29.06.2020
Calculation of the capacitor capacitance for an LED lamp. Capacitors for LED bulbs. There are several significant disadvantages
Sometimes in electrical engineering, power supplies are used that do not contain a transformer. This raises the problem of lowering the input voltage. For example, lowering the alternating mains voltage (220 V) at a frequency of 50 hertz to the required voltage value. An alternative to a transformer is a capacitor, which is connected in series with the voltage source and the load (for more information on the use of capacitors, see section "). Such a capacitor is called a quenching capacitor.
To calculate a quenching capacitor means to find the capacitance of such a capacitor, which, when connected to the circuit described above, will lower the input voltage to the required voltage at the load. Now we get the formula for calculating the capacitance of the quenching capacitor. A capacitor operating in an alternating current circuit has a capacitance (), which is related to the frequency of the alternating current and its own capacitance () (moreover), more precisely:
By condition, we included resistance (resistive load ()) and a capacitor in the alternating current circuit. The total resistance of this system () can be calculated as:
Since the connection is serial, using , we write:
where is the voltage drop across the load (device supply voltage); - mains voltage, - voltage drop across the capacitor. Using the above formulas, we have:
If the load is small, then using a capacitor, including it in series in the circuit, is the easiest way to reduce the mains voltage. In the event that the voltage at the power output is less than 10-20 volts, then the capacitance of the quenching capacitor is calculated using the approximate formula:
It is more profitable and easier to power low-voltage electrical and radio equipment from the mains. For this, transformer power supplies are most suitable, since they are safe to operate. However, interest in transformerless power supplies (BTBP) with a stabilized output voltage is not weakening. One of the reasons is the complexity of manufacturing a transformer. But for BTBP, it is not needed - only the correct calculation is needed, but this is exactly what scares inexperienced novice electricians. This article will help make the calculation and facilitate the design of a transformerless power supply.
A simplified diagram of the BPTP is shown in fig. 1. The diode bridge VD1 is connected to the network through a quenching capacitor C gas, connected in series with one of the diagonals of the bridge. The other diagonal of the bridge works on the load of the block - resistor R n. In parallel with the load, a filter capacitor C f and a zener diode VD2 are connected.
The calculation of the power supply begins with setting the voltage U n at the load and the current strength I n. consumed by the load. The greater the capacitance of the capacitor C gas, the higher the energy capabilities of the BPTP.
Capacitance calculation
The table shows data on the capacitance X from the capacitor C gas at a frequency of 50 Hz and the average value of the current I cf passed by the capacitor C gas, calculated for the case when R n \u003d 0, that is, with a short circuit of the load. (After all, the BTBP is not sensitive to this anomalous mode of operation, and this is another huge advantage over transformer power supplies.)Other values of capacitive resistance X c (in kiloohms) and the average current value I cf (in milliamps) can be calculated using the formulas:
C gas - the capacitance of the quenching capacitor in microfarads.
If we exclude the zener diode VD2, then the voltage U n at the load and the current I n through it will depend on the load R n. It is easy to calculate these parameters using the formulas:
U n - in volts, R n and X n - in kiloohms, I n - in milliamps, C gas - in microfarads. (The formulas below use the same units.)
With a decrease in load resistance, the voltage across it also decreases, moreover, according to a nonlinear dependence. But the current passing through the load increases, however, very slightly. So, for example, a decrease in R n from 1 to 0.1 kOhm (exactly 10 times) leads to the fact that U n decreases by 9.53 times, and the current through the load increases by only 1.05 times. This "automatic" current stabilization favorably distinguishes BTBP from transformer power supplies.
Power P n at the load, calculated by the formula:
with a decrease in R n decreases almost as intensively as U n. For the same example, the power consumed by the load is reduced by a factor of 9.1.
Since the load current I n at relatively small values of resistance R n and voltage U n changes very little on it, in practice it is quite acceptable to use approximate formulas:
Restoring the zener diode VD2, we get voltage stabilization U n at the level U st - the value is almost constant for each particular zener diode. And with a small load (high resistance R n), the equality U n \u003d U st will be fulfilled.
Load resistance calculation
To what extent can R n be reduced so that the equality U n \u003d U st is true? Until the inequality is fulfilled:
Therefore, if the load resistance turns out to be less than the calculated R n, the voltage across the load will no longer be equal to the stabilization voltage, but will turn out to be somewhat less, since the current through the zener diode VD2 will stop.
Calculation of the allowable current through the zener diode
Now let's determine what current I n will flow through the load R n and what current - through the zener diode VD2. It is clear that
As the load resistance decreases, the power consumed by it P n =I n U n =U 2 st /R n increases. But the average power consumed by the BPTP is equal to
remains unchanged. This is explained by the fact that the current I cf branches into two - I n and I st - and, depending on the load resistance, is redistributed between R n and the zener diode VD2, and so that the lower the load resistance R n, the less current goes through stabilizer and vice versa. This means that if the load is small (or completely absent), the VD2 zener diode will be in the most difficult conditions. That is why it is not recommended to remove the load from the BPTP, otherwise all the current will go through the zener diode, which can lead to its failure.
The amplitude value of the mains voltage is 220·√2=311(V). The pulse value of the current in the circuit, if we conditionally neglect the capacitor C f, can reach
Accordingly, the Zener diode VD2 must reliably withstand this pulsed current in the event of an accidental load disconnection. We should not forget about possible voltage overloads in the lighting network, which are 20 ... 25% of the nominal value, and calculate the current passing through the zener diode with the load turned off, taking into account the correction factor 1.2 ... 1.25.
If there is no powerful zener diode
When there is no zener diode of suitable power, it can be fully replaced with a diode-transistor counterpart. But then the BTBP should be built according to the scheme shown in Fig. 2. Here, the current flowing through the zener diode VD2 decreases in proportion to the static current transfer coefficient of the base of the powerful n-p-n transistor VT1. The analog voltage UCT will be approximately 0.7V higher than U st of the lowest power zener diode VD2 if the transistor VT1 is silicon, or by 0.3V if it is germanium.The transistor of the p-n-p structure is also applicable here. However, then the circuit shown in Fig. 3.
Half-wave block calculation
Along with a full-wave rectifier in BTBP, the simplest single-wave rectifier is also sometimes used (Fig. 4). In this case, its load R n is fed only by positive half-cycles of alternating current, and the negative ones pass through the VD3 diode, bypassing the load. Therefore, the average current I cf through the diode VD1 will be half as much. This means that when calculating the block, instead of X c, one should take 2 times the resistance equal to
and the average current with a short-circuited load will be equal to 9.9 πC gas \u003d 31.1 C gas. Further calculation of such a variant of the BPTP is carried out in exactly the same way as in the previous cases.
Calculation of the voltage on the quenching capacitor
It is generally accepted that at a mains voltage of 220V, the rated voltage of the quenching capacitor C gas must be at least 400V, that is, with approximately a 30% margin in relation to the amplitude network, since 1.3 311 = 404 (V). However, in some of the most critical cases, its rated voltage should be 500 or even 600V.And further. When choosing a suitable capacitor C gas, it should be borne in mind that capacitors such as MBM, MBPO, MBGP, MBGTS-1, MBGTS-2 cannot be used in BTBP, since they are not designed to work in AC circuits with an amplitude voltage value exceeding 150V.
The most reliable BTBP capacitors are MBGCH-1, MBGCH-2 for a rated voltage of 500V (from old washing machines, fluorescent lamps, etc.) or KBG-MN, KBG-MP, but for a rated voltage of 1000V.
filter capacitor
It is difficult to calculate the capacity of the filter capacitor C f analytically. Therefore, it is selected experimentally. Approximately, it should be considered that for each milliamp of the average current consumed, it is required to take at least 3 ... 10 microfarads of this capacitance if the BTBP rectifier is full-wave, or 10 ... 30 microfarads if it is single-half-wave.The rated voltage of the used oxide capacitor C f must be at least U st A if there is no zener diode in the BTBP, and the load is constantly on, the rated voltage of the filter capacitor must exceed the value:
If the load cannot be turned on permanently, and there is no zener diode, the nominal voltage of the filter capacitor should be more than 450V, which is hardly acceptable due to the large size of the capacitor C f. By the way, in this case, the load should be connected again only after the BTBP is disconnected from the network.
And that is not all
It is desirable to supplement any of the possible options for the BTBP with two more auxiliary resistors. One of them, the resistance of which can be in the range of 300 kΩ ... 1 MΩ, is connected in parallel with the capacitor C quenched. This resistor is needed to accelerate the discharge of the capacitor C dies after the device is disconnected from the network. Another - ballast - with a resistance of 10 ... 51 Ohm is included in the break of one of the network wires, for example, in series with the capacitor C extinguished. This resistor will limit the current through the diodes of the VD1 bridge at the time the BTBP is connected to the network. The dissipation power of both resistors must be at least 0.5 W, which is necessary to guarantee against possible surface breakdowns of these resistors by high voltage. Due to the ballast resistor, the zener diode will be loaded somewhat less, but the average power consumed by the BTBP will increase markedly.Which diodes to take
The function of the BTBP full-wave rectifier according to the circuits in fig. 1 ... 3 can perform diode assemblies of the KTs405 or KTs402 series with letter indices Zh or I, if the average current does not exceed 600 mA, or with indices A, B, if the current value reaches 1 A. Four separate diodes connected according to bridge circuit, for example, KD105 series with indices B, C or D, D226 B or C - up to 300 mA, KD209 A, B or C - up to 500 ... 700 mA, KD226 V, G or D - up to 1.7 A .Diodes VD1 and VD3 in BTBP according to the circuit in fig. 4 can be any of the above. It is also permissible to use two diode assemblies KD205K V, G or D for current up to 300 mA or KD205 A, V, Zh or I - up to 500 mA.
And the last. The transformerless power supply, as well as the equipment connected to it, are directly connected to the AC mains! Therefore, they must be securely insulated from the outside, say, placed in a plastic case. In addition, it is strictly forbidden to "ground" any of their outputs, as well as open the case when the device is turned on.
The proposed method for calculating the BPTP has been tested by the author in practice for a number of years. The entire calculation is based on the fact that the BPTP is essentially a parametric voltage stabilizer, in which the quenching capacitor acts as a current limiter.
Magazine "SAM" №5, 1998
Something often began to ask me how to connect a microcontroller or what kind of low-voltage circuit directly to 220 without using a transformer. The desire is quite obvious - the transformer, even if it is pulsed, is very bulky. And pushing it, for example, into the control circuit of a chandelier placed directly in the switch will not work with all your desire. Is it possible to hollow out a niche in the wall, but this is not our method!
Nevertheless, there is a simple and very compact solution - this is a divider on a capacitor.
True, capacitor power supplies do not have decoupling from the network, so if something suddenly burns out in it, or goes wrong, then it can easily shock you with current, or burn your apartment, but ruining a computer is generally a nice thing, in general, a technique security here must be honored more than ever - it is painted at the end of the article. In general, if I didn’t convince you that transformerless power supplies are evil, then Pinocchio is evil to himself, I have nothing to do with it. Okay, closer to the topic.
Remember the usual resistive divider?
It would seem that what is the problem, I chose the necessary denominations and got the desired voltage. Then he straightened and Profit. But not everything is so simple - such a divider can and will be able to give the right voltage, but it will not give the right current at all. Because resistance is very high. And if the resistances are proportionally reduced, then a large current will flow through them, which at a voltage of 220 volts will give very large heat losses - the resistors will heat like a stove and eventually either fail or start a fire.
Everything changes if one of the resistors is replaced with a capacitor. The bottom line is - as you remember from the article about capacitors, the voltage and current on the capacitor are out of phase. Those. when the voltage is at its maximum, the current is at its minimum, and vice versa.
Since we have an alternating voltage, the capacitor will constantly discharge and charge, and the peculiarity of the discharge-charge of the capacitor is that when it has the maximum current (at the moment of charging), then the minimum voltage and set. When it is already charged and the voltage on it is maximum, then the current is zero. Accordingly, in this scenario, the heat loss power released on the capacitor (P=U*I) will be minimal. Those. he doesn't even sweat. And the reactive resistance of the capacitor is Xc \u003d -1 / (2pi * f * C).
Theoretical digression
There are three types of resistance in a circuit:
Active - Resistor (R)
Reactive - capacitor (X c) and coil (X L)
The total circuit resistance (impedance) Z \u003d (R 2 + (X L + X s) 2) 1/2
Active resistance is always constant, while reactance depends on frequency.
X L \u003d 2pi * f * L
Xc=-1/(2pi*f*C)
The reactance sign of an element indicates its nature. Those. if greater than zero, then these are inductive properties, if less than zero, then capacitive. From this it follows that inductance can be compensated by capacitance and vice versa.
f is the current frequency.
Accordingly, at direct current at f \u003d 0 and X L of the coil becomes equal to 0 and the coil turns into an ordinary piece of wire with only active resistance, and Xc of the capacitor goes to infinity, turning it into a break.
We get the following scheme:
Everything, in one direction the current flows through one diode, in the other through the second. As a result, on the right side of the circuit we no longer have a change, but a pulsating current - one half-wave of a sinusoid.
Add a smoothing capacitor to make the voltage quieter, microfarads by 100 and volts by 25, electrolyte:
In principle, it’s already ready, the only thing is that you need to put the zener diode on such a current so that it doesn’t die when there is no load at all, because then he will have to take the rap for everyone, dragging all the current that the PSU can give.
And you can help him lightly. Install a current limiting resistor. True, this will greatly reduce the load capacity of the power supply, but this is enough for us.
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The current that this circuit can give can be, EMNIP, approximately calculated by the formula:
I \u003d 2F * C (1.41U - Uout / 2).
- F is the mains frequency. We have 50hz.
- C - capacity
- U - voltage in the outlet
- Uout - output voltage
The formula itself is derived from eerie integrals of the form of current and voltage. In principle, you can google it yourself using the keyword “quenching capacitor calculation”, there is plenty of material.
In our case, it turns out that I = 100 * 0.46E-6 (1.41 * U - Uout / 2) = 15mA
Not extravaganza, but for the operation of the MK + TSOP + some kind of optointerface is more than enough. And more is usually not required.
Add a couple of conduits for additional power filtering and you can use:
Then, as usual, I etched everything and soldered it:
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The scheme has been repeatedly tested and works. I once shoved it into the thermal glass heating control system. There were places with a matchbox, and safety was guaranteed by the total vitrification of the entire block.
SAFETY
In this scheme there is no voltage decoupling from the supply circuit, which means the circuit VERY DANGEROUS in terms of electrical safety.
Therefore, it is necessary to take an extremely responsible approach to its installation and selection of components. And also carefully and very carefully handle it when setting it up.
First, notice that one of the pins goes to GND directly from the socket. And this means that there may be a phase, depending on how the plug was inserted into the outlet.
Therefore, strictly observe a number of rules:
- 1. Ratings should be set with a margin for as much voltage as possible. This is especially true for the capacitor. I have 400 volts, but this is the one that was available. It would be better in general for 600 volts, because. in the power grid sometimes there are voltage surges much higher than the nominal value. Standard power supplies, due to their inertia, will easily survive it, but the capacitor can break through - imagine the consequences for yourself. Well, if there is no fire.
- 2. This circuit must be carefully insulated from the environment. Sturdy housing so that nothing sticks out. If the circuit is mounted in a wall, then it should not touch the walls. In general, we pack the whole thing tightly in plastic, vitrify and bury it at a depth of 20 meters. :)))))
- 3. When adjusting, in no case do not climb with your hands to any of the elements of the chain. Let it not calm you down that there is 5 volts at the output. Since five volts are there exclusively relative to itself. But in relation to the environment, there are still the same 220.
- 4. After disconnection, it is highly desirable to discharge the quenching capacitor. Because there remains a charge of 100-200 volts in it, and if you carelessly poke your head somewhere in the wrong place, it will hurt your finger. It is unlikely that it is fatal, but it is not pleasant, and from surprise you can do trouble.
- 5. If a microcontroller is used, then flash it ONLY when completely turned off from the network. And you need to turn it off by pulling it out of the socket. If this is not done, then with a probability close to 100% the computer will be killed. And most likely all.
- 6. The same applies to communication with the computer. With such a power supply, it is forbidden to connect via USART, it is forbidden to combine grounds.
If you still want to communicate with a computer, then use potentially separated interfaces. For example, a radio channel, infrared transmission, at worst, the division of RS232 by optocouplers into two independent parts.
Online calculation of the quenching capacitor of a transformerless power supply (10+)
Transformerless power supplies - Online calculation of the quenching capacitor of a transformerless power supply
But the scheme (A1) will not work, since current flows through the capacitor in only one direction. It will quickly charge the capacitor. After that, the voltage will no longer be applied to the circuit. It is necessary that the capacitor, having been charged in one half-cycle, could be discharged in the other. For this, in the scheme (A2) introduced the second diode.
The mains voltage is applied between the terminal marked 220V and the common wire. Resistor R2 needed to limit the current surge. When the circuit is in steady state with a good quality mains voltage, no current surges occur. But at the moment of switching on, we can get not to the zero value of the input voltage (which would be optimal), but to any, up to the amplitude one. The capacitor is then discharged, so that the low-voltage part will be connected directly to the 310V amplitude of the mains voltage. It is necessary that at this moment the diodes do not burn out. For this:
[Resistor R2, Ohm] = 310 / [The maximum allowable one-time current pulse through the diode, A]
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Ask a Question. Article discussion. messages.
Good evening. No matter how hard I tried, I could not use the above formulas for Fig. 1.2 to learn the values of the capacitances of capacitors C1 and C2 with the given data values \u200b\u200bin your table (Uin ~ 220V, Uout 15V, Iout 100mA, f 50Hz). I have a problem, turn on the coil of a small-sized DC relay for an operating voltage of -25V to the network ~ 220V, the operating current of the coil is I = 35mA. Maybe I'm not something
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After reading this headline, someone may ask: “Why?” Yes, if you just plug it into an outlet, even turning it on according to a certain scheme, it has no practical significance, it will not bring any useful information. But if the same LED is connected in parallel to a heating element controlled by a thermostat, then you can visually control the operation of the entire device. Sometimes such an indication allows you to get rid of many minor problems and troubles.
In light of what has already been said, the task seems trivial: just put a limiting resistor of the desired value, and the issue is resolved. But all this is good if you feed the LED with a rectified constant voltage: as the LED was connected in the forward direction, it remained the same.
When working on alternating voltage, everything is not so simple. The fact is that, in addition to the forward voltage, the LED will also be affected by the reverse polarity voltage, because each half-cycle of the sinusoid changes sign to the opposite. This reverse voltage will not illuminate the LED, but it can become unusable very quickly. Therefore, it is necessary to take measures to protect against this "harmful" voltage.
In the case of mains voltage, the calculation of the quenching resistor should be based on a voltage value of 310V. Why? Everything is very simple here: 220V is, the amplitude value will be 220 * 1.41 = 310V. The amplitude voltage at the root of two (1.41) times greater than the current one, and this must not be forgotten. This is the forward and reverse voltage applied to the LED. It is from the value of 310V that the resistance of the quenching resistor should be calculated, and it is from this voltage, only of reverse polarity, that the LED is protected.
How to Protect an LED from Reverse Voltage
For almost all LEDs, the reverse voltage does not exceed 20V, because no one was going to make a high-voltage rectifier on them. How to get rid of such misfortune, how to protect the LED from this reverse voltage?
It turns out that everything is very simple. The first way is to turn on the usual one with a high reverse voltage (not lower than 400V) in series with the LED, for example, 1N4007 - reverse voltage 1000V, forward current 1A. It is he who will not miss the high voltage of negative polarity to the LED. The scheme of such protection is shown in Fig. 1a.
The second method, no less effective, is to simply shunt the LED with another diode connected back-to-back - in parallel, Fig. 1b. With this method, the protective diode does not even have to be with a high reverse voltage, any low-power diode, for example, KD521, is enough.
Moreover, you can simply turn on two LEDs in parallel - opening one by one, they will protect each other themselves, and both will emit light, as shown in Figure 1c. This is already the third way to protect. All three protection schemes are shown in Figure 1.
Figure 1. LED Reverse Voltage Protection Circuits
The limiting resistor in these circuits has a resistance of 24KΩ, which at an operating voltage of 220V provides a current of the order of 220/24 = 9.16mA, can be rounded up to 9. Then the power of the quenching resistor will be 9 * 9 * 24 = 1944mW, almost two watts. This is despite the fact that the current through the LED is limited to 9mA. But prolonged use of the resistor at maximum power will not lead to anything good: at first it will turn black, and then it will completely burn out. To prevent this from happening, it is recommended to put two 12KΩ resistors in series with a power of 2W each.
If you set the current level to 20mA, then it will be even more - 20 * 20 * 12 = 4800mW, almost 5W! Naturally, no one can afford a stove of such power for heating a room. This is based on one LED, but what if there is a whole one?
Capacitor - wattless resistance
The circuit shown in Figure 1a, with a protective diode D1, “cuts off” the negative half-cycle of the alternating voltage, therefore, the power of the quenching resistor is halved. But, still, the power remains very significant. Therefore, it is often used as a limiting resistor: it will limit the current no worse than a resistor, but it will not generate heat. After all, it is not for nothing that a capacitor is often called a wattless resistance. This switching method is shown in Figure 2.
Figure 2. Scheme for switching on an LED through a ballast capacitor
Everything seems to be fine here, there is even a protective diode VD1. But two details are missing. First, the capacitor C1 after the circuit is turned off can remain in a charged state and store a charge until someone discharges it with their own hand. And this, believe me, will definitely happen someday. The electric shock is, of course, not fatal, but quite sensitive, unexpected and unpleasant.
Therefore, in order to avoid such a nuisance, these quenching capacitors are shunted with a resistor with a resistance of 200 ... 1000 KΩ. The same protection is also installed in transformerless power supplies with a quenching capacitor, in optocouplers, and in some other circuits. In Figure 3, this resistor is labeled R1.
Figure 3. Scheme of connecting the LED to the lighting network
In addition to the resistor R1, another resistor R2 appears on the circuit. Its purpose is to limit the inrush current through the capacitor when voltage is applied, which helps to protect not only the diodes, but also the capacitor itself. It is known from practice that in the absence of such a resistor, the capacitor sometimes breaks, its capacitance becomes much less than the nominal one. Needless to say, the capacitor must be ceramic for an operating voltage of at least 400V or special for operation in AC circuits for a voltage of 250V.
Another important role is assigned to the resistor R2: in the event of a breakdown of the capacitor, it operates as a fuse. Of course, the LEDs will also have to be replaced, but at least the connecting wires will remain intact. In fact, this is how the fuse works in any - the transistors burned out, and the printed circuit board remained almost untouched.
The circuit shown in Figure 3 shows only one LED, although in fact they can be connected in series with several pieces. The protective diode will quite cope with its task alone, but the capacitance of the ballast capacitor will have to be calculated, at least approximately, but still.
In order to calculate the resistance of the quenching resistor, it is necessary to subtract the voltage drop across the LED from the supply voltage. If several LEDs are connected in series, then simply add their voltages, and also subtract from the supply voltage. Knowing this residual voltage and the required current, according to Ohm's law, it is very simple to calculate the resistance of a resistor: R \u003d (U-Ud) / I * 0.75.
Here U is the supply voltage, Ud is the voltage drop across the LEDs (if the LEDs are connected in series, then Ud is the sum of the voltage drops across all the LEDs), I is the current through the LEDs, R is the resistance of the quenching resistor. Here, as always, - voltage in Volts, current in Amperes, result in Ohms, 0.75 - a coefficient to increase reliability. This formula has already been cited in the article.
The amount of forward voltage drop for LEDs of different colors is different. At a current of 20mA, red LEDs are 1.6 ... 2.03V, yellow 2.1 ... 2.2V, green 2.2 ... 3.5V, blue 2.5 ... 3.7V. White LEDs with a wide emission spectrum of 3.0 ... 3.7 V have the highest voltage drop. It is easy to see that the spread of this parameter is quite wide.
Here are the voltage drops of just a few types of LEDs, simply by color. In fact, there are many more of these colors, and the exact value can only be found in the technical documentation for a particular LED. But often this is not required: to get a result acceptable for practice, it is enough to substitute some average value (usually 2V) into the formula, of course, if it is not a garland of hundreds of LEDs.
To calculate the capacitance of the quenching capacitor, the empirical formula C \u003d (4.45 * I) / (U-Ud) is used,
where C is the capacitance of the capacitor in microfarads, I is the current in milliamps, U is the peak voltage of the network in volts. When using a chain of three series-connected white LEDs, Ud is about 12V, U is the amplitude voltage of the network is 310V, to limit the current at 20mA, you need a capacitor with a capacitance
C \u003d (4.45 * I) / (U-Ud) \u003d C \u003d (4.45 * 20) / (310-12) \u003d 0.29865 μF, almost 0.3 μF.
The nearest standard capacitor value is 0.15uF, so two capacitors connected in parallel will have to be used in this circuit. Here it is necessary to make a remark: the formula is valid only for an alternating voltage frequency of 50 Hz. For other frequencies, the results will be incorrect.
The capacitor needs to be checked first.
Before using the capacitor, it must be checked. To begin with, just turn on 220V to the network, it is better through a 3 ... 5A fuse, and after 15 minutes check by touch, is there any noticeable heating? If the condenser is cold, then you can use it. Otherwise, be sure to take another, and also pre-check. After all, 220V is no longer 12, everything is a little different here!
If this check was successful, the capacitor did not heat up, then you can check if there was an error in the calculations, whether the capacitor is of the same capacity. To do this, you need to turn on the capacitor as in the previous case in the network, only through an ammeter. Naturally, the ammeter must be AC.
This is a reminder that not all modern digital multimeters can measure alternating current: simple cheap devices, for example, very popular with radio amateurs, can only measure direct current, which no one knows what such an ammeter will show when measuring alternating current. Most likely it will be the price of firewood or the temperature on the moon, but not alternating current through the capacitor.
If the measured current is approximately the same as it turned out when calculating according to the formula, then you can safely connect the LEDs. If instead of the expected 20 ... 30mA it turned out 2 ... 3A, then here, either an error in the calculations, or the marking of the capacitor was incorrectly read.
Illuminated switches
Here you can focus on another way to turn on the LED in the lighting network used. If such a switch is disassembled, then it can be found that there are no protective diodes there. So, everything that is written just above is nonsense? Not at all, you just need to carefully look at the disassembled switch, or rather, the value of the resistor. As a rule, its nominal value is at least 200KΩ, maybe even a little more. In this case, it is obvious that the current through the LED will be limited to about 1mA. The circuit of the illuminated switch is shown in Figure 4.
Figure 4. Wiring diagram for an LED in a backlit switch
Here, with one resistor, several “hares” are killed at once. Of course, the current through the LED will be small, it will glow weakly, but it is quite bright to see this glow on a dark night in the room. But during the day this glow is not needed at all! So let yourself shine imperceptibly.
In this case, the reverse current will also be weak, so weak that in no way will it be able to burn the LED. Hence the savings on exactly one protective diode, which was described above. With the release of millions, and maybe even billions, of switches per year, the savings are considerable.
It would seem that after reading articles about LEDs, all questions about their application are clear and understandable. But there are still many subtleties and nuances when turning on LEDs in various circuits. For example, parallel and series connection or, in other words, good and bad circuits.
Sometimes you want to assemble a garland of several dozen LEDs, but how to calculate it? How many LEDs can be connected in series if there is a power supply with a voltage of 12 or 24V? These and other questions will be discussed in the next article, which we will call "Good and bad LED switching schemes."
