So with two 3/4"x 1/8" aluminum strips for the tracks/heat sinks I use the same thermal
paste used for a computer CPU between the bulbs and the strips for the heat issue.
The rest I'll pick up a little at a time till I'm comfortable/w it.
I was a late bloomer to computers/w my first being in 06. It lasted me 8 years and due
to my dissatisfaction/w the limits of it I built my second one which I'm now on/using. I think you can see where that is going.
Remember an LED is a diode first.. Basically a solid state on/off switch..but mostly directional.
When you apply a voltage (which is really nothing more than a slope of a hill.. a "potential" difference)
the switch blocks electrons (amps) from traveling "downhill" until you reach its threshold (V(f)). At that point the current is allowed to flow.
As the current flows photons are generated :
The charge carriers recombine in a forward P-N junction as the electrons cross from the N-region and recombine with the holes existing in the P-region. Free electrons are in the conduction band
of energy levels, while holes are in the valence energy band
. Thus the energy level of the holes will be lesser than the energy levels of the electrons. Some part of the energy must be dissipated in order to recombine the electrons and the holes. This energy is emitted in the form of heat and light. The electrons dissipate energy in the form of heat for silicon and germanium diodes but in gallium arsenide phosphide
(GaAsP) and gallium phosphide
(GaP) semiconductors, the electrons dissipate energy by emitting photons
. If the semiconductor is translucent, the junction becomes the source of light as it is emitted, thus becoming a light-emitting diode, but when the junction is reverse biased no light will be produced by the LED and, on the contrary, the device may also get damaged.
As w/ any diode the amount of current allowed to flow is exponentially increased as the "potential" or voltage changes by a small amount..Of course heat will exponentially increase as well..
soo the important thing to know is what is the safety range of the diode in terms of voltage and at what current will you need to be able to supply
The rest is "tinker toy" parts.. Drivers, power supplies and control circuits..
To eliminate a lot of potential problems, using constant current drivers simplifies any build.
All you need to know is the V(f) of the diodes at a pre-determined current (there are complications, mostly minor of the V(f) actually changing under load (heating) but it is mostly irreverent w/ enough slack built in)
Next step voltages add in series so w/ a "string" of LED's at a set current all you do is add the voltages (V(f)) at that current and get a stable (switching) power supply to supply it. A driver (or a contained circuit in a power supply driver combo) will regulate the current in the string and conversely maintain the voltage through the entire string, regardless of ind. voltage needs of each diode.
now take the simplest (well more or less the cheapest) amp "governor" a Meanwell LDD. now the circuitry to function requires some in/out differential (don't ask why) and must be added to the power supply equation.
So say you have 5 diodes of various colors..
From the spec sheet at 500mA (a combination of your choice and the diodes physical properties) the V(f) is 3.6, 2.2,3.6, 3.4, and 3V.
15.8V is what you need to supply to 5 all set +..-..+..- ect..
For the LDD about 3V in/out differential is needed.. Add that to the 15.8 and you get 18.8V. so any ps greater than 18.8V (the LDD "governor" will adj. the voltage to the right "out" of 15.8V) is sufficient.
And since the current required is only 500mA (1/2A) a ps capable of not burning out at 1/2 amp "draw" is necessary..
Adding a bit (well a big bit) of fudge factor 1A PS at say 24V is perfect.
so a 24 Watt ps..
Say you double the diodes but run them 2 parallel strings, one driver each..
So your voltage stays the same but your current needs double.
If instead you just used one driver but attached the 2 series strings in parallel, the PS stays the same BUT the current into each row of 5 is 1/2..
[COLOR=Red]Amps add in parallel, voltage adds in series.
[COLOR=Black]For the most part, that is really all you need to know electrically.
Control is another whole set of tinker-toys..
Oh and I suppose you need to consider exactly what a 3W diode really means.. Though there are infinite combinations possibe the diode is what determines the reality..Say at 1000mA the voltage across the diode is 3V
1Ax3V =3W. Consider it more of a maximum recommended than a real number..Say you supply 3.2V to it and have a ps capable of supplying more than 1A the amps flowing will increase (remember this is all exponential) the current flowing (will increase (as will heat AND photons out) to say..1300mA.. So now your 3W diode is really a 1.3X 3.2V= 4.16W diode..
you may now see why regulating current is easier than regulating voltage, and using voltage is practically impossible in a mixed V(f) array.
In other words the watt output of a diode is what you make it to be, by choosing voltage and/or drive current.
At least this is about as easy as I make it for myself..
The confusion will set in when you read that LED's are "current controlled" devices.. which is true but the concept is a bit confusing in a practical sense..
voltage and current sort of go hand in hand..
You can have a 1000A power supply that is at 2V and it won't light a single diode w/ a V(f) of 3V
You could also have a ps of 1000V yet not capable of more than a 0.1A output and your diode will light but generally quite feebly..
And your power supply will need to dissipate say 1000-3V of potential 999V 99.9Watts of waste and heat..
AFAICT.. and yes it is a bit more complicated, but not much..
any corrections/additions to this soapbox more than welcome..
Can't have a post w/out a pretty picture: