Led bulb / lights
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
4 days ago by 🇮🇹
AlessandroSPQR (
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Captain)
RC & Electrics