From: Bill Paxton <goawaypleaseus@ya...> - 2002-06-15 21:10:58
Going thru the archives I could not find this one
answered although it was asked multiple times. I'll
try. (It can be done many ways and I apologize if this
is answered repeatedly and I'm just blind)
Please remember this is a serial port. Like always if
you fry something - whoops. Your bad. Besides this
could be a trick and I want you to destroy your serial
1 - low current (~2mA) LED of your choice
1 - 3.3k+ resistor of your choice
Wire LED+ to the resistor. Wire other side of resistor
to detector input pin. Wire LED- to detector output
This is really not that bright, and I mean
intelligence not millicandelas.
It /must be/ a low current LED. We shouldn't use more
than 1mA. Already there is a Vreg(~3mA) and a detector
(~3mA) plus if you use the 4.7k pullup resistor,
another ~1mA. Measuring one on the table right now
using an L78L05, PNA4602, and a 4k7 pullup it's taking
6.2mA from RTS.
3.3k ohm would allow 1mA if it were constant and the
LED was 1.7Vfd so it's less really. If you don't need
the brightness by all means increase the resistor.
PLEASE do NOT take advantage of the fact that it is
pulsed and lower the resistor. You probably can, some.
But for the record "don't do it." (if you make it too
low or omit it, your detector will stop working
because it's 'stealing' all the output)
Unless you want to use transistors and external power
this is the best I can suggest. Hmm, although you
could do a modified version of Enrique's capacitor
reservoir transmitter... ah, nevermind. :-)
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