LEDs birghtness

Hi, it's stupid questions time=)

I didn't upload any program to the Arduino Uno board. It's used only as a power source

I have two schemas, they have difference only in LEDs count
First: 5V pin => 220 OHM resistor => push button => LED => ground pin
Second: 5V pin => 220 OHM resistor => push button => 1 LED => 2 LED => ground pin

In the second schema LEDs' bright less intensive than LED bright in the first.
Both schemas have the same voltage and resistance, so the current should be also the same.
Why than LEDs' brightness is less in the second schema? Does it depend not only on current?

Your post was MOVED to its current location as it is more suitable.

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If you want the same brightness, you have to change the value of the resistor. It depends if the LEDs are in series or in parallel

For example, supposing a voltage source of 5V, and the LED(s) forward voltage is 2V and you want to limit the LED(s) current to 10mA each :

  • With one LED :

    • the voltage drop in the resistor is : 5V - 2V = 3V
    • the resistor must be : 3V / 0.01A = 300 ohms
  • With two LEDs in series:

    • the voltage drop in the resistor is : 5V - (2 * 2V) = 1V
    • the resistor must be : 1V / 0.01A = 100 ohms
  • With two LEDs in parallel:

    • the voltage drop in the resistor is : 5V - 2V = 3V
    • the resistor must be : 3V / (2 * 0.01A) = 150 ohms

@al-rd, your topic has been moved to a more suitable location on the forum (again); I suggest that you leave it there.

The introductory tutorials section is for tutorials, not for questions.

You mean posters can actually move their posts? :astonished:

Yes, one can change title and location with the black pencil next to the title.

And you can /should be able to view the edit history of a post by clicking on the orange pencil above a post (not the topic).

This is going to be interesting ... :roll_eyes:

LEDs have internal resistance, or did you not measure it. Also called "voltage drop".

I'm sorry to be an argumentative nit-picker, but for a newbie it's better not to talk about internal resistance for an LED because it changes over orders of magnitude with the voltage applied. It's a much better mental model to just think of it as a fixed voltage drop, which is broadly independent of current.*

*I realise it's not fully independent, but it's a good model to start out with.

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OP: Yes, the brightness does depend upon the current and the current did go down when you added added a second led in series.

You should try three led’s in series - this may add to your new found understanding of the forward voltage drop (Vf) rating of led’s.

Perhaps it’s time to invest $5 in a basic DMM (digital multi-meter).

Incorrect, the voltage is definitely different. Having things in series means the voltage is split between them.

In the first case: 5V = IR + Vled - therefore Vled = 5-IR
in the second case: 5V = IR + Vled + Vled - therefore Vled = 0.5(5-IR)

Clearly the voltage seen by the leds is different in the second case, and
the current will be different too as a consequence (much less as less voltage
available to the resistor).

Lets assume red LEDs with approx 1.4V forward voltage - first case about 3.6V across resistor, so I = 16mA. Second case 2.2V for R, so I = 10mA.

In reality the second case LED forward voltage is likely to be more like 1.35V
than 1.4V as the current is lower - LED voltage does depend on current, but