Hi,
Second Edit I've just realised the 6x9 is for one lot of two, so double the current and power requirements I give below! So, you need total 360mA current at 12V and 4.32 watts power.
First, the brighness does not go up linearly with current - so 50mA *won't* give 2.5 times the brightness of 20mA. It will be significantly brighter, but will definetely stress the components more (probably 3 times the heat).
As to the voltage overhead, yes, you are on the *only just* range with 0.9V. The problem is that if the supply is not well regulated, a small increase in voltage will cause an increase in current. With a larger overhead, and therefore a larger dropping resistor, the increase in current is smaller.
On the other hand, a small decrease in voltage will cause a large drop in current, causing the LED's to dim.
If your supply is well regulated by a proper regulator, like a 7812, or at least a power transistor and zener diode circuit, then you should be fine That means that your transformer must be at least about 15 volts, not 12, to allow for the drop on the regulator.
If your supply is just a transformer, rectifier and capacitor, you might run into problems if the mains is not well regulated, or if the circuit is run in different locations with slightly different mains voltages. For the low cost of a regulator, it makes sense to use one, then you have no problems.
As to the current needed: Current flows *through* components, for this, think of it as water. The *same* 20mA will flow through each series LED, the power the LED gets is this 20mA *combined with* the voltage drop across the LED. Just like water going through a pipe - the same amount of water flows at every point of the pipe, but its pressure (like voltage) will get less as it passes obstructions (components) and uses up some of its energy.
So, each series string of LED's only has 20mA (or whatever you decide to feed it) flowing through. But, to get the energy into them, you must have enough *volts* available for each one.
Each string (9 of them) gets 20mA and since these 9 are in parallel, they draw 9 lots of 20mA from the supply lines. So, you need 180mA total from the supply, at 11.1 volts across the strings plus the 0.9V dropping resistor.
You cannot just feed them 11.1 volts, as the slightest variation in voltage will cause large variations in current though the LED's (LED's are not linear). You *must* use a resistor, which sort of 'absorbs' a little voltage variation. The larger the resistor (and thus the 'voltage overhead' mentioned earlier) the more voltage variation can be tolerated without stressing the LED's. Even with a perfect voltage regulator, LED's are different to each other, so you still need a resistor. Higher resistors will increase the total power required and wasted though.
You could feed the LED's from a constant current, just as easy to do with a regulator, but then they really should be all in series. Otherwise, if an LED failed in one series string, the other strings would be forced to have a much higher current through them, leading to quick failure. Also, you would need to supply a high enough voltage to run them all. So this is only done with small numbers of series'ed LED's.
Finally, note that however you connect the LED's, you will need to supply the same *power* (not including any small waste in the resistors or regulator).
Assuming 12V and 180mA, you need to supply 2.16 watts of power. If you were to use 6V, therefore strings of 3 LED's and 18 strings in parallel, you would still need 2.16 watts ( 6V x 20mA x 18 ).
Cheers,
FoxyRick.
EDIT:
I just though, if you give a link to the datasheet, I'll have a quick check (probably tomorrow) and see if 20mA is OK. If you want to drive a different current, remember the voltage drop will change.