# 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?

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Welcome,

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?

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

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This is going to be interesting ...

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

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 ledâ€™s in series - this may add to your new found understanding of the forward voltage drop (Vf) rating of ledâ€™s.

Perhaps itâ€™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
logarithmically.

Hi
Thanks for helping!

1. From your comment I can see the next formulas to calculate voltage:
1.1 In series: V = I * R + (LEDForwardVoltage * LEDsCount)
1.2 In parallel: V = R * I * LEDsCount - LEDForwardVoltage
Are they correct?
2. If first is yes. They both against Omn's low where just V = R * I. Is Omn's low isn't really practical?

Hello=)

I feel like I'm missing something((
We've Ohm's low, where V = I * R. I'm not sure how voltage drop goes here...
Does R depend on it and the actual formula is V = I * (R + (VolrageDrop * LEDsCount))?

Hi, thanks!

There's Ohm's low that say: V = I * R. I assume that V equals voltage of pin, R equals voltage of resistance. Obviously, that's wrong. How can I see dependency between LEDs count and current in Ohm's low?

Hi, thanks for so detailed reply !
Looks like there's a formula: V = I * R + (LEDForwardVoltage * LEDsCount). But Ohm's law doesn't have LEDForwardVoltage * LEDsCount part. How is it ?

LedForwardVoltage * NumberOfLeds still results in a voltage that you can plug into Ohm's law.
Typically when calculating the resistor you need for a led, you subtract the forward voltage of the led(s) from the supply voltage and divide the outcome by the current to get your resistor value.
So e.g. a red led with Vf=1.7V, a supply voltage of 5v and a desired current of 15mA (0.015A) gives (5-1.7)/0.015 = 220 Ohm. Conveniently that's a standard resistor value.

Ohm's law applies ONLY to a single resistor and allows you to calculate the current through it if you know the voltage across it.

For more complicated circuits you need different formulas.

Kirchhoffs Voltage Law or KVL, states that â€śin any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loopâ€ť which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero.

FIg 1: - you dont know the voltage across the resistor so you CAN NOT work out the current.
Fig 2: however if you "pretend" the LED is a voltage source of 2.0V that leaves 5v-2v=3V across the resistor. Now Ohms law gives you i2 = 3V / 220 ohms = 13mA.

(A forward biased LED GENERALLY has a voltage drop of about 2V) @SteveThackery

Fig 3 shows your circuit with two LED's. The voltage remaining across the resistor is now just 5v - 2V -2V = ABOUT 1V

So the current i3 = 5V - 2V - 2v (=1V) / 220 ohms = 4.5mA.

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Provided it's red, and its forward voltage isn't actually 1.5-1.7V, which actually is more common for red leds. With green sitting somewhere around 2V and blue 0.4-0.5V above that. Long story short: look up forward voltage of the specific led used, because there's just no single 'correct' voltage and it really does depend (on color and power rating, mostly).

LEDs are not a linear resistance, like common resistors are. Your LED specifications will have a graph that shows the voltage drop(resistance) changes as the applied voltage changes.
Paul

@anon35827816

.. depending on
the current .. and ..
the temperature .. and ...

thats all too complicated for the OP.