When I do have a serie circuit with a power source of 5V then a resistor (for instance 220 Ohm) en 1 red LED. This LED does have a voltage drop of ±1.9V. So I would expect (conform Ohm's law) a LED has also a resistance which transforms into light. But this isn't true..?

I should say: Current in circuit = 5/(220+resistance LED)

Can someone explane why the resistance of the LED is ignored? (page 30 projectbook)
Thank you in advance.

The LED is a current device. With that in mind a good LED with a data sheet will reflect the rated forward current (Ifwd) and the forward voltage drop (Vfwd). So to get the current limited for asn applied voltage we get Vsupply - Vfwd / Ifwd. So in your example it would be 5V - 1.9V = 3.1V / 0.227 Amp = 13.65 Ohms. In other words 13.65 Ohms * .227 Amps = 3.1 Volt.

Since 13.65 Ohms is an odd number just choose the next highest resistance value and make sure the resistor will dissipate the required power. In this case .227 Amp * 3.1 Volts = 0.703 Watt and with a margin I would look for a 2 watt resistor.

Thanks for your quick answer. Sorry I formulated my question a bit wrong.
I ment to ask: what is the current in this circuit?
Normally I should calculate the "Replacement resistor" Rv= R1 + R2.
When I have Rv, I calculate the current using I=U/R but R2 is 0 when it is a LED?
I don't understand the resistance is 0 while there is a voltage drop.

Please note I did adjust my main post a bit to make it more clear.

You are correct that the current is the same through all series components.

The voltage divides (or sums-up) across series components, so right, you have 3.1V across the resistor.

3.1V across 220 Ohms gives you 0.014 Amps (14mA).

LEDs are non-linear. Like all diodes the resistance changes with voltage. We can calculate the resistance of the LED at the particular voltage if we know the current but we rarely bother with that because the resistance changes depending on the conditions.

This highly non-linear characteristic makes the LED voltage “fall magically into place” with correct current through it. And, it’s reason you need a series resistor (or some kind of current limiting/control).

a LED has also a resistance which transforms into light. But this isn’t true…?

I don’t understand the physics, but power (Watts, milliwatts, energy) is calculated as Power = Voltage x Current. So 1.9V x 14mA is 27mW.

Some of that energy gets converted to light energy. The remaining energy is “lost” as heat.

But the project book says: 5/220= 0.023 A So calculated without the voltage drop of the LED. Why?

I don't know what book you have.

That calculation is correct if you have 5V across the resistor. If 5V is the total applied voltage and there is an LED in series, the calculation is wrong.

I do have the Arduino ProjectsBook. The calculation is on mid page 30 and the circuit is on page 29.
5V is the main power supply and the 220 ohm resistor and red LED are in serie.

I see what you are saying. Here is a quote from the book which I found online.

"In the circuit shown in Fig. 5, you’re supplying 5 volts. The resistor ofers 220 ohms
resistance. To find the amperage used by the LED, replace the values in the equation. You should have 5=I*220. Dividing both sides of the equation by 220, you’ll
find that I = .023. That’s 23 thousandths of an amp, or 23 milliamps (23 mA) used
by the LED. That value is just about the maximum you can safely use with these
LEDs, which is why you used a 220-ohm resistor".

That is incorrect as the circuit reflects a 220 Ohm resistor, a LED in series between 5 volts and ground. They do not consider the Vfwd of the LED in fact they seem to ignore the presence of the LED. I am never too quick to say the book is wrong but to me it seems the book is wrong. Also as mentioned the resistance of an LED is not even close to a linear function, it is totally not even close to linear as was mentioned. So I can't say if a LED has a Vfwd of 2.0 volts and Ifwd of 0.020 (20 mA) Amp that the resistance is 100 Ohms and if I apply 1.0 volt across the same LED the current will be 0.010 Amp and the resistance the same 100 Ohms.

Anyway, as I read it between Figure 5 and the text on page 30 something is wrong and yes, you are reading it correctly. Unless I am missing something?

I think the problem here is you are treating a led as if it were linear, which it is not.

A led is a non-linear device. It has a rated forward voltage specified for some specific current (typ 20mA)
so a led with a V_{f} of 1.9V would give you 5V-1.9V = 3.1V drop across the 220 ohm resistor,
which would probably yield a current of 14mA based on the V_{f}.

The V_{f} tells you what the led voltage will be so you can adjust the resistor value to obtain the
desired current.
If for example you wanted 20mA led current you do the following:

raschemmel:
I think the problem here is you are treating a led as if it were linear, which it is not.

A led is a non-linear device. It has a rated forward voltage specified for some specific current (typ 20mA)
so a led with a V_{f} of 1.9V would give you 5V-1.9V = 3.1V drop across the 220 ohm resistor,
which would probably yield a current of 14mA based on the V_{f}.

The V_{f} tells you what the led voltage will be so you can adjust the resistor value to obtain the
desired current.
If for example you wanted 20mA led current you do the following:

3.1V/0.020A = 155 Ohms.

Take a look at this with a focus on Figure 5 & 6 page 24 and page 30 with the text I quoted from understanding Ohm’s law. As written and shown does it make sense to you? To me they omit the forward voltage drop of the LED.

Actually , they are assuming that parameters
like V forward ate too advanced for someone
who just learning to connect a led. This is is very
common in tutorials for beginners with no
electronics experience. Ask yourself this:
did they tell you how to increase the led brightness
or did the just assume that if it lights with a 220 ohm resistor you'll be happy and not ask any questions ?

I understand what you are saying. Maybe my knowledge is a bit to much for the first projects. But if you make a calculation it should be correct in my opinion.
Maybe they just ignore the resistance of the LED, because it is not much compared to the resistor.
Thanks for your reaction guys.

For the calculation to be correct in an absolute sense you'd need to have other things in there, such as temperature, LED I-V characteristic as a function of temperature, wire resistances, power supply internal resistance...

In teaching its common to simplify a calculation to the bare essentials, to give a quick back-of-the-envelope
value. However here the LED voltage probabaly should not be simplified away, its not negligible, and for
a blue LED it might be more than 3V in fact.

OK while I can understand making things simple but when teaching while simple is a good thing leaving out important information is not so good. The paragraph in question clearly states the following:

"In the circuit shown in Fig. 5, you’re supplying 5 volts. The resistor offers 220 ohms
resistance. To find the amperage used by the LED, replace the values in the equation. You should have 5=I*220. Dividing both sides of the equation by 220, you’ll find that I = .023. That’s 23 thousandths of an amp, or 23 milliamps (23 mA) used
by the LED. That value is just about the maximum you can safely use with these
LEDs, which is why you used a 220-ohm resistor".

That could not be any further wrong. Here are the figures 5 and also 6 with figure 5 being referenced.

The LED is going to have a forward voltage drop, be it a volt, two volts, or three volts the simple fact is it will have a drop. Since I have a pen in my hand I claim Vfwd is 1.9 volts and I fwd is 20 mA and that's at ambient temperature on a lovely sunny day at sea level right on the beach. Now if my LED is dropping 1.9 volts that is subtracted from my 5 volt supply. I think we can all agree on that? Therefore 5 volts - 1.9 Volts leaves me only 3.1 volts across my remaining 220 Ohm resistor making it 3.1 Volts / 220 Ohms = 14.09 mA which is a far cry from what is quoted above. In the above paragraph there is no consideration given to the forward voltage drop of the LED and if all that was wanted was to convey a little Ohm's Law then the LED should have been omitted but with it included it makes a game changer. That's all there really is to it.

I call it a failure. If you're going to grossly simplify a detailed technical explanation like this, you can at least include a short sentence explaining the simplification.

"I call it a failure. If you're going to grossly simplify a detailed technical explanation like this, you can at least include a short sentence explaining the simplification."

Not sure who that is directed to.
Where is your explanation ?