Connecting an LED - Do you have a question?

Question

Electronics

We all know that a current-limiting resistor is needed to connect an LED to a circuit.
1 If I connect the LED through one 1.5V battery + resistor, then it will not glow - this is understandably not enough voltage.
2 If you connect without a resistor to one 1.5V battery, then it will burn out - it will not glow, but the current will go through the diode and since the current will be large (for a diode), it will burn out.
3 If the battery is 3V (two AA 1.5V each) - then you can just take a 220 Ohm resistor and an LED and everything will work fine. At the same time, the resistor will limit the current from the battery in the entire network, this is understandable - but what about the voltage? When you connect a resistor, the voltage should also drop (and fall!), And it doesn’t matter in what sequence. But the LED will shine - why doesn't it shine in the first case?

Answer 1

Here it would be first to read what an LED is and why it glows at all. If very generally, the LED has a minimum voltage (minimum potential difference) at which it can generally glow and a maximum (breakdown voltage). But the brightness of the glow does not depend on what voltage, but on what current flows through it . For an abstract LED in a vacuum, for example, it may be necessary to glow about 23 mA, which at 220 ohms and 5V, if I'm not mistaken, by changing the voltage, we will also change the current (see Ohm's law).
1. It depends on which LED and which resistor, it is clear that if a 500KΩ resistor, no LED will start to glow even if it is connected to 220. Although if it explodes, it can and will glow :)
2. Not necessarily, only if the battery can give enough current to burn the LED, for example, a typical BIOS battery, at 1.2-1.5 volts, the LED will glow as much as necessary, because they cannot give enough current to burn it
3. The LED is needed connect in parallel with the load, not in series.

Answer 2

Look at the VAC of the LED and it will immediately become clear to you.
for example WAH .

Answer 3

you can roughly estimate the required resistance of the resistor.
The voltage drop of an "open" LED is about half a volt. It burns well at 20 mA. Accordingly, Rlimit = (Upit - 0.5V) / 20mA.
That is, at Upit=3V Rogr=125Ohm.
Power sources have some more internal resistance (for a battery it depends on the degree of its discharge), which must be added Rlimit. And the voltage drop across the diode is not quite 0.5V. There is a non-linear dependence of the current on the voltage across the diode.
It is possible, of course, to calculate everything beautifully, but it is hardly necessary.