An LED is a light emitting diode.
A diode is a semiconductor junction which allows the passage of current when its forward voltage is exceeded.
LEDs emit light, of varying frequencies and hence colour, depending on the material used for the diode junction.
The two important values needed when selecting an LED are:
1) the forward voltage Vf. This is the minimum voltage across the LED necessary to make the diode conduct and hence the LED to produce light.
2) the forward current If. This is the current which will pass when the diode is conducting, up to a maximum, at which point the magic smoke escapes and it stops working.
The Benedini TBS5 is run from a 5 volt supply and has a 270 ohm resistor in series with the main gun and MG outputs on connector 6.
Making the assumption that the LED has a fairly standard forward voltage of 2 volts, we can use Ohm's Law to calculate that the forward current will be the supply voltage less the forward voltage and all divided by the resistance. This works out to be:
( 5V - 2V ) / 270R, which equals 0.011A, or 11mA.
This is fine for a standard LED with a maximum forward current of 20mA or so.
However, to use higher powered LEDs, so that the main gun muzzle flash will be visible in daylight, much more current is needed, which is why I designed the original recoil unit around a high powered transistor switching chip, as the Benedini by itself cannot provide the current necessary to illuminate a high powered LED.
It should also be noted that it will only work when connected the right way round. The picture below shows the ways to identify the positive and negative pins on an LED.
As Mark stated, the centre pin on the TBS5 is the positive supply, so needs to be connected to the Anode on the LED. The top and bottom pins connect to the negative supply when activated by a sound, so need to be connected to the Cathode of the LED.
If you run two LEDs in parallel, the voltage drop will be the same, but each LED will only receive half of the 11mA, so will be dimmer and may not even light. Also, if the voltage drops across the two LEDs are slightly different, one will take a much higher proportion of the available current than the other, so the subordinate one may not light at all.
If you run them in series, the voltage drop across the pair doubles, so the available current becomes:
(5v -2v -2v) / 270R, which equals 0.0037A, or 3.7mA, which may
just light the LEDs. But probably not.
Adrian.
I have spent a lot of time on this to no avail 
