Calculation of the quenching capacitor in the transformer power supply. Transformerless power supplies, voltage converters without transformer. Calculation. Calculate online, online

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, downgrading AC voltage 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 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. Capacitor working in a circuit alternating current, 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:

In such a power supply, a series-connected capacitor and a load are connected to an alternating voltage network. Let us first consider the operation of a source with a purely resistive load (Fig. 1a).

From the course of electrical engineering, it is known that the impedance of a series-connected capacitor C1 and a resistor Rn is equal to:

where X c 1 \u003d 1 / 2n * f * C1 is the capacitance of the capacitor at frequency f. Poet-

Fig.1

mu effective alternating current in the circuit Ieff \u003d Uс / Z (Uc is the supply voltage). The load current is related to the capacitance of the capacitor, the output voltage of the source and the mains voltage as follows

For small output voltages

Ieff \u003d 2l * f * C1 * Uc.

As an example useful in practice, we will calculate a quenching capacitor for connecting a 127 V soldering iron with a power of 40 W to a 220 V network. The required effective value of the load current Ieff \u003d 40/127 \u003d 0.315 A. Estimated capacitance of the quenching capacitor

For the operation of heating devices, the value of the effective current is important. However, if the load is, for example, a battery included in the diagonal of the rectifier bridge (Fig. 1, b), it will be charged by an average rectified (pulsating) current, the numerical value of which is less than Ieff:

In amateur radio practice, a source is often used in which a quenching capacitor is connected to the network in series with a diode bridge, and the load, shunted by another capacitor, is powered from the output diagonal of the bridge (Fig. 2). In this case, the circuit becomes sharply non-linear and the shape of the current flowing through the bridge and quenching the capacitor will differ from sinusoidal. Because of this, the above calculation is incorrect.

What are the processes taking place in a source with a smoothing capacitor C2 with a capacity sufficient to consider the output voltage ripple as negligible? For the quenching capacitor C1, the diode bridge (together with C2 and Rн) in the steady state is a kind of equivalent of a symmetrical zener diode. When the voltage on this equivalent is less than a certain value (it is practically equal to the voltage Uout on the capacitor C2), the bridge is closed and does not conduct current, at a higher voltage, current flows through the open bridge, preventing the voltage at the bridge input from increasing.

Consideration will start from the moment ti, when the mains voltage is maximum (Fig. 3). Capacitor C1 is charged to the amplitude voltage of the network Uc.amp minus the voltage on the diode bridge um, approximately equal to Uout. The current through the capacitor C1 and the closed bridge is zero. The voltage in the network decreases according to the cosine law (graph 1), it also decreases on the bridge (graph 2), and the voltage across the capacitor C1 does not change.

The capacitor current will remain zero until the voltage on the diode bridge, changing sign to the opposite, reaches the value -Uout (moment t2). At this moment, the current lei will appear abruptly through the capacitor C1 and the bridge. Starting from the moment t2, the voltage on the bridge does not change, and the current is determined by the rate of change of the mains voltage and, therefore, will be exactly the same as if only capacitor C1 was connected to the network (graph 3).

When the mains voltage reaches a negative amplitude value (moment t 3), the current through the capacitor C1 will again become zero. The process is then repeated every half cycle.

The current through the bridge flows only in the time interval from t 2 to t 3, its average value can be calculated as the area of ​​​​the shaded part of the sinusoid in graph 3. Simple calculations, which, however, require knowledge of differential and integral calculus, give such a formula for the average current Iav through the load Rн:

(2)

At low values ​​of the output voltage, this formula and the previously obtained (1) give the same result. If in (2) the output current is equated to zero, we get Uvyx=Uc*2 ^1/2, i.e., with a load current equal to zero (in case of accidental disconnection of the load, say, due to an unreliable contact), the output voltage of the source becomes equal to the amplitude voltage of the network. This means that all elements of the source must withstand such a voltage. When the load current decreases, for example, by 10%, the output voltage will increase so that the expression in brackets also decreases by 10%, i.e. by about 30 V (at Uout = 10 V). Conclusion - the inclusion of a zener diode in parallel with the load Rn (as shown by dashed lines in Fig. 2) is almost mandatory.

For a half-wave rectifier (Fig. 4), the current is calculated using the following formula:

Naturally, at low values ​​of the output voltage, the load current will be half as much as for a full-wave rectifier, and the output voltage at zero load current will be twice as much - after all, this is a voltage-doubling rectifier!

The procedure for calculating sources according to the scheme in fig. 2 next. Initially, they are set by the output voltage Uout, the maximum In max and the minimum I n min values ​​of the load current, the maximum Uc max and the minimum Uc min values ​​of the mains voltage. It has already been indicated above that with a changing load current, a zener diode is required, connected in parallel with the load Rn. How to choose it? At the minimum mains voltage and maximum load current, a current of at least the permissible minimum stabilization current 1st min must flow through the zener diode. You can set the value within 3...5 mA. Now determine the capacitance of the quenching capacitor C1 for a full-wave rectifier:

C1 \u003d 3.5 (Ist min + ln max) / (Uc min-0.7 Uvyx). (3)

The formula is obtained from (2) by substituting the corresponding values. The current in it is in milliamps, the voltage is in volts; capacitance is in microfarads. The result of the calculation is rounded up to the nearest higher denomination; you can use a battery of several capacitors connected in parallel.

I st max \u003d (U c max -0.7 Uout) C 1 / 3.5-I n min (4)

In the absence of a zener diode for the required voltage Uout, allowing the calculated maximum stabilization current, you can connect several zener diodes for a lower voltage in series or use an analogue of a powerful zener diode.

The minimum load current In mm should be substituted into formula (4) only when this current is long - a few seconds or more. With a short-term minimum load current (fractions of a second), it must be replaced by an average (in time) load current. If the zener diode allows a current greater than that calculated by formula (4), it is advisable to use a quenching capacitor of a slightly larger capacity to reduce the requirements for the accuracy of its selection.

I needed a power supply for a homemade mini drill made from a 17 volt motor. I reviewed many schemes of various PSUs, but all of them used a transformer, which I don’t have, but I’m somehow reluctant to buy. Then I decided to do it easier and collect power for a given voltage - 17 Volts. The circuit is quite simple, you need to supply 220 volts of alternating voltage to such a ready-made power supply, in short, power the circuit from the outlet, and at the output we get 17 volts of direct voltage. Typically, power supplies of this type are used in all sorts of small household items, for example, in a flashlight with a battery, as a charger, where a small current is needed, up to 150 mA, or in electric shavers.

So, the details for the circuit. This is what the high-voltage metal-film capacitors (the ones that are red) look like, and to the left of them there is a 100 uF electrolytic capacitor.

Instead of a microcircuit 78l08 You can use voltage stabilizers such as KR1157EN5A (78l08) or KR1157EN5A (7905).

If there is no rectifier diode 1N4007 , then it can be replaced by 1N5399 or 1N5408 , which are designed for more high current. The gray circle on the diode indicates its cathode.

Resistor R1 took 5W, and R2 - 2W, for insurance, although both could be used at 0.5 watts.

zener diode BZV85C24 (1N4749), designed for a power of 1.5 W, and for voltages up to 24 volts, it can be replaced with domestic 2S524A .

This transformerless PSU was assembled without adjusting the output voltage, but if you want to organize such a function, then simply connect a variable resistor of about 1 kOhm to pin 2 of the 78L08 microcircuit, and its second output to the minus circuit.

Of course, there is a board for the transformerless power supply circuit, the lay format, you can download it. I think you understand that unmarked diodes are 1n4007 .

The finished structure must be placed in a plastic case, due to the fact that the circuit included in the network is under voltage of 220 volts and in no case should you touch it!

In these photos you can see the input voltage, that is, the voltage in the outlet, and how many volts we get at the PSU output.

Video of the operation of the transformerless power supply circuit

The big advantage of this scheme we can consider the very modest dimensions of the finished device, because due to the lack of a transformer, this PSU can be made small, and the cost of parts for the circuit is relatively inexpensive.

Minus the scheme it can be considered that there is a danger of accidentally touching a working source and getting an electric shock. Article author - egoruch72.

Discuss the article TRANSFORMER-LESS POWER SUPPLY

At the very beginning of the topic, regarding the selection of a quenching capacitor, consider a circuit consisting of a resistor and a capacitor connected in series to the network. The total resistance of such a circuit will be equal to:

The effective value of the current, respectively, is found according to Ohm's law, the mains voltage is divided by the impedance of the circuit:

As a result, for the load current and input and output voltages, we obtain the following relationship:

And if the output voltage is small enough, then we have the right to consider approximately equal to:

However, let's consider from a practical point of view the issue of selecting a quenching capacitor for inclusion in the AC network of a load designed for a voltage less than the standard network voltage.

Let's say we have a 100 watt incandescent lamp rated at 36 volts, and for some incredible reason we need to power it from a 220 volt household network. The lamp requires an effective current equal to:

Then the capacitance of the required quenching capacitor will be equal to:

Having such, we gain hope to get a normal glow of the lamp, we expect that it will at least not burn out. This approach, when we start from the effective value of the current, is acceptable for resistive loads, such as a lamp or a heater.

But what if the load is non-linear and connected through a diode bridge? Let's say you need to charge a lead-acid battery. What then? Then the charging current will be pulsating for the battery, and its value will be less than the effective value:

Sometimes it can be useful for a radio amateur to use a power source in which a quenching capacitor is connected in series with a diode bridge, at the output of which there is, in turn, a filter capacitor of significant capacity, to which a load is connected. direct current. It turns out a kind of transformerless power supply with a capacitor instead of a step-down transformer:

Here, the load as a whole will be non-linear, and the current will no longer be sinusoidal, and it will be necessary to conduct calculations in a slightly different way. The fact is that a smoothing capacitor with a diode bridge and a load will externally manifest itself as a symmetrical zener diode, because the ripples with a significant filter capacitance will become negligible.

When the voltage on the capacitor is less than a certain value, the bridge will be closed, and if it is higher, the current will flow, but the voltage at the output of the bridge will not increase. Consider the process in more detail with graphs:

At time t1 the mains voltage has reached its amplitude, the capacitor C1 is also charged at this moment to the maximum possible value minus the voltage drop across the bridge, which will be approximately equal to the output voltage. The current through the capacitor C1 is equal to zero at this moment. Further, the voltage in the network began to decrease, the voltage on the bridge too, but on the capacitor C1 it does not change yet, and the current through the capacitor C1 is still zero.

Further, the voltage on the bridge changes sign, tending to decrease to minus Uin, and at that moment a current rushes through the capacitor C1 and through the diode bridge. Further, the voltage at the output of the bridge does not change, and the current in the series circuit depends on the rate of change of the supply voltage, as if only capacitor C1 was connected to the network.

When the network sinusoid reaches the opposite amplitude, the current through C1 again becomes equal to zero and the process will go in a circle, repeating every half period. Obviously, the current flows through the diode bridge only between t2 and t3, and the average current can be calculated by determining the area of ​​the patch under the sinusoid, which will be equal to:

If the output voltage of the circuit is small enough, then this formula approaches the one obtained earlier. If the output current is set equal to zero, then we get:

That is, when the load is interrupted, the output voltage will become equal to the amplitude of the mains !!! This means that such components should be used in the circuit so that each of them would withstand the amplitude of the supply voltage.

By the way, when the load current decreases by 10%, the expression in brackets will decrease by 10%, that is, the output voltage will increase by about 30 volts, if initially we are dealing with, say, 220 volts at the input and 10 volts at the output. Thus, the use of a zener diode in parallel with the load is strictly necessary!!!

What if the rectifier is half-wave? Then the current must be calculated using the following formula:

At low values ​​of the output voltage, the load current will become half as much as with full bridge rectification. And the voltage at the output without load will be twice as large, since here we are dealing with a voltage doubler.

So, a power supply with a quenching capacitor is calculated in the following order:

The first step is to choose what the output voltage will be.

Then determine the maximum and minimum load currents.

If the load current is assumed to be non-constant, a zener diode in parallel with the load is required!

Finally, the capacitance of the quenching capacitor is calculated.

For a full-wave rectified circuit, for a mains frequency of 50 Hz, the capacitance is given by the following formula:

The result obtained by the formula is rounded up towards the capacity of a larger denomination (preferably no more than 10%).

The next step is to find the stabilization current of the zener diode for the maximum supply voltage and minimum current consumption:

For a half-wave rectification circuit, the quenching capacitor and the maximum current of the zener diode are calculated using the following formulas:

When choosing a quenching capacitor, it is better to focus on film and metal-paper capacitors. Film capacitors of small capacity - up to 2.2 microfarads for an operating voltage of 250 volts work well in these circuits when powered by a 220 volt network. If you need a large capacitance (more than 10 microfarads) - it is better to choose a capacitor for an operating voltage of 500 volts or more.

Andrey Povny ( Google Plus ,

Several schemes and calculation of transformerless power supplies with a quenching capacitor

Mains power supply with quenching capacitor(Fig. 1), in fact, there is a voltage divider, in which the upper arm is a capacitor, and the lower one is a complex nonlinear diode-resistor-capacitor circuit. This determines the disadvantages (and advantages, of course) of such devices.

Picture 1:

In order for the source to operate in a wide range of load current with high efficiency, it is enough to make the input voltage divider purely reactive, for example, capacitor (Fig. 2).

Figure 2:

It allows you to additionally stabilize the output voltage of the source with a series-connected compensation or switching regulator, which cannot be done in a conventional source with a quenching capacitor. As shown in the article by S. Biryukov “ Calculation of a network power supply with a quenching capacitor” – “Radio”, 1997, N 5, p. 48-50, - a serial stabilizer can only be used if the voltage at its input is limited, which again significantly reduces efficiency.

It is advisable to use a source with a capacitor voltage divider for joint operation with switching regulators. It is ideal for a device that consumes low current for a long time, but at some point requires a sharp increase in it. An example is an apartment watchdog device based on “MOS” microcircuits with an actuating unit on a relay and an audible signaling device.

The current drawn by the capacitor divider will have a phase shift of 90 degrees. relative to the mains voltage, so the voltage divider on the reactive elements does not require cooling. Based on the foregoing, it seems that the current through the divider can be chosen arbitrarily large. However, an unjustified increase in the divider current will lead to active losses in the wires and to an increase in the mass and volume of the device. Therefore, it is advisable to take the current through the voltage divider within 0.5 ... 3 of the maximum load current.

The calculation of a source with a capacitive divider is simple. As follows from f-ly (2) in the mentioned article, the output voltage Uout and total output current (zener diode and load Iout) of the source according to scheme 1, but are related as follows:

Iout = 4fC1(2Uc-Uout)

This formula is also suitable for calculating a source with a capacitor divider, in it you just need to replace C1 with the total capacitance of capacitors C1 and C2 connected in parallel, shown in Fig. 2. a Uc - on Uc2x (voltage on capacitor C2 at RH = ° °), i.e.

Uc2x = Uc-C1/(C1+C2)

Then
Iout = 4f(C1+C2)x x
or after obvious transformations
Iout = 4f-C1 .

Since the voltage drop across the diodes of the bridge Ud at small values ​​of Kout becomes noticeable, we finally get

Iout = 4f-C1 .

It can be seen from the formula that at Рн=0 (i.e., at Uout=0), the current Iout, if we neglect the voltage drop across the diodes, remains the same as for the power source assembled according to scheme 1,a. The voltage at the output without load decreases:
Uaux = =Uc-C1^/2/(C1+C2)-2Un.

The capacitance and operating voltage of the capacitor C2 are selected based on the required output voltage - the ratio of the capacitance values ​​C1 / C2 is inversely proportional to the values ​​​​of the voltage falling on C1 and C2. For example, if C1″ = 1 μF, and C2 = 4 μF, then the voltage Uc1 will be equal to 4/5 of the mains voltage, and Uc2=Uc/5, which corresponds to 186 and 44 V with a mains voltage of Uc = 220 V. It must be taken into account that that the amplitude value of the voltage is almost 1.5 times higher than the current one, and select capacitors for the appropriate rated voltage.

Despite the fact that theoretically capacitors in an AC circuit do not consume power, in reality, some heat can be generated in them due to the presence of losses. You can check in advance the suitability of the capacitor for use in the source by simply connecting it to the mains and estimating the case temperature after half an hour. If the capacitor C1 has time to noticeably warm up, it should be considered unsuitable for use in the source.

Special capacitors for industrial electrical installations practically do not heat up - they are designed for high reactive power. These capacitors are used in fluorescent lamps, in ballasts of asynchronous electric motors, etc.

Below are two practical power supply circuits with a capacitor divider: a five-volt general purpose one (Fig. 3) for a load current of up to 0.3 A and an uninterruptible power supply for quartz electronic-mechanical watches (Fig. 4).

Figure 3:

Figure 4:

The voltage divider of a five-volt source consists of a paper capacitor C1 and two oxide capacitors C2 and C3, forming a non-polar shoulder with a capacity of 100 microfarads. The polarizing diodes for the oxide pair are left-handed bridge diodes according to the scheme. With the ratings of the elements indicated in the diagram, the closing current (at Rn = 0) is 600 mA, the voltage across the capacitor C4 in the absence of load is 27 V.

Electronic-mechanical clocks are usually powered by a single galvanic cell with a voltage of 1.5 V. The proposed source generates a voltage of 1.4 V at an average load current of 1 mA. The voltage removed from the C1C2 divider rectifies the node on the elements VD1, VD2. SZ. Without load, the voltage across the capacitor C3 does not exceed 12V.