Beautiful place, Mike, an enviable little lake.
I love your dock and I can finally see it in the water.

Thanks for sharing, bravo hermanK.

I like how you're showcasing your work.
I think it's very important and helpful to detail every step of the process.
Well done, Chugalone.

I wanted to add two more tips for those who intend to build a circuit themselves and use resistors.

MEASURING RESISTANCE

Resistance measurement (all multimeters, even low-cost ones, have this feature) should be performed on a de-energized circuit.
Be careful if the resistor is not isolated but connected to other components, the measurement may be distorted.
For example, if it is connected in parallel with other resistors, you will have the value in Ohms of the equivalent resistance of the resistors in parallel and not of the individual resistor.

DISPOSITION OF POWER INVOLVED

For small circuits with low currents, this information may be useless, but it's good to know.
When purchasing a resistor, only the resistance value in Ohms is considered.
In reality, we must also check its power, or rather, its ability to dissipate heat in Watts without breaking.

A small resistor, like those used in electronics, is typically 250 mW (1/4 Watt).
Well, to know if it's suitable for our use and won't get damaged, just do a little calculation.
In direct current, according to Ohm's third law, electrical power is given by the product of voltage and current.
P = V x I
For example, if a resistor is subjected to a voltage of 12 volts and a current of 40 mA, it must support a power of at least 0.48 W.
480 mW is greater than 250 mW, so one of the resistors mentioned above cannot be used. The resistor was 300 Ohms, of course.
So, in summary, the advice is: always calculate the power involved based on the voltage and current.

Hi EdW, as I was saying to Dave, everyone has their own needs, interests, and hobbies. I agree with you.
I agree with you about the difficulties of learning Arduino (which requires knowledge of computer science, not just electronics) and building a DC/DC yourself.
I agree that building electronic products, now sold cheaply on Aliexpress, is uneconomical.
Furthermore, people often want to save time. I could have made the anchors myself because I knew how to do it, but I chose to buy pre-made ones.
Some time ago, I started sourcing components to build an ESC myself, just for fun, certainly not to achieve better performance than ESCs sold online. I stopped because I couldn't find everything and didn't have time to finish a time-consuming project. I don't rule out picking it up again in the future, along with many other things.
Personally, I don't need to adjust the brightness of the lights on a boat. Once I've found the right brightness (combined with the right current, thus choosing the right resistor), I don't think it's necessary to change them, but that's my personal opinion.
In any case, I could place a variable resistor in series with a fixed resistor for fine adjustments.
As for me, I respect everyone's opinions and choices, so if anyone needs information on basic electronics, to the extent I can, I'm available and happy to share my limited knowledge.

Hi Dave, in electronics, there are many ways to achieve the same result.
I didn't say using resistors is the best solution.
The advantages of a small LED circuit with resistors are these:
You can understand what you're doing, and once you've learned it, you can repeat it over and over again, applying it to various situations.
It's fun to learn and apply the basics of electronics.
In fact, JSS4 is doing it very well with Ronald, congratulations to both of you.
Other systems, like DC-DC Buck Step Down or Arduino, are much more difficult to DIY. Impossible for a novice to learn easily.
There are those who want to learn more, and those who are content to buy what they need and assemble it.
Basically, with discrete components (resistors, capacitors, transistors, etc., etc.), it's simply fun.
This is obviously not a criticism; everyone is free to do as they prefer, and from your point of view, you've certainly made the best choice in achieving the result you wanted.
Furthermore, resistors are very cheap.

Sorry Ron, I'll reply to your PM as soon as possible.

Hi Ron, I counted at least eight lights, great challenge. I'll be rooting for you, of course.


I will reply to your PM as soon as possible.

Thanks for the reply Ron, I was really curious. Will you use this power supply (5.88 volts) for the LEDs or directly from the battery (7.2 volts)?

Great work JSS4, the photo with the boat illuminated at night is very suggestive.

That's fine, Ron.
Since there are only two LEDs, there's no chance of confusion.
You need to tidy up when you have a lot of LEDs to manage.
What were you measuring that gives you 5.88 volts?

Hi Ron, as I was saying, I've added the last three sketches to post number 9 of this topic.

I did the same for the voltage and current measurement sketch.
I preferred to put them all where the main explanation was given.

Anyway, I'm attaching them in this post as well.

The first one is about voltage and current measurements (later I'd like to give you some helpful tips on resistance measurements, which any multimeter can perform). The second, third, and fourth ones show an example of moving from a circuit diagram to practical wiring.
Generally, the connection logic is on a board (as a beginner, it's best to buy a punch board because it's very easy to use). From there, each LED will be connected to two wires, a positive and a negative, which will run through the boat as best as possible.
Some people, in fact, prefer to proceed as follows:
1. Position the LEDs and wiring.
2. Prepare the board with the resistors, connected to the battery with a manual or remote-controlled switch (or, better yet, both).
3. Solder the LED wires in the correct positions on the board.

In order to facilitate research and understanding of a topic that is widely discussed in dynamic naval models, i.e. LED lighting, I am inserting the following links from this same forum.

forum/135645

forum/138041

forum/162486#162586

Hi Ron, I've added the last three sketches to post number 9 of this topic.

After addressing the issue of voltage and current measurements, there's another aspect to consider:
Transitioning from an electrical diagram to a real circuit with actual connections and wiring isn't easy for all beginners.

So I've considered a practical example.
Let's say a modeler wants to position the regulation position lights on a boat over 50 meters (according to COLREG).
I've included both the electrical diagram and the possible practical connection.
In practice, there are many possible approaches; the one illustrated is one of many.

Fantastico!

Oh, okay, this connection is very nice.

I added the seventh diagram in message #9. It's for current and voltage measurements.

I hope it's clear, but if you're not sure about something, just ask.

Ron, you're lucky to have a guide like JSS4.
I agree with him on resistors and LEDs.

If I understand correctly, you live nearby and can meet in person. Is that right?
Well, that's the best thing; explaining things remotely isn't always easy.


P.S. Sorry I haven't posted all the schematics yet, but I'll do it soon (it's one thing to write on your cell phone while standing, but it's another to find a surface to draw on. So far, I've had a few opportunities).

Hi Edw, you're absolutely right, but with an LED driver, beginners and those just starting out with electronics wouldn't understand much.
They'd just buy the pre-built circuit and connect it.
With basic circuits, you learn to do something yourself by thinking.
You're an electronics engineer and don't need to learn things you've known for a long time and very well, but imagine that many modelers buy kits and circuits (for lighting, sound, or other things) without fully understanding how they work.
Of course, you're right about efficiency and speed (and that's not a criticism); buying an LED driver saves you time doing calculations and soldering.
For those who've never done them before, it can be a lot of fun and satisfying.

Ron, if at 76 you want to learn some basic electronics, you have all my admiration, you're great.

Hi Chugalone, thanks for the compliments.

I think that to understand electronics, you have to start with the simple things.
There were many more things to say, but I risked confusing Ron.
Maybe I'll add them later.
By applying Ohm's law, you can manage LEDs well. Mathematically, they are very simple operations.

JSS4 mentioned MOSFETs and Arduino, but there things get complicated; it's better to take it one step at a time. It's clear that he has in-depth knowledge of electronics.
For those starting from scratch, it's a bit too difficult.

I recommend proceeding as follows:
1. Determine how many LEDs to use, what type, and how to power them.
2. Then draw a small circuit with resistors and LEDs (choosing the best solution).
3. Reproduce it on a breadboard for verification and possible measurements and tests.
4. Apply it to the model by running the wires and positioning the resistors (even all on one board).

If you need clarification, advice, or help, I'd be happy to help.

Now I still have to upload the other schematics I promised and check for errors. But in the meantime, you can view the videos and photos I took previously on other topics.

With simple, discrete components (small space and minimal cost), you can make a circuit with flashing LEDs instead of solid ones, if you like.

Hi Ron, sorry for the delay, I'll post the text first and then try to attach the diagrams (I'll correct any errors as they arise).
Very good, let's talk about a specific case and make practical assumptions.

With a voltage of 7.2 volts, you can choose from several solutions:

1. PARALLEL CONNECTION (ONE RESISTOR FOR EACH LED)
With one resistor for each LED (and then all the LED-resistor branches connected in parallel to the battery), the calculation is easy (see diagram no. 1).
To find the correct resistance value, simply apply the inverse formula of Ohm's law, and then:
R=V/I
The unknowns are the current intensity and the voltage.

Regarding the current, the value is given by the LED's factory data.
If I remember correctly (but I recommend always checking the datasheet for each component you use, it's important), the values ​​for 3 mm electronic bulb LEDs range from 15 mA to 20 mA (i.e., 0.015 A and 0.018 A). With 15 mA, you'll get a dimmer light but less heat loss and a longer LED lifespan; vice versa, with 20 mA.

The value I'll consider for the following calculations is 18 mA (0.018 A).

Regarding voltage, the value to consider won't be the battery voltage, because the LED voltage drop must be taken into account. So, battery voltage minus LED voltage drop.
Again, the LED voltage drop is a factory setting.
For these types of LEDs, the typical voltage drops are: 2 volts for red, 2.2 volts for yellow and green, and 3 volts for blue. Check the datasheet to be sure before proceeding.

With all the data available, you can apply Ohm's law:
R = (Vbatt - Vled) / I
With a red LED, we have:
R = (7.2 - 2) / 0.018 = 5.2 / 0.018 = 289 Ohms
At this point, you'll need to find the most suitable commercial resistor for this resistance value.

1.1 CHOOSING THE MOST SUITABLE COMMERCIAL RESISTOR FOR THE THEORETICAL VALUE
With the most common commercial series, you have two available values: 270 Ohm and 300 Ohm.
If you choose the 270 Ohm commercial resistor, you will have higher current (19 mA or 0.019 A) almost at the limit, and greater brightness.
If you choose the 300 Ohm commercial resistor, you will have lower current (17 mA or 0.017 A) almost at the limit, and greater brightness.
There are precision series that are much closer to the theoretical value, but I don't think there's any need; you can easily choose between them.

2. PARALLEL CONNECTION (EACH BRANCH HAS TWO LEDS IN SERIES WITH A RESISTOR)
(See diagram number 2).

In this case, the same calculations apply, but since there are two LEDs in series, the voltage drop to be considered will be the sum of the two, so 2 + 2 = 4 Volts.
R = (Vbatt - Vled1 - Vled2) / I
With a red LED, we have:
R = (7.2 - 4) / 0.018 = 3.2 / 0.018 = 178 Ohms
At this point, you'll need to find the most suitable commercial resistor for this resistance value, which, luckily, is 178 Ohms.

3. PARALLEL CONNECTION (EACH BRANCH HAS THREE LEDS IN SERIES WITH A RESISTOR)
(SEE diagram number 3).

In this case, the same calculations apply, but since there are three LEDs in series, the voltage drop to be considered will be the sum of the three, so 2 + 2 + 2 = 6 Volts.
R = (Vbatt-Vled1 –Vled2 – Vled3)/I
With a red LED, we have:
R = (7.2 -6) / 0.018 = 1.2/0.018 = 67 Ohms
At this point, you'll need to find the most suitable commercial resistor for this resistance value, which is 68 Ohms.

For obvious reasons, we can't connect four LEDs in series because the sum of their voltage drops is greater (8 Volts) than the supply voltage (7.2).


4. SUMMARY OF RESISTORS TO BE USED

With a single LED per resistor, use a 270 Ohm resistor.
With two LEDs per resistor, use a 178 Ohm resistor.
With three LEDs per resistor, use a 68 Ohm resistor.

There is a slight difference between red and green LEDs.

5. CONSIDERATIONS ON TOTAL CURRENT DRAWN BY THE CIRCUIT

For simplicity's sake, let's assume you want to power twelve LEDs with a 7.2 Volt battery (I'm using twelve LEDs for example only).

If you connect them all in parallel, you'll have a total current draw of 216 mA or 0.216 A (i.e., 0.18 mA x 12).
(See diagram 4.)

If you connect them in groups of two in parallel, you'll have a total current draw of 108 mA or 0.108 A (i.e., 0.18 mA x 6, because there are six branches in parallel).
(See diagram 5.)

If you connect them in groups of four in parallel, you'll have a total current draw of 72 mA or 0.072 A (i.e., 0.18 mA x 4, because there are four branches in parallel).
(See diagram 6.)

When you connect an LED in series with another, one of the two that breaks will prevent the current from flowing to the other, which will therefore appear off even if it's healthy.

Hi Ron, I think your friend JSS4 is right. The best approach is to know how to use resistors (individual components) to operate and understand how an LED circuit works.
As soon as I can, I'll send you some practical instructions applied to the specific case of a 7.2 volt power supply I'm preparing (which you can show to JSS4).

As of today, how many of the participants in today's question are still unaware of the existence and function of a bulbous bow?
AI is keen for this number to approach zero, hahaha.
It wants everyone to understand the concept.

We'd already talked about it, not in the AI ​​question of the day, but when we asked the questions of the day.
The question posed was the difference between a great circle [ortodromia] and a rhumb line [lossodromia].
Then I think the AI ​​also touched on the topic.
However, it's true that a route like the one described in today's question, on a map using the Mercator projection, might appear to be the longest route.

Beautiful pictures Ron!

Guys, I chose the Proxxon, thanks to everyone who gave me advice and information.

Hi Ron, great job. The new recruit gave you great advice.
If you'd like an overview of how to control LEDs with a resistor, adjust the current, and thus the correct brightness, you can see these posts from older threads.
However, remember: when choosing a (correct) parallel connection, as in your case, the currents are added for each LED you add (LED-resistor branch), so if you add additional LEDs, be careful about the total current in the circuit. The batteries (or a BEC or other type of power supply) must be able to supply it adequately. Logically, in your case, with a current draw of 18-20 mA per LED, and therefore a total of 40 mA, you shouldn't have any problems.

forum/135645

forum/138041

So glad I could help, Duncan!

Hi Mike, laser cutting is an amazing new tool. Kudos to you for using CAD so well.

Hi Robbob, congratulations on the great ideas and truly original additions to the list of useful tools and utensils.
A true professional.