P3D Viewer Circuitry

[Nick Merz <merz@xxxxxxxxx>]

>
>I see no reason why a pair of white LEDs would not be sufficient for a
>viewer. That would require a 9vdc battery, 2 white LEDs, a 90 ohm
>resistor, and a switch wired in series. The current drain from the
>battery would be 20ma. Power consumed by the LEDs and resistor would be
>180mw.

There was a good discussion a while ago regarding optics and details 
of building a viewer based on white LEDs.  I have a few things to add 
regarding the circuitry which might be helpful.  I can think of three 
desirable features that could be achieved in the circuit:

o  Easily identify burned out bulbs.
o  Drive the viewer from wall juice in addition to batteries.
o  Get a constant light output regardless of how low the battery 
charge is becoming.

For the first feature, I will use one white LED per eye, and wire 
them in parallel so when one burns, the other will still shine.  For 
the second, I will power the system with 9v batteries and wire a 
standard jack in parallel with them.  120vAC to 9vDC power plugs are 
quite standard, and can be bought for around $12.  For the third, I 
will add a voltage regulator to the circuit.  This is a simple device 
that works as a constant current source by regulating a voltage 
output.

The white LEDs I got need 50mA each to maximize their brightness, 
which means the circuit needs to run at 100mA.  That's fairly high, 
so two 9v batteries in parallel is what I intend to try.  It may even 
be the case that if I use rechargeable 9v, they will juice-up when I 
plug the viewer into the wall.  (I'm not sure if this is really true.)

The circuit, with voltages, looks like...
(two 9v batteries and power jack, all in parallel) in series with, +9
(momentary push-button switch) in series with, +9
(voltage regulator) in series with, +7
(25ohm current-limiting resistor) in series with, +4.5
(two white LEDs in parallel), +0

The observed behavior will be that the LEDs will put out as much 
light as they can for as long as possible, then will suddenly get 
dramatically dimmer or go off altogether when the batteries are spent 
and the regulator shuts itself down.  I will attempt to use a 
National Semiconductor LM1086 for my project which is analog.  I have 
also just found an LM2578A which is a switching regulator and may be 
much more power efficient, easier on batteries.  These regulators run 
about $2.50

If anyone on the list has tried something similar to this, let me 
know how it went.

Credit where it's due:  much of this power circuitry knowledge comes 
from James Blanc, the electrical engineer who designs all the 
charging and power circuits for Apple Computer's laptops.

Ciao,
Nick Merz