Electronic – Voltage Divider Supply (Unloaded and Loaded)

Edits to this answer have subtly but significantly changed things I was trying to say. I've changed some back, but it gets hard and time consuming and it's easy to make a mess unawares. I could have just done a "revert" but that would be a sad thing to do as there were numerous worthwhile changes.
If editing please try to retain the sense of each point being edited. Fixing typos and spill chucking is fine. Adding poncy formula renderings is fine. If you feel you wish to change phraseology or grammar rather than technical content I'll hazard that you'll find that there are many more worthy candidates which warrant prior attention.

A major problem is that you have not told us what you are trying to achieve but rather how you are hoping to achieve whatever it is that you are doing. A good overall description is liable to result in a superior solution.

Things to consider:

• Why are you using those particular LEDs?

• Do you need a high CRI?

• How long do you want these LEDs to last?

• Do you care or know anything about this manufacturer?

These LEDs MAY be extremely good and have a long lifetime.
The webpage and site look good and the published specifications are good...
However, experience has shown that for other than well known manufacturers, it is highly likely that the LEDs will be be sub-standard, having a short lifetime and low efficiency.

LEDs of this power level and this cost essentially must be driven with constant current(CC). While it is possible to get CC power supplies which control a resistive pass element these dissipate excessive energy in high power applications. What you want is a CC power supply that adjusts adjusts its voltage as required to control current so that energy dissipation in the driver is minimised. What the LED manufacturer terms the "largest value" in his spec sheet (which is why I used the term here but it was edited out) = 20 V is the maximum value = V_{f, max} that may appear across the LED at rated current, but usually actual values will usually be lower and will vary between devices.
If you parallel LEDs the differences in V_f will at best result in different light output from each LED, and at worst may cause a cascade of failures over time. (Some LEDs having low Vf hog the current and starve high Vf LEDs. The low Vf LEDs dissipate excessive power and die early leaving fewer LEDs to carry the same current and the process accelerates. )

It is not obvious that you have considered LED heat sinking or other thermal management. [My reference to LED heatsinking was edited out but I was specifically meaning that as other heat dissipatioon had been mentioned, making the LED aspect more noticeable by its ommission. /. Most of the almost 2000 W input will be dissipated in the LEDs as heat and will require substantial heat sinks.

The LEDs are specified as having 100-120 lumens per watt (l/W) or thereabouts. If they achieve this then about 25% of the input power is converted to light and 75% as heat you must deal with. Cree is selling 200 l/W LEDs and they have achieved 350 l/W the lab. However, most LEDs are far lower. You can buy 120-150 l/W LEDs from top manufacturers, but many lesser-known manufacturers claim much but deliver less. High CRI will likely reduce l/W. Looking at their graphs the methods they use to (claim to) achieve include bandwidth notching (where part of the spectrum is suppressed) which reduces l/W if it is done by adsorbing emitted light. Even Cree appear to do this rather than solely using phosphor spectrum tailoring.

Anyways, if you budget for about 1500 W of heat dissipation and say 80 °C their T_opr,max) LED at the manufacturer's external measurement point, and keep an ambient temperature of 30 °C (optimistic) you need a 0.03 degree C per Watt heatsink [just try and buy one!]. This requires an extremely good (or large) heat sink.

Liquid cooling or well-designed fan cooling in conjunction with large heat sinks is required for such power levels. If you were to pack the 13.5 mm square LEDs at, say, 20 mm centres (6.5 mm = 1/4" between each LED) they would occupy about 220 mm square (~9" square). Evacuating 1.5 kW of heat with a lower than 50 °C rise (given no hot spots) will be "interesting".

Simple test:

• Take a 1 kW heater element.
• Enclose in a say 9"x 9" metal box - say filled with oil. (Do not short electrically).
• Attach cooling methods of your choice.
• Operate element at 1 kW.
• Monitor oil temperature. MUST stay <= 80C.

Success?
OK. Try 1.5 kW.

You have an N channel MOSFET connected as a source follower. A source follower (just like an emitter follower for BJTs) can't have as it's output a voltage any higher than what is put on the gate. In fact, for a MOSFET, the source voltage will be a couple of volts less than what is on the gate - this is why you see 3V. The act of connecting your load to the source ensures that you cannot achieve full-turn-on unless the gate rises higher than 17 or 18 volts from ground. But all is not lost....

Ask yourself - what turns the FET on - answer - the gate to source voltage has to be a couple of volts or even higher on some fets. This can only mean that the fet is partially turned on.

You ought to consider grounding the source and putting the train lights in the drain. Like this: -

Rd is where you'd connect the lights. Now with the mosfet gate at 5V with respect to grounded source the fet turns on much better.

Another way is to use a P ch MOSFET driven from a BJT: -

Because the P channel FET's source is up to the positive supply, pulling the gate down to GND ensures it turns on to the lowest resistance. Don't run this circuit any higher than about 15 volts though because the maximum FET gate voltage might be limited to 15 volts - read the data sheets - a lot of mosfets are OK for +/-20 volts gate-source but some are only rated at max 12 volts.