P3D Re: Viewer Circuitry

[Tom Hubin <thubin@xxxxxxxxx>]

Tom Hubin wrote:

> >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.

to which Nick Merz replied and Tom Hubin comments:
  
> 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.  

If you use a battery with sufficient voltage for series wiring then the
battery will last twice as long if you do wire the LEDs in series. If
you also wire a 10k ohm resistor in parallel with each LED then you can
tell which one is dead. When one burns out the other will glow faintly
with about 1% power. The bad LED won't glow at all.

If you must wire them in parallel then do so with a battery that is only
slightly greater than what you need. A white LED usually needs about 3.6
volts. So 4.5 volts would be a good battery value. Maybe 3 AAA batteries
in series. Or some other small 4.5v or 6v battery for watches or
something.

> 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. 

This is a fine idea. Use the same voltage as the battery you choose.

> 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 regulator needs to be about 2v to 3v less than your battery to work.
You could use a 6v battery and regulate to 4v. Then a series resistor to
limit the current.

> 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. 

Do you mean two in parallel or one for each LED?

> 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.)

Seems reasonable. But I do not know much about battery recharging. Some
current limiting may be needed. Just a thought.

The regulator probably won't do anything that a single series resistor
won't do here. A regulator is needed when you expect great variations in
supply voltage. A 9vdc battery puts out 9vdc for about as long as it
can. I would bet that by the time it drops to putting out 8vdc that it
is ready for the trash anyway. If that is so then just use and
appropriate series resistor. When you notice the LEDs getting dimmer
then the battery is nearly dead anyway.

If you must regulate for a single LED then regulate for as low as you
can. 5vdc regulators (78L05) are common as dirt. The series resistor is
(5v-3.6v)/0.050a = 28 ohms for each LED. 

Do not wire the LEDs in parallel. If you wire them in parallel then
whichever one requires less voltage will dominate the circuit and be
much brighter than the other. Wire a series resistor with each LED. Then
wire the LED/resistor combos in parallel if you like.

> 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
> 
<snip>
> 
> Ciao,
> Nick Merz

Tom Hubin
thubin@xxxxxxxxx