Selection of quenching capacitor. Calculation of a capacitor for LEDs. Capacitance calculation

The article provides a method for calculating capacity quenching capacitor and voltage at its terminals in the active load circuit,in particular, a soldering iron, which can significantly reduce the amount of calculations, reducing them to a minimum, which simplifies calculations and reduces time, necessary to select a quenching capacitor of the required capacity and the corresponding rated voltage.

The material presented suggests method for calculating capacitor capacity and the voltage on it when it is connected in series with a soldering iron, and two options are considered. In the first option, it is necessary to reduce the power of the soldering iron by the required amount using a quenching capacitor, and in the second, turn on low voltage soldering iron into a 220 V network, extinguishing excess voltage with a capacitor.

Implementation of the first option(Fig. 1) involves two calculations with initial data (the current consumed by the soldering iron from the network I and the resistance of the soldering iron R1), then two intermediate calculations (the current consumed by the soldering iron at less power by the required value II and the capacitance of the capacitor Rc) and, finally , the last two calculations that give the required

Fig.1

values ​​of the capacitance of the capacitor C at a frequency of 50 Hz and the voltage at the terminals of the capacitor Uc). Thus, to solve the problem according to the first option, it is necessary to carry out 6 calculations.

According to the second option (Fig. 2),to solve the problem, it is necessary to perform two calculations with the initial data, as in the first option, namely: find the current

I, consumed by the soldering iron from the network, and the resistance of the soldering iron R, then follows one intermediate calculation, from which, as in the first option, the capacitance of the capacitor Rc is found and, finally, the last two calculations, from which the capacitance of the capacitor C is determined at a frequency of 50 Hz and on-

Fig.2

voltage at the capacitor terminals Uc. Thus, to solve the problem using the second option, it is necessary to carry out five calculations.

Solving problems using both options requires a certain amount of time. The technique does not allow one to immediately determine the capacitance of the quenching capacitor and, accordingly, the voltage at its terminals in one step, bypassing the initial and intermediate calculations.

We were able to find expressions that allow us to immediately, in one step, calculate the capacitance of the quenching capacitor, and then the voltage at its terminals for the first option. In a similar way, an expression was obtained to determine the capacity of the quenching capacitor for the second option.

Option 1. We have a 100 W 220 V soldering iron and want to operate it at a power of 60 W, using a quenching capacitor connected in series with it. Initial data: rated power of the soldering iron P = 100 W; rated network voltage U = 220 V; required soldering iron power P1 = 60 W. It is required to calculate the capacitance of the capacitor and the voltage at its terminals according to Fig. 1. The formula for calculating the capacity of the quenching capacitor is:

C = P∙10 6 /2πf 1 U 2 (P/P 1 - 1) 0.5 (µF).

At mains frequency = 50 Hz, the formula takes the form:

C = 3184.71 R/U 2 (R/R 1 - 1) 0.5 =

3184.71-100 /220 2 (100/60-1) = 8.06 µF.

In the test example, the capacitance of the capacitor is 8.1 µF, i.e. we have a complete coincidence of results. The voltage at the capacitor terminals is

Uc = (PP 1) 0.5 ∙10 6 /2πf 1 CU (V).

At network frequency f 1 = 50 Hz, the formula simplifies:

Uc = 3184.71 (PP 1) 0.5 /CU =

3184,71(60∙100) 0,5 /8,06 220 =

139.1 V.

In the test example, Uc = 138 V, i.e. practical coincidence of the result. Thus, to solve the problem according to the first option, instead of six calculations, you need to make only two (without intermediate calculations). If necessary, the capacitance of the capacitor can be immediately calculated using the formula:

Rc = U 2 (P/P, - 1) 0.5 /P =

220 2 (100/60 - 1) 0.5 /100 = 395.2 Ohm.

In the test example, Rc = 394 Ohm, i.e. practical coincidence.

Option 2. We have a soldering iron with a power of 25 W, a voltage of 42 V and we want to connect it to a 220 V network. It is necessary to calculate the capacitance of the quenching capacitor connected in series to the soldering iron circuit and the voltage at its terminals according to Fig. 2. Initial data: nominal soldering iron capacity P = 25 W; rated voltage Ur = 42 V; mains voltage U = 220 V. The formula for calculating the capacitor capacity is:

C = Р∙10 6 /2πf 1 Ur(U 2 - Ur 2) 0.5 μF.

At network frequency f 1 = 50 Hz, the formula takes the form:

C = 3184.71 P/Ur(U 2 - Ur 2) 0.5 =

3184,71 -25/42(220 2 - 42 2) =

8.77 µF.

The voltage at the capacitor terminals can be easily determined using the initial data using the Pythagorean theorem:

Uc = (U 2 - Ur 2) 0.5 = (220 2 - 42 2) =

216 V.

Thus, to solve the problem using the second option, instead of five calculations, it is necessary to carry out only two. If necessary, the capacitance value of the capacitor for this option can be determined by the formula:

Rc = Ur(U 2 - Ur 2) 0.5 /P =

42(220 2 - 42 2)/25 = 362.88 Ohms.

According to the test example, Rc = 363 Ohm. It is advisable to bypass the quenching capacitor C in the given figures with a discharge resistor MLT-0.5 with a nominal value of 300...500 kOhm.

Conclusions. The proposed method for calculating the capacitance of a quenching capacitor and the voltage at its terminals makes it possible to significantly reduce the amount of calculations, reducing them to a minimum.

K. V. Kolomoytsev.

Circuit voltage, Ua

Circuit frequency, f

Capacity of the step-down capacitor, C

Load voltage, Ub

Current flowing through the load, I

Load power, P

If you have ever had the task of reducing the voltage to any level, for example from 220 Volts to 12V, then this article is for you.

There are many ways to do this using available materials. In our case, we will use one part - a container.

In principle, we can use ordinary resistance, but in this case, we will have the problem of overheating of this part, and then a fire is not far away.

In the case when a capacitance is used as a reducing element, the situation is different.

A capacitance connected to an alternating current circuit has (ideally) only reactance, the value of which is found according to the well-known formula.

In addition, in our circuit we include some kind of load (light bulb, drill, washing machine), which also has some kind of resistance R

Thus, the total resistance of the circuit will be as

Our circuit is in series, and therefore the total voltage of the circuit is the sum of the voltages on the capacitor and on the load

Using Ohm's law, we calculate the current flowing in this circuit.

As you can see, knowing the parameters of the circuit, it is easy to calculate the missing values.

And remembering how power is calculated, it is easy to calculate the parameters of the capacitor based on the power consumption of the load.

Keep in mind that in such a circuit you cannot use polar capacitors, that is, those that are included in electronic circuit in strict accordance with the indicated polarity.

In addition, it is necessary to take into account the network frequency f. And if in Russia we have a frequency of 50Hz, then for example in America the frequency is 60Hz. This also affects the final calculations.

Calculation examples

It is necessary to power a 36W light bulb designed for a voltage of 12V. What capacity of the step-down capacitor is needed here?

If we're talking about about electrical networks in Russia, the input voltage is 220 Volts, frequency 50 Hz.

The current passing through the light bulb is 3 Amperes (36 divided by 12). Then the capacity according to the above formula will be equal to:

The obtained parameters of the step-down capacitor

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 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 = 0, that is, when short circuit loads. (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 milliamperes, 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 it consumes 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 large sizes 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. Transformerless unit power supply, as well as the equipment connected to it, are connected to the AC network directly! 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

Why did I order these capacitors? The answer is banal. To “collective farm” LED lightening. Where else can they be used? I'll tell you how to calculate the ballast capacity for an LED light bulb. Control review. For those who are not afraid to use such drivers, let's go. For those who do not respect such schemes, there is no need to enter.

First, as usual, let's see what was in the package

And the parcel contains two bags of Conders, exactly 50 pcs each. in everyone. I also ordered these condensers
$7.85 (50 pieces) from the same seller.

I chose it not only by voltage and capacity, but also by size. They should be minimal, otherwise they won’t be applicable everywhere.



I also ordered diodes.


$8.21 (1000pcs.)


Of course, I went overboard with the diodes. 1000 pieces is a lot. But the price difference between 100 and 1000 is simply ridiculous. Diodes 1N4007 (1A 1000V) are widely used in imported household appliances. We can say that not a single product can do without them. It can be used in ours too. Let them sit, if anything happens, I’ll give some of them to my friends.

Well, now let's get down to business.
We take a standard Chinese light bulb. Here is its diagram (slightly improved).


Added R4, it will be instead of a fuse, and will also soften the starting current. The current through the LEDs determines the rating of capacitance C1. Depending on what current we want to pass through the LEDs, we calculate its capacity using formula (1).


For calculations, we need to know the voltage drop across the LEDs. It's easy to calculate. The LED behaves in the circuit like a zener diode with a stabilization voltage of about 3V (there are exceptions, but very rare). At serial connection LEDs, the voltage drop across them is equal to the number of LEDs multiplied by 3V (if there are 5 LEDs, then 15V, if 10 - 30V, etc.). Let's say we want to make a light bulb with ten 5730smd LEDs. According to the passport data, the maximum current is 150mA. I am not a supporter of violence. Therefore, we calculate the light bulb at 100mA. There will be a power reserve. And the supply, as they say, is not enough for the pocket.
Using formula (1) we get: C=3.18*100/(220-30)=1.67 μF. The industry does not produce such a capacity, not even the Chinese one. We take the nearest convenient one (we have 1.5 μF) and recalculate the current using formula (2).
(220-30)*1.5/3.18=90mA. 90mA*30V=2.7W. This is the rated power of the light bulb. It's simple. In life, of course, it will be different, but not much. It all depends on the actual voltage in the network, the exact capacity of the ballast, the actual voltage drop across the LEDs, etc. By the way, using formula (2) you can calculate the power of light bulbs already purchased. The voltage drop across R2 and R4 can be neglected; it is insignificant. You can connect quite a lot of LEDs in series, but the total voltage drop should not exceed half the mains voltage (110V). If this voltage is exceeded, the light bulb reacts painfully to all changes in the network voltage. The more it exceeds, the more painfully it reacts (this is friendly advice).
And yet, let’s check how accurate the capacity ratings are. First 2.2 µF.


Now 1uF.

The errors are small, no more than 2%. You can safely take it.
Let's move on to practical application. If anyone is interested, look where I applied it. This was already in one of the previous reviews, so I hid it under a spoiler.

Excerpt from panel overview

In one of my reviews I connected panels to a driver on a Conder. This is the light bulb made from an energy-saving lamp. Let me remind you that the module consists of five parallels. Each parallel contains 18 2835smd LEDs. Voltage drop 51V.

Let's calculate the current from formula (2):
We get current =(220-51)*2.2/3.18=117mA. 51V*117mA=6W LED power (66.7mW for each LED - 33% of the nominal) - the calculated power of the lamp. We assemble and turn on. WORKS!

But such light bulbs cannot be used without protective glass or a plastic diffuser. All LEDs are under phase and cannot be touched in operating mode. Now let's see what the instruments show. Where would I be without them?


The device showed 5.95W.
Of course, such a light bulb can only be used in a barn.
And people have sheds and garages. And something needs to be screwed in there too (the village version, I’ll explain why). In the summer I often go to the village. But in the countryside, the voltage does not rise above 200V; sometimes it can be lower. Now let’s calculate the power of our light bulb at 180V in the network. Using the same formula, we first find the current that flows through the LEDs. Only instead of 220V in the formula we will put 180V. Total 110mA*51V=5.6W. As you can see, the power has hardly changed. But incandescent light bulbs smoke at this voltage.
Option with garage. In the garage, on the contrary, I don’t have time to change light bulbs - at least 240V. Let's calculate the current and power at 260V, all using the same formula. We have: 145mA*51V=7.4W (41% of maximum power). Burnout is too far away. Conclusion: it will light at 180V and will not burn out at 260V.
Now I’ll try to evaluate the quality characteristics of the light. Tried to light the wall

It shines very brightly, with a warm, pleasant light, brighter than a 60W incandescent lamp (picture below). You can compare brightness and color tone. Everything was filmed under the same conditions, at the same distance from the wall.

I also measured the power of the incandescent lamp for the purity of the experiment, with the same device under the same conditions.
Incandescent lamp – 56.5 W.
LED lamp – 5.95W.
I inserted both light bulbs one by one into a table lamp with a reflector. You saw him.

Now a clipping from my last review. True, I added dimensions.

Excerpt from the review About 1W LED Bulbs High power diodes

Using these LEDs I decided to remake the lamp.


The light bulbs have already deteriorated, and the new ones are of low quality.


I decided to connect the lamp via condensers; I don’t need more power, but electronic driver I'll save it for something more worthwhile. And here is the diagram.


I connect all the diodes in series.


I also made the driver board from what I had (quickly)






There was even a pin for fastening. I did not remove the throttle. I left it for weight, otherwise the lamp will fall.




I did it according to all electrical safety rules. Not a single energized element comes out. The board is secured with printed conductors inside.
Let's calculate the power of the resulting light bulb. First, using formula (2), we find the current through the LEDs with a ballast capacity of 3.2 μF. (220-18)*3.2/3.18=203.2mA. 203.2mA*18V=3.66W – calculated power (at a network voltage of 220V).
We look at the device


The device shows 3.78W. But the socket is 232V, not 220V. The error is minimal.
And, as usual, let's see how it shines.

This is a 40W light bulb. Naturally, all light bulbs are in equal conditions (shutter speed on the handbrake, the distance to the wall is the same).

This is my LED light. The photo exposure meter tells you that the light is brighter than forty.

And finally, the third device where they (conders) can be used. I have been using a homemade charger for many years.

Additional Information

It also contains a current driver based on capacitors.


It was made long before I received the condensers and diodes from China. Therefore, all parts are domestic.


The circuit is standard, like in Chinese light bulbs.


It was for this charging that I derived the formula for calculating the ballast capacity. So, if anyone wants, they can calculate both the current and charging time with other capacitors in the ballast.

Now let's try to summarize. I will try to highlight all the pros and cons of such schemes.
-During operation, you categorically cannot touch the circuit elements, they are under phase.
-It is impossible to achieve high currents of LEDs, since this requires large capacitors.
- Large pulsations of the light flux with a frequency of 100 Hz require large filter capacities at the output.
+The circuit is very simple and does not require any special skills in manufacturing.
+ Does not require special material costs during production. Most parts can be found in any shed or garage (old TVs, etc.).
+Indispensable as an initial LED experience, as the first step in mastering LED lighting.
I wrote my vision, my attitude to such schemes. It may differ from yours. But I expressed it. And as always, it’s up to you to draw the conclusion.
That's all. More to detailed analysis I will not return to such schemes. Worn them out from top to bottom.
And at the end for those who keep track of the tracks.

Some radio amateurs, when designing network power supplies, use capacitors instead of step-down transformers. ballast, dampening excess voltage (Fig. 1).

A non-polar capacitor included in an AC circuit behaves like a resistance, but, unlike a resistor, does not dissipate the absorbed power in the form of heat, which makes it possible to design a compact power supply that is lightweight and cheap. The capacitance of a capacitor at frequency f is described by the expression:

The capacitance value of the ballast capacitor Cb is determined with sufficient accuracy using the formula:

where U c is the network voltage, V;

I N - load current, A;

U H is the load voltage, V. If U H is in the range from 10 to 20 V, then the following expression is quite acceptable for calculation:

Substituting the values ​​of U c = 220 V and U H = 15 V, at I n = 0.5 A we obtain the values ​​Sb = 7.28 μF (1) and Sb = 7.27 μF (2). For both expressions, a very decent agreement is obtained, especially considering that the capacity is usually rounded to the nearest greater value. It is better to select capacitors from the K73-17 series with an operating voltage of at least 300 V.

When using this circuit, you must always remember that it is galvanically connected to the network and you risk being hit electric shock with potential mains voltage. In addition, measuring equipment or any additional devices should be connected very carefully to a device with a transformerless power supply, otherwise you may end up with fireworks that are not at all festive.

To power even low-power devices, it is better to use step-down transformers. If the voltage of its secondary winding does not correspond to the required (exceeds), then it is quite safe to use a quenching capacitor in the circuit of the primary winding of the transformer to reduce the voltage or to connect a transformer with a low-voltage primary winding to the network (Fig. 2) Ballast capacitor in this case, it is selected so that at the maximum load current, the output voltage of the transformer corresponds to the specified one.

Literature

1. Biryukov S.A. Devices on microchips. - M., 2000.

I. SEMENOV,

Dubna, Moscow region.