```From: Jim Large <jml@nomos.com>
Date: Wed, 13 Jan 1999 19:06:23 -0500

Donald,

I found your web page, http://www.cs.rutgers.edu/~watrous/ledlite.html, in a
search engine while I was looking for something else, and it caught my eye
because I've built several LED flashlights by the same method.

The plan calls for a 150 ohm resistor to limit current through the LED to
the recommended, 20 milliamps, but 150 ohms is too high.  It will limit the
current to about 7.6 mA, and the LED will not shine as brightly as it could.

The plan fails to account for the forward voltage drop of the diode.  If you
build the circuit, and measure the voltages across each component, you
should find 3V across the battery stack (as you would expect if the
batteries are fresh), 1.85 V across the diode, and (3 - 1.85) or 1.15 V
across the resistor.  The 1.85 V drop across the LED is a characteristic of
the diode, and remains nearly constant over the normal range of operating
currents.  You will probably find it listed on the back of the package that
the LED came in.

The current through each component is the same because they are wired in
series.  To compute it, we use Ohm's law, but note!  The LED and the battery
do not obey Ohm's law.  The only component that obeys it is the resistor.
Ohm's law says that
I = E/R
where I is the current through the resistor, E is the voltage across the
resistor, and R is the resistance.  Plugging in the numbers we get,
I = 1.15/150 = 0.00767 A == 7.67 mA.

I'm guessing that the value, 150, was chosen based on the erroneous
assumption that there is no voltage drop across the LED, and that the
resistor drops the full three volts of the battery stack.

To get the recommended 20mA, you need to use a lower valued resistor.  57.5
ohms would be ideal, but you will probably need to substitute a different,
more readily available value.  A 47ohm (+/- 20%) resistor should give you
somewhere between 20.4mA and 30.6mA. -- more than you need, but probably
within the absolute limits of the LED.

Even if the maximum current rating of the LED is 30mA (I forget whether it's
30 or 50), a 47 ohm resistor should be safe because there's another
resistance in the circuit that I haven't accounted for.  The batteries have
internal resistance.  I don't know how much, but it will be higher for
smaller batteries.  It might be significant if you're using AA or AAA size
batteries.  I'll leave the method of measuring battery resistance as an
exercise for the reader (Hint: don't try simply hooking one up to an ohm
meter).

-- Jim Large