Ok the first thing to understand is most current DIY designs use constant current circuits in the driver (unless using strips which are constant voltage).. Basically a Buck voltage regulator w/ feedback control of the amp output..
Put simply.. a power supply (constant voltage switching power supply) spec-d to the output needs of the LED string (more on this later) feeds a "driver" (buck voltage converter w/ feedback loop and reference) that lowers or raises the voltage to supply a set current.
If this is an all in one unit.. fine.. As separate parts you have to consider the design of the driver As an example a Meanwell LDD-H constant current driver ($7) will "lose" (don't ask me how) 3V from the orig power supply. So a 12V PS will only be able to output 9V ect.
LED's have a V(f) at which, at minimum they will light.. them Max before burn out..
Each one (types) are different (as ell as lot to lot variation) but using constant current in the string this becomes unimportant, as the driver will output a constant current, allowing each diode its V(f) at that current..
All that would be important is knowing the V(f)'s of the diode at that amperage , and adding the diodes V(f) together in series..
Say you have a 24V ps and a meanwell at 500mA you have 5 diodes that at 500mA have a voltage drop of 3V, 3.2V 2.1v 3.5v, 3.5v for a total of 15.3V.. Well within the ps -driver voltage (24-3= 19)
IF the diodes can take it and you want more output, changing the driver to say 1000mA will increase the V(f) of each diode, and you must recalculate.
But back to your orifg. question:
You could (if the PS allows it) trim the ps voltage down to more closely match the driver/led circuit to save the ps a bit of work.
The drivers themselves normally don't heat that much dissipating the excess voltage (part of the magic I don't understand but it has to do w/ the driver circuitry)
Of course the PS needs the amps (watts) but in larger ps this will usually exceed the simple setups..In the above example the ps needs to be capable of outputting 7.65W (15.3x.5) or to add a decent fudge factor 1A @ 24V (actually 1/2A)
It is simpler than I'm probably making it sound.;
steves drivers are a bit of an oddity since they are not true buck drivers.. you DO need to trim the ps to match the REAL V drop of any string.. and therefore "I" do not recommend them for a beginner.. It isn't hard, there data shhets have a good "setup" protcol but.. why??
Oh and instead of doing a full blown adruino Steves Typhon is a fair "clone" at a fair price.. Though simplistic.. BUT reprogrammable..
If a PSU is 25V, can you wire only one 12V LED into it?
Maybe but it depends on the actual v(f) at the current you want to drive it at and the type of driver (lossy or not).
Will it simply let the rest of the voltage return to the PSU? Or do I need to connect a load that takes up the entire 24V provided?
no and no..
Does it overdrive the LEDs if I don't dim them?
W/ constant current this is not possible.. You could decide to over drive them..Say max current is recommended at 700mA but you use a 1000mA driver..
As to optics.. Simple geometry..
Tan 45 degrees (1/2 of 90 degree optic) is 1 so 1" off the water and your cone is 2" in diameter..
So for full surface coverage of say 12" tank (front to back) each diode needs to be about 6" off the water surface.
I'm leaving strips out for now, but one obvious problem is at 120 degrees (native optics for most strip diodes) At 6" you have a 21" cone of light.. so light spill is a problem the higher the strip from the tank surface. W/out a good reflective shield they are fairly ineffective at distances of more than an inch or so..
re: 12V Luxeon-m's.. From the data sheet the V(f) at 700mA is
Sooo running 2 in series w/ a 500mA driver (v(f) will be lower than 10.5-11.7) w/ a 24V PS "should" be possible IF
you can tweak the ps a bit.. though it is close..
24(ps)-3(driver loss)=21V @ 500mA assuming a slight v drop you should be close to 21v... 2X11.05= 22.1)...PRETTY close..
A 350mA driver would be better ..or 2 @500 1 per LED.. You only need to double the ps output (2X.5) which would be easy enough to do (2A ps )