0
\$\begingroup\$

Thanks to the help I've received on here the last few days I have been able to put together a schematic for a project I am planning.

The project itself is lighting up a 1/400 scale model of the Cunard Queen Mary 2.

enter image description here

I will be running this from a 3amp 12v power source using a range of mostly 3mm LEDs along with some 0402SMDs. They will mostly be in series chains of 2 or 3, with each chain having its own direct connection. I've grouped the LEDs into like for like, so no chain is made up using different colours or types/makes.

I've broken this all up into 5 layers, depending on where the lighting sits within the model vertically.

Below are 6 images. The first 5 are each layer individually, with the final one a combination of all 5. (Please Note: There may look like there are some broken or random wires, there isn't that is just me making crude use of the software to create an outline of the model)

enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

(The software states I will require 1.89A but my calculations come closer to 2.5A)

Should I fit a fuse? If so, what value?

I'd appreciate any input, but please keep in mind my knowledge is absolute minimum, especially where terminology is concerned, thanks.

Edit: I've corrected the resistor value on the single red LED on layer 3, to 560ohm.

\$\endgroup\$
  • \$\begingroup\$ Beware that different LEDs have different voltage drops. You may need a different resistor value for a chain of red LEDs than for a chain of white LEDs. Moreover, I’d suggest you use chains as long as possible (probably 3 for white LEDs, perhaps 4 or even 5 for yellow or red ones), to reduce the power loss in resistors. \$\endgroup\$ – user2233709 Nov 19 '17 at 14:14
  • \$\begingroup\$ Thanks, I think I've already taken into account the differences in drops, Almost all of the LEDs I will be using come from the same seller, and their information regarding the different ones are here: link \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 14:30
  • \$\begingroup\$ The only exception to this are the white ones in Layer 3 (shown external to the ship). I've used the LED series parallel array wizard to find out what value resistors I need, and have selected the closes higher value I have been able to find, again from the same seller. Fingers crossed I've taken into account everything correctly. Regarding grouping into longer chains, I've no way of being able to calculate the resistors I'd need, as that wizard always defaults to 3 or less. And I am damn useless at math lol \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 14:30
  • \$\begingroup\$ No idea what wizard you’re talking about, but does it allow you to specify the forward voltage of the LED you use? It’s surprising in layer 3 that you have the same resistor value for a single red LED and for a single green one. \$\endgroup\$ – user2233709 Nov 19 '17 at 14:43
  • 1
    \$\begingroup\$ It's not built yet LOL. I'll add a photo to my OP. \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 17:01
1
\$\begingroup\$

The reason you'd fit a fuse is to protect the wiring.

You state that you'll be running this from a 3A 12v power source. If the source is current limited, and cannot deliver more than 3A, then as long as your wiring will handle 3A, so at least 0.3 \$mm^2\$, then there's no need for a fuse.

If you substitute a car battery for your PSU, then you should fit a fuse.

\$\endgroup\$
  • \$\begingroup\$ It is limited. Basically a mains 12v adaptor @ 3amps \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 14:31
0
\$\begingroup\$

I think a more clever design for lighting on this is to replace the wires with optical fibre and choose emitters for each level or zone you wish to control intensity of. Choose the aperture size of fibre to suit your scale.

This requires using optically clear epoxy ( like that used in 5mm LEDs) to create an adapter or manifold to capture the light from a large area LED chip and neatly routing the optics. I would choose the warmest LED to simulate tungsten bulbs.

Designing it so the LED is replaceable requires mechanical-optical skills than you can hone from web-searching. I expect about 10 sq.cm./watt heatsink if it has convection flow and a few watts is then all you need if your losses are minimized. Voltage options are wide ranging on large area chips, so match Voltage to power source and use an active current limiter for dimming.

Glass fibre is better but plastic is cheaper. For the easiest solution , get a fibre-optic lamp and change the LEDs.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Thanks. The white LEDs shown in the centre of layer 3 and all of the white ones in layers 4 and 5 will all be fibre optic drivers. I couldn't possibly have done this with just LEDs alone, the count would have been near a 1000, and would have to be absolutely tiny ones. A lot of the fibre optics I will be using are 0.25mm, with the rest made up of 0.75mm - 2.00mm. They will be used to simulate the balcony cabin porch lamps, pool lighting, flood lights for the name on the side of the smoke stack, and so on. \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 15:14
  • \$\begingroup\$ Do you have a schematic and BOM? I have written many times how to compute series R for arrays based on p/n and Pd, Vf of each LED ( more accurate than your wizard) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '17 at 16:50
  • \$\begingroup\$ Say what now? Perhaps you missed the part of my post where I said "I'd appreciate any input, but please keep in mind my knowledge is absolute minimum, especially where terminology is concerned, thanks." LOL I understood R (resistor) and Vf (forward voltage) but the rest...not a clue lol \$\endgroup\$ – Dominic James Sibthorp Nov 19 '17 at 17:04
  • \$\begingroup\$ @DominicJamesSibthorp BOM = bill of materials (parts list), p/n = part number, Pd = power dissipation. \$\endgroup\$ – Tom Carpenter Nov 19 '17 at 17:07
  • 1
    \$\begingroup\$ Photos look good, but I didn't zoom (arg) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '17 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.