> On Sunday 02 March 2014 18:12:42 Peter C. Wallace did opine:
> 
> > On Sun, 2 Mar 2014, Gene Heskett wrote:

> > >> If you have a 5V push-pull output, another way would be to connect
> > >> this output to Cathode1 and Anode2 and connect Anode1 to 5V through
> > >> a resistor and Cathode2 to ground through another resistor.
> > > 
> > > But the i limiting R can also be in the common signal input line I'd
> > > assume?  Or will that common voltage drop eat my lunch? OTOH,
> > > resistors in 500 packs from the shack are cheap. :)
> > > 
> > 
> > Sure, one resistor is fine for the second method. Slight advantage of
> > two is that the LED drives can be individualy adjusted.

Wait a minute.  It sounds like you are suggesting anode 1 to +5V.
Cathode 1 to anode 2, with a resistor to the signal source.  And 
cathode 2 to ground. That won't work, there is a path from +5V
thru both LEDs to ground, with no current limiting resistor.  That
will let the smoke out of both LEDs.

If you are using a single ended source, you want two limiting
resistors.  +5V, thru resistor 1, to anode 1.  Cathode 1 to anode 2
and input signal.  Cathode 2 thru resistor to ground.

You can put the resistors on the inside if you want:  +5 V to
anode 1.  Cathode 1 thru resistor to input, then another resistor
from input to anode 2.  Cathode 2 to ground.  But you still need
two resistors.

If the input signal is differential, then you can hook the LEDs 
back-to-back, and use a single limiting resistor.

-- 
  John Kasunich
  [email protected]

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