From: "Gary & Tena Strong" <gstrong@uswest.net>
To: "Andrew Plunkett" <aplunkett@home.com>
Cc: "lancair list" <lancair.list@olsusa.com>
Subject: RE: 10 LED position sensor
Date: Sun, 19 Mar 2000 11:20:24 -0600
Message-ID: <000901bf91c7$695598c0$d290e1cf@oemcomputer>
In-Reply-To: <000901bf9173$afb981e0$80b10c18@bvrtn1.or.home.com>
Importance: Normal
Mime-Version: 1.0

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Andrew,
A quick look at the schematic tells me that the 7805 should easilly handle
the leds without any trouble.  A typically LED pulls maybe 20 milliamps
times 10 would equal .2 of an amp.  A 7805 usually with any sort of heatsink
will handle around 3 amps without getting warm, so the .2 amp load of the
leds is inconsequential. Although I have an electrical engineering
background, I've been in computers for years so please excuse any 'fading"
of the memory if what I describe here doesn't sound correct.

I'd guess that you've possibly got a small short circuit causing the
excessive load.  The short is most likely in a wire to wire (or trace to
trace) rather than a wrong pin on a chip  (if a wrong pin on a chip was the
case, the chip would have instantly exploded at this amperage).  In the fear
of being too basic (please forgive me if these next few questions insult
you, I just don't know your skill level in electronics).

1.  when a schematic shows wires crossing, you only attach them if their is
a circle around the cross over point.
2.  the leds have a shorter leg and this difference in length determines
polarity and the anode or cathode leg of the led.
3.  if capacitor c1 is shorted (you can easily check this with an ohm meter,
or if you don't have one, throw it away and put in a new one.  with the
spikes the circuit has experienced from what you describe, its probably bad.
Also capacitors are polarized as well, you'll need to make sure + is on +,
etc. or it will instantly short out causing you the problems your
experiencing.

I'd be glad to help you sort this out, just let me know how I can be of
help.

Andrew
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