Capacitors for LED light bulbs. Calculation of mixed connection of capacitors online

Sometimes in electrical engineering power supplies that do not contain a transformer are used. In this case, the task of reducing the input voltage arises. For example, reducing the alternating voltage of the network (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 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 a capacitor that, when connected to the circuit as described above, will reduce the input voltage to the required voltage at the load. Now we get the formula for calculating the capacity of the quenching capacitor. Capacitor operating in a circuit alternating current, has a capacitance (), which is related to the frequency of the alternating current and its own capacitance () (and), more precisely:

According to the condition, we included a 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 mains voltage. If the voltage at the power output is less than 10-20 volts, then the capacity of the quenching capacitor is calculated using the approximate formula:

(5.4.4)

More often in practice, smaller units of capacitance are used: 1 nF (nanofarad) = 10 –9 F and 1 pkF (picofarad) = 10 –12 F.

There is a need for devices that accumulate charge, and isolated conductors have low capacity. It was experimentally discovered that the electrical capacity of a conductor increases if another conductor is brought close to it - due to electrostatic induction phenomena.

Capacitor - these are two conductors called linings, located close to each other .

The design is such that the external bodies surrounding the capacitor do not affect its electrical capacity. This will be done if the electrostatic field is concentrated inside the capacitor, between the plates.

Capacitors are flat, cylindrical and spherical.

Since the electrostatic field is inside the capacitor, the electric displacement lines start on the positive plate, end on the negative plate, and do not disappear anywhere. Therefore, the charges on the plates opposite in sign, but equal in magnitude.

The capacitance of a capacitor is equal to the ratio of the charge to the potential difference between the plates of the capacitor:

(5.4.5)

In addition to capacitance, each capacitor is characterized U slave (or U etc . ) – the maximum permissible voltage, above which a breakdown occurs between the plates of the capacitor.

Connection of capacitors

Capacitive batteries– combinations of parallel and series connections of capacitors.

1) Parallel connection of capacitors (Fig. 5.9):

In this case, the common voltage is U:

Total charge:

Resulting capacity:

Compare with parallel connection resistance R:

Field strength inside the capacitor (Fig. 5.11):

Voltage between plates:

where is the distance between the plates.

Since the charge is

.

2. Capacitance of a cylindrical capacitor

The potential difference between the plates of a cylindrical capacitor shown in Figure 5.12 can be calculated using the formula:

Transformerless power supplies with a quenching capacitor are convenient in their simplicity, have small dimensions and weight, but are not always applicable due to the galvanic connection of the output circuit with a 220 V network.

In a transformerless power supply, a series-connected capacitor and load are connected to an alternating voltage network. A non-polar capacitor connected to an alternating current circuit behaves like a resistance, but, unlike a resistor, does not dissipate the absorbed power as heat.

To calculate the capacity of the quenching capacitor, the following formula is used:

C is the capacitance of the ballast capacitor (F); Ieff - effective load current; f - input voltage frequency Uc (Hz); Uс - input voltage (V); Un - load voltage (V).

For ease of calculations, you can use an online calculator

The design of devices powered from them must prevent the possibility of touching any conductors during operation. Special attention attention should be paid to insulating the controls.

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  • The need to connect an LED to the network is a common situation. This includes an indicator for turning on devices, a backlit switch, and even a diode lamp.

    There are many schemes for connecting low-power indicator LEDs through a resistor current limiter, but such a connection scheme has certain disadvantages. If you need to connect a diode with a rated current of 100-150mA, you will need a very powerful resistor, the dimensions of which will be significantly larger than the diode itself.

    This is what the connection diagram for a tabletop LED lamp would look like. And powerful ten-watt resistors at low room temperatures could be used as an additional heating source.

    The use of conductors as a current limiter allows one to significantly reduce the dimensions of such a circuit. This is what the power supply for a 10-15 W diode lamp looks like.

    The principle of operation of circuits using a ballast capacitor


    In this circuit, the condenser is a current filter. Voltage is supplied to the load only until the condenser is fully charged, the time of which depends on its capacity. In this case, no heat generation occurs, which removes restrictions on load power.

    To understand how this circuit works and the principle of selecting a ballast element for an LED, let me remind you that voltage is the speed of electrons moving along the conductor, and current is the electron density.

    For a diode, it is absolutely indifferent at what speed electrons will “fly” through it. The calculation of the conductor is based on the current limitation in the circuit. We can apply at least ten kilovolts, but if the current is several microamps, the number of electrons passing through the light-emitting crystal will be enough to excite only a tiny part of the light emitter and we will not see the glow.

    At the same time, at a voltage of several volts and a current of tens of amperes, the electron flux density will significantly exceed the throughput of the diode matrix, converting the excess into thermal energy, and our LED element will simply evaporate in a cloud of smoke.

    Calculation of a quenching capacitor for an LED

    Let's look at the detailed calculation; below you can find the online calculator form.

    Calculation of capacitor capacity for an LED:

    C(uF) = 3200 * Isd) / √(Uin² - Uout²)

    With uF– condenser capacity. It should be rated at 400-500V;
    ISD– rated current of the diode (see the passport data);
    Uin– amplitude network voltage - 320V;
    Uout– rated LED supply voltage.

    You can also find the following formula:

    C = (4.45 * I) / (U - Ud)

    It is used for

    Hi all! I surfed the site a lot, and especially in my thread, and found a lot of interesting things. In general, in this article I want to collect all kinds of amateur radio calculators so that people don’t search too hard when the need arises for calculations and circuit design.

    1. Inductance Calculator- . We thank you for the presented program. crab

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    Light indication is an integral part of electronics, with the help of which a person can easily understand the current state of the device. In household electronic devices The indication role is performed by an LED installed in the secondary power circuit, at the output of the transformer or stabilizer. However, in everyday life there are also many simple electronic designs that do not have a converter, in which an indicator would be a useful addition. For example, an LED built into a wall switch key would be an excellent reference for the location of the switch at night. And the LED in the body of the extension cord with sockets will signal that it is connected to a 220 V power supply.

    Below are several simple circuits, with the help of which even a person with minimal knowledge of electrical engineering can connect an LED to an alternating current network.

    Connection diagrams

    An LED is a type of semiconductor diode with a supply voltage and current much lower than in a household electrical network. If connected directly to a 220 volt network, it will instantly fail. Therefore, the light-emitting diode must be connected only through a current-limiting element. The cheapest and easiest to assemble are circuits with a step-down element in the form of a resistor or capacitor.

    An important point that you need to pay attention to when connecting an LED to an AC network is the reverse voltage limitation. This task can easily be accomplished by any silicon diode designed for a current no less than that flowing in the circuit. The diode is connected in series after the resistor or with reverse polarity in parallel with the LED.

    There is an opinion that it is possible to do without limiting the reverse voltage, since electrical breakdown does not cause damage to the light-emitting diode. However, reverse current may cause overheating p-n junction, resulting in thermal breakdown and destruction of the LED crystal.

    Instead of a silicon diode, you can use a second light-emitting diode with a similar forward current, which is connected in reverse polarity in parallel with the first LED.

    The downside of current-limiting resistor circuits is the need for dissipation high power. This problem becomes especially relevant when connecting a load with high current consumption. This problem is solved by replacing the resistor with a non-polar capacitor, which in such circuits is called ballast or quenching.

    A non-polar capacitor connected to an AC network behaves like a resistance, but does not dissipate the power consumed in the form of heat.

    In these circuits, when the power is turned off, the capacitor remains undischarged, which creates a risk of electric shock. This problem is easily solved by connecting a 0.5-watt shunt resistor with a resistance of at least 240 kOhm to the capacitor.

    Calculation of a resistor for an LED

    In all of the above circuits with a current-limiting resistor, the resistance is calculated according to Ohm's law: R = U/I, where U is the supply voltage, I is the operating current of the LED. The power dissipated by the resistor is P = U * I. This data can be calculated using.

    Important. If you plan to use the circuit in a low convection package, it is recommended to increase maximum value power dissipated by the resistor by 30%.

    Calculation of a quenching capacitor for an LED

    Calculation of the capacity of the quenching capacitor (in μF) is carried out using the following formula: C = 3200*I/U, where I is the load current, U is the supply voltage. This formula is simplified, but its accuracy is sufficient for serial connection 1-5 low current LEDs.

    Important. To protect the circuit from voltage surges and impulse noise, a quenching capacitor must be selected with an operating voltage of at least 400 V.

    It is better to use a ceramic capacitor of the K73–17 type with an operating voltage of more than 400 V or its imported equivalent. Electrolytic (polar) capacitors must not be used.

    You need to know this

    The main thing is to remember safety precautions. The presented circuits are powered by 220 V AC, and therefore require special attention during assembly.

    Connecting the LED to the network must be carried out in strict accordance with circuit diagram. Deviation from the diagram or negligence can lead to a short circuit or failure of individual parts.

    You should carefully assemble transformerless power supplies and remember that they do not have galvanic isolation from the network. Ready scheme must be reliably isolated from adjacent metal parts and protected from accidental contact. It can only be dismantled with the power supply switched off.

    A little experiment

    To lighten up the boring diagrams a little, we suggest that you familiarize yourself with a small experiment that will be of interest to both novice radio amateurs and experienced professionals.

    Read also

    It is more profitable and easier to power low-voltage electrical and radio equipment from the mains. Transformer power supplies are most suitable for this, since they are safe to use. However, interest in transformerless power supplies (BTBP) with stabilized output voltage does not wane. One of the reasons is the complexity of manufacturing the transformer. But for the BTBP it is not needed - only the correct calculation is required, but this is precisely what scares inexperienced novice electricians. This article will help you make calculations and facilitate the design of a transformerless power supply.

    A simplified diagram of the BPTP is shown in Fig. 1. 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 for the load of the block - resistor R n. A filter capacitor C f and a zener diode VD2 are connected in parallel to the load.

    The calculation of the power supply begins with setting the voltage U n on the load and the current strength I n. consumed by the load. The greater the capacitance of the capacitor C, the higher the energy capabilities of the BPTP.

    Capacitance calculation

    The table shows data on the capacitance X c of the capacitor C extinguished at a frequency of 50 Hz and the average value of the current I cf passed by the capacitor C extinguishing, calculated for the case when R n = 0, that is, with a short circuit of the load. (After all, the BTBP is not sensitive to this abnormal operating mode, and this is another huge advantage over transformer power supplies.)

    Other values ​​of capacitance X s (in kilo-ohms) and the average current value I sr (in milliamps) can be calculated using the formulas:


    C extinguisher is the capacitance of the quenching capacitor in microfarads.

    If we exclude the zener diode VD2, then the voltage U n on 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 kilo-ohms, I n - in milliamps, C gas - in microfarads. (The formulas below use the same units of measurement.)

    As the load resistance decreases, the voltage on it also decreases, and according to a nonlinear dependence. But the current passing through the load increases, although 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 distinguishes BTBP from transformer power supplies.

    Power Рн at the load, calculated by the formula:



    with a decrease in Rn, it decreases almost as intensely as Un. For the same example, the power consumed by the load is reduced by 9.1 times.

    Since the current I n of the load at relatively small values ​​of resistance R n and voltage U n on it changes extremely little, in practice it is quite acceptable to use approximate formulas:



    By restoring the zener diode VD2, we obtain stabilization of the voltage U n at the level of U st - a value that is practically constant for each specific zener diode. And with a small load (high resistance R n), the equality U n = U st.

    Load resistance calculation

    To what extent can R n be reduced so that the equality U n = U st is valid? As long as the inequality holds:



    Consequently, if the load resistance turns out to be less than the calculated Rn, the voltage on the load will no longer be equal to the stabilization voltage, but will be somewhat less, since the current through the zener diode VD2 will stop.


    Calculation of permissible current through a zener diode

    Now let’s determine what current I n will flow through the load R n and what current will flow 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 flows through Zener diode, and vice versa. This means that if the load is small (or completely absent), the zener diode VD2 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 network voltage is 220·√2=311(V). The pulse value of the current in the circuit, if we neglect the capacitor C f, can reach



    Accordingly, the zener diode VD2 must reliably withstand this pulse current in case of accidental disconnection of the load. We should not forget about possible voltage overloads in the lighting network, amounting to 20...25% of the nominal value, and calculate the current passing through the zener diode when the load is off, taking into account a correction factor of 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 analogue. 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 transfer coefficient of the powerful base current npn transistor VT1. The voltage of the UCT analogue will be approximately 0.7V higher than Ust of the lowest-power zener diode VD2 if the transistor VT1 is silicon, or by 0.3V if it is germanium.

    A transistor is also applicable here. p-n-p structures. However, then the circuit shown in Fig. is used. 3.

    Half-wave block calculation

    Along with a full-wave rectifier, the simplest half-wave rectifier is sometimes used in BTBP (Fig. 4). In this case, its load Rn is powered only by positive half-cycles of alternating current, and the negative ones pass through the diode VD3, bypassing the load. Therefore, the average current I cf through diode VD1 will be half as much. This means that when calculating the block, instead of X c, you should take 2 times the resistance equal to



    and the average current with a short-circuited load will be equal to 9.9 πС extinguisher = 31.1 С extinguishing. Further calculation of this version of the BPTP is carried out completely similarly to the previous cases.

    Calculation of voltage on the quenching capacitor

    It is generally accepted that with a network voltage of 220V, the rated voltage of the quenching capacitor C should be at least 400V, that is, with approximately a 30 percent margin in relation to the amplitude network voltage, 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 selecting a suitable capacitor C, it should be taken into account that it is impossible to use capacitors of the types MBM, MBPO, MBGP, MBGTs-1, MBGTs-2 in BTBP, since they are not designed to operate in alternating current circuits with an amplitude voltage value exceeding 150V.

    Capacitors MBGCh-1, MBGCh-2 with a rated voltage of 500V work most reliably in BTBP (from old washing machines, fluorescent lamps etc.) or KBG-MN, KBG-MP, but for a rated voltage of 1000V.

    Filter capacitor

    The capacitance of the filter capacitor C f is difficult to calculate analytically. Therefore, it is selected experimentally. Approximately, it should be assumed that for each milliamp of average current consumed, it is required to take at least 3...10 μF of this capacitance if the BTBP rectifier is full-wave, or 10...30 μF if it is half-wave.

    The rated voltage of the oxide capacitor used C f must be at least U st And 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 constantly and there is no zener diode, the rated 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 reconnected only after disconnecting the BTBP from the network.

    And that is not all

    It is advisable to supplement any of the possible BTBP options with two more auxiliary resistors. One of them, the resistance of which can be in the range of 300 kOhm...1 MOhm, is connected in parallel with the capacitor C extinguisher. This resistor is needed to speed up the discharge of capacitor C after disconnecting the device from the network. The other - ballast - with a resistance of 10...51 Ohms is connected to the break of one of the network wires, for example, in series with the capacitor C extinguisher. This resistor will limit the current through the diodes of the VD1 bridge when 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 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 noticeably.

    What diodes to take

    The function of the full-wave rectifier BTBP according to the circuits in Fig. 1...3 can be made by diode assemblies of the KTs405 or KTs402 series with letter indices Ж or И, 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, V or G, D226 B or V - up to 300 mA, KD209 A, B or V - up to 500...700 mA, KD226 V, G or D - up to 1.7 A .

    Diodes VD1 and VD3 in the BTBP according to the diagram in Fig. 4 can be any of the above. It is also permissible to use two diode assemblies KD205K V, G or D for a current of up to 300 mA or KD205 A, V, Zh or I - up to 500 mA.

    And one last thing. The transformerless power supply, as well as the equipment connected to it, are connected directly to the AC network! Therefore, they must be reliably insulated from the outside, say, placed in a plastic case. In addition, it is strictly forbidden to “ground” any of their terminals, as well as to open the case when the device is turned on.

    The proposed methodology for calculating BPTP has been tested by the author in practice for a number of years. The entire calculation is carried out based on the fact that the BPTP is essentially a parametric voltage stabilizer, in which the role of a current limiter is performed by a quenching capacitor.

    Magazine "SAM" No. 5, 1998