Re: TI-H: NS Illuminator

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Re: TI-H: NS Illuminator

Here's a small explanation of resistance and voltage in series circuits:
(assuming I remember last year's physics correctly)

First, let's pretend we have a 4V circuit with two 1 ohm resistors. 
Each resistor would eat up 2V.  If you had a 3 ohm resistor and a 1 ohm
resistor, the 3 ohm would use 3V and the 1 ohm 1V.  So, the voltage used
by each load is the load's fraction of the total resistance; i.e.
Vload=V*Rload/(Rload+Rload2+Rload3+...).

Well, find out the resistance of your LED (I'll call it R) and then
multiply it by 3V (total voltage) and divide it by 2.4V (voltage for the
LED).  Take that quotient and subtract the resistance of your LED, and
you have the resistance of the resistor that you need.  Example, if your
LED used 4 ohms, you take (4*3/2.4)-4=1 ohm and you use a 1 ohm
resistor. 

Even if the diode works, a resistor has a couple advantages.  First, a
constant .6V drop yields too little voltage if your batteries drop below
1.35V each.  With resistors, when the batteries lose voltage, the
resistor takes less voltage too, so your batteries can drop to 1.31V
before the LED gets less power than it needs.

NDStein@aol.com wrote:
> 
> I know a resistor is better and more efficient.  However, this is as far as
> I've gone so far.  I don't know how to calculate the resistance I need, as I
> thought you need to know amperage to calculate the resistance you should
> have.  I don't know the amperage of 2 AA batteries, and I assumed it varied.
>  Maybe you can explain this to me in #ticalc some day.  Also, I thought
> silicon diodes (normal ones, not the rectifier diodes, etc.) always gave a .6
> v drop.  This is what I read in the radio shack electronics book ;-).  In
> this case, what kind of drop would you expect, and why would I burn something
> out?  However, I was planning on substituting a resistor when I found out
> what to use.
> 
>                                                  -Noah

-- 
Jonathan Anderson
sarlok@geocities.com

"I can't be wrong - my modem is error correcting."
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