (which is better).
What I read in the post I replied to was the suggestion that old style two lead bipolar LEDs could be created by taking three lead bicolor LEDs and connecting the two uncommon leads together. The point of my post was that it wouldn't create a two lead bipolar because the three lead package starts with a common cathode or anode.
What confuses me in reading it now is what the purpose of that was. The original poster wants an easy way to switch an led between two colors. A two lead bipolar requires both sides to be driven, one by a buffer (or directly if the original output is strong enough) and the other side through an inverter. The three lead LEDs would just need that same inverter, just arranged differently.
But it really seams that you want something that does away with an inverter.
Whether the LED has two leads or three leads, you need the inverter (unless you go with the scheme of resistors that someone suggested).
With the two lead LED, you drive one side directly (or via a buffer if more current is needed) and the other side via an inverter. When the side directly connected to the buffer is high, that pin will be high while the other pin will be low (because of the inverter). Switch the buffer output to low, and that pin of the LED will be low while the other side is high (because of the inverter). You are changing the direction of the voltage flow, and hence can switch the LED between red and green.
With the three lead LED, you connect the common anode to the positive supply, and then connect one of the non-common leads to your buffer. Then also connect an inverter to that buffer, and connect the output of the inverter to the other non-common lead.
When that buffer is low, it will switch on the first LED (by grounding it), while the inverter output will be high and thus the second LED can't turn on. Drop the buffer output to high, and the first LED will have both sides at the same potential, and hence not be on, while the second LED will see a low, and thus turn on.
(It would be a lot easier with a diagram, but I can't be bothered trying to draw it here).
"Carrie" wrote in news:IDJPg.23160$ snipped-for-privacy@tornado.texas.rr.com:
I already explained a way. I since looked up PIC chip outputs, they can source AND sink, up to 50 mA either way. All you need to do is supply the PIC with 5 volts, and make a 2.5 volt rail to use as LED common connection. Feed each two-lead bipolar LED through a resistor selected for a 2.5V supply. High level gives one colour, low level gives the other.
Take care to work out the power dissipation fpr the op-amp you'll need to make the 2.5V rail, use a quad op-amp with all 4 stages controlled as voltage followers by one pair of series resistors of equal value. If you don't need much LED current, you might get by with an 8-pin dual op-amp IC.
snipped-for-privacy@FreeNet.Carleton.CA (Michael Black) wrote in news:eeo0l1$svm$ snipped-for-privacy@theodyn.ncf.ca:
That 'someone' was me. There was a tad more to it than resistors though.
I think Carrie's first post left no room for doubt that she knew how the inverters worked for this. What she wants is to avoid them, they take too much space on the board.
I'm not sure if have it figured out by now, but here is a simple way to figure it.
Consider the led's red or green has a forward drop of about 2 volt, and its equivalent resistance at 20 ma will be 100 ohms. Now take 2) 100 ohm resistors to make a voltage divider ( + 5 volt to resistor a with resistor b in series then connecting to ground). now with the back to back led connected to the center of your divider being 2.5 volt and the other end of the led to the Pic output, A high or a low will give roughly 2.5 volt difference on the led. You may or may not even need a resistor to drop the extra .5 volt but if you do, you end up with 3 resistors and the led instead of the extra opamp's. I would still consider at least a transistor buffer on the Pic outputs to drive the leds. JTT
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