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bi-color LEDs with mixed common lead

L

Lostgallifreyan

Jan 1, 1970
0
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

Do you really think that 24 transistors are easier to handle than an op-amp
IC to make the 2.5V rail?
 
J

James Thompson

Jan 1, 1970
0
Lostgallifreyan said:
Do you really think that 24 transistors are easier to handle than an
op-amp
IC to make the 2.5V rail?
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
 
L

Lostgallifreyan

Jan 1, 1970
0
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.

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.
 
L

Lostgallifreyan

Jan 1, 1970
0
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.

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.
 
J

James Thompson

Jan 1, 1970
0
Lostgallifreyan said:
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.

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.
 
L

Lostgallifreyan

Jan 1, 1970
0
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.

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.
 

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