Theory VS Experimental : Love-O-Meter

Hello everyone,
Today, I test the Love-O-Meter project of Arduino Project Book.
There are using red leds with a 220Ohm resistor in serie.
With 5V input, there is a 20mA current in the circuit (almost).
I was telling myself, well, I want to reduce the luminosity of the LED.

I want to have a circuit with 1mA instead of 20mA.
So, with a reference of 2V@20mA for a LED, we can say that "there is a resistor of 100Ohm in the LED".

Now I want 1mA so the voltage on the LED should be V = RI = 1001*10^(-3) = 0.1 V = 100mV
So, with an input of 5V, my resistor will consumes 5-0,1=4.9V.

So, the value of my resistor should be R = V/I = 4.9/(1*10^(-3)) = 4,9kOhm so let's take a 4,7kOhm resistor.

I realized the circuit and take some measures with my multimeter and that is my question :
I really have a current of 1mA in my circuit.
However, I found that my LED consumes 1.6V (normal if it's want to have enough voltage for luminosity) and my resistor consumes 3.4V so the resistor can be considered more about 3kOhm...

I don't understand why with a wrong reflexion I have the good result...
I hope to be clear enough and can someone help me to understand

Thank you.

Now I want 1mA so the voltage on the LED should be V = RI = 1001*10^(-3) = 0.1 V = 100mV
So, with an input of 5V, my resistor will consumes 5-0,1=4.9V.

No.... Like all diodes, LEDs are non-linear. Their resistance changes when voltage changes... Ohm's Law is a law of nature and it's always true, but it's not useful since we don't know the effective resistance.

They are considered "constant voltage". Under normal operating conditions the voltage is always (approximately) 2.2V. (If you've measured 1.6V, that's OK.)

With a 5V supply and 2.2V across the LED, that leaves 2.8V across the resistor. So, in order to calculate the current (which is the same through all series components) you have to calculate the current through the resistor.

DVDdoug:
No.... Like all diodes, LEDs are non-linear. Their resistance changes when voltage changes... Ohm's Law is a law of nature and it's always true, but it's not useful since we don't know the effective resistance.

They are considered "constant voltage". Under normal operating conditions the voltage is always (approximately) 2.2V. (If you've measured 1.6V, that's OK.)

With a 5V supply and 2.2V across the LED, that leaves 2.8V across the resistor. So, in order to calculate the current (which is the same through all series components) you have to calculate the current through the resistor.

Thank you for your response. I understand now that it's not linear.
However, I don't understand why when I put a 4.7kOhm, only 3.4kOhm is really taken .... ?

Did I have to consider that 2V will be consumed by the LED, calculate the resistor, make the circuit and check the real value to know if it's enough ? I don't know the behaviour to have ....

I don't know what project you are referring to nor what kind of LEDs do you have but I'd say 4.7KOhm on 5 V would not light up your leds. Double or triple the 220 OHMs.

Oscarko:
I don't know what project you are referring to nor what kind of LEDs do you have but I'd say 4.7KOhm on 5 V would not light up your leds. Double or triple the 220 OHMs.

I'm okay with that but how I have to reflecting to have a 1mA in a circuit where there is a red LED, a resistor (to calculate) and a voltage of 5V ?

If you are talking about LEDs , then you should be talking about FORWARD VOLTAGE@SPECIFIED CURRENT rating
ie:
2.2V@20 mA (from LED datasheet)

If you want 1 mA, then 5V - 2.2V = 2.8V/0.001 A = 2.8k ohm

Most people use 220 ohms.

R =2.8V /220 ohms = 0.01277 A (12.7 mA)

I often use 1K when I want an LED as a low-brightness indicator, works pretty well for many LEDs.
High brightness LEDs, like 8000 to 12000mcd parts, will still be really bright. Experiment with a resistor & LED and solderless breadboard before soldering things together.

I'm sorry guys but I can't understand well ...
Let's take 2.2V consumes by the LED so 2.8V for the resistor.
If I want 5mA, it seems a resistor of : 2.8/(5*10^(-3)) = 560
How can I check amperage in the circuit ? because I can't take it on the LED (otherwise the current go through my multimeter rather than the LED)

LetMeR00t:
How can I check amperage in the circuit ?

You have to break the circuit with the meter. Have a look at my photos in reply #7 here.

LetMeR00t:
Let's take 2.2V consumes by the LED so 2.8V for the resistor.

The LED consumes current not voltage.

The LED is fixed voltage so effectiveley the resistor is also.

Resistor is 2,8 v so use V=IR to set the required current.

Voltage is measured across a component.
Current is measured through a component by breaking the circuit.

JimboZA:
You have to break the circuit with the meter. Have a look at my photos in reply #7 here.

Thank you very much, I found 5.7mA so it's what I expected ^^

Boardburner2:
The LED consumes current not voltage.

The LED is fixed voltage so effectiveley the resistor is also.

Resistor is 2,8 v so use V=IR to set the required current.

Well, the objective is to consider (anytime) that the LED consumes 2V (with the reference 2V@20mA).

LetMeR00t:
Thank you very much, I found 5.7mA so it's what I expected ^^

Well, the objective is to consider (anytime) that the LED consumes 2V (with the reference 2V@20mA).

No the LED consumes 5.7 mA, you have already measured it.

The 2V is the Potential Difference (PD) across it which for a diode is fairly constant.

Is English your first language ? (No slur intended).

You should read a bit further in this post: up to #11.

Boardburner2:
No the LED consumes 5.7 mA, you have already measured it.

The 2V is the Potential Difference (PD) across it which for a diode is fairly constant.

Is English your first language ? (No slur intended).

I'm french, my english is so bad ? ^^

JimboZA:
You should read a bit further in this post: up to #11.

Thank you, I will learn it completely

JimboZA:
You should read a bit further in this post: up to #11.

Your explanation (with images) is very clear, thank you now it's clear for me.

LetMeR00t:
Your explanation (with images) is very clear, thank you now it's clear for me.

LetMeR00t:
I'm french, my english is so bad ? ^^

No,

The word consumption normally refers to current or power not voltage, at least in the electronics world.
Understanding the usage could save problems later on with discussion.

Boardburner2:
No,

The word consumption normally refers to current or power not voltage, at least in the electronics world.
Understanding the usage could save problems later on with discussion.

Ok, but is it wrong to say that a a dipole consumes voltage ? Well we start with 5V and finish with 0 to it means that it's consumes it some way no ?

I think this is a language or understanding problem.
Context is important.

I do not understand your post.
EDIT

By dipole did you mean diode?

A dipole is a word normally used for an aerial.(or other physical stuff)

Boardburner2:
I think this is a language or understanding problem.
Context is important.

I do not understand your post.
EDIT

By dipole did you mean diode?

A dipole is a word normally used for an aerial.(or other physical stuff)

Sorry for my language I don't use the correct words.

I would like to know if we can say that an element in the circuit such as a led, a resistor or something else consumes voltage in other words, can we imagine voltage as a chocolate bar (yes, that's my example ) the chocolate bar represents the 5V, each component of the system will consumes the chocolate bar such as resistor that consumes 3/5 of the chocolate bar and the led 2/5. At the end of the circuit the chocolate bar is totally consumes.
Consumes seems to be a non correct word so well it's probably me who can't understand the concept of voltage