Mark:
: The LED method looks fairly easy, and the parts don't appear that
expensive. I know how to solder. What I am not as familiar with is all
the symbols used in circuits. Some like the switches are not that hard to
figure out. Others are more difficult, but it basically is just a plan for
the path of electrons. I just need a reference on how to read the plan.
grin.
omg sorry! I am so often in my own crazy world.. :)
reposting the design again with a better explanation
switch ___ 68 ohm r
\ | |
+9VDC-- \-------------| A |------/\/\--
| |___| | |
| | | _|_
| | | \ / LED
_|_ | _|_ __V__
100uF --- | --- |
| | 0.1uF| |
| | | |
___________|_____________|______|______|
|
0 Volts DC
A= 7805 voltage regulator
100uF = capacitor (polarity is important on this one)*
0.1uF = other capacitor
68 ohm is a resistor
LED is the white LED - polarity is important*
The polarity of the pins on the voltage rglator will be determined by the
manufacturer though in the standard configuration, laying the transistory
lookking thing flat on it's back with the pins toward you, the pin on the
left (pin one) is the output voltage (5V which goes through the resistor to
the LED), pin 2 goes as per the diagram to 0 volts and pin 3 on the right
is the input voltage of something around 9 volts (it has to be at least 1.5
volts higher than the output voltage for the regulator to work)
the 100uf capacitor doesn't need to be a polarized one but the link I
included was for such a beast, in which case the fly lead marked (-) goes
to 0 volts
the LED has a short lead (cathode or -) and an anode (which connects to
the + side of the circuit) - so the short lead as per the diagram goes to 0
volts
What's happening in the circuit: 9V battery provides the power which is
switched on using a switch of your choosing, the voltage is stabilized by
the 100uF capacitor which sits between the 9V and 0 volts and provides
power to the voltage regulator..
The voltage regulator does its magic and outputs a stable 5 volts which is
further stabilized by the 0.1 uF capacitor. The 5 volts then pass through
a resistor to drop the voltage to the 3.3 volts needed to power the LED.
Without the resistor the voltage would be too high and the LED would fry,
without the capacitor there might be little spikes in the power which have
the potential to spike the light levels as the LED is turned on or off.
it's straightforward and simple, no globes to burn out and drawing such a
low current should last for years of use in a densitometer circuit without
the need to change the battery (as long as you turn it off when it's not in
use! ;)
You'll also need a battery clip which I neglected to mention to connect the
battery to the circuit (or do as I often do, ply the head off a dead 9V
battery and solder on leads to use that as the clip) and a few bits of
wire - if you're neat you can make this without any circuit board and once
you've tested it, simply hot glue it in place inside the case
have fun =)
karl
