Many have seen miniature flashlights powered by a single 1.5 volt battery. Theoretically, this voltage is not enough to light up a white LED. This means that there is some kind of device hidden under the case that increases the voltage to the required level. This device can be made with your own hands within half an hour, using inexpensive and accessible parts. This article will tell you in detail how to connect an LED to a 1.5V battery.
Scheme and principle of its operation
The LED power supply circuit from a 1.5V battery is shown in the figure. The main functional elements are a single-stage transistor amplifier and pulse transformer, due to which deep positive feedback is achieved. The base current of the transistor is limited by resistor R1, and to optimize the output parameters, a diode VD1 and a capacitor C1 are installed, which will be discussed a little later.
The LED power supply circuit from one battery operates on the principle of a blocking generator. The formation of pulses is carried out by unlocking the transistor and transitioning it to saturation mode using positive feedback. Exit from saturation occurs due to a decrease in the base current. The transistor turns off and the transformer energy is dumped into the load. As a result, the LED flashes for a short period of time.
Now let's take a closer look at the operation of the circuit shown in the figure. It is known that the current in an inductor cannot change instantly. First, when voltage is applied from the battery, the transistor is in the closed state. A gradual increase in current in the collector and then in the base winding leads to a smooth unlocking of the transistor. This leads to an increase in the collector current, which also flows through the collector winding. This increase in current is transformed into the base winding and further increases the base current.
As a result of such an avalanche-like process, saturation enters the transistor. In saturation mode, the collector current stops increasing, which means the voltage on the base winding will become zero. This will lead to a decrease in the base current and the transistor will exit saturation. The voltage on the base winding changes polarity, which contributes to almost instantaneous blocking of the transistor. As a result, all the accumulated energy flows into the load. The LED flashes and passes a current through itself, which decreases from the collector current value to zero. At this time interval, a reverse blocking process occurs in the transformer, which leads to the next unlocking of the transistor. Then the cycle repeats.
The circuit operates at a frequency of several tens of kilohertz. Therefore, thousands of flashes per second are perceived by the human eye as a constant glow. But the circuit can be slightly modified by eliminating current dips through the LED to zero, and adding a smoothing capacitor and a diode to it. Capacitor C1 is connected in parallel to the LED, observing polarity, and diode VD1 is connected in series, into the load current flow circuit. VD1 prevents the capacitor from discharging to an open transistor.
Connecting an LED to a battery, according to this diagram, requires compliance with one rule: you cannot turn on the assembled device without a load (the transistor may burn out).
Calculation and assembly details
All radio components necessary for practical implementation are inexpensive or are available in the stocks of radio amateurs. The exception is the transformer, which will require a little work.
The transformer is made by hand from a ferrite ring dismantled from a faulty compact fluorescent lamp or pulse block nutrition. The outer diameter of the ring is approximately 10 mm with possible tolerance in both directions. For winding, two single-core wires of the same length with a cross section of 0.5 mm 2 are used. Twisted pair cable used in a LAN network connection is ideal.
Both wires (preferably different colors) are folded towards each other and wound around the ring, laying the turns around the circumference. There should be 20 turns in total. In this case, the beginnings of the wires come out on one side, and the ends on the other. After this, the beginning of a wire of one color is connected to the end of a wire of a different color and connected to the positive of the battery. The two remaining ends are connected to the transistor collector and resistor.
The transistor is selected based on the highest collector current with a double margin to avoid overheating. In this case, KT315V or KT3102A is suitable. Instead, you can install the imported BC547A with the following parameters:
- maximum collector current – 100 mA;
- maximum collector-emitter voltage – 45V;
- gain h 21E – 100-220.
It is advisable to choose a transistor with a h21E value close to 100.
Having set the maximum operating collector current of 25 mA, you can calculate the base current: I B = I K / h 21E = 25/100 = 0.25 mA.
Theoretically, the resistance of resistor R1 can be calculated using the formula: R1=(U BAT -U BE)/I B =(1.5-0.6)/0.00025=3600 Ohm.
However, in practice, a resistor with a nominal value of 1 kOhm is sufficient, since the calculation does not take into account the input resistance of the power source and the high-frequency operating mode and the magnetizing current, which is the ballast component of the collector current. It should also be taken into account that as the emf of the battery decreases, a resistor with a lower resistance will be more effective. With a 1kOhm-0.125W±5% resistor, the amplitude value of the LED current does not exceed 26 mA.
The circuit can be powered not only from a 1.5V battery, but also from a 1.2V AA battery.
Diode VD1 in this case must have a low voltage drop in the open state. Schottky diodes of type 1N5817-1N5819 are suitable for this purpose, in which the voltage drop at low currents is 0.2-0.4V. Capacitor C1 is electrolytic at 10 uF-6.3V. This capacity is enough to smooth out the current ripples on the LED.
During operation, the battery loses capacity and the voltage at its terminals decreases. In this case, the LED will continue to glow as long as the condition is met: U BAT >U BE (on average 0.6V). Thus, the LED power supply circuit from one battery allows you to use a AA battery with maximum efficiency.
Printed circuit board
The printed circuit board of the simplest blocking generator can be downloaded. This is a single-sided board measuring 10 by 20 mm, which easily fits into the flashlight body. It is advisable to place the finished board with parts and wiring to the LED in a thermal tube and place it next to the battery. If you use an SMD transistor and a resistor, excluding the diode with a capacitor, you can make an even smaller board for the smallest flashlight.
Afterword
The considered circuit solution is effective in the case of switching on 1-3 LEDs of any color with a maximum current of up to 30 mA. To power a more powerful LED from a single battery, some adjustments will need to be made. In the above circuit, you can reduce the resistance of the resistor, thereby increasing the amplitude of the collector current (but not more than the maximum rated value).
To connect a 1W LED, you will have to replace all the parts of the circuit with more powerful ones: a transformer with a larger core and a transistor with a collector current of at least 500 mA. When setting up a circuit for a flashlight on one battery, you need to use an oscilloscope to monitor the LED current.
On the Internet you can find many diagrams for connecting an LED to a battery. At the same time, the authors do not hesitate to show photos of their measurements, where the current in the load exceeds the permissible value for a low-power LED (30 mA). Why doesn't the LED burn out? The fact is that most multimeters measure alternating voltage and current only in the range of 40-400 Hz, and this is stated in the instructions. But many radio amateurs do not know this nuance. Naturally, the multimeter cannot measure the LED current, pulsating at a frequency of tens of kHz, and displays a random number on the screen.
Read also
LEDs have long replaced incandescent light bulbs in almost all areas. This is understandable: LEDs are brighter than lamps, given their energy consumption.
But LEDs also have a number of disadvantages. Of course, we won’t talk about all of them, but we will discuss one. This is a high initial power threshold - it is about 1.8-2.2 volts. Naturally, you can’t power it from one battery...
To overcome this shortcoming, we will build a simple converter using an absolute minimum of parts.
Thanks to this converter, you can connect an LED (or several LEDs) to one battery and make a small flashlight.
We will need:
- Light-emitting diode.
- 2N3904 or BC547 silicon transistor, or any other n-p-n structure.
- Wire.
- Resistor 1 kOhm.
- Ring cores or ferrite cores.
Converter circuit
I will give you two diagrams. One for winding a ring transformer, the other for those who do not have a ring core on hand.
This is the simplest blocking generator, with a free excitation frequency. The idea is as old as time. The device will have a high efficiency.
Winding the inductor
Regardless of whether you use a ring core or a regular ferrite core, wind 10 turns of each winding. Your inductor is ready for this.
Generator check
We assemble according to the diagram and check. The generator should work and does not need adjustment.If suddenly, even though the elements are working properly, the LED does not light up, try changing the ends of one of the windings of the induction transformer.
Now the LED shines very brightly even with a dead battery. The lower limit of the power supply for the entire device is now somewhere around 0.6 volts.
The efficiency of a transformer with a ring core is slightly higher. Not critical of course, but just keep it in mind.
The availability and relatively low prices of ultra-bright light-emitting diodes (LEDs) allow them to be used in various amateur devices. Beginning radio amateurs who are using LEDs in their designs for the first time often wonder how to connect an LED to a battery? After reading this material, the reader will learn how to light an LED from almost any battery, what LED connection diagrams can be used in this or that case, how to calculate the circuit elements.
What batteries can the LED be connected to?
In principle, you can simply light the LED using any battery. Electronic circuits developed by radio amateurs and professionals make it possible to successfully cope with this task. Another thing is how long the circuit will operate continuously with a specific LED (LEDs) and a specific battery or batteries.
To estimate this time, you should know that one of the main characteristics of any battery, be it a chemical cell or a battery, is capacity. Battery capacity – C is expressed in ampere-hours. For example, the capacity of common AAA AA batteries, depending on the type and manufacturer, can range from 0.5 to 2.5 ampere-hours. In turn, light-emitting diodes are characterized by an operating current that can be tens and hundreds of milliamps. Thus, you can approximately calculate how long the battery will last using the formula:
T= (C*U baht)/(U work led *I work led)
In this formula, the numerator is the work that the battery can do, and the denominator is the power consumed by the light-emitting diode. The formula does not take into account the efficiency of the specific circuit and the fact that it is extremely problematic to fully use the entire battery capacity.
When designing battery-powered devices, they usually try to ensure that their current consumption does not exceed 10–30% of the battery capacity. Guided by this consideration and the above formula, you can estimate how many batteries of a given capacity are needed to power a particular LED.
How to connect from a AA 1.5V AA battery
Unfortunately, it doesn't exist simple way Power the LED from one AA battery. The fact is that the operating voltage of light-emitting diodes usually exceeds 1.5 V. For this value lies in the range of 3.2 - 3.4V. Therefore, to power the LED from one battery, you will need to assemble a voltage converter. Below is a diagram of a simple voltage converter with two transistors that can be used to power 1 – 2 super-bright LEDs with an operating current of 20 milliamps.
This converter is a blocking oscillator assembled on transistor VT2, transformer T1 and resistor R1. The blocking generator produces voltage pulses that are several times higher than the voltage of the power source. Diode VD1 rectifies these pulses. Inductor L1, capacitors C2 and C3 are elements of the anti-aliasing filter.
Transistor VT1, resistor R2 and zener diode VD2 are elements of a voltage stabilizer. When the voltage across capacitor C2 exceeds 3.3 V, the zener diode opens and a voltage drop is created across resistor R2. At the same time, the first transistor will open and lock VT2, the blocking generator will stop working. This ensures stabilization of the converter output voltage at 3.3 V.
It is better to use Schottky diodes as VD1, which have a low voltage drop in the open state.
Transformer T1 can be wound on a ferrite ring of grade 2000NN. The diameter of the ring can be 7 – 15 mm. As a core, you can use rings from converters of energy-saving light bulbs, filter coils of computer power supplies, etc. The windings are made of enameled wire with a diameter of 0.3 mm, 25 turns each.
This scheme can be painlessly simplified by eliminating stabilization elements. In principle, the circuit can do without a choke and one of the capacitors C2 or C3. Even a novice radio amateur can assemble a simplified circuit with his own hands.
The circuit is also good because it will operate continuously until the power supply voltage drops to 0.8 V.
How to connect 3V batteries
You can connect a super-bright LED to a 3V battery without using any additional parts. Since the operating voltage of the LED is slightly higher than 3 V, the LED will not shine at full strength. Sometimes it can even be useful. For example, using an LED with a switch and a 3 V disk battery (popularly called a tablet), used in computer motherboards, you can make a small flashlight keychain. This miniature flashlight can be useful in different situations.
From such a battery - 3 Volt tablets you can power an LED
Using a pair of 1.5 V batteries and a purchased or homemade converter to power one or more LEDs, you can make a more serious design. The diagram of one of these converters (boosters) is shown in the figure.
The booster based on the LM3410 chip and several attachments has the following characteristics:
- input voltage 2.7 – 5.5 V.
- maximum output current up to 2.4 A.
- number of connected LEDs from 1 to 5.
- conversion frequency from 0.8 to 1.6 MHz.
The output current of the converter can be adjusted by changing the resistance of the measuring resistor R1. Despite the fact that from the technical documentation it follows that the microcircuit is designed to connect 5 LEDs, in fact you can connect 6 to it. This is due to the fact that the maximum output voltage of the chip is 24 V. The LM3410 also allows LEDs to glow (dimming) . The fourth pin of the chip (DIMM) is used for these purposes. Dimming can be done by changing the input current of this pin.
How to connect 9V Krona batteries
“Krona” has a relatively small capacity and is not very suitable for powering high-power LEDs. The maximum current of such a battery should not exceed 30 - 40 mA. Therefore, it is better to connect 3 light-emitting diodes connected in series with an operating current of 20 mA to it. They, as in the case of connecting to a 3 volt battery, will not shine at full power, but the battery will last longer.
Krona battery power supply circuit
It is difficult to cover in one material all the variety of ways to connect LEDs to batteries with different voltages and capacities. We tried to talk about the most reliable and simple designs. We hope that this material will be useful to both beginners and more experienced radio amateurs.
From a battery with a voltage of 1.5 volts or lower, it’s simply not realistic. This is due to the fact that most LEDs have a voltage drop exceeding this figure.
How to light an LED from a 1.5 volt battery
A way out of this situation may be to use a simple one transistor and inductance. In essence, it is peculiar. The circuit is a simple blocking generator, powered by a 1.5 volt battery, generating fairly powerful pulses as a result of pumping energy into the inductor. The circuit is simple and can be assembled in literally 10 minutes.
The T1 inductor is made on a ferrite ring with a diameter of 7 millimeters (its dimensions are K7x4x3). The winding contains 21 turns, made of double-folded enameled PEV copper wire with a diameter of 0.35 millimeters.
After winding is completed, the end of one of the wires must be connected to the beginning of the other wire. The result is a tap from the center of the winding. By selecting the resistance, you can achieve better light output.
So we have a Panasonic RF-800UEE-K radio receiver; there is plenty of information on the Internet about all its advantages and disadvantages. On the plus side, I would like to note the very good quality of the tuner, wooden (plywood) case, decent sound quality for this segment of receivers. It is very easy to disassemble, no latches, five screws on the back panel and two more screws securing the front panel to the plywood body.
Disadvantages include mono sound and lack of normal bass. But there is an input and output; those who don’t have enough bass can connect it to external speakers.
The receiver is so successful that, in order not to put this device into the class of multimedia centers, the manufacturer cut down some of the functionality of the MP3 player and did not install backlighting on the receiver scale, although judging by the configuration of the front panel it was supposed to be there. The body is glued together from pressed shavings and is quite loose, but this is easy to fix.
We glue all the seams with joinery PVA with a “slide” until completely dry.
Then we impregnate the ends and insides with polyurethane varnish, it penetrates very well, so you will have to apply three or four generous layers.
After drying, the body stretches and begins to “sound” like the front soundboard of a guitar :-)
We measure the seat for installing the light, in our case it is a socket 90 mm long and 7 mm wide.
We cut the foil PCB into panels of the required size.
The receiver is powered by a voltage of 6V; for illumination I want to try orange and yellow LEDs with a direct voltage of 2.1V. I will put them in pairs, the excess voltage with such a circuit will be 1.8V, we will deposit it on a resistor. The resistor value is calculated according to Ohm's law R=U/I. In our case, U=1.8 V, and current I=20 mA (the maximum permissible forward current for this type of LED), it turns out that at R=90 Ohm everything should work, but we will go further and limit the current to 10-9mA, while There is no significant decrease in brightness. We get R=220 Ohm. The calculation can be made using the link provided at the bottom of this post.
I assemble two strips of yellow and orange on different types LEDs. In order not to make a fuss, I use one side of the foiled PCB as a minus, the other as a plus. 
Orange SMD LEDs gave a more saturated glow.
This plank went into action. I glue it with double-sided tape, and the LEDs shine strictly at the end of the scale, there is a technological gap there.
Magic scale.
Plus output to the power knob (volume control)
Minus on the central core of the power connector. With this switching scheme, the backlight will only work when operating from an external power supply; in battery mode, it will not illuminate, saving batteries. I think the manufacturer deliberately separated the two power circuits through a diode.
