I'm wiring in a bunch of 3V LED's to use as lighting on my model train layout and as I've normally done, I have a 12V wire bus running under the layout and just tap into it (essentially wiring them as parallel) with each 3V LED which is wired to an appropriate resistor (600-750 Ohm). I've never had any problems doing this before. I found this diagram from one of the sellers of the LEDs and they make it sound as though you must use shunt resistors up to an amount equal to draw enough current to equal that provided by the 12V power supply. What's the purpose of doing that and why is it necessary? Since each LED already has it's own resistor to reduce the voltage and current to it I don't understand why this would need to be done. I'm not doing this now and it doesn't seem to be a problem? Should I install these shunt resistors or not worry about it?

(In before the imitation crabmeat jokes).

Their explanation:
3) Approx. 50 pcs 3V LED made Lights, the lights must be prewired a 620 ohm resistor in series. You may connect 620 ohm 1/4W resistors in parallel to share the exceed current if you use less than 50 lights. for example, you must use 20 pcs resistors if you have 30 lights only.

Their diagram:

Attached File

Short answer, they're wrong.

current is drawn by a load, not pushed by a supply.

That's a very simplistic explanation, but it should suffice for this purpose.

Utter garbage.

Holy crap!
That explanation in the OP is just plain wrong!
First post got it right.

Yea, just do the resistor in series with each LED.

Let's hope that person also isn't a plumber, otherwise someone's water bill is gonna be HUUUUUGE.

ETA:  "Gotta have an open pipe of the right diameter at the end to drain off the excess water."

Treat each LED individually.

This page may help: https://en.m.wikipedia.org/wiki/LED_circuit

Yeahhh, that thing is the text book of a electrical “vampire”, it will draw more current than necessary to run the circuit.

When circuits are run in parallel with identical resistance on each leg, each leg of the circuit will draw the same current, by putting a resistor on the end of the circuit you create another path that doesn’t need to be there and is serving no real purpose other than drawing more current and wasting energy. If that resistor on the end has less resistance than the other parallels then it will reduce voltage output at each other path.

Their explanation is wrong, with a possible explanation of why they do it wrong.

Some kinds of (mostly cheap) power supplies need a constant minimal current draw in order to function correctly. Or they may actually be using an unregulated power supply where the voltage is constant only if they keep a constant current draw. In your case that does not sound like an issue.

It sounds like either a constant current supply, or just no or very poor regulation.

Is the vendor describing connection to a “12V” power supply they are providing, or just the LEDs? What does the output voltage measure with no LEDs connected?

Glad to know I'm not the only one scratching my head over that diagram & explanation. So in other words, ignore the shunt resistor nonsense and just keep doing what I have been doing, adding LED's as my layout progresses up to the point that the total amount doesn't exceed the rating of the power supply?

This is the one that has the explanation that I posted Transformer

I'm currently using a 12V 1A switching Power Supply that I believe came off of a cable TV box or something. I plan to swap that out with a 2.5A switching PS. Right now, the total current draw off of the PS is about 450ma when I last measured it.

Correct.

Thank you.

And I will second the "correct"!

Unregukated supply would be my guess. The voltage swings around with the current, so they may be expecting a consitant load on the supply. We used to have contant current chargers we used on the railroad. They were a big PITA. Constant voltage chargers are much better!

First reply is right.  The shunt resistor is just a waste of power.  You don't need it.

Now, what I'd do, because I'm that kind of obsessive-compulsive freak, is I'd put a 1K variable resistor in series with each LED and trim each LED to be the exact intensity I want it to be.

REEEEEEEEE made my head hurt!

You'd still need a fixed resistor in series with the trimmer to protect the LED when the trimmer gets too low.\

ETA, I'd probably go for a 4.7k ohm trimmer also. The LED's may still be fairly bright at 1k ohm.

That writing looks like poorly translated Chinese to me.

It's an unregulated power supply circuit.

In the case shown, it delivers ~12V at a load current of 1000mA (1A). At lighter loads the voltage will rise, maybe up to 20V or more, and the LED's would be overdriven. Too bright or smoke.

Very, very common in cheap Chinese stuff.

A shunt resistor is unecessary with a regulated power supply.

Yet. This answer is completely accurate..

OP, just don't let the overall current drain by the load exceed the capacity of the power supply, and you'll be good to go.

Also, the resistors in series with the KED are comminly referred ti as "dropping resistors". The diode limits the current on each branch oh the circuit, the resistor is of a value calculated  so that at the given current, turns any more voltage that the supply is delivering to the circuit than what the LED can handke into heat..

Without the dropping resistors, LEDs can be real bright, but only briefly.  :(

Where are the pictures, OP?

The PS is "capable" of supplying a certain load, that doesn't mean it HAS to be max loaded.

That's what I was thinking but linear circuits was 35 years ago so some things have faded.

Here's a few to show what I'm trying to achieve. Ignore all the junk and ammo boxes, it's still a work in progress.

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Wow!  The answers and discussion in this thread are almost frightening!

This makes no sense. There is no reason to have an extra resistor using up the full capacity of a power supply.

The rule of thumb I've used for years is to subtract the forward voltage drop of the diode from the power supply voltage and then calculate what size resistor will provide the rated current to the diode.  For example a "12-volt" power supply providing 14.4V to a 20mA blue diode with a 2.7V drop would be

V=IR
R=V/I=(14.4-2.7)/0.020=267 Ohm

A 300 Ohn resistor with +/- 10% tolorance would never exceede 20mA.

Mike

PS. If this is an unregulated supply get a better supply.

PPS.  Hope you have a lot of LEDs you're better off putting them in series with a current regulator circuit.

OP,  you need to buy a regulated supply for this project. They are a little expensive, but provide clean, voltage regulated current.

Another option is to buy a few cheap parts and work with what you'be already got. For instance, you could buy a voltage regulator in the TO-220 package, say a LM7809, which is a nine volt linear regulator. You would power that from the power supply you have and it would provide 9 volts output to your LED's. All you would have to do is change the values of the resistors. You would need to heatsink the LM7809, but that is cheap and easy to do.

As for mirrormirror's idea about the trimmer pots, that is a good way to "personalize" each light and make it burn with whatever intensity you like. For probably around $10, you could build a nice little set up. You would need the 7809, a heatsink, a couple of capacitors for smoothing and filtering the 7809, some cheap trimpots, some fixed value resistors a little solder, some wire, a piece of proto (perf) board and of course your LED's. Then you would have a decent set up that you can easily customize and you won't need to worry about any shunt resistors! It could easily be built in a couple of hours. If you are interested in going this route, let me know and I will be glad to help you get the parts together.

ETA, are you using the supply shown in the OP, or something else?

I'm not using the power supply shown, I'm using another one like that, but it says it's a switching power supply. I plan to switch that to a larger 2.5A, what should be a switching supply, which looks like one of the chargers that's typical of what is used for laptop computers. It says "Comcast" on it and was used for some sort of their cable boxes. I'm guessing that would probably be regulated.
I have several of these that I use to step down to 5V to control some servos, Would something like this, since it's adjustable, be adequate to use to provide a filtered supply? Is there something pre made like this available that could be used to filter my existing supply? Is there any way to test if my current power supplies are filtered by using a multimeter (or some other way), or is a switching power supply considered to be filtered?

power supply converter

Use your multimeter to look at the voltage of any supply, unloaded.

Any regulated supply will be within +/- 10% (usually better) of the specified output voltage.

Any unregulated supply, like a super cheap chicom wall wart, will be higher than the spec output, which is specified as X volts AT Y current. That's why the original schematic showed an additonal load (the extra resistor not feeding one of the LED's)

You can always measure ripple by switching your multimeter to AC and looking at the output of any power supply.

Caveat: switching supplies may have an AC ripple component that is beyond the bandwidth of your meter, especially if it's like one of the Harbor Freight freebies.

Switching supplies are regulated.

The cheapest chicom wall wart unregulated supplies may have little more than a transformer and diodes (half or full wave rectifier configuration). Most have some filter capacitor.

Since many people in this thread do not understand why unregulated power supply voltage changes, and why one would need an additional resistor to load the supply to bring the voltage down, here is a tutorial.

.

Neat

Is it wrong that I imagine Godzilla stomping all of that into bits?

Merrel"s post is pretty good for determinig what you have. How do adjust the voltage on it? As for the filtering caps, if it is a regulated supply, don't worry about it, I was just suggesting them if you went the add-on regulator route. The trimpots will still work if you want to customize the brightness of each LED.

Depends some power supplies may have voltage regulation issues at very low currents, so the extra loads are there to ensure that the power supply is providing a minimum current that provides appropriate voltage regulation.

Mike

Shunt resistor is to get the voltage across the LED in the appropriate range. Do the math to do that, size your resistors appropriately. This full power things is bullshit.

Uh, no. The shunt resistor is for voltage regulation by putting the proper load on the circuit to balance the output from the unregulated supply for what you need. The LED resistors are there to limit the current on the LEDs.

The worst thing about the light kit the OP shows is that you can't customize it. Say you get your LED's hooked up and your shunt resistor dialed in right for the circuit. You have to burn all the LED's all the time. You want to shut five of them off? @ 20mA a piece, you have now thrown your circuit off by 100mA, or 10% of what the supply is rated for. Want to turn more off, or do some dimming? Now you have thrown everything out of balance even more. Who knows what the voltage output is now? It's probably more than the current limiting resistors were figured for. Will you burn out your LED's? Probably not, but could dramatically shorten their lifespan.

Yep. And the problem with the unregulated supply / load resistor is adding or subtracting more LED's will affect the PS voltage, and thus brightness of all the LED's.

Additionally, an unregulated PS will change its output depending on changes in AC line voltage. It's not a consistent 110-120V. Anyone.who has watched regular room lights dim when an AC or fridge compressor has seen this.

A regulated supply eliminates this. Also, if using say 12V, one can put two or three LED's in series for each tap off the 12V supply. It's important to use matched LED's doing this, usually from the same lot is sufficient, otherwise brightness can and will vary.

If trying to recreate the "look" of an incandescent light with white LED's, select ones with a color temperature of ~3000°K ("warm" LED color can cover 3500°K down to 2700 and lower). 3000 is closest, and if you don't mind soldering small surface mount LED's, get selected ones from Mouser or Digikey with a color temperature of 3000°K and a CRI (color rendering index) of 80 or better. These will look very close to incandescent. CREE is very reliabls and outputs are consistent. Ebay ones are all over the map.

Use this search

For example, this CREE 3000K 95CRI LED 941-JE2835AWTA0ZF730 is inexpensive, very efficient and highly accurate for simulating an incandescent light. 10 cents each in singles, well under a nickel in quantities of 100 and up.

Data sheet - run this LED at a couple of mA and it will be plenty bright. It can run much higher, but that would be way too much for an HO structure. Another benefit is at 5 or 10 mA, the voltage drop will be on the order of 2.5 to 2.6V, so you could easily run 3 in series with one current limiting resistor from a regulated 12V supply. These are very efficient and also have a wide distribution lattern of light. Use a small (10W is plenty) soldering iron with a fine tip. Use small gauge (28-30 is fine) wire. Tin both and touch them together and hit very briefly with the iron. At a nickel each or less, if you melt a few practicing it's no big deal.

Have fun!

I was thinking of hooking up in series. These are some of the lights I'm using. They have the resistors already on them and are made for 12V (but they are 1K Ohm which does make them too dim). I'm also using some others that don't have the resistors on them. Since they are 2 LEDs to a lamppost, I was thinking of cutting off the resistors and hooking them up in series, and them hooking them up in series to another lamppost configured the same way. Then I would hook up the pair of lampposts (4 LED's in total) to the 12V. There should be no problem doing this right? And I would actually be saving on current used since there is none being wasted through the resistors, is that correct?

Lampposts

Thanks for the tutorial link. Interesting. I may play around with different Ohm loads like in that one chart and see what voltage & current reading I get.

Just don't use a piece of shit power supply and life is fine and you don't have to worry about this kind of silliness.

OP, if you 're willing to do just a little soldering look up "constant current source with two transistors".

This super-simple circuit will keep your LED at a rock steady brightness over a very wide range of voltages.

LED's with integrated resistors are kind of limited. If you had one which was designed for 12V, with an integrated 1kO resistor, the lamp would be driven at about 9mA (12V-3V drop over the LED ~=9V.

9V/1kO = 9mA.

If you put two of the "12V LED's" in series, you'd have 12V - 3V (for LED1) - 3V (for LED2) = 6V. Since both have 1kO resistors, the current through the LED string in series would be 6V/2kO, or 3mA.

Once you play around a bit with loose (no integrated resistor) LED's, you won't look back. The integrated ones are much more expensive and have inherent limitations.

By all means, experiment. Use an old variable trainset DC power supply like an MRC or even Tyco. You can adjust the output, and thus brightness you want. Measure that voltage, and/or current, and you'll know where to set the current for hooking up to a layout-wide system supply.

You can always pick up cheap LED's on ebay to practice with, often for pennies apiece. The reason I noted CREE LED's from Mouser or Digikey is the reliability is much much higher. You can incorporate them into structures being comparatively sure they won't burn out.

See also https://www.ngineering.com/ho_lighting_kits.htm & https://www.ngineering.com/lightng.htm for some ideas and also a taste of the markup vs buying components direct from a distributor.

ETA: if the integrated resistor is a discrete component soldered to the leg of the LED, then yes, you can cut them off and treat it as a normal LED. Watch overheating. The junctions in LED's are tiny. Forceps make a handy heatsinking clamp to keep the hot solder joint away from the sensitive LED chip.