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<p class=MsoNormal><font size=2 color=navy face=Arial PTSIZE=10
FAMILY=SANSSERIF><span style='font-size:10.0pt;font-family:Arial;color:navy'>Hi
Robot,</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>Some real basic LED info (my apologies if
you already know this). The battery life will depend on how much current you’re
drawing from it. To figure out how much current you’ll need to light your
LED, you need to find out a couple of things: 1) the forward voltage drop
across the LED and 2) the maximum current the LED will take. Both of these can
usually be found in a spec sheet for the LED. There are a couple of ‘general’
assumptions you can make to get you in the ballpark: The forward voltage drop
is usually around 1.7V. Some LEDs drop as much as 2.2V, so again, check the
spec sheet if at all possible. Second assumption – most LEDs will take a MAXimum
of 20mA (0.020A) current. With that in mind, you can figure out what resistor
you’ll need to run the LED.</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>R = Vbatt-Vled / Iled (this is just Ohm’s
law R = V / I, but remember we have to subtract the voltage drop across the LED)</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>For a 3V source (two batteries in series),
and running 10mA (usually gives good brightness), we get</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>R = 3 – 1.7 / 0.010</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>R = 1.3 / 0.010</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>R = 130</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>So a 130 Ohm resistor and a three volt
supply will give you 10mA through the LED. You could probably get away with
even less current. It depends on the LED. Some high efficiency LEDs will have a
lower voltage drop and will give very good light levels with at very low
currents. </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>After you know how much current you need
to run through the LED, THEN you can check the mAH rating of the batteries. This
is where I’m a little unclear – so perhaps a battery guru can step
in. Basically, mAH is a rating of how much current the battery can supply over
time. I don’t know how to apply this at this point, but this can probably
be found out with a little Googling (funny how that word has become a verb).</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>Tim (needs more verbs in his life) Servo</span></font></p>
<div>
<p style='margin-bottom:12.0pt'><em><i><font size=2 color=blue face=Arial><span
style='font-size:10.0pt;font-family:Arial;color:blue'>"Imagination is more
important than knowledge." - Albert Einstein</span></font></i></em></p>
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<p class=MsoNormal><font size=2 face=Tahoma><span style='font-size:10.0pt;
font-family:Tahoma'>-----Original Message-----<br>
<b><span style='font-weight:bold'>From:</span></b> Robotboy8@aol.com
[mailto:Robotboy8@aol.com] <br>
<b><span style='font-weight:bold'>Sent:</span></b> Monday, January 17, 2005
9:06 AM<br>
<b><span style='font-weight:bold'>To:</span></b> synth-diy@dropmix.xs4all.nl<br>
<b><span style='font-weight:bold'>Subject:</span></b> [ot] [sdiy] Battery
questions</span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Two questions. Firstly, how long can I expect two AA
batteries to last when connected to a single LED (no other circuitry)?
And will the color of the LED matter? Secondly, what the heck is a UM-2
battery? I have an old tape recorder that calls for four um-2's, they
seem to be the same size as "C" cells, are they the same thing?<br>
<br>
-eric</span></font></p>
<p class=MsoNormal><font size=3 face=Arial><span style='font-size:12.0pt;
font-family:Arial'> </span></font></p>
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