Power led from 220. How to connect the LED to the lighting network. Scheme for connecting devices to a computer

Schemes for the simplest connection of LEDs to 220V

Because you need to correctly solve two problems at once:

1. Limit the forward current through the LED so it doesn't burn out.
2. Protect the LED from reverse current breakdown.

If you ignore any of these items, the LED will instantly be covered with a copper basin.

In the simplest case, you can limit the current through the LED with a resistor and / or a capacitor. And to prevent breakdown from reverse voltage, you can use a conventional diode or another LED.

Therefore, the simplest scheme for connecting an LED to 220V consists of only a few elements:

The protective diode can be almost anything, because. its reverse voltage will never exceed the forward voltage of the LED, and the current is limited by a resistor.

The resistance and power of the limiting (ballast) resistor depends on the operating current of the LED and is calculated according to Ohm's law:

R = (U in - U LED) / I

And the power dissipation of the resistor is calculated as follows:

P = (U in - U LED) 2 / R

where U in = 220 V,
U LED - direct (working) voltage of the LED. Usually it lies in the range of 1.5-3.5 V. For one or two LEDs, it can be neglected and, accordingly, the formula can be simplified to R \u003d U in / I,
I - LED current. For conventional indicator LEDs, the current will be 5-20 mA.

Ballast Resistor Calculation Example

Let's say we need to get the average current through the LED = 20mA, hence the resistor should be:

R \u003d 220V / 0.020A \u003d 11000 Ohm(we take two resistors: 10 + 1 kOhm)

P \u003d (220V) 2 / 11000 \u003d 4.4 W(we take with a margin: 5 W)

The required resistor value can be taken from the table below.

Table 1. Dependence of the LED current on the resistance of the ballast resistor.

Other connection options

In the previous circuits, the protective diode was connected in anti-parallel, but it can also be placed like this:

This is the second circuit for switching on 220 volt LEDs without a driver. In this circuit, the current through the resistor will be 2 times less than in the first variant. And, therefore, 4 times less power will be allocated on it. This is a definite plus.

But there is also a minus: the full (amplitude) mains voltage is applied to the protective diode, so any diode will not work here. You will have to pick up something with a reverse voltage of 400 V and above. But these days, that's not a problem at all. Perfect, for example, is the ubiquitous 1000 volt diode - 1N4007 (KD258).

Despite the common misconception, during the negative half-cycles of the mains voltage, the LED will still be in a state of electrical breakdown. But due to the fact that the resistance of the reverse-biased p-n junction of the protective diode is very high, the breakdown current will not be enough to disable the LED.

Attention! All the simplest circuits for connecting 220 volt LEDs have a direct galvanic connection with the network, so touching ANY point in the circuit is EXTREMELY DANGEROUS!

To reduce the value of the touch current, you need to halve the resistor into two parts, so that it turns out as shown in the pictures:

Thanks to this solution, even by swapping the phase and zero, the current through a person to the "ground" (in case of an accidental touch) cannot exceed 220/12000 = 0.018A. And it's not so dangerous anymore.

What about pulsations?

In both circuits, the LED will only glow during the positive half-cycle of the mains voltage. That is, it will flicker at a frequency of 50 Hz or 50 times per second, and the ripple span will be 100% (10 ms on, 10 ms off, and so on). It will be visible to the eye.

In addition, when flashing LEDs illuminate any moving objects, such as fan blades, bicycle wheels, etc., a stroboscopic effect will inevitably occur. In some cases, this effect may be unacceptable or even dangerous. For example, when working at a machine tool, it may seem that the cutter is stationary, but in fact it rotates at breakneck speed and is just waiting for you to stick your fingers into it.

To make the ripple less noticeable, you can double the LED turn-on frequency using a full-wave rectifier (diode bridge):

Note that compared to circuit #2, with the same resistor value, we got twice the average current. And, accordingly, four times the power dissipation of the resistors.

At the same time, there are no special requirements for the diode bridge, the main thing is that the diodes of which it consists can withstand half the operating current of the LED. The reverse voltage on each of the diodes will be quite negligible.

Still, as an option, you can organize an anti-parallel connection of two LEDs. Then one of them will burn during the positive half-wave, and the second - during the negative.

The trick is that with this inclusion, the maximum reverse voltage on each of the LEDs will be equal to the forward voltage of the other LED (a few volts maximum), so each of the LEDs will be reliably protected from breakdown.

LEDs should be placed as close to each other as possible. Ideally, try to find a dual LED, where both crystals are placed in the same package and each has its own conclusions (although I have never seen such ones).

Generally speaking, for LEDs that perform an indicator function, the magnitude of the pulsations is not very important. For them, the most important thing is the most noticeable difference between the on and off states (on / off indication, playback / recording, charge / discharge, normal / accident, etc.)

But when creating fixtures, you should always try to minimize pulsations. And not so much because of the dangers of the stroboscopic effect, but because of their harmful effects on the body.

What ripples are considered acceptable?

It all depends on the frequency: the lower it is, the more noticeable the ripple. At frequencies above 300 Hz, the ripples become completely invisible and are not normalized at all, that is, even 100% are considered the norm.

Despite the fact that light pulsations at frequencies of 60-80 Hz and above are not visually perceived, however, they can cause increased eye fatigue, general fatigue, anxiety, decreased visual performance, and even headaches.

To prevent the above consequences, the international standard IEEE 1789-2015 recommends the maximum level of brightness ripple for a frequency of 100 Hz - 8% (guaranteed safe level - 3%). For a frequency of 50 Hz, these will be 1.25% and 0.5%, respectively. But this is for perfectionists.

In fact, in order for the brightness pulsations of the LED to stop at least somehow annoying, it is enough that they do not exceed 15-20%. This is exactly the level of flickering of medium-power incandescent lamps, and no one has ever complained about them. Yes, and our Russian SNiP 23-05-95 allows light flickering at 20% (and only for particularly painstaking and responsible work, the requirement is increased to 10%).

In accordance with GOST 33393-2015 "Buildings and structures. Methods for measuring the pulsation coefficient of illumination" to assess the magnitude of pulsations, a special indicator is introduced - the coefficient of pulsations (K p).

Coeff. ripple is generally calculated using a complex formula using an integral function, but for harmonic oscillations the formula is simplified to the following:

K p \u003d (E max - E min) / (E max + E min) ⋅ 100%,

where E max is the maximum illumination value (amplitude), and E min is the minimum.

We will use this formula to calculate the capacitance of a smoothing capacitor.

You can very accurately determine the pulsations of any light source using a solar panel and an oscilloscope:

How to reduce pulsations?

Let's see how to turn on the LED in a 220 volt network to reduce ripple. To do this, the easiest way is to solder a storage (smoothing) capacitor in parallel with the LED:

Due to the non-linear resistance of LEDs, calculating the capacitance of this capacitor is a rather non-trivial task.

However, this task can be simplified by making a few assumptions. First, imagine the LED as an equivalent fixed resistor:

And secondly, to pretend that the brightness of the LED (and, consequently, the illumination) has a linear dependence on the current.

Calculation of the capacitance of the smoothing capacitor

Let's say we want to get the coefficient. ripple 2.5% at a current through the LED 20 mA. And let us have an LED at our disposal, on which 2 V drops at a current of 20 mA. The network frequency, as usual, is 50 Hz.

Since we decided that the brightness depends linearly on the current through the LED, and we presented the LED itself as a simple resistor, we can safely replace the illumination in the formula for calculating the ripple coefficient by the voltage on the capacitor:

K p \u003d (U max - U min) / (U max + U min) ⋅ 100%

We substitute the initial data and calculate U min:

2.5% = (2V - Umin) / (2V + Umin) ⋅ 100% => Umin = 1.9V

The period of voltage fluctuations in the network is 0.02 s (1/50).

Thus, the voltage waveform on the capacitor (and therefore on our simplified LED) will look something like this:

We recall trigonometry and calculate the charge time of the capacitor (for simplicity, we will not take into account the resistance of the ballast resistor):

t charge = arccos(U min /U max) / 2πf = arccos(1.9/2) / (2 ⋅ 3.1415⋅ 50) = 0.0010108 s

The rest of the period of the conder will be discharged. Moreover, the period in this case must be halved, because. we use a full-wave rectifier:

t razr \u003d T - t charge \u003d 0.02 / 2 - 0.0010108 \u003d 0.008989 s

It remains to calculate the capacity:

C=I LED ⋅ dt/dU = 0.02 ⋅ 0.008989/(2-1.9) = 0.0018 F (or 1800 uF)

In practice, it is unlikely that anyone will install such a large conder for the sake of one small LED. Although, if the task is to get a ripple of 10%, then only 440 microfarads are needed.

We increase efficiency

Have you noticed how much power is dissipated in the quenching resistor? Wasted power. Is there any way to reduce it?

It turns out that it is still possible! It is enough to take a reactive resistance (capacitor or inductor) instead of active resistance (resistor).

We, perhaps, will immediately discard the inductor because of its bulkiness and possible problems with the self-induction EMF. And what about capacitors, you can think.

As you know, a capacitor of any capacitance has an infinite resistance for direct current. But the resistance to alternating current is calculated by this formula:

Rc = 1 / 2πfC

that is, the larger the capacitance C and the higher the frequency f- the lower the resistance.

The beauty is that on the reactance and the power is also reactive, that is, not real. It's like it's there, but it's like it's not there. In fact, this power does not do any work, but simply returns back to the power source (to the outlet). Household meters do not take it into account, so you do not have to pay for it. Yes, it creates an additional load on the network, but you, as the end user, are unlikely to be very worried =)

Thus, our do-it-yourself LED power supply circuit from 220V takes the following form:

But! It is in this form that it is better not to use it, since in this circuit the LED is vulnerable to impulse noise.

Turning on or off powerful inductive loads located on the same line as you (air conditioner motor, refrigerator compressor, welding machine, etc.) leads to very short voltage surges in the network. Capacitor C1 represents almost zero resistance for them, therefore a powerful impulse will go straight to C2 and VD5.

Another dangerous moment occurs if the circuit is turned on at the moment of the antinode of the voltage in the network (that is, at the very moment when the voltage in the outlet is at its peak value). Because C1 is completely discharged at this point, then there is too much current inrush through the LED.

All this over time leads to progressive degradation of the crystal and a decrease in the brightness of the glow.

In order to avoid such unfortunate consequences, the circuit must be supplemented with a small quenching resistor of 47-100 ohms and a power of 1 watt. In addition, the resistor R1 will act as a fuse in case of breakdown of the capacitor C1.

It turns out that the scheme for connecting the LED to the 220 volt network should be like this:

And there remains one more small nuance: if you pull this circuit out of the socket, then some charge will remain on the capacitor C1. The residual voltage will depend on the moment at which the power circuit was broken and in some cases may exceed 300 volts.

And since the capacitor has nowhere to discharge, except through its internal resistance, the charge can be stored for a very long time (a day or more). And all this time, the conder will be waiting for you or your child, through which it will be possible to discharge properly. Moreover, in order to get an electric shock, you do not need to climb into the bowels of the circuit, just touch both pins of the plug.

To help the conder get rid of unnecessary charge, we will connect any high-resistance resistor (for example, 1 MΩ) in parallel with it. This resistor will have no effect on the design mode of the circuit. It won't even warm up.

Thus, the finished scheme for connecting an LED to a 220V network (taking into account all the nuances and improvements) will look like this:

The value of the capacitance of the capacitor C1 to obtain the desired current through the LED can be immediately taken from, or you can calculate it yourself.

Calculation of the quenching capacitor for the LED

I will not give tedious mathematical calculations, I will immediately give a ready-made capacitance formula (in Farads):

C \u003d I / (2πf√ (U 2 in - U 2 LED))[F],

where I is the current through the LED, f is the frequency of the current (50 Hz), U in is the effective value of the mains voltage (220V), U LED is the voltage on the LED.

If the calculation is carried out for a small number of series-connected LEDs, then the expression √ (U 2 in - U 2 LED) is approximately equal to U in, therefore the formula can be simplified:

C ≈ 3183 ⋅ I LED / U in[µF]

and, since we are doing calculations under U in = 220 volts, then:

C≈ 15⋅I LED[µF]

Thus, when the LED is turned on at 220 V, for every 100 mA of current, approximately 1.5 μF (1500 nF) of capacitance will be required.

Who is at odds with mathematics, pre-calculated values ​​can be taken from the table below.

Table 2. Dependence of the current through the LEDs on the capacitance of the ballast capacitor.

A little about the capacitors themselves

It is recommended to use noise-suppressing capacitors of class Y1, Y2, X1 or X2 for a voltage of at least 250 V as quenching capacitors. They have a rectangular case with numerous certificate designations on it. They look like this:

In short, then:

• X1- used in industrial devices connected to a three-phase network. These capacitors are guaranteed to withstand a 4 kV surge;
• X2- the most common. They are used in household appliances with a rated mains voltage of up to 250 V, withstand surges up to 2.5 kV;
• Y1- operate at rated mains voltage up to 250 V and withstand impulse voltage up to 8 kV;
• Y2- a fairly common type, can be used with a mains voltage of up to 250 V and can withstand impulses of 5 kV.

It is permissible to use domestic film capacitors K73-17 for 400 V (or better - for 630 V).

Today, Chinese "chocolates" (CL21) are widely used, but in view of their extremely low reliability, I highly recommend resisting the temptation to use them in your schemes. Especially as ballast capacitors.

Attention! Polar capacitors should never be used as ballast capacitors!

So, we looked at how to connect an LED to 220V (schemes and their calculation). All the examples given in this article are well suited for one or more low-power LEDs, but are completely impractical for powerful fixtures, such as lamps or spotlights - it is better to use for them, which are called drivers.

When designing radio equipment, the question of power indication often arises. The age of incandescent lamps for indication has long passed, a modern and reliable radio element of indication at the moment is the LED. This article will propose a diagram for connecting an LED to 220 volts, that is, the possibility of powering the LED from a household AC mains - a socket that is in any comfortable apartment.
If you need to power several LEDs at the same time, then we will also mention this in our article. In fact, such schemes are used for LED garlands or lamps, this is a little different. In fact, here it is necessary to implement the so-called driver for LEDs. So, let's not lump everything together. Let's try to figure it out in order.

The principle of lowering the supply voltage for the LED

Two power paths can be selected to power a low voltage load. The first is, so to speak, the classic version, when the power is reduced due to the resistor. The second option, which is often used for chargers, is a quenching capacitor. In this case, the voltage and current flow like pulses, and these same pulses must be precisely matched so that the LED, the load does not burn out. Here you need a more detailed calculation than with a resistor. The third option is a combined power supply, when both methods of lowering the voltage are used. Well, now about all these options in order.

Scheme for connecting an LED to a voltage of 220 volts (quenching capacitor)

The scheme for connecting an LED to 220 volts does not look complicated, the principle of its operation is simple. The algorithm is the following. When voltage is applied, capacitor C1 begins to charge, while in fact it is charged directly on one side, and on the other through a zener diode. The zener diode must match the voltage of the LED. As a result, the capacitor is fully charged. Next comes the second half-wave, when the capacitor starts to discharge. In this case, the voltage also goes through the zener diode, which is now operating in its normal mode and through the LED. As a result, a voltage equal to the stabilization voltage of the zener diode is applied to the LED at this time. Here it is important to choose a zener diode with the same rating as the LED.

Here everything seems to be simple and theoretically implemented normally. However, exact calculations are not so simple. After all, in fact, it is necessary to calculate the capacitance of the capacitor, which in this case will be quenching. This is done with a formula.

Let's estimate: 3200 * 0.02 / √ (220 * 220-3 * 3) \u003d 0.29 mF. Here's what the capacitor should be with a voltage for the LED of 3 volts, and a current of 0.02 A. You can substitute your values ​​​​and calculate your own version.

Radio components for connecting an LED to 220 volts

The power of the resistor can be minimal, 0.25 W is quite suitable (the value on the diagram is in ohms).
It is better to choose a capacitor (capacity is indicated in microfarads) with a margin, that is, with an operating voltage of 300 volts.
The LED can be any, for example, with a glow voltage of 2 volts AL307 BM or AL 307B and up to 5.5 volts - this is KL101A or KL101B.
The zener diode, as we have already mentioned, must correspond to the supply voltage of the LED, so for 2 volts it is KS130D1 or KS133A (stabilization voltage is 3 and 3.3 volts, respectively), and for 5.5 volts KS156A or KS156G

This method has its drawbacks, since with a slight voltage surge or deviation in the operation of the capacitor, we can get voltages much higher than 3 volts. The LED will burn out at once. The advantage is the economy of the circuit, as it is pulsed. Let's just say, not high reliability, but efficiency. Now about the combined option.

Scheme for connecting an LED to a voltage of 220 volts (quenching capacitor + resistor)

Everything is the same here, except that a resistor was added to the chain. In general, the influence of the resistor can make the whole circuit more predictable, more reliable. There will be less impulse currents with high voltage. This is good!

(... as in the diagram above, a quenching capacitor + resistor is used)

All the pros and cons are akin to the option with a quenching capacitor, but there is no reliability here either. Even more so, using a diode rather than a zener diode will affect the protection of the LED when the capacitor is discharged. That is, all the current will flow through the LED, and not, as in the previous case, through the LED and the zener diode. This option is the same. And here is the last case, using a resistor.

Scheme for connecting an LED to a voltage of 220 volts (resistor)

It is these schemes that we recommend you to assemble. Everything here is according to classical principles, Ohm's law and the formula for calculating power. First, let's calculate the resistance. When calculating the resistance will neglect the internal resistance of the LED and the voltage drop across it. In this case, we get a small margin, since the actual voltage drop across it will allow it to operate in a slightly more gentle mode than prescribed by the characteristics. So, let's say we have a LED current of 0.01 A and 3 volts.

R \u003d U / I \u003d 220 / 0.01 \u003d 22000 Ohm \u003d 22 kOhm. In the circuit, 15 kOhm, that is, the current was 0.014666 A, which is quite acceptable. This is how resistors are calculated for these cases. The only thing here will depend on how many resistors you use. If two are as in the first diagram, then we divide the resulting result in half.

If one, then by itself all the voltage will fall only on it.

Well, as expected, let's talk about the pros and cons. Plus one and very large, the circuit is very reliable. There is also one minus, that all the voltage will drop on 1-2 resistors, which means it will dissipate more power. Let's guess. P=U*I=220*0.02=4.4 Watts. That is, there should be a resistor as much as 4 watts if the current is 0.02 A. In this case, it is worth choosing a resistor carefully, it should be at least 3-4 watts. Well, you yourself understand that in this case there is no question of efficiency when 4 watts are dissipated on the resistor, and the LED can be neglected. In fact, it is almost like a small LED lamp, but only 1 LED is lit.

Connecting multiple LEDs to 220 volts

When you need to connect several LEDs at once, this is a somewhat different story. In fact, such circuit variations, or rather stabilizer circuits for LEDs, are called a driver. Apparently from the word drive (English) in motion. That is, it seems like a circuit that starts a group of LEDs. We will not talk about the correctness of the use of this word and about new words that we constantly borrow from other languages. Let's just say that this is a slightly different option, which means that we will analyze it in our other article "

One of the important issues when working with LEDs is its connection to AC and high voltage. It is known that the LED cannot be directly powered from the 220 V network. How to properly assemble the circuit and provide power to solve the problem?

Electrical Properties

To answer the above question, it is necessary to study the electrical properties of the LED.

Its current-voltage characteristic is a steep line. This means that with an increase in voltage, even by a very small amount, the current through the radiating semiconductor increases sharply. An increase in current leads to heating of the LED, as a result of which it can simply burn out. This problem is solved by including a limiting resistor in the circuit.

The LED has a low reverse breakdown voltage (about 20 volts), so it cannot be connected to a 220 volt AC network. To prevent the flow of current in the opposite direction, it is necessary to include a diode in the circuit or turn on the second one towards the first LED. The connection must be in parallel.

So, we know that any scheme for connecting an LED to a 220 volt network must contain a resistor and a rectifier, otherwise power will not be possible.

Why is such a scheme needed? First of all, for the design of the network indicator. An LED bulb can be an excellent indicator to help you determine if an electrical appliance is plugged in or not. It is added to the circuit of switches and sockets in order to easily find them in the dark.

Such an indicator begins to glow at a voltage of only a few volts. At the same time, it consumes a minimum amount of electricity due to a small (several miles of amperes) current.

What resistor to use?

To select the optimal resistance of the resistor, you must use Ohm's law.

R \u003d (Unetwork-Ur.) / Ir.nom.

Suppose we took a red LED for the indicator with a nominal current value of 18mA and a forward voltage of 2.0 Volts.

(311-2) / 0.018 \u003d 17167 Ohm \u003d 17 kOhm

Let's explain where the number 311 came from. This is the peak of the sinusoid, along which the voltage in our network changes. Without going into the realm of mathematics with all its calculations, we can simply say that the peak voltage is 220 * √2.

Sometimes there are circuits in which there is no rectifying diode. In this case, the resistance must be increased several times in order to make the current smaller and protect the indicator light from burning out.

Elementary circuit of the current indicator

What is needed to make the simplest indicator, which is powered by a 220 volt network? Here is the list:

• ordinary indicator LED of any color you like;
• resistor from 100 to 200 kOhm (the greater the resistance, the less brightly the bulb will glow);
• diode with a reverse voltage of 100 volts or more;
• a low-power soldering iron so as not to overheat the LED.

Since the number of parts is minimal, the board is not used in the installation. The indicator is connected in parallel with the electrical appliance.

For those who do not have the desire to run around in search of a diode, manufacturers have come up with a ready-made two-color indicator in the form of two LEDs of different colors built into one case. Usually it is red and green. In this case, the number of circuit details is further reduced.

There are other connection schemes in which the resistor is replaced with a capacitor or diode bridges, transistors, etc. are used. But no matter what design features are introduced, the main task is to rectify the current and lower it to a safe value.

Quite often we have to deal with such a question - how to connect an LED to 220 V, or simply to an AC electrical network. As such, direct connection of the diode directly to the network does not carry any semantic load. Even when using certain schemes, we will not get the desired effect.

If we need to connect the LED to a constant voltage network, then this task is solved very simply - we put a limiting resistor and forget it. The LED as it worked "in the forward direction" will continue to work.

If we need to use a 220 V network to connect an LED, then it will already be affected by reverse polarity. This is clearly seen by looking at the graph of the sinusoid, where each half-cycle of the sinusoid tends to change its sign to the opposite.

In this case, we will not get a glow in this half-cycle. In principle, it's okay))), but the LED will fail very quickly.

In general, a quenching resistor should be chosen from the condition of a design voltage of 310 V. Explaining why this is so is a dreary task, but you just need to remember it, because. the effective voltage value is 220 V, and the amplitude already increases by the root of two from the current one. Those. this way we get the applied forward and reverse voltage to the LED. The resistor is selected for 310V reverse polarity in order to protect the LED. How you can protect, we will see below.

How to connect LEDs to 220V in a simple way using resistors and a diode - option 1

The first circuit works on the principle of damping the reverse half-cycle. The vast majority of semiconductors have a negative attitude towards reverse voltage. To block it, we need a diode. As a rule, in most cases, diodes of the IN4004 type are used, designed for voltages greater than 300 V.

Connecting LED according to a simple circuit with a resistor and a diode - option 2

Another simple circuit for connecting LEDs to a 220 V AC network is not much more complicated and can also be classified as simple circuits.

Consider the principle of operation. With a positive half-wave, current flows through resistors 1 and 2, as well as the LED itself. In this case, it is worth remembering that the voltage drop across the LED will be reversed for a conventional diode - VD1. As soon as the negative half-wave of 220 V "gets" into the circuit, the current will go through a conventional diode and resistors. In this case, the direct voltage drop across VD1 will be reversed with respect to the LED. Everything is simple.

With a positive half-wave of the mains voltage, the current flows through the resistors R1, R2 and the HL1 LED (in this case, the forward voltage drop on the HL1 LED is the reverse voltage for the VD1 diode). With a negative half-wave of the mains voltage, the current flows through the diode VD1 and resistors R1, R2 (in this case, the forward voltage drop across the VD1 diode is the reverse voltage for the HL1 LED).

Calculation part of the scheme

Rated mains voltage:

U S.NOM = 220 V

The minimum and maximum mains voltage is accepted (experimental data):

U S.MIN = 170 V

U S.MAX = 250 V

The HL1 LED is accepted for installation, having the maximum allowable current:

I HL1.DOP = 20 mA

Maximum calculated amplitude current of the LED HL1:

I HL1.AMPL.MAX = 0.7*I HL1.DOP = 0.7*20 = 14 mA

Voltage drop across the HL1 LED (experimental data):

Minimum and maximum operating voltage across resistors R1, R2:

U R.MIN REAL = U S.MIN = 170 V

U R.REAL MAX = U S.MAX = 250 V

Estimated equivalent resistance of resistors R1, R2:

R EQ.CALC = U R.AMPL.MAX /I HL1.AMPL.MAX = 350/14 = 25 kΩ

P R.MAX = U R.REAL MAX 2 /R EQ. CALC = 2502/25 = 2500 mW = 2.5 W

Estimated total power of resistors R1, R2:

P R.CALC = P R.MAX /0.7 = 2.5/0.7 = 3.6W

A parallel connection of two resistors of the MLT-2 type is accepted, having a total maximum allowable power:

P R.DOP \u003d 2 2 \u003d 4 W

Estimated resistance of each resistor:

R CALC = 2*R EQ. CALC = 2*25 = 50 kOhm

The nearest larger standard resistance of each resistor is taken:

R1 = R2 = 51 kΩ

Equivalent resistance of resistors R1, R2:

R EKV \u003d R1 / 2 \u003d 51/2 \u003d 26 kOhm

Maximum total power of resistors R1, R2:

P R.MAX = U R.REAL MAX 2 /R EQ = 2502/26 = 2400 mW = 2.4 W

Minimum and maximum amplitude current of the HL1 LED and VD1 diode:

I HL1.AMPL.MIN = I VD1.AMPL.MIN = U R.AMPL.MIN /R EQ = 240/26 = 9.2 mA

I HL1.AMPL.MAX = I VD1.AMPL.MAX = U R.AMPL.MAX /R EQ = 350/26 = 13 mA

Minimum and maximum average current of HL1 LED and VD1 diode:

I HL1.AVERAGE MIN = I VD1.AVERAGE MIN = I HL1.REAL MIN / K F = 3.3 / 1.1 = 3.0 mA

I HL1.MED.MAX = I VD1.MED.MAX = I HL1.REAL MAX /K F = 4.8/1.1 = 4.4 mA

Reverse voltage diode VD1:

U VD1.OBR = U HL1.PR = 2 V

Design parameters of the diode VD1:

U VD1.CALC = U VD1.OBR / 0.7 = 2 / 0.7 = 2.9 V

I VD1.CALC = U VD1.AMPL.MAX /0.7 = 13/0.7 = 19 mA

A VD1 diode of the D9V type is adopted, which has the following main parameters:

U VD1.DOP = 30 V

I VD1.DOP = 20 mA

I 0.MAX = 250uA

Cons of using the scheme for connecting LEDs to 220 V according to option 2

The main disadvantages of connecting LEDs according to this scheme are the low brightness of the LEDs, due to the low current. I HL1.SR \u003d (3.0-4.4) mA and high power on resistors: R1, R2: P R.MAX \u003d 2.4 W.

Option 3 for connecting LEDs to an AC 220 V electrical network

With a positive half-cycle, current flows through the resistor R1, the diode and the LED. With a negative current does not flow, because. the diode in this case turns on in the opposite direction.

The calculation of the circuit parameters is similar to the second option. To whom it is necessary - will count and compare. The difference is small.

Cons of connecting according to option 3

If the "inquisitive minds" themselves have already calculated, they can compare the data with the second option. To whom laziness - it is necessary to take a word. The disadvantage of such a connection is also the low brightness of the LED, because. the current flowing through the semiconductor is only I HL1.CP = (2.8-4.2) mA.

But with such a scheme, we get a noticeable decrease in the power of the resistor: P R1.MAX \u003d 1.2 W instead of 2.4 W obtained earlier.

Connecting a 220 V LED using a diode bridge - option 4

As you can see in the graphic picture, in this case, we use resistors and a diode bridge to connect to 220.

In this case, the current through 2 resistors and the LED current will flow both with a positive and a negative half-wave of the sinusoid due to the use of a rectifier bridge on diodes VD1-VD4.

U VD.CALC = U VD.OBR / 0.7 = 2.6 / 0.7 = 3.7 V

I VD.CALC = U VD.AMPL.MAX /0.7 = 13/0.7 = 19 mA

Diodes VD1-VD4 type D9V are accepted, having the following main parameters:

U VD.DOP = 30 V

I VD.DOP = 20 mA

I 0.MAX = 250uA

Disadvantages of the connection scheme for option 4

However, with this scheme, we will get a noticeable increase in the brightness of the LED: HL1: I HL1.CP = (5.9-8.7) mA instead of (2.8-4.2) mA

In principle, these are the most common schemes for connecting any LED to a 220 V network using a conventional diode and resistors. For ease of understanding, calculations have been given. Not for everyone, maybe understandable, but who needs it, he will find, read and understand. Well, if not, then a simple graphic part will suffice.

How to connect an LED to 220V using a capacitor

Above, we looked at how easy it is, using only diodes and resistors, to connect any LED to a 220 V network. They were simple diagrams. Now let's look at more complex, but better in terms of implementation and durability. For this we need a capacitor.

The current limiting element is a capacitor. On the diagram - C1. The capacitor must be designed to operate with a voltage of at least 400 V. After charging the latter, the current through it will limit the resistor.

Connecting an LED to a 220 V network using the example of a backlit switch

Now you will not surprise anyone with a switch with an integrated LED backlight. Having disassembled it and figured it out, we will get another way, thanks to which we can connect any LED to a 220 V network.

All illuminated switches use a resistor of at least 200 kΩ. The current in this case is limited to about 1A. When connected to the network, such an LED will glow. At night it can be easily seen on the wall. The reverse current in this case will be very small and cannot damage the semiconductor. In principle, such a circuit also has the right to exist, but the light from such a diode will still be negligible. And whether the game is worth the candle is not clear.