How to solder an LED lamp?

Question

Do it yourself

Good afternoon. Just don't laugh.

I want to make a lamp for my beloved with my own hands. but all my experience in electrical engineering and soldering came down to soldering soldered wires.

I know that for a LED lamp it is necessary to use components such as a driver.

A hint is needed:

1. Circuit diagram
2. Name of the element with a description of the characteristics (or instructions on how to select these characteristics, if you rely on the fact that the lamp will be powered by a 9V Krona battery.
3. Tips for soldering
4. Links where you can read in detail about all this, but not in such a way as to climb into the wilds.

It is possible that I will be very interested in this topic and study the material in detail, but now I am interested in solving an applied problem. Thank you!

Answer 1

So, on the fingers :))
There is an LED, it has two key parameters:
Voltage drop and operating current.
What is voltage drop?
Imagine a battery running on a closed circuit of two simple resistors (load resistance). A battery always has such a parameter as EMF (those very volts), which depend on the chemistry of the battery. This is a constant. There is internal resistance (the fresher the battery, the lower it is).
And then the circuit turns out like this: EMF source - internal resistance - load resistance1 - load resistance2 - looped to the other end of the EMF source.
We have one branch. Therefore, the current in it is also equal on all elements. According to Ohm's law, U \u003d I * R. This very U is the voltage drop. And the circuit equation must be such that the sum of all drops is equal to the same EMF.
E \u003d I * Rvn battery resistance + I * R1 + I * R2
E - battery emf, constant, always equal to a certain value, depending on the battery reagents. A typical battery has 1.5 volts.
Those. EMF, as it were, is divided by all the resistances of the circuit, in proportion to these resistances. The greater the resistance of the section, the greater the piece of voltage it will drag on itself.
When the battery dies, the EMF is still = 1.5 volts, but the internal resistance of the battery increases. This means that the current drops, which means that the distribution of drops changes and most of the EMF settles on the internal resistance, and not on the load resistances, a minuscule remains.
Let's go back to the LED. Its feature is that it is a non-linear element. Those. its resistance depends on the current along a tricky curve. And the essence of this curve is that its drop (I diode * R diode) = const and its value is in the documentation for the diode. What does it threaten. And here's what:
Let's say a diode drop according to the passport is 1.5 volts. And you hang it on a crown, whose EMF = 9 volts:
You get an EMF circuit = I * Rvn battery resistance + U diode pad.
U diode drop = const and equal to one and a half volts. Even if you kill it, it will be so. Where to put the remaining volts? They can only be landed on the internal resistance of the load. It, at the moment, is also a certain constant (we assume that our battery is infinitely charged). Let it be in 1OM
What happens? 9 volts - 1.5 = 7.5 volts should land on a resistance of 1 ohm. How? Yes, easily - the current in our circuit (and through the diode too) will be only 7.5A. Weak, huh? The diode from such an akhtung will instantly evaporate (After all, its operating current is thousands of times less!) Scattering splashes of burnt plastic around the area.
How to be? There are several options:
Sculpt so many LEDs in series so that their total voltage drop balances the EMF of the battery minus a small drop in the internal resistance of the battery (this drop will be Rin * Iworking LED)
Or add one more additional resistance to the circuit. Which will be such that I \u003d 0.03A (let it be the operating current of the LED) And then R of this resistance can be easily calculated from the equation:
9 \u003d R * 0.03 - 1 * 0.03 - 1.5
Now about the drivers:
What does an LED driver do? And he just adjusts his output voltage in such a way that it is consistent with the voltage drop across the LED, providing a strictly specified operating current on it. If you put two lights in series, the driver will double the voltage while holding the current. You put three - three times and so on until the driver has enough margin for regulation (depending on the input voltage and the driver circuit).

Answer 2

Theory: The
LED is powered by a certain current. The value of this current depends on the type of LED used and is sought in its passport characteristics. The required current is provided by a current stabilizer, one of the varieties of which is what you called a “driver”.
Practice:
To select the desired driver, you need to decide on the type of LED used. In the passport data, find the value of the optimal current for the LED. Next, look for a driver on sale so that its input voltage includes your 9 volts from the crown, and the output current is equal to the current of your LED. For example, - input voltage 5-12 Volts, output current 30 mA (substitute the data of your diode here). Several LEDs can be switched on in series. The supply current will remain the same. But the number of diodes in the circuit will be limited by your supply voltage (that is, your 9 volts - actually a little less, given the losses on the driver). The voltage drop across your diode is also found out from the passport data, then they just add up. In your case, it will be possible to turn on two or three diodes in series (depending on the diodes).
A cheaper option is to set the current to the diodes with a current-limiting resistor. It's just a resistor in series with a diode circuit. The resistance of the resistor is calculated based on the data of the voltage source, the voltage drops across the diodes and the required current using Ohm's law from a school textbook. The disadvantage of this option is the decrease in brightness when the battery is discharged.
The option with “winding on a ferrite ring” does not suit you - in fact, this is the same “driver” from the first option, but assembled on the knee. Based on what you wrote, it is obvious that you do not have the skills to calculate, manufacture and configure this node. And even if it were enough (but then you would not have asked this question), then in most cases, the laboriousness of its manufacture in single copies is higher than the cost of the finished unit, and indeed higher than reasonable.
As for the directivity of the radiation of the diodes (which look like a droplet), this is treated by grinding this very “droplet”, which is a focusing lens, until the desired directivity pattern is given (even sand with a fine sandpaper so that the surface becomes matte for greater uniformity).

Answer 3

For a simple lamp, an LED and a resistor are enough. What specific lamp do you want to make?

Answer 4

According to paragraph 3, I recommend http://mightyohm.com/files/soldercomic/translations/Soldering%20is%20easy%20(RUS).pdf (or in the original mightyohm.com/files/soldercomic/FullSolderComic_EN.pdf ).
For the rest of the points: it all depends on what exactly you want to do.

Answer 5

If you need simplicity - and more or less quality - I would recommend buying an LED strip (there are different options) + a driver with the desired input voltage.
Cons:
for fairly large fixtures
, not the cheapest option ~ 700-1000r
Pros
are no more difficult than soldering a couple of wires

Answer 6

You chose the Krona battery unsuccessfully for the lamp. The capacitance is small, the internal resistance is large. The light will be bad and not for long. And something relatively powerful and will not pull at all.