How to make a powerful flashlight. Do-it-yourself repair and modernization of Lentel, Foton, Smartbuy Colorado and RED LED lights. LED lamp

We make a flashlight on LEDs with our own hands

LED flashlight with 3V converter for LED 0.3-1.5V 0.3-1.5 VLEDflash light

Usually, a blue or white LED requires 3 - 3.5v to operate, this circuit allows you to power a blue or white LED with low voltage from a single AA battery.Normally, if you want to light up a blue or white LED you need to provide it with 3 - 3.5 V, like from a 3 V lithium coin cell.

Details:
Light-emitting diode
Ferrite ring (~10 mm diameter)
Winding wire (20 cm)
1kΩ resistor
N-P-N transistor
Battery

Parameters of the used transformer:
The winding going to the LED has ~45 turns wound with 0.25mm wire.
The winding going to the base of the transistor has ~30 turns of 0.1mm wire.
The base resistor in this case has a resistance of about 2K.
Instead of R1, it is desirable to put a tuning resistor, and achieve a current through the diode ~ 22mA, with a fresh battery, measure its resistance, then replacing it with a constant resistor of the received value.

The assembled circuit must work immediately.
There are only 2 reasons why the scheme will not work.
1. the ends of the winding are mixed up.
2. too few turns of the base winding.
Generation disappears, with the number of turns<15.

Put the pieces of wire together and wind around the ring.
Connect the two ends of different wires together.
The circuit can be placed inside a suitable enclosure.
The introduction of such a circuit into a flashlight operating from 3V significantly extends the duration of its operation from one set of batteries.

Variant of execution of a lamp from one battery 1,5v.

The transistor and resistance are placed inside the ferrite ring

White LED powered by a dead AAA battery

Upgrade option "flashlight - pen"

The excitation of the blocking generator shown in the diagram is achieved by a transformer connection at T1. The voltage pulses that occur in the right (according to the scheme) winding are added to the voltage of the power source and fed to the VD1 LED. Of course, it would be possible to exclude the capacitor and resistor in the base circuit of the transistor, but then VT1 and VD1 may fail when using branded batteries with low internal resistance. The resistor sets the operating mode of the transistor, and the capacitor passes the RF component.

The circuit used a KT315 transistor (as the cheapest, but any other with a cutoff frequency of 200 MHz or more), an ultra-bright LED. For the manufacture of a transformer, a ferrite ring is required (approximate size 10x6x3 and a permeability of about 1000 HH). The wire diameter is about 0.2-0.3 mm. Two coils of 20 turns each are wound on the ring.
If there is no ring, then a cylinder similar in volume and material can be used. You just have to wind 60-100 turns for each of the coils.
Important point : you need to wind the coils in different directions.

Flashlight photos:
the switch is located in the "fountain pen" button, and the gray metal cylinder conducts current.

We make a cylinder according to the size of the battery.

It can be made from paper, or a piece of any rigid tube can be used.
We make holes along the edges of the cylinder, wrap it with tinned wire, pass the ends of the wire into the holes. We fix both ends, but leave a piece of conductor at one of the ends: so that you can connect the converter to the spiral.
A ferrite ring would not fit into a lantern, so a cylinder of similar material was used.

Cylinder from an inductor from an old TV.
The first coil is about 60 turns.
Then the second, winds in the opposite direction again 60 or so. The threads are held together with glue.

We assemble the converter:

Everything is located inside our case: We unsolder the transistor, the resistor capacitor, solder the spiral on the cylinder, and the coil. The current in the coil windings must go in different directions! That is, if you wound all the windings in one direction, then swap the conclusions of one of them, otherwise generation will not occur.

It turned out the following:

We insert everything inward, and use nuts as side plugs and contacts.
We solder the coil leads to one of the nuts, and the VT1 emitter to the other. Glue. we mark the conclusions: where we will have an output from the coils, we put “-”, where the output from the transistor with the coil we put “+” (so that everything is like in a battery).

Now you should make a "lamp diode".

Attention: on the base should be minus the LED.

Assembly:

As is clear from the figure, the converter is a "substitute" for the second battery. But unlike it, it has three points of contact: with the plus of the battery, with the plus of the LED, and the common body (through the spiral).

Its location in the battery compartment is specific: it must be in contact with the positive of the LED.

Modern flashlightwith the mode of operation of the LED powered by constant stabilized current.

The current stabilizer circuit works as follows:
When power is applied to the circuit, transistors T1 and T2 are locked, T3 is open, because an unlocking voltage is applied to its gate through resistor R3. Due to the presence of an inductor L1 in the LED circuit, the current increases smoothly. As the current in the LED circuit increases, the voltage drop across the R5-R4 chain increases, as soon as it reaches about 0.4V, transistor T2 opens, followed by T1, which in turn closes the current switch T3. The increase in current stops, a self-induction current arises in the inductor, which begins to flow through the diode D1 through the LED and the chain of resistors R5-R4. As soon as the current decreases below a certain threshold, transistors T1 and T2 will close, T3 will open, which will lead to a new cycle of energy accumulation in the inductor. In normal mode, the oscillatory process occurs at a frequency of the order of tens of kilohertz.

About details:
Instead of the IRF510 transistor, you can use the IRF530, or any n-channel field-effect key transistor for a current of more than 3A and a voltage of more than 30 V.
Diode D1 must necessarily be with a Schottky barrier for a current of more than 1A, if you put an ordinary even high-frequency type KD212, the efficiency will drop to 75-80%.
The inductor is homemade, it is wound with a wire no thinner than 0.6 mm, better with a bundle of several thinner wires. About 20-30 turns of wire on the B16-B18 armor core are required with a non-magnetic gap of 0.1-0.2 mm or close to 2000NM ferrite. If possible, the thickness of the non-magnetic gap is selected experimentally according to the maximum efficiency of the device. Good results can be obtained with ferrites from imported inductors installed in switching power supplies, as well as in energy-saving lamps. Such cores have the form of a thread spool, do not require a frame and a non-magnetic gap. Coils on toroidal cores made of pressed iron powder, which can be found in computer power supplies (they are wound with output filter inductors), work very well. The non-magnetic gap in such cores is evenly distributed in volume due to the production technology.
The same stabilizer circuit can also be used in conjunction with other batteries and batteries of galvanic cells with a voltage of 9 or 12 volts without any change in the circuit or cell ratings. The higher the supply voltage, the less current the flashlight will consume from the source, its efficiency will remain unchanged. The stabilization current is set by resistors R4 and R5.
If necessary, the current can be increased up to 1A without the use of heat sinks on the parts, only by selecting the resistance of the setting resistors.
The charger for the battery can be left "native" or assembled according to any of the known schemes, or even use an external one to reduce the weight of the flashlight.

LED flashlight from calculator B3-30

The converter is based on the B3-30 calculator circuit, in the switching power supply of which a transformer with a thickness of only 5 mm is used, which has two windings. Using a pulse transformer from an old calculator made it possible to create an economical LED flashlight.

The result is a very simple circuit.

The voltage converter is made according to the scheme of a single-cycle generator with inductive feedback on a transistor VT1 and a transformer T1. The impulse voltage from the windings 1-2 (according to the B3-30 calculator circuit diagram) is rectified by the VD1 diode and fed to the super-bright HL1 LED. Capacitor C3 filter. The design is based on a Chinese-made flashlight designed to install two AA batteries. The transducer is mounted on a printed circuit board made of one-sided foil-coated fiberglass with a thickness of 1.5 mmfig.2sizes that replace one battery and inserted into the flashlight instead of it. A contact made of double-sided foil fiberglass with a diameter of 15 mm is soldered to the end of the board marked with a “+” sign, both sides are connected by a jumper and soldered.
After installing all the parts on the board, the “+” end contact and the T1 transformer are filled with hot glue to increase strength. The layout of the lantern is shown infig.3and in a particular case depends on the type of lamp used. In my case, no modification of the lamp was required, the reflector has a contact ring, to which the negative output of the printed circuit board is soldered, and the board itself is attached to the reflector with hot glue. The printed circuit board assembly with the reflector is inserted instead of one battery and clamped with a cover.

The voltage converter uses small parts. Resistors of the MLT-0.125 type, capacitors C1 and C3 are imported, up to 5 mm high. Diode VD1 type 1N5817 with a Schottky barrier, in its absence, you can use any rectifier diode that is suitable for the parameters, preferably germanium due to the lower voltage drop across it. A properly assembled converter does not need to be adjusted if the transformer windings are not reversed, otherwise swap them. In the absence of the above transformer, you can make it yourself. Winding is carried out on a ferrite ring of size K10 * 6 * 3 with a magnetic permeability of 1000-2000. Both windings are wound with PEV2 wire with a diameter of 0.31 to 0.44 mm. The primary winding has 6 turns, the secondary 10 turns. After installing such a transformer on the board and checking its performance, it should be fixed on it with hot glue.
Flashlight tests with an AA battery are presented in Table 1.
The test used the cheapest AA battery costing only 3 rubles. The initial voltage under load was 1.28 V. At the output of the converter, the voltage measured on a superbright LED was 2.83 V. The brand of the LED is unknown, the diameter is 10 mm. The total current consumption is 14 mA. The total operating time of the flashlight was 20 hours of continuous operation.
When the voltage on the battery drops below 1V, the brightness drops noticeably.

Time, h	V batteries, V	V conversion, V
0	1,28	2,83
2	1,22	2,83
4	1,21	2,83
6	1,20	2,83
8	1,18	2,83
10	1,18	2.83
12	1,16	2.82
14	1,12	2.81
16	1,11	2.81
18	1,11	2.81
20	1,10	2.80

Homemade flashlight with LEDs

The basis is a flashlight "VARTA" powered by two AA batteries:
Since diodes have a highly non-linear IV characteristic, it is necessary to equip the flashlight with a circuit for working with LEDs, which will provide a constant brightness of the glow as the battery is discharged and will remain operational at the lowest possible supply voltage.
The heart of the voltage regulator is the MAX756 micropower DC/DC boost converter.
According to the declared characteristics, it works when the input voltage drops to 0.7V.

Switching scheme - typical:

Mounting is carried out in a hinged way.
Electrolytic capacitors - tantalum CHIP. They have a low series resistance, which improves efficiency somewhat. Schottky diode - SM5818. Chokes had to be connected in parallel, because. there was no suitable value. Capacitor C2 - K10-17b. LEDs - superbright white L-53PWC "Kingbright".
As you can see in the figure, the whole circuit easily fit into the empty space of the light emitting node.

The output voltage of the stabilizer in this switching circuit is 3.3V. Since the voltage drop across the diodes in the nominal current range (15-30mA) is about 3.1V, the extra 200mV had to be extinguished by a resistor connected in series with the output.
In addition, a small series resistor improves load linearity and circuit stability. This is due to the fact that the diode has a negative TCR, and when it is heated, the direct voltage drop decreases, which leads to a sharp increase in current through the diode when it is powered from a voltage source. It was not necessary to equalize the currents through the diodes connected in parallel - no difference in brightness was observed by eye. Moreover, the diodes were of the same type and taken from the same box.
Now about the design of the light emitter. As you can see in the photos, the LEDs in the circuit are not tightly soldered, but are a removable part of the structure.

The native light bulb is gutted, and 4 cuts are made in the flange from 4 sides (one was already there). 4 LEDs are arranged symmetrically in a circle. The positive leads (according to the diagram) are soldered to the base near the cuts, and the negative leads are inserted from the inside into the central hole of the base, cut off and also soldered. "Lamp diode", inserted in place of a conventional incandescent light bulb.

Testing:
The stabilization of the output voltage (3.3V) continued until the supply voltage dropped to ~1.2V. The load current in this case was about 100mA (~ 25mA per diode). Then the output voltage began to gradually decrease. The circuit has switched to a different mode of operation, in which it no longer stabilizes, but outputs everything it can. In this mode, it worked up to a supply voltage of 0.5V! The output voltage at the same time dropped to 2.7V, and the current from 100mA to 8mA.

A little about efficiency.

The efficiency of the circuit is about 63% with fresh batteries. The fact is that the miniature chokes used in the circuit have an extremely high ohmic resistance - about 1.5 ohm
The solution is a µ-permalloy ring with a permeability of about 50.
40 turns of PEV-0.25 wire, in one layer - it turned out about 80 μG. The active resistance is about 0.2 Ohm, and the saturation current, according to calculations, is more than 3A. We change the output and input electrolyte to 100 microfarads, although without prejudice to efficiency it can be reduced to 47 microfarads.

Scheme of the LED lampon DC/DC converter from Analog Device - ADP1110.

Standard typical connection diagram of ADP1110.
This converter chip, according to the manufacturer's specifications, is available in 8 versions:

Model	Output voltage
ADP1110AN	Adjustable
ADP1110AR	Adjustable
ADP1110AN-3.3	3.3V
ADP1110AR-3.3	3.3V
ADP1110AN-5	5V
ADP1110AR-5	5V
ADP1110AN-12	12V
ADP1110AR-12	12V

Microcircuits with indices "N" and "R" differ only in the type of package: R is more compact.
If you bought a chip with an index of -3.3, you can skip the next paragraph and go to the "Details" item.
If not, I present to your attention another scheme:

It adds two parts to get the required 3.3 volts output to power the LEDs.
The circuit can be improved by taking into account that the LEDs need a current source, not a voltage source, to operate. Changes in the circuit so that it would give out 60mA (20 for each diode), and the diodes will automatically set the voltage to us, the same 3.3-3.9V.

resistor R1 is used to measure the current. The converter is designed in such a way that when the voltage at the FB (Feed Back) pin exceeds 0.22V, it will finish increasing the voltage and current, which means that the value of the resistance R1 is easy to calculate R1 = 0.22V / In, in our case 3.6Ω. Such a circuit helps to stabilize the current, and automatically select the required voltage. Unfortunately, voltage will drop across this resistance, which will lead to a decrease in efficiency, however, practice has shown that it is less than the excess that we chose in the first case. I measured the output voltage and it was 3.4 - 3.6V. The parameters of the diodes in such an inclusion should also be as similar as possible, otherwise the total current of 60mA was not distributed equally between them, and again we will get different luminosity.

Details
1. A choke will fit any 20 to 100 microhenry with a small (less than 0.4 ohm) resistance. The diagram indicates 47 μH. You can make it yourself - wind about 40 turns of PEV-0.25 wire on a µ-permalloy ring with a permeability of about 50, size 10x4x5.
2. Schottky diode. 1N5818, 1N5819, 1N4148 or equivalent. Analog Device DOES NOT RECOMMEND the use of the 1N4001
3. Capacitors. 47-100 microfarads at 6-10 volts. It is recommended to use tantalum.
4. Resistors. A power of 0.125 watts with a resistance of 2 ohms, possibly 300 kΩ and 2.2 kΩ.
5. LEDs. L-53PWC - 4 pieces.

Voltage converter for powering a white LED DFL-OSPW5111P with a brightness of 30 cd at a current of 80 mA and a radiation pattern width of about 12°.

The current consumed from a battery with a voltage of 2.41V is 143mA; in this case, a current of about 70 mA flows through the LED at a voltage of 4.17 V on it. The converter operates at a frequency of 13 kHz, the electrical efficiency is about 0.85.
Transformer T1 is wound on an annular magnetic circuit of size K10x6x3 made of ferrite 2000NM.

The primary and secondary windings of the transformer are wound simultaneously (i.e., in four wires).
The primary winding contains - 2x41 turns of wire PEV-2 0.19,
The secondary winding contains - 2x44 turns of wire PEV-2 0.16.
After winding, the winding leads are connected in accordance with the diagram.

Transistors KT529A of the p-n-p structure can be replaced with KT530A of the n-p-n structure, in this case it is necessary to change the polarity of connecting the GB1 battery and the HL1 LED.
Details are placed on the reflector using hanging mounting. Pay attention to the fact that the contact of the parts with the tin plate of the flashlight, which supplies the “minus” of the GB1 battery, is excluded. The transistors are fastened together with a thin brass clamp, which provides the necessary heat removal, and then glued to the reflector. The LED is placed instead of the incandescent lamp so that it protrudes 0.5 ... 1 mm from the socket for its installation. This improves heat dissipation from the LED and simplifies its installation.
When you first turn on the battery power is supplied through a resistor with a resistance of 18 ... 24 ohms so as not to damage the transistors if the terminals of the transformer T1 are connected incorrectly. If the LED does not shine, it is necessary to swap the extreme terminals of the primary or secondary winding of the transformer. If this does not lead to success, check the serviceability of all elements and the correct installation.

Voltage converter for powering an industrial design LED lamp.

Voltage converter for powering the LED lamp

The circuit is taken from the Zetex manual for the use of ZXSC310 microcircuits.
ZXSC310- LED driver chip.
FMMT 617 or FMMT 618.
Schottky diode- almost any brand.
Capacitors C1 = 2.2uF and C2 = 10uFfor surface mounting, 2.2 uF is the value recommended by the manufacturer, and C2 can be set from about 1 to 10 uF

Inductor 68 microhenries at 0.4 A

The inductance and resistor are installed on one side of the board (where there is no print), all other parts are on the other. The only trick is making a 150 milliohm resistor. It can be made from 0.1 mm iron wire, which can be obtained by unwinding the cable. The wire should be annealed on a lighter, carefully wiped with a fine sandpaper, tinned the ends and soldered a piece about 3 cm long into the holes on the board. Further, in the process of tuning, it is necessary, by measuring the current through the diodes, to move the wire, while heating the place of its soldering to the board with a soldering iron.

Thus, something like a rheostat is obtained. Having achieved a current of 20 mA, the soldering iron is removed, and an unnecessary piece of wire is cut off. The author came out with a length of about 1 cm.

Flashlight on power source

Rice. 3.A flashlight on a current source, with automatic current equalization in the LEDs, so that the LEDs can be with any spread of parameters (the VD2 LED sets the current that the transistors VT2, VT3 repeat, so the currents in the branches will be the same)
Transistors, of course, should also be the same, but the spread of their parameters is not so critical, so you can take either discrete transistors, or if you can find three integrated transistors in one package, their parameters are as close as possible. Play with the placement of the LEDs, you need to choose a pair of LED-transistor so that the output voltage is minimal, this will increase the efficiency.
The introduction of transistors evened out the brightness, but they have resistance and voltage drops on them, which forces the converter to increase the output level to 4V, to reduce the voltage drop across the transistors, you can propose a circuit in Fig. 4, this is a modified current mirror, instead of the reference voltage Ube = 0.7V in the circuit in Fig. 3, you can use the 0.22V source built into the converter, and maintain it in the VT1 collector using an op-amp, also built into the converter.

Rice. 4.Flashlight on a power source, with automatic current equalization in the LEDs, and with improved efficiency

Because the output of the opamp is of the “open collector” type; it must be “pulled up” to the power supply, which makes the resistor R2. Resistors R3, R4 act as a voltage divider at point V2 by 2, so the opamp will maintain a voltage of 0.22 * 2 = 0.44V at point V2, which is 0.3V less than in the previous case. It is impossible to take a divider even less in order to lower the voltage at point V2. the bipolar transistor has a resistance Rke and during operation, the voltage Uke will drop on it, so that the transistor works correctly V2-V1 must be greater than Uke, for our case 0.22V is enough. However, bipolar transistors can be replaced with field-effect transistors, in which the drain-to-source resistance is much less, this will make it possible to reduce the divider, so that the difference V2-V1 is completely insignificant.

Throttle.The inductor must be taken with a minimum resistance, special attention should be paid to the maximum allowable current, it should be of the order of 400 -1000 mA.
The rating doesn't matter as much as the maximum current, so Analog Devices recommends something between 33 and 180uH. In this case, theoretically, if you do not pay attention to the dimensions, then the larger the inductance, the better in all respects. However, in practice this is not entirely true, because. we have a non-ideal coil, it has active resistance and is not linear, in addition, the key transistor at low voltages will no longer give out 1.5A. Therefore, it is better to try several coils of different types, designs and different ratings in order to choose a coil with the highest efficiency and the smallest minimum input voltage, i.e. the coil with which the flashlight will glow for as long as possible.

Capacitors.C1 can be anything. C2 is better to take tantalum because. it has a small resistance, which increases the efficiency.

Schottky diode.Any for current up to 1A, preferably with minimal resistance and minimal voltage drop.

Transistors.Any with collector current up to 30 mA, coefficient current amplification of the order of 80 with a frequency of up to 100 MHz, KT318 is suitable.

LEDs.You can white NSPW500BS with a glow of 8000mCd from Power Light Systems.

Voltage transformerADP1110, or its replacement ADP1073, to use it, the circuit in Fig. 3 will need to be changed, take a 760μG inductor, and R1 = 0.212 / 60mA = 3.5Ω.

Lantern on ADP3000-ADJ

Options:
Power supply 2.8 - 10 V, efficiency approx. 75%, two brightness modes - full and half.
The current through the diodes is 27 mA, in half brightness mode - 13 mA.
In order to obtain high efficiency, it is desirable to use chip components in the circuit.
A properly assembled circuit does not need to be configured.
The disadvantage of the circuit is the high (1.25V) voltage at the FB input (pin 8).
Currently, DC / DC converters with an FB voltage of about 0.3V are being produced, in particular, by Maxim, on which it is realistic to achieve an efficiency above 85%.

Scheme of a lantern on Kr1446PN1.

Resistors R1 and R2 - current sensor. Operational amplifier U2B - amplifies the voltage taken from the current sensor. The gain = R4 / R3 + 1 and is approximately 19. The gain is required so that when the current through the resistors R1 and R2 is 60 mA, the output voltage opens the transistor Q1. By changing these resistors, you can set other stabilization current values.
In principle, an operational amplifier can be omitted. It’s just that instead of R1 and R2 one 10 Ohm resistor is placed, from it the signal through the 1kOhm resistor is fed to the base of the transistor and that’s it. But. This will lead to a decrease in efficiency. On a 10 ohm resistor at a current of 60 mA, 0.6 volts - 36 mW is wasted in vain. In the case of using an operational amplifier, the losses will be:
on a 0.5 Ohm resistor at a current of 60 mA = 1.8 mW + the consumption of the op-amp itself is 0.02 mA, let at 4 Volts = 0.08 mW
= 1.88 mW - significantly less than 36 mW.

About components.

In place of KR1446UD2, any low-power op-amp with a low minimum supply voltage can work, OP193FS would be better, but it is quite expensive. Transistor in SOT23 package. The polar capacitor is smaller - type SS at 10 Volts. Inductance CW68 100uH for 710mA. Although the cutoff current of the converter is 1 A, it works normally. It has the best efficiency. I selected the LEDs for the most identical voltage drop at a current of 20 mA. Assembled a flashlight in a case for two AA batteries. I shortened the place for the batteries to fit the size of AAA batteries, and in the freed space I assembled this circuit by surface mounting. A case for three AA batteries will work well. You will need to install only two, and place the scheme in place of the third.

The efficiency of the resulting device.
Input U I P Output U I P Efficiency
Volt mA mW Volt mA mW %
3.03 90 273 3.53 62 219 80
1.78 180 320 3.53 62 219 68
1.28 290 371 3.53 62 219 59

Replacing the light bulb of the flashlight “Zhuchok” with a module from the companyLuxionLumiledLXHL-NW 98.
We get a dazzlingly bright flashlight, with a very light press (compared to a light bulb).

Modification scheme and module parameters.

StepUP DC-DC converters ADP1110 from Analog devices.

Power supply: 1 or 2 batteries 1.5V operability is maintained up to Uin.=0.9V
Consumption:
*with open switch S1 = 300mA
*with switch closed S1 = 110mA

LED electronic flashlight
Powered by just one AA or AAA finger-type battery on a microcircuit (KR1446PN1), which is a complete analogue of the MAX756 microcircuit (MAX731) and has almost identical characteristics.

A flashlight is taken as a basis, in which two AA batteries (accumulators) are used as a power source.
The converter board is placed in the lantern instead of the second battery. On one end of the board, a tinned sheet contact is soldered to power the circuit, and on the other, an LED. A circle of the same tin is put on the conclusions of the LED. The diameter of the circle should be slightly larger than the diameter of the reflector base (by 0.2-0.5 mm), into which the cartridge is inserted. One of the terminals of the diode (negative) is soldered to the mug, the second (positive) passes through and is insulated with a piece of PVC or fluoroplastic tubing. The purpose of the circle is twofold. It provides the structure with the necessary rigidity and at the same time serves to close the negative contact of the circuit. A lamp with a cartridge is removed from the lantern in advance and a circuit with an LED is placed instead. Before installation on the board, the LED leads are shortened in such a way as to ensure a tight, play-free fit “in place”. Typically, the length of the leads (excluding soldering to the board) is equal to the length of the protruding part of the fully screwed lamp base.
The connection diagram of the board and the battery is shown in fig. 9.2.
Next, the lantern is assembled and its performance is checked. If the circuit is assembled correctly, then no settings are required.

The design uses standard installation elements: capacitors of the K50-35 type, EC-24 chokes with an inductance of 18-22 μH, LEDs with a brightness of 5-10 cd with a diameter of 5 or 10 mm. Of course, it is also possible to use other LEDs with a supply voltage of 2.4-5 V. The circuit has a sufficient power reserve and allows you to power even LEDs with a brightness of up to 25 cd!

On some test results of this design.
The lantern modified in this way worked with a “fresh” battery without interruption, in the switched on state, for more than 20 hours! For comparison, the same flashlight in the "standard" configuration (that is, with a lamp and two "fresh" batteries from the same batch) worked for only 4 hours.
And one more important point. If rechargeable batteries are used in this design, it is easy to monitor the state of their discharge level. The fact is that the converter on the KR1446PN1 chip starts up stably at an input voltage of 0.8-0.9 V. And the glow of the LEDs is consistently bright until the battery voltage reaches this critical threshold. The lamp will still burn at this voltage, of course, but it is hardly possible to speak of it as a real light source.

Rice. 9.2Figure 9.3

The printed circuit board of the device is shown in fig. 9.3, and the location of the elements - in fig. 9.4.

Turning the flashlight on and off with one button

The circuit is assembled on a CD4013 D-trigger chip and an IRF630 field effect transistor in the "off" mode. the current consumption of the circuit is practically 0. For stable operation of the D-flip-flop, a filter resistor and a capacitor are connected to the input of the microcircuit, their function is to eliminate contact bounce. It is better not to connect unused microcircuit pins anywhere. The microcircuit operates from 2 to 12 volts; any powerful field-effect transistor can be used as a power switch, because. the drain-source resistance of the field-effect transistor is negligible and does not load the output of the microcircuit.

CD4013A in SO-14 package, analogue to K561TM2, 564TM2

Simple generator circuits.

Allow to feed the LED with ignition voltage 2-3V from 1-1.5V. Short pulses of increased potential open the p-n junction. The efficiency of course decreases, but this device allows you to "squeeze out" almost all of its resource from an autonomous power source.
Wire 0.1 mm - 100-300 turns with a tap from the middle, wound on a toroidal ring.

Dimmable LED flashlight with beacon mode

The power supply of the microcircuit - a generator with an adjustable duty cycle (K561LE5 or 564LE5) that controls the electronic key, in the proposed device is carried out from a step-up voltage converter, which allows you to power the lamp from one galvanic cell 1.5.
The converter is made on transistors VT1, VT2 according to the transformer oscillator circuit with positive current feedback.
The oscillator circuit with an adjustable duty cycle on the K561LE5 chip mentioned above has been slightly modified in order to improve the linearity of current regulation.
The minimum current consumption of the flashlight with six parallel-connected super-bright L-53MWC white LEDs from Kingbnght is 2.3 mA. The dependence of the current consumption on the number of LEDs is directly proportional.
The "Beacon" mode, when the LEDs flash brightly at a low frequency and then go out, is implemented by setting the brightness control to maximum and turning on the flashlight again. The desired frequency of light flashes is regulated by the selection of the capacitor C3.
The flashlight remains operational when the voltage drops to 1.1v, although the brightness decreases significantly
A field-effect transistor with an insulated gate KP501A (KR1014KT1V) was used as an electronic key. In terms of the control circuit, it is in good agreement with the K561LE5 microcircuit. The KP501A transistor has the following limiting parameters, the drain-source voltage is 240 V; gate-source voltage - 20 V. drain current - 0.18 A; power - 0.5 W
It is permissible to connect transistors in parallel, preferably from the same batch. Possible replacement - KP504 with any letter index. For field-effect transistors IRF540, the supply voltage of the DD1. generated by the converter must be increased to 10 V
In a lamp with six L-53MWC LEDs connected in parallel, the current consumption is approximately equal to 120 mA when the second transistor is connected in parallel to VT3 - 140 mA
Transformer T1 is wound on a ferrite ring 2000NM K10-6 "4.5. The windings are wound in two wires, and the end of the first winding is connected to the beginning of the second winding. The primary winding contains 2-10 turns, the secondary - 2 * 20 turns Wire diameter - 0.37 mm. brand - PEV-2. The inductor is wound on the same magnetic circuit without a gap with the same wire in one layer, the number of turns is 38. The inductance of the inductor is 860 μH

Converter circuit for LED from 0.4 to 3V- powered by one AAA battery. This flashlight increases the input voltage to the required voltage with a simple DC-DC converter.

The output voltage is approximately 7 watts (depending on the voltage of the installed LEDs).

Building the LED Head Lamp

As for the transformer in the DC-DC converter. You must make it yourself. The image shows how to assemble the transformer.

Another version of converters for LEDs _http://belza.cz/ledlight/ledm.htm

Flashlight on a lead-acid sealed battery with a charger.

Lead acid sealed batteries are currently the cheapest. The electrolyte in them is in the form of a gel, so the batteries allow operation in any spatial position and do not produce any harmful fumes. They are characterized by great durability, if you do not allow deep discharge. Theoretically, they are not afraid of overcharging, but this should not be abused. Batteries can be recharged at any time without waiting for them to be completely discharged.
Lead-acid sealed batteries are suitable for use in portable flashlights used in the household, in summer cottages, and in production.

Fig.1. Diagram of an electric lantern

The electrical circuit diagram of a flashlight with a charger for a 6-volt battery, which allows in a simple way to prevent deep discharge of the battery and thus increase its service life, is shown in the figure. It contains a factory-made or self-made transformer power supply and a charger-switching device mounted in the lamp housing.
In the author's version, a standard block designed to power modems is used as a transformer unit. The output AC voltage of the block is 12 or 15 V, the load current is 1 A. There are also such blocks with built-in rectifiers. They are also suitable for this purpose.
The alternating voltage from the transformer unit is supplied to the charging and switching device, which contains a plug for connecting the charger X2, a diode bridge VD1, a current stabilizer (DA1, R1, HL1), a GB battery, a toggle switch S1, an emergency power button S2, an incandescent lamp HL2. Each time the toggle switch S1 is turned on, the battery voltage is supplied to the relay K1, its contacts K1.1 close, supplying current to the base of the transistor VT1. The transistor turns on by passing current through the lamp HL2. The lamp is turned off by switching the toggle switch S1 to its original position, in which the battery is disconnected from the winding of relay K1.
The allowable battery discharge voltage is selected at the level of 4.5 V. It is determined by the turn-on voltage of relay K1. You can change the allowable value of the discharge voltage using the resistor R2. With an increase in the value of the resistor, the allowable discharge voltage increases, and vice versa. If the battery voltage is below 4.5 V, then the relay will not turn on, therefore, voltage will not be applied to the base of the transistor VT1, which turns on the HL2 lamp. This means that the battery needs to be charged. At a voltage of 4.5 V, the illumination created by the flashlight is not bad. In case of emergency, you can turn on the flashlight at low voltage with the S2 button, provided that the S1 toggle switch is first turned on.
A constant voltage can also be applied to the input of the charging-switching device, without paying attention to the polarity of the connected devices.
To transfer the flashlight to the charge mode, it is necessary to dock the X1 socket of the transformer unit with the X2 plug located on the lamp body, and then plug the plug (not shown in the figure) of the transformer unit into the 220 V network.
In the above embodiment, a 4.2 Ah battery is used. Therefore, it can be charged with a current of 0.42 A. The battery is charged with direct current. The current stabilizer contains only three parts: an integrated voltage regulator DA1 type KR142EN5A or imported 7805, an HL1 LED and a resistor R1. The LED, in addition to working in a current stabilizer, also performs the function of an indicator of the battery charge mode.
Setting up the electrical circuit of the flashlight is reduced to adjusting the current of the battery charge. The charging current (in amperes) is usually chosen ten times less than the numerical value of the battery capacity (in ampere-hours).
For tuning, it is best to assemble the current stabilizer circuit separately. Instead of a battery load, connect an ammeter for a current of 2 ... 5 A to the connection point of the cathode of the LED and resistor R1. By selecting resistor R1, set the calculated charge current using the ammeter.
Relay K1 - reed switch RES64, passport RS4.569.724. The HL2 lamp consumes a current of approximately 1A.
The KT829 transistor can be used with any letter index. These transistors are composite and have a high current gain of 750. This should be taken into account in case of replacement.
In the author's version, the DA1 chip is installed on a standard ribbed heatsink with dimensions of 40x50x30 mm. Resistor R1 consists of two 12W wirewound resistors connected in series.

Scheme:

REPAIR OF LED FLASHLIGHT

Part ratings (C, D, R)
C = 1 uF. R1 = 470 kOhm. R2 = 22 kOhm.
1D, 2D - KD105A (admissible voltage 400V limit current 300 mA.)
Provides:
charging current = 65 - 70mA.
voltage = 3.6V.

LED Treiber PR4401 SOT23

Here you can see what the results of the experiment led to.

The circuit offered to your attention was used to power an LED flashlight, recharge a mobile phone from two metal hydrite batteries, when creating a microcontroller device, a radio microphone. In each case, the operation of the circuit was flawless. The list where you can use the MAX1674 can be continued for a long time.

The easiest way to get a more or less stable current through the LED is to connect it to the unregulated power circuit through a resistor. Keep in mind that the supply voltage must be at least twice the operating voltage of the LED. The current through the LED is calculated by the formula:
I led \u003d (Umax. supply - U working diode) : R1

This scheme is extremely simple and in many cases justified, but it should be used where there is no need to save electricity, and there are no high requirements for reliability.
More stable circuits - based on linear stabilizers:

As stabilizers, it is better to choose adjustable, or fixed voltage, but it should be as close as possible to the voltage on the LED or a string of LEDs connected in series.
Stabilizers like LM 317 are very suitable.
German text: iel war es, mit nur einer NiCd-Zelle (AAA, 250mAh) eine der neuen ultrahellen LEDs mit 5600mCd zu betreiben. Diese LEDs benötigen 3,6V/20mA. Ich habe Ihre Schaltung zunächst unverändert übernommen, als Induktivität hatte ich allerdings nur eine mit 1,4mH zur Hand. Die Schaltung lief auf Anhieb! Allerdings ließ die Leuchtstärke doch noch zu wünschen übrig. Mehr zufällig stellte ich fest, dass die LED extrem heller wurde, wenn ich ein Spannungsmessgerät parallel zur LED schaltete!??? Tatsächlich waren es nur die Messschnüre, bzw. deren Kapazität, die den Effekt bewirkten. Mit einem Oszilloskop konnte ich dann feststellen, dass in dem Moment die Frequenz stark anstieg. Hm, also habe ich den 100nF-Condensator gegen einen 4.7nF Typ ausgetauscht und schon war die Helligkeit wie gewünscht. Anschließend habe ich dann nur noch durch Ausprobieren die beste Spule aus meiner Sammlung gesucht... Das beste Ergebnis hatte ich mit einem alten Sperrkreis für den 19KHz Pilotton (UKW), aus dem ich die Kreiskapazität entfernt habe. Und hier ist sie nun, die Mini-Taschenlampe:

Sources:
http://pro-radio.ru/
http://radiokot.ru/

LED strips are now used everywhere and sometimes pieces of such strips, strips with LEDs burned out in places, fall into the hands. And there are a lot of whole, working LEDs and it’s a pity to throw away such goodness, I want to use them somewhere. There are also various types of batteries. In particular, we will consider the elements of a "dead" Ni-Cd (nickel-cadmium) battery. From all this rubbish, you can build a solid home-made lantern, with a high probability better than the factory one.

LED strip how to check

As a rule, LED strips are rated for 12 volts and consist of many independent segments connected in parallel to form a strip. This means that if any element fails, only the corresponding element loses its functionality, the remaining segments of the LED strip continue to work.

Actually, you just need to apply a supply voltage of 12 volts to special contact points that are on each piece of tape. In this case, the voltage will go to all segments of the tape and it will become clear where the non-working sections are.

Each segment consists of 3 LEDs and a current limiting resistor connected in series. If you divide 12 volts by 3 (number of LEDs), you get 4 volts per LED. This is the supply voltage of one LED - 4 volts. I emphasize, since the resistor limits the entire circuit, a voltage of 3.5 volts is enough for the diode. Knowing this voltage, we can directly test any LED on the tape individually. This can be done by touching the leads of the LED with probes connected to a power supply with a voltage of 3.5 volts.

For these purposes, you can use a laboratory, regulated power supply or mobile phone charger. The charger is not recommended to be connected directly to the LED, because its voltage is about 5 volts and theoretically the LED can burn out from a large current. To prevent this from happening, you need to connect the charger through a 100 ohm resistor, so we will limit the current.

I made myself such a simple device - charging from a mobile with crocodiles instead of a plug. It is very convenient for turning on cell phones without a battery, recharging batteries instead of a "frog" and other things. Good for testing LEDs too.

For the LED, the polarity of the voltage is important, if you confuse the plus with the minus, the diode will not light up. This is not a problem, the polarity of each LED is usually indicated on the tape, if not, then you need to try this and that. From the confused pluses or minuses, the diode will not deteriorate.

LED lamp

For a flashlight, it is necessary to make a light-emitting unit, a lamp. Actually, you need to dismantle the LEDs from the tape and group them according to your taste and color, by quantity, brightness and supply voltage.

To remove from the tape, I used a utility knife, carefully cutting off the LEDs directly with pieces of the conductive wires of the tape. I tried to solder, but something I did badly succeeded. Having picked 30-40 pieces, I stopped, more than enough for a flashlight and other crafts.

Connect the LEDs according to a simple rule: 4 volts per 1 or several diodes in parallel. That is, if the assembly is powered from a source of no more than 5 volts, no matter how many LEDs there are, they must be soldered in parallel. If you plan to power the assembly from 12 volts, you need to group 3 consecutive segments with an equal number of diodes in each. Here is an example of an assembly that I soldered from 24 LEDs, dividing them into 3 consecutive sections of 8 pieces. It is rated for 12 volts.

Each of the three sections of this element is designed for a voltage of about 4 volts. The sections are connected in series, so the entire assembly is powered by 12 volts.

Someone writes that LEDs should not be connected in parallel without an individual limiting resistor. Maybe this is correct, but I do not focus on such trifles. For a long service life, in my opinion, it is more important to choose a current-limiting resistor for the entire element and it should be selected not by measuring the current, but by feeling the working LEDs for heating. But more on that later.

I decided to make a flashlight powered by 3 nickel-cadmium cells from a used screwdriver battery. The voltage of each element is 1.2 volts, therefore 3 elements connected in series give 3.6 volts. We will focus on this tension.

By connecting 3 battery cells to 8 parallel diodes, I measured the current - about 180 milliamps. It was decided to make a light-emitting element of 8 LEDs, just as it fits successfully into a reflector from a halogen spot lamp.

As a base, I took a piece of foil fiberglass about 1cmX1cm, it will fit 8 LEDs in two rows. I cut 2 separating strips in the foil - the middle contact will be "-", the two extreme ones will be "+".

For soldering such small parts, my 15-watt soldering iron is too much, or rather too big a sting. You can make a tip for soldering SMD components from a piece of 2.5mm electrical wire. To keep the new tip in place in the large hole in the heater, you can bend the wire in half or add additional pieces of wire to the large hole.

The base is tinned with rosin solder and the LEDs are soldered with polarity. Cathodes ("-") are soldered to the middle strip, and anodes ("+") to the extreme ones. The connecting wires are soldered, the extreme strips are connected by a jumper.

You need to check the soldered structure by connecting it to a 3.5-4 volt source or through a resistor to the phone charger. Do not forget about the polarity of the inclusion. It remains to come up with a flashlight reflector, I took a reflector from a halogen lamp. The light element must be securely fixed in the reflector, for example with glue.

Unfortunately, the photo cannot convey the brightness of the glow of the assembled structure, I’ll say from myself: it doesn’t blind very well!

Battery

To power the flashlight, I decided to use rechargeable batteries from a "dead" screwdriver battery. I took out all 10 elements from the case. The screwdriver worked on this battery for 5-10 minutes and sat down, according to my version, the elements of this battery may well be suitable for the flashlight to work. After all, a flashlight needs currents that are much smaller than for a screwdriver.

I immediately unhooked three elements from the common bundle, they will just give a voltage of 3.6 volts.

I measured the voltage on each element individually - all were about 1.1 V, only one showed 0. Apparently this is a faulty bank, it's in the trash. The rest will still work. Three cans will be enough for my LED assembly.

Having studied the Internet, I deduced important information for myself about nickel-cadmium batteries: the nominal voltage of each cell is 1.2 volts, the bank should be charged to a voltage of 1.4 volts (the voltage on the bank without load), it should be discharged at least 0.9 volts - if several cells are composed in series, then not less than 1 volt per element. You can charge with a current of a tenth of the capacity (in my case 1.2A / h = 0.12A), but in fact it can be large (the screwdriver is charged for no more than an hour, which means the charging currents are at least 1.2A). For training / recovery, it is useful to discharge the battery to 1 V with some load and charge again, so several times. At the same time, estimate the approximate operating time of the flashlight.

So, for three elements connected in series, the parameters are as follows: charging voltage 1.4X3=4.2 volts, nominal voltage 1.2X3=3.6 volts, charge current - which will give a mobile charger with a stabilizer of my manufacture.

The only not clear moment: how to measure the minimum voltage on discharged batteries. Before connecting my lamp, there was a voltage of 3.5 volts on three elements, when connected - 2.8 volts, the voltage is quickly restored when disconnected again to 3.5 volts. I decided this: at the load, the voltage should not fall below 2.7 volts (0.9 V per element), without load it is desirable that there be 3 volts (1 V per element). However, it will take a long time to discharge, the longer you discharge, the more stable the voltage, it stops dropping quickly on the lit LEDs!

I discharged my already discharged batteries for several hours, sometimes turning off the lamp for several minutes. As a result, it turned out 2.71 V with a connected lamp and 3.45 V without a load, I did not dare to discharge further. I note that the LEDs continued to shine, albeit dimly.

Charger for nickel-cadmium batteries

Now you should build a charger for a flashlight. The main requirement is that the output voltage should not exceed 4.2 V.

If you plan to power the charger from any source of more than 6 volts, a simple circuit on KR142EN12A is relevant, this is a very common microcircuit for regulated, stabilized power. Foreign analogue of LM317. Here is a diagram of the charger on this chip:

But this scheme did not fit into my idea - versatility and maximum convenience for charging. After all, for this device you will need to make a transformer with a rectifier or use a ready-made power supply. I decided to make it possible to charge batteries from a mobile phone charger and a computer USB port. For implementation, a more complicated scheme is required:

The field effect transistor for this circuit can be taken from a faulty motherboard and other computer peripherals, I cut it off from an old video card. There are plenty of such transistors on the motherboard near the processor and not only. To be sure of your choice, you need to drive the transistor number into the search and make sure from the datasheets that this is a field transistor with an N-channel.

As a zener diode, I took the TL431 chip, it is found in almost every charger from a mobile phone or in other switching power supplies. The outputs of this microcircuit must be connected as in the figure:

I assembled the circuit on a piece of textolite, immediately provided a USB socket for connection. In addition to the circuit, I soldered one LED near the socket to indicate charging (that voltage is being supplied to the USB port).

A few explanations for the diagram Since the charging circuit will be connected to the battery all the time, the VD2 diode is necessary so that the battery does not discharge through the stabilizer elements. By selecting R4, you need to achieve a voltage of 4.4 V at the specified control point, you need to measure it with the battery unhooked, 0.2 volts is a margin for drawdown. And in general, 4.4 V does not go beyond the recommended voltage for three battery cans.

The charger circuit can be greatly simplified, but it will only have to be charged from a 5 V source (the USB port of the computer meets this requirement), if the charger of the phone produces a higher voltage, it cannot be used. According to a simplified scheme, theoretically, batteries can be recharged, but in practice, batteries are charged this way in many factory products.

LED current limit

To prevent overheating of the LEDs, and at the same time reduce the current consumption from the battery, you need to select a current-limiting resistor. I picked it up without any devices, estimating the heat by touch and controlling the brightness of the glow by eye. The selection must be made on a charged battery, you should find the optimal value between heating and brightness. I got a 5.1 ohm resistor.

Working hours

I made several charges and discharges and got the following results: charging time - 7-8 hours, with the lamp continuously on, the battery is discharged to 2.7 V in about 5 hours. However, when turned off for a few minutes, the battery recovers a little and can work for another half an hour, and so on several times. This means that the flashlight will work for a long time if it does not shine all the time, but in practice it does. Even if you use it practically without turning it off, it should be enough for a couple of nights.

Of course, a longer time of work without interruption was expected, but do not forget that the batteries were taken from a "dead" screwdriver battery.

housing for lantern

The resulting device needs to be placed somewhere, to make some kind of convenient case.

I wanted to place batteries with an LED flashlight in a polypropylene water pipe, but the cans did not even fit into a 32 mm pipe, because the inner diameter of the pipe is much smaller. As a result, I settled on couplings for 32 mm polypropylene. I took 4 couplings and 1 plug, glued them together with glue.

By gluing everything into one structure, we got a very massive lantern, about 4 cm in diameter. If you use any other pipe, you can significantly reduce the size of the lantern.

Having wrapped the whole thing with electrical tape for a better look, we got this lantern:

Afterword

In conclusion, I would like to say a few words about the resulting review. Not every USB port of a computer can charge this flashlight, it all depends on its load capacity, 0.5 A should be enough. For comparison, cell phones, when connected to some computers, may show charging, but in fact there is no charge. In other words, if the computer charges the phone, then the flashlight will also charge.

The FET circuit can be used to charge 1 or 2 battery cells from USB, you just need to adjust the voltage accordingly.

Somehow I ordered SMD 5630 LEDs from China for a future robot, which I have been assembling for half a year, and now there are a lot of diodes, a whole bay, and the surplus must be used somewhere 🙂 I decided to assemble a backlight for the door at the entrance to the house. Having started experimenting, it turned out that you can make good flashlights for lighting in various places in the house, and most importantly, everything can be made from improvised materials! 🙂

The first step is to collect the necessary materials, namely:

A lid from kefir or milk - the basis of the flashlight body
SMD 5630 or 5730 LEDs
Resistors 3.3 - 12 ohms (depending on power supply)
Mounting or PCB
wires
Plexiglas - as case cover
3.7 Volt battery or 5 Volt power supply

In this article, I used SMD 5630 LEDs with an operating voltage of 3.3 Volts and a current of 150 milliamps. The power source is a cell phone battery with a capacity of 5000 mAh and a voltage of 3.8 volts. At this voltage, 3.3 Ohm resistors are needed, but in the absence of these, 2.2 Ohm had to be used.

When the battery is discharged, its voltage drops and generally does not exceed 3.6 volts, which is consistent with the resistance rating of 2.2 ohms.

A small piece of circuit board is suitable for mounting LEDs and resistors.

We solder diodes, resistors and power wires according to the diagram.

The diagram shows the resistor values for 3.7 and 5 Volts. For a brighter glow, you can add additional LEDs - 3, 4 or more pieces, depending on the size of the housing cover and the required brightness.

After that, you should check the operability of the circuit by applying power to the appropriate wires.

Now you can fix the board in the cover with hot glue.

We pass the wires through the side opening of the cover, also fixing them with hot glue.

Now we fix the transparent plexiglass cover with the help of second super glue.

I cut out the cover using a 44 mm crown and a screwdriver from a sheet of plexiglass.

Apply glue to the edges of the glass. It can be dots, or it can be a solid line.

Firmly press the body of the flashlight and hold for a few seconds.

Lid in place. The flashlight is almost ready.

The hole in the center of the flashlight, obtained by drilling a circle of plexiglass, can be closed with a furniture plug.

The body of the flashlight is ready. If desired, you can sand the plexiglass with sandpaper to obtain a matte surface. In the photo below, on the left, a flashlight with transparent glass, and on the right, with a matte glass obtained with sandpaper.

Connect both flashlights to a power source.

This is what the finished product looks like.

The brightness of such lanterns is enough to illuminate the whole room.

For example, you can make a backlight on a bookshelf.

Or on the shelf with clothes in the closet.

Light sources of new generation - light-emitting diodes - despite still high cost become more and more popular.

Due to their low power consumption, they are successfully used not only in stationary lighting fixtures, but also in stand-alone, battery-powered ones.

In this article we will talk about how you can make an LED flashlight with your own hands and what advantages it will have in comparison with the usual one.

An LED (foreign name - Light Emitting Diode or LED), like a conventional diode, consists of two semiconductors with electronic and hole conductivity.

But in this case, such materials are used, for which the glow in the pn-junction zone is characteristic.

Generally speaking, LEDs have been used in electronics for a long time.

But before they glowed barely, and therefore were used only as indicators, for example, indicating that the device was turned on.

With the development of technology, LEDs have learned to make much brighter, so they have become full-fledged light sources. At the same time, their cost is constantly decreasing, although, of course, they are still very far from an ordinary light bulb.

But many buyers are willing to overpay, because LEDs have a number of advantages:

They consume 10-15 times less electricity than incandescent lamps of the same brightness.
They simply have a huge resource, which is expressed in 50 thousand hours of work. Moreover, manufacturers back up their promises with a warranty period of 2 or even 3 years.
They emit white light, very similar to natural.
They are much less afraid of shocks and vibrations than other light sources.
They have high resistance to voltage drops.

Thanks to all these qualities, LEDs today are confidently replacing other light sources from almost everywhere. They are used in everyday life, and in car headlights, and in advertising, and in portable flashlights, one of which we will now learn how to make.

Required elements for manufacturing

First of all, you need to get all the components that will make up the device.

There are not many of them:

Light-emitting diode.
Ferrite ring with a diameter of 10 - 15 mm.
Wire for winding with a diameter of 0.1 and 0.25 mm (pieces of 20 - 30 cm).
Resistor 1 kOhm.
NPN transistor.
Battery.

Well, if you can get the case from a purchased flashlight. If it is not there, any base can be used to fasten the components.

Assembly diagram

If everything is ready, we can start:

We make a transformer: a ferrite ring will act as a magnetic circuit of a home-made transformer. First, 45 turns of winding wire with a diameter of 0.25 mm are wound on it, forming a secondary winding. In the future, an LED will be connected to it. Next, from a wire with a diameter of 0.1 mm, you need to make a primary winding with 30 turns, which will be connected to the base of the transistor.
Resistor selection: The base resistor should be approximately 2 kΩ.

But the value of the second resistor must be selected. It is done like this:

a tuning (variable) resistor is installed in its place.
Having connected the flashlight to a new battery, set such a resistance on the variable resistor that a current of 22 - 25 mA flows through the LED.
Measure the resistance value on a variable resistor and install a constant resistor with the same rating instead.

As you can see, the circuit is extremely simple and the probability of error can be considered minimal.

Do-it-yourself LED flashlight - diagram

If the flashlight still turned out to be inoperative, the reason may be as follows:

In the manufacture of the windings, the condition of multidirectional currents was not observed. In this case, the generation of current in the secondary winding will not occur. For the circuit to be working, you must either wind the windings in different directions, or swap the conclusions of one of the windings.
The winding contains too few turns. It should be borne in mind that the required minimum is 15 turns.

If they are present in the winding in a smaller amount, the generation of current will again be impossible.

DIY 12 volt LED flashlight

Those who need not a flashlight, but a whole spotlight in miniature, can assemble a device with a more powerful power source. As the latter, a 12-volt battery will be used. This product will have a slightly larger size, but it will still be easy enough to carry.

To create a high power light source, you need to prepare the following:

polymer pipe with a diameter of about 50 mm;
glue for gluing PVC parts;
a pair of threaded fittings for PVC pipes;
screw cap;
toggle switch;
12 V LED;
12-volt battery;
auxiliary elements for the installation of electrical wiring - heat shrink tubes, electrical tape, plastic clamps.

As a power source, you can use several batteries from broken radio-controlled toys, which are combined into one 12 V battery. Batteries, depending on their type, will need from 8 to 12.

A 12-volt LED flashlight is assembled like this:

We solder pieces of wire to the contacts of the LED, which are a couple of centimeters longer than the battery. In this case, it is necessary to ensure reliable isolation of the connections.
The wires connected to the battery and the LED are equipped with special connectors that allow you to make quick connections.
When assembling the circuit, the toggle switch is installed so that it is on the opposite side with respect to the LED. The electronic filling is ready, and if the tests have shown that it works properly, you can start manufacturing the case.

The case is made of a polymeric pipe. It is done like this:

The pipe is cut to the desired length, after which all the electronics are placed inside it.
We put the battery on glue so that it remains motionless during carrying and manipulating the flashlight. Otherwise, a heavy battery may hit the LED element and disable it.
Glue the threaded fitting to the pipe at both ends. Glue does not need to be saved - the connection should be tight. Otherwise, water may seep into the housing at this point.
We fix the toggle switch inside the fitting installed on the side opposite to the LED. We put the switch on the glue, while it should not protrude outward so that the plug can be screwed onto the fitting.

To switch the toggle switch, the plug will need to be unscrewed, then reinstalled. This is somewhat inconvenient, but this solution ensures complete tightness of the case.

A question of price and quality

Of all the flashlight components, the 12-volt LED is the most expensive. You will have to pay 4 - 5 USD for it.

Everything else can be obtained for free: batteries, as already mentioned, are removed from radio-controlled toys, plastic pipes and parts very often remain as waste after installing plumbing or heating in a house.

If absolutely all the components have to be purchased in a store, then the cost of the lighting device will result in about 10 USD.

A homemade lamp from an LED strip can be built quickly and easily. - see the manufacturing instructions and make your own unique product.

Read about how to properly install the LED strip with your own hands.

Conclusion

A handy flashlight that gives a bright light and at the same time is able to work for a long time without recharging the battery is always needed in the household. As you can see, you can easily make it yourself, which will save you some money. The main thing is to be careful and strictly adhere to all the recommendations set out in the article.

Step 1:

The first thing to do is to assemble the battery pack.

Step 2:

Install the LED and connect the wires. The wiring was mounted according to the diagram shown in the video.

Step 3: Preparing the Lantern Body

Due to the fact that a significant amount of heat is generated during the operation of a high-power light source, it is necessary to cut ventilation holes in the housing. We will close them with ventilation grilles.