Good morning to all the naval modelers of this forum.

One of the recurring questions is this: is the resistor in series with the LED necessary or not?
The other is this: is it better to connect the LEDs in series or in parallel?

I forgot to complete the answer.
I link to the fourth message of this topic and the following ones.



EXPLANATION OF THE VIDEOS

In the videos you can see a parallel circuit powered by a 9 volt battery.
As you can see there is a resistor in series with each LED. In this way there will be the right voltage on the LED to turn it on.
A variable resistor has also been added to the third LED to show that too high a resistance will cause too little current to flow and the LED will be too dim (not very bright) or will not light up.
I avoided showing how too low a resistance burns the LED.



WHEN TO APPLY THE RESISTANCE AND WHAT VALUE

Powering a LED of this type with 9 volts means burning it.
If I don't want to put a resistor I have to power it with the correct voltage which, in this case, is 2 volts.
Two volts corresponds to the voltage drop that the LED causes in the circuit.

When powering a circuit with a LED and a resistor with a 9 volt battery, I have to understand which resistor to put so that the right current passes through the circuit and so that there is a voltage of 2 volts at the ends of the LED.
So if the voltage is 9 volts and there must be 2 volts on the LED, it is logical that I must have 7 volts at the ends of the resistor.
By applying Ohm's law I find the value of the resistance, knowing what the operating current is (for example 18 mA).
R = (Vcc-Vled)/I = (9-2)/0.018 = 7/0.018 = 389 Ohms.

By measuring the voltage at the ends of the LED with a voltmeter we will in fact find a value of around 2 volts and at the ends of the resistor around 7 volts, with a logically closed circuit (look at the seventh attached image).

With lower voltage batteries we will have to use resistors with lower ohmic values until we reach the point where no resistance is needed.
In this case, powering at 2 volts we will not need any resistance.
With higher voltage batteries we will have to use resistors with higher ohmic values.
Attention, when increasing the voltage, the electrical power involved must be taken into consideration, as common 250 mW (like the ones in the photo) resistors may no longer be suitable.

If we power an LED with a 6 volt battery, the resistor must be 222 Ohm.
But if we put three LEDs in series we won't need any resistors, because the sum of the voltage drops (2 2 2) of each LED corresponds precisely to the 6 volts of power (see the fourth attached image).
With a 6 volt battery and only two LEDs in series we will need a 111 Ohm resistor.


BETTER SERIES OR PARALLEL?

None is better than the other, there are pros and cons to evaluate.

The advantage of parallel connection is that if one LED breaks, the others will continue to work normally.
Instead, in series connection, if one LED breaks, all the others will turn off.

The advantage of series connection is that less current flows (less battery consumption and less power absorbed) compared to parallel. In parallel connection, for each LED added to the circuit, the current of that branch will be added to the total current of the circuit.
For example, ten LEDs in series will absorb 18 mA while ten LEDs in parallel will absorb 180 mA.

Attention, so far we have talked about signaling LEDs for electronics.
The situation changes for high-brightness LEDs.
Look at the eighth photo, these LEDs have a voltage drop of 3 volts (therefore 3 volts of direct power supply) and an operating current of 300 mA.
Look at the ninth photo, these LEDs have a voltage drop of 3 volts (therefore 3 volts of direct power supply) and an operating current of 700 mA.

In this case ten LEDs of this type (700 mA) in parallel will absorb 7 Amperes.

Furthermore, even with a single high brightness LED (3 volt; 300 mA) powered by a 9 volt battery, the power involved will be too high for the small resistors shown in the photo.
Don't use these resistors because you will definitely burn them.
Either use resistors with adequate powers (but you will waste a lot of energy dispersed in heat, due to the Joule effect) or you will have to power the LEDs without a resistor and therefore with exact power supplies (voltage regulators where or 3 volt battery packs).

I hope this clears up any doubts but any questions will not be unwelcome.

I also include the YouTube link for those who are unable to open the video file directly.


https://studio.youtube.com/video/xLmrpiLY2hM/edit

Ciao Hello Hola Bombero.

I'm going to see your latest construction: Finnmarken blog.

just to let you know my latest build (Finnmarken blog) has in excess of 100 lights ( i am experimenting with the wiring) and at present they are in parallel with a resistor on each deck(5 plus running lights) and are working OK. I would not run in series because if 1 fails I would have a major problem finding the faulty light!

Just my 2 cents worth but if you wire bulbs in series if one of the bulbs goes out then the entire string of bulbs goes out and it is a real pain to find the offending bulb and then replace it. Just my humble opinion. Len

You're welcome PhilipH (not to be confused with PhilH), "head under your feet", as that famous Italian actor said, ahahahahahah.

Hi alessandrospqr
Or commander to his friends
Could you draw the a diagram for so I can see it to be clear. This site is great .
I love it all of you who have helped give yourself a clap 👏.
I can just picture people all round the world clapping 🤣🤣🤣
Thanks all
Phil uk

Hi PhilH

In this way you have made the connection in parallel (first and second image of the fourth message of this topic).

The three LEDs are powered at the same voltage, the currents add up but only three LEDs are not a problem.

If powered at their direct voltage they do not require resistors.
Logically you can manage them with any DC-DC circuit instead of a simple resistor (if you have space and want to spend more). Alternatively, you can make a manual voltage regulator yourself with the LM317.

However you solved the problem and that's what matters.
Often the right commercial advice is more effective than doing it yourself.

Hi all thanks to everyone who has answered my question ❓ well I have finally managed to get the lights working on the mast I don't know what you call it but I joined all the reds on one side and all the black on the other side so 3 reds & 3 blacks on other sides, like I said I can't get my head around it ( dyslexia) 😔

A bit down the list now, and answered by AlessandroSPQR, but yes, if in series connect LED negative to next LED positive, all the way to the end.

Hi Phil
Re your less
I have used these a lot on models
I use 2 1.5v AA batteries that means you can use them without resistors job done

Regards
Dave

I use these lights for all my lighting needs and i don't have to worry about resistors and the leads are super thin so you can fit more wire through smaller areas.

then I add a adjustable voltage regulator
5Pcs LM2596S DC-DC LM2596 with LED Display Voltmeter Buck Converter Step-Down Regulator 4.0-40V to 1.25-37V DC 36V to 24V to 12V to 5V Power Supply Module Compatible with Car Motor Buck
EDGELEC 20pcs 0402 0603 0805 1206 Pre Wired SMD LED Light Emitting Diode Micro Litz Length 7.8in. Soldered White Red Green Yellow Blue Orange Pink UV (20pcs DC 3V 0402 White Prewired Mini Lights)



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Sorry, I always travel a little late.
When I finish reviewing the translation, which for me is the most difficult thing, I always realize that there are other messages that have already answered the question.
I haven't read it yet (because I should translate it) but I really think that all the answers are already in the attached link.
This time too I probably wrote in vain.

Wow Simplesailor e Mike, if you liked it it means you read my message.
Since I realized it was very long I figured no one would want to read it, ahahahahahahah.
Congratulations to you. Tenacious and patient.

Hi thanks for replying, I have seen that link before but like I say maybe it's my dyslexia or maybe as my sister always told me "your just thick" let's go right back to basics, battery+long leg positive so in on positive and out with neg next bulb is it positive in negative out then positive in negative out. Then resistor battery-
Phil uk👍

Good morning PhilipH

I will try to be of help to you.

I don't know what your basic knowledge is about electricity so I will try not to assume anything.
If I say things that you already know you will want to excuse me and please do not be offended, but if I have not been clear about some things just ask.

LEDs are not real light bulbs (like incandescent or neon ones) even if today they are completely replacing them.


GENERAL NOTIONS
The LED, as the English acronym says, is a diode that emits light.
L.E.D. (Light Emitting Diod).
In fact, the electrical symbol is the same as the diode but with two arrows (outwards) which signify the emission of light.

Like all LEDs it only conducts in one direction. If you get the polarity wrong it won't be damaged, it just won't emit light.

Many years ago LEDs were used in electronics for signaling, today in addition to this task they are used for lighting.
The difference is the amount of light they can emit and the current they absorb. A high brightness LED absorbs 100-300 mA and above.
I think you want to use classic LEDs for electronics, so I'll limit the discussion to these.

The signaling LEDs (white, yellow, green, red) with the classic bulb shape absorb approximately 18 mA.
You have to look at the datasheet but I can tell you that their range varies from a minimum of 15 mA to a maximum of 20 mA.

An LED with too much current flowing through it burns out (irreparable damage).

To turn on an LED you must make sure to power it at the correct voltage at its ends, so that the right current flows through it.
Each LED, in addition to the operating current, is characterized by its voltage drop, or we could say by the direct supply voltage, i.e. the voltage that can be supplied directly to the LED without the need for resistors.
Also in this case you need to read the datasheet, but typically the voltage drop of an LED is 2 or 3 volts depending on the type (yellow, green, red, white, etc.).

From now on, to give practical examples, I will consider 18 mA as the operating current and 2 volts as the voltage drop.
If during the tests you see that the LED is dimmer (not very bright) you can consider 20 mA as the operating current.

SERIES OR PARALLEL CIRCUITS

The LEDs can be connected in series or in parallel (or series-parallel logically).
From now on I will only do the examples with three LEDs (because this is your specific case).

The parallel connection is what you see in the first and second images.
The first image represents the simple electrical diagram, the second is how the wires can actually be connected (one of the possible ways)-
At the ends of each branch (consisting of a LED and a resistor in series with it) there is always the same voltage.
The current suitable for the LED circulates in each branch (we said 18 mA).
The total current, being a parallel, will be given by the sum of the three currents.
So in the specific case 18 18 18 = 54 mA.



The series connection is what you see in the third and fourth images.
The third image represents the simple electrical diagram, the fourth is how the wires can actually be connected (one of the possible ways).
The voltage drops of each LED add up while the total current is always the same (i.e. 18 mA in the specific case).

In parallel, if one LED breaks, the others continue to work.
In series, the breakage of one LED will lead to the opening of the circuit (the other two will also turn off)
In series, at the same voltage, there is less current absorption and less dispersion due to the Joule effect.

HOW TO SIZE THE OHMIC VALUES OF THE RESISTORS

PARALLEL CIRCUIT
Let's first look at the parallel circuit (the first image attached).
If you supply the right voltage (2 or 3 volts) you don't need to put any resistance.
Since the typical voltages are higher (4.7 – 5 – 6 – 7.2 – 9 and 12 volts) it is necessary to create a voltage drop, so that there is the right one at the ends of the LED.

I will give some examples on typical cases, that is: 6, 9 and 12 volts (the formulas are always the same).

Let's assume that:
the supply voltage is 6 volts,
the voltage drop of the LED is 2 volts,
the operating current must be 18 mA, i.e. 0.018 A.

It should be considered a single branch.
To obtain a 2 volt drop on the LED it means that 4 volts must drop on the resistor (6-2 =4).
To have a voltage drop of 4 volts, with a flow of 0.018 A, what resistance should we put?
Just apply Ohm's first law (V= R * I).
R=V/I
R= 4/0.018 = 222 Ohms
The resistance to have a current flow of 18 mA with these voltages is 222 Ohm.
We will choose the commercial resistor closest to this value which is a 220 Ohm resistor.


Let's assume that:
the supply voltage is 6 volts,
the voltage drop of the LED is 3 volts,
the operating current must be 18 mA, i.e. 0.018 A.

R = (6-3)/0.018 = 167 Ohm (commercial value 150 or 180 Ohm)


Let's assume that (fifth photo attached):
the supply voltage is 9 volts,
the voltage drop of the LED is 2 volts,
the operating current must be 18 mA, i.e. 0.018 A.
R = (9-2)/0.018 = 389 Ohm (commercial value 390 Ohm)

with a 3 volt drop on the LED
R = (9-3)/0.018 = 333 Ohm (commercial value 330 Ohm)


Let's assume that:
the supply voltage is 12 volts,
the voltage drop of the LED is 2 volts,
the operating current must be 18 mA, i.e. 0.018 A.
R = (12-2)/0.018 = 556 Ohm (commercial value 560 Ohm)

with LEDs powered directly at 3 volts
R = (12-3)/0.018 = 500 Ohm (commercial value 470 or 560 Ohm)



You understand that the more the voltage increases, the more there is a loss of power in heat, due to the Joule effect.

SERIES CIRCUIT

In the series circuit, if the voltage drop is 2 volts for each LED, by powering the circuit with a 6 volt battery it will not be necessary to add any resistance.

If we power the circuit with a 9 volt battery we will have to calculate the value of a resistor that has a voltage drop of 3 volts (9 - 6).
In this case it will be:
R= 3/0.018 = 167 Ohm (commercial value 150 or 180 Ohm)
(sixth image attached)

If we power the circuit with a 12 volt battery we will have to calculate the value of a resistor that has a voltage drop of 6 volts (12-6).
In this case it will be:
R= 6/0.018 = 333 Ohm (commercial value 330 Ohm)



WITH COMMON FACTOR
If you know how to measure voltage and current you can size resistors much more precisely.

To respect the polarity of the LED you can recognize the anode and the cathode inside, but it is simpler to do some tests (even with a tester) or observe the pins, the longest one is on the positive (if I remember correctly, otherwise it will be the opposite, ahahahahahaha)

A quick look at the powers.
If you use small quarter-watt resistors, do not exceed this value.
Let's look at the riskiest case:
12 volt with LED at 18 mA.
By Ohm's third law: P=V*I
P=12 * 0.018 = 0.216W
We are within the 250 mW limits of these small resistors, everything is ok, otherwise we would have had to use larger ones (capable of supporting greater powers).

Hi phil, component shop do a catalogue on led and do a very simple diagram on calculating ohms and resistors , hope that helps , ian

Here's a good article that may help, below.

Series seems to be the answer - all LEDs get the same current then. You would need a resistor in the circuit, which will vary depending on what voltage you are supplying. There are plenty of LED calculators available to guide - and the resistor is not too critical if you are unsure of your LED specs

And because you said you cant get them to work - and without knowing your experience are you aware that LEDs have to be wired the correct way round. The LED anode (generally the longer wire) is connected to the positive, then the cathode is connected to the next anode, and so on.

Good luck!
https://www.ledsupply.com/blog/wiring-leds-correctly-series-parallel-circuits-explained/.