is it not the case that, in the water wheel scenario, should something go wrong, all LED's tied to the same resistor can go *poof*? And is this not why it is often recommended to have one resistor per LED?
![](http://forum.arcadecontrols.com/index.php?PHPSESSID=lnnio2871bb05p21lb118h4co6&action=dlattach;topic=140466.0;attach=314040;image)
The big advantages to wiring the LEDs in series like the left half of the water wheel diagram/schematic are the increased efficiency and greater fault tolerance.
To wire the 12 IR LEDs for this project, you can either do 4 of the groups of three LEDs on the left
or 12 of the individual LEDs on the right.
Both configurations require current limiting resistors that dissipate the voltage drop as heat
-- Left needs 4 (1 per group of 3 LEDs) that drop 0.5v (4 * 0.5 = 2)
-- Right needs 12 (1 per LED) that drop 3.5v (12 * 3.5 = 42)
-- The right option has to dissipate 21 times the heat that the left option does.
Next consider how much current each configuration uses
-- These LEDs draw 100mA IIRC
-- Left uses 100mA flowing through each of the 4 groups of three LEDs (400mA total)
-- Right uses 100 mA flowing through each of the 12 LEDs (1200mA total)
-- The right draws three times the current that the left option does
That brings us to the question of circuit protection if the current limiting resistor goes bad -- let's assume that it shorts out. (worst case scenario)
-- The LEDs are rated for 1.5v IIRC
-- Left now has 5v applied to 3 LEDs in series. This results in a 1.67v drop across each LED = 0.17v over rated voltage.
-- Right now has 5v applied to 1 LED = 3.5v over rated voltage
-- The right LED will be VERY bright for 1 or 2 seconds before it
definitely burns out, the left LEDs might survive
![I dunno! :dunno](http://newforum.arcadecontrols.com/Smileys/default/dunno.gif)
This leads us to the question of what happens if an LED goes bad.
- A burned out LED stops current flow (most likely scenario)
-- Left would have 3 LEDs go dark (1 burns out like a fuse, protecting the other 2)
-- Right would have 1 LED go dark = slightly easier troubleshooting (
finally a win for the right side configuration
![LOL :lol](http://newforum.arcadecontrols.com/Smileys/default/lol.gif)
)
- A shorted LED (HIGHLY unlikely, almost certainly caused on the wiring end, not the light end)
-- Left will burn out the other two LEDs in the group and slightly dim the other groups (less current flowing through them)
-- Right will slightly dim the other LEDs (less current flowing through them)
The answer to your "one resistor per LED" question depends on the configuration.
If *only* one resistor is used to protect a bunch of LEDs wired in parallel, it will have to be large enough to handle all the current flow and heat dissipation for the whole setup.
When a carbon resistor has too much current flowing through it, the resistor starts to cook itself and break down.
As it breaks down, the resistance decreases, causing more current to flow through it, causing it to break down further . . . . until it becomes a short circuit.
Less resistance = less voltage drop across the resistor.
All the LEDs will be over-voltaged and burned out, probably before the resistor breaks down completely.
For protection puprposes, the three 1.5v LEDs on the left half of the water wheel diagram are effectively a single, much more robust, 4.5v LED.
![Grin ;D](http://newforum.arcadecontrols.com/Smileys/default/grin.gif)
Scott