A method for limiting the inrush current when connecting a capacitor. Limiting the charging current of the capacitor of the SMPS mains rectifier. Scheme, description. Beijing crashes Wall Street

Power supply circuits

M. DOROFEEV, Moscow
Radio, 2002, No. 10

One of important issues in network switching power supplies - charging current limitation a large capacity smoothing capacitor installed at the output of the mains rectifier. Its maximum value, determined by the resistance of the charging circuit, is fixed for each specific device, but in all cases it is very significant, which can lead not only to blown fuses, but also to failure of input circuit elements. The author of the article offers a simple way to solve this problem.

A lot of works have been devoted to solving the problem of limiting the starting current, in which so-called “soft” switching devices are described. One of the widely used methods is the use of a charging circuit with a nonlinear characteristic. Typically, the capacitor is charged through a current-limiting resistor to operating voltage, and then this resistor is closed with an electronic key. The simplest way to obtain such a device is to use a thyristor. The figure shows a typical input node diagram pulse source nutrition. The purpose of elements not directly related to the proposed device (input filter, mains rectifier) ​​is not described in the article, since this part is made in a standard manner.

Smoothing capacitor C7 is charged from the mains rectifier VD1 through the current-limiting resistor R2, in parallel with which the thyristor VS1 is connected. The resistor must meet two requirements: firstly, its resistance must be sufficient so that the current through the fuse during charging does not lead to its burnout, and secondly, the power dissipation of the resistor must be such that it does not fail before fully charging capacitor C7.

The first condition is satisfied by a resistor with a resistance of 150 Ohms. The maximum charging current in this case is approximately equal to 2 A. It has been experimentally established that two resistors with a resistance of 300 Ohms and a power of 2 W each, connected in parallel, meet the second requirement.

The capacitance of capacitor C7 660 μF was selected from the condition that the amplitude of the rectified voltage pulsations at a maximum load power of 200 W should not exceed 10 V. The values ​​of elements C6 and R3 are calculated as follows. Capacitor C7 will be charged almost completely through resistor R2 (95% of the maximum voltage) in time t=3R2 C7=3 150 660 10 -6 ≈0.3 s. At this moment, the thyristor VS1 should open.

The thyristor will turn on when the voltage at its control electrode reaches 1 V, which means that capacitor C6 must charge to this value in 0.3 s. Strictly speaking, the voltage on the capacitor grows nonlinearly, but since the value of 1 V is about 0.3% of the maximum possible (approximately 310 V), this initial section can be considered almost linear, therefore the capacitance of capacitor C6 is calculated using a simple formula: C = Q /U, where Q=l t - capacitor charge; I - charging current.

Let's determine the charging current. It should be slightly greater than the control electrode current at which the thyristor VS1 turns on. We select the KU202R1 thyristor, similar to the well-known KU202N, but with a lower turn-on current. This parameter in a batch of 20 SCRs ranged from 1.5 to 11 mA, and for the vast majority its value did not exceed 5 mA. For further experiments, a device with a switching current of 3 mA was selected. We select the resistance of resistor R3 equal to 45 kOhm. Then the charging current of capacitor C6 is 310 V/45 kOhm = 6.9 mA, which is 2.3 times greater than the turn-on current of the thyristor.

Let's calculate the capacitance of capacitor C6: C=6.9 10 -3 0.3/1≈2000 μF. The power supply uses a smaller capacitor with a capacity of 1000 μF for a voltage of 10 V. Its charging time has been halved, to approximately 0.15 s. I had to reduce the time constant of the charging circuit for capacitor C7 - the resistance of resistor R2 was reduced to 65 Ohms. In this case, the maximum charging current at the moment of switching on is 310 V/65 Ohm = 4.8 A, but after a time of 0.15 s the current will decrease to approximately 0.2 A.

It is known that a fuse has significant inertia and can pass short pulses without damage, much exceeding its rated current. In our case, the average value over a time of 0.15 s is 2.2 A and the fuse tolerates it “painlessly”. Two resistors with a resistance of 130 Ohms and a power of 2 W each, connected in parallel, also cope with such a load. During the charging time of capacitor C6 to a voltage of 1 V (0.15 s), capacitor C7 will be charged to 97% of the maximum.

Thus, all conditions for safe operation are met. Long-term operation of a switching power supply has shown high reliability of the described unit. It should be noted that a gradual increase in voltage over smoothing capacitor C7 over 0.15 s has a beneficial effect on the operation of both the voltage converter and the load.

Resistor R1 serves to quickly discharge capacitor C6 when the power supply is disconnected from the network. Without it, this capacitor would take much longer to discharge. If in this case you quickly turn on the power supply after turning it off, then the thyristor VS1 may still be open and the fuse will burn out.

Resistor R3 consists of three connected in series, with a resistance of 15 kOhm and a power of 1 W each. They dissipate about 2 W of power. Resistor R2 is two parallel-connected MLT-2 with a resistance of 130 Ohms, and capacitor C7 is two, with a capacity of 330 microns for a rated voltage of 350 V, connected in parallel. Switch SA1 - toggle switch T2 or push-button switch PKN 41-1. The latter is preferable because it allows you to disconnect both conductors from the network. The KU202R1 thyristor is equipped with an aluminum heat sink with dimensions of 15x15x1 mm.

LITERATURE
1. Secondary power sources. Reference manual. - M.: Radio and communication, 1983.
2. Eranosyan S. A. Network power supplies with high-frequency converters. - L.: Energoatomizdat, 1991.
3. Frolov A. Limitation of the capacitor charging current in a network rectifier. - Radio, 2001, No. 12, p. 38, 39, 42.
4. Mkrtchyan Zh. A. Power supply of electronic computers. - M.: Energy, 1980.
5. Integrated circuits of foreign household video equipment. Reference manual. - St. Petersburg: Lan Victoria, 1996.

The circuit is designed to protect against surge charging current when an uncharged capacitor is connected to the on-board network. Anyone who hasn't tried to connect an uncharged capacitor to a network without a limiting resistor - it's better not to... At a minimum, the contacts will burn out.

When the discharged capacitance is connected to the network, capacitance C1 is discharged, T1 (n-MOS switch with low channel resistance) is closed. Capacity C2 (the same farad) is charged through low-resistance R5. T2 opens almost instantly, I shunt C1 to ground and T1 gate. When the potential of the negative terminal of C2 drops below 1V (charging to Ubattery - 1V), T2 closes, C1 smoothly charges to approximately 9/10 Ubattery, opening T1. The time constant R2C1 is large enough so that the current jump T1 (recharging C2 by +1V to Uakb) does not exceed the permissible limit for T1.

In the future, the negative terminal C2 is constantly shorted to ground through T1, INDEPENDENT OF THE DIRECTION OF CURRENT T1 (both in the forward direction - from drain to source, and in the reverse direction). There is nothing wrong with “turning over” an OPEN MOS transistor. When choosing a sufficiently well-conducting transistor, the entire reverse current will flow through the channel, and the built-in reverse diode will not open, since the voltage drop across the channel is several times less than the 0.5-0.8 V required for opening. By the way, there is a whole class of MIS devices (the so-called FETKY) designed specifically for operation in the reverse direction (synchronous rectifiers), their built-in diode is shunted by an additional Schottky power diode.

Calculation: for transistor IRF1010 (Rds=0.012 Ohm), a voltage drop of 0.5 Ohm will be achieved only with a channel current of 40A (P=20W). For four such transistors in parallel and the same discharge current of 40A, each transistor will dissipate 0.012*(40/4)^2 = 1.2 W, i.e. they will not need radiators (especially since 1.2W will be dissipated only when the current consumption changes, but not constantly).

For dense installation (do you have a lot of space for an extra radiator?) - it is advisable to parallel small-sized (TO251, DIP4 package) transistors that do not provide radiators at all, based on the ratio of current (power) consumption of the amplifier - Rds - maximum power dissipation. Since Pds max is usually 1W (800mW for DIP4), the amount n transistors (with Rds each) for an amplifier with output power Pout must be at least n > 1/6 * Pout * sqrt(Rds) at 12V power supply (I omitted the dimensions in the formula). In fact, taking into account the short duration of current pulses, n can be safely halved compared to this formula .

The charging resistor R5 is selected based on a compromise between thermal power and charging time. At the specified 22 Ohms, the charging time is about 1 minute with a power dissipation of 7 W. Instead of R5, you can turn on a 12V light bulb, say, from a turn signal. Resistors R1, R3 are safety resistors (discharge capacitors when disconnected from the network).

To indicate switching on, we connect an additional inverter (reducing R2). Attention! The circuit is operational when using npn transistors T2, T3 with h21e > 200 (KT3102). Depending on the brightness of the LED, select R1 in the range of 200 Ohm - 1 kOhm.

And here is a version of the circuit in which the gate switch is controlled by the REMOTE signal (transistor AND). When REMOTE is not connected or turned off, the key transistor is guaranteed to be closed. LEDs D3-D4 indicate charging of C1, D5-D6 - open state of the key.

Accurate indication of the network voltage threshold is most easily provided by the TL431 (KR142EN19) IC in the typical voltage comparator mode (with a corresponding divider in the input circuit and a current-limiting R in the cathode circuit).

Circuit losses largely depend on installation. It is necessary to ensure a minimum resistance (and wire thicknesses corresponding to the current) in the power circuit (terminal + / C2 / T1 / terminal -). In amateur practice, I think it is not advisable to make external terminals - it is better to immediately solder the short AWG8 wires that connect the circuit to the amplifier terminal block.

Often in various sources power supply, the task arises of limiting the starting current surge when turned on. The reasons may be different - rapid wear of relay contacts or switches, reduced service life of filter capacitors, etc. I recently had a similar problem. I use a good server power supply in my computer, but due to the unsuccessful implementation of the standby section, it overheats severely when the main power is turned off. Because of this problem, I had to repair the standby board twice already and change some of the electrolytes located next to it. The solution was simple - turn off the power supply from the outlet. But it had a number of disadvantages - when turned on, there was a strong surge of current through the high-voltage capacitor, which could damage it, in addition, after 2 weeks the power plug of the unit began to burn out. It was decided to make an inrush current limiter. In parallel with this task, I had a similar task for powerful audio amplifiers. The problems in amplifiers are the same - burning of switch contacts, current surge through the bridge diodes and filter electrolytes. You can find quite a lot of surge current limiter circuits on the Internet. But for a specific task, they may have a number of disadvantages - the need to recalculate circuit elements for the required current; for powerful consumers - selection of power elements that provide the necessary parameters for the calculated allocated power. In addition, sometimes it is necessary to provide a minimum starting current for the connected device, which increases the complexity of such a circuit. To solve this problem, there is a simple and reliable solution - thermistors.

Fig.1 Thermistor

A thermistor is a semiconductor resistor whose resistance changes sharply when heated. For our purposes, we need thermistors with a negative temperature coefficient - NTC thermistors. When current flows through the NTC thermistor, it heats up and its resistance drops.

Fig.2 TKS thermistor

We are interested in the following thermistor parameters:

Resistance at 25˚C

Maximum steady current

Both parameters are in the documentation for specific thermistors. Using the first parameter, we can determine the minimum current that will pass through the load resistance when connecting it through a thermistor. The second parameter is determined by the maximum power dissipation of the thermistor and the load power must be such that the average current through the thermistor does not exceed this value. For reliable operation of the thermistor, you need to take the value of this current less than 20 percent of the parameter specified in the documentation. It would seem that it would be easier to select the right thermistor and assemble the device. But you need to consider some points:

1. The thermistor takes a long time to cool down. If you turn off the device and immediately turn it on again, the thermistor will have low resistance and will not perform its protective function.
2. You cannot connect thermistors in parallel to increase the current - due to the spread of parameters, the current through them will vary greatly. But it is quite possible to connect the required number of thermistors in series.
3. During operation, the thermistor becomes very hot. The elements next to it also heat up.
4. The maximum steady-state current through the thermistor should be limited by its maximum power. This option is specified in the documentation. But if the thermistor is used to limit short current surges (for example, when the power supply is initially turned on and the filter capacitor is charging), then impulse current maybe more. Then the choice of thermistor is limited by its maximum pulse power.

The energy of a charged capacitor is determined by the formula:

E = (C*Vpeak²)/2

where E is the energy in joules, C is the capacitance of the filter capacitor, Vpeak is the maximum voltage to which the filter capacitor will be charged (for our networks you can take the value 250V*√2 = 353V).

If the documentation indicates the maximum pulse power, then based on this parameter you can select a thermistor. But, as a rule, this parameter is not specified. Then the maximum capacity that can be safely charged with a thermistor can be estimated from the already calculated tables for thermistors of standard series.

I took a table with the parameters of NTC thermistors from Joyin. The table shows:

Rnom- nominal resistance of the thermistor at a temperature of 25°C

Imax- maximum current through the thermistor (maximum steady-state current)

Smax- maximum capacity in the test circuit that is discharged onto the thermistor without damaging it (test voltage 350v)

You can see how the test is carried out on page seven.

A few words about the parameter Smax– the documentation shows that in the test circuit the capacitor is discharged through a thermistor and a limiting resistor, which releases additional energy. Therefore, the maximum safe capacity that a thermistor can charge without such resistance will be less. I searched for information in foreign thematic forums and looked at typical circuits with limiters in the form of thermistors, for which data is given. Based on this information, you can take the coefficient for Smax in a real scheme 0.65, by which to multiply the data from the table.

Table of parameters of NTC thermistors from Joyin

By connecting several identical NTC thermistors in series, we reduce the requirements for the maximum pulse energy of each of them.

Let me give you an example. For example, we need to select a thermistor to turn on the computer power supply. The maximum power consumption of the computer is 700 watts. We want to limit the starting current to 2-2.5A. The power supply contains a 470 µF filter capacitor.

We calculate the effective current value:

I = 700W/220V = 3.18A

As I wrote above, for reliable operation of the thermistor, we will select the maximum steady-state current from the documentation that is 20% greater than this value.

Imax = 3.8A

We calculate the required thermistor resistance for a starting current of 2.5A

R = (220V*√2)/2.5A = 124 Ohm

From the table we find the required thermistors. 6 pieces of JNR15S200L thermistors connected in series suit our needs Imax, general resistance. The maximum capacity that they can charge will be 680 µF * 6 * 0.65 = 2652 µF, which is even more than we need. Naturally, with a decrease Vpeak, the requirements for maximum pulse power thermistor. Our dependence is on the square of the voltage.

AND last question regarding the choice of thermistors. What if we have selected the thermistors required for maximum pulse power, but they are not suitable for us? Imax(the constant load is too high for them), or do we not need a source of constant heating in the device itself? To do this, we will use a simple solution - we will add another switch to the circuit in parallel with the thermistor, which we will turn on after charging the capacitor. Which is what I did in my limiter. In my case, the parameters are as follows: the maximum power consumption of the computer is 400W, the starting current limitation is 3.5A, the filter capacitor is 470uF. I took 6 pieces of 15d11 (15 ohm) thermistors. The diagram is shown below.

Rice. 3 Limiter circuit

Explanations for the diagram. SA1 disconnects the phase wire. LED VD2 serves to indicate the operation of the limiter. Capacitor C1 smoothes out ripples and the LED does not flicker at the mains frequency. If you don’t need it, then remove C1, VD6, VD1 from the circuit and simply connect the LED and diode in parallel in the same way as the elements VD4, VD5. To indicate the charging process of the capacitor, LED VD4 is connected in parallel with the thermistors. In my case, when charging the capacitor of the computer power supply, the entire process takes less than a second. So, let's collect.

Fig.4 Assembly kit

I assembled the power indicator directly in the cover of the switch, throwing out a Chinese incandescent lamp, which would not have lasted long.

Rice. 5 Power indicator

Fig.6 Thermistor block

Rice. 7 Assembled limiter

This could have been completed if all the thermistors had not failed after a week of work. It looked like this.

Rice. 8 Failure of NTC thermistors

Despite the fact that the margin for the permissible capacitance value was very large - 330 µF * 6 * 0.65 = 1287 µF.

I bought the thermistors from a well-known company, with different values ​​- all defective. Manufacturer unknown. Either the Chinese pour thermistors of smaller diameters into large cases, or the quality of the materials is very poor. As a result, I bought an even smaller diameter - SCK 152 8mm. The same China, but already branded. According to our table, the permissible capacitance is 100 µF * 6 * 0.65 = 390 µF, which is even slightly less than needed. However, everything works fine.

Capacitor charge

In order to charge a capacitor, it must be connected to a DC circuit. In Fig. Figure 1 shows a capacitor charging diagram. Capacitor C is connected to the generator terminals. Using the key, you can close or open the circuit. Let us consider in detail the process of charging a capacitor.

The generator has internal resistance. When the key is closed, the capacitor will charge to a voltage between the plates equal to e. d.s. generator: Uc = E. In this case, the plate connected to the positive terminal of the generator receives a positive charge (+q), and the second plate receives an equal negative charge (-q). The amount of charge q is directly proportional to the capacitance of the capacitor C and the voltage on its plates: q = CUc

P is. 1

In order for the capacitor plates to charge, it is necessary that one of them gains and the other loses a certain number of electrons. The transfer of electrons from one plate to another is carried out along an external circuit by the electromotive force of the generator, and the process of moving charges along the circuit itself is nothing more than electricity, called charging capacitive current I charge

The charging current usually flows in thousandths of a second until the voltage across the capacitor reaches a value equal to e. d.s. generator The graph of the voltage increase on the capacitor plates during its charging is shown in Fig. 2a, from which it is clear that the voltage Uc gradually increases, first quickly, and then more and more slowly until it becomes equal to e. d.s. generator E. After this, the voltage across the capacitor remains unchanged.

Rice. 2. Graphs of voltage and current when charging a capacitor

While the capacitor is charging, a charging current flows through the circuit. Schedule charging current shown in Fig. 2, b. At the initial moment, the charging current is the greatest value, because the voltage on the capacitor is still zero, and according to Ohm’s law io charge = E/ Ri, since all e. d.s. generator is applied to resistance Ri.

As the capacitor charges, that is, the voltage across it increases, the charging current decreases. When there is already voltage on the capacitor, the voltage drop across the resistance will be equal to the difference between e. d.s. generator and the voltage on the capacitor, i.e. equal to E - U s. Therefore i charge = (E-Uс)/Ri

It can be seen from this that with an increase in Uс, i charge decreases and at Uс = E the charging current becomes equal to zero.

The duration of the capacitor charging process depends on two values:

1) from the internal resistance of the generator Ri,

2) from the capacitance of the capacitor C.

In Fig. Figure 2 shows graphs of charged currents for a capacitor with a capacity of 10 μF: curve 1 corresponds to the charging process from a generator with e. d.s. E = 100 V and with internal resistance Ri = 10 Ohm, curve 2 corresponds to the charging process from a generator with the same e. d.s, but with lower internal resistance: Ri = 5 Ohm.

From a comparison of these curves it is clear that with a lower internal resistance of the generator, the strength of the charge current at the initial moment is greater, and therefore the charging process occurs faster.

Rice. 2. Graphs of charging currents at different resistances

In Fig. Figure 3 compares the graphs of charging currents when charging from the same generator with e. d.s. E = 100 V and internal resistance Ri = 10 ohm of two capacitors of different capacities: 10 μF (curve 1) and 20 μF (curve 2).

The value of the initial charging current io charge = E/Ri = 100/10 = 10 A is the same for both capacitors, but since a capacitor with a larger capacity accumulates a larger amount of electricity, its charging current must take longer, and the charging process is longer.

Rice. 3. Graphs of charging currents at different capacities

Capacitor discharge

Let's disconnect the charged capacitor from the generator and connect a resistance to its plates.

There is a voltage U c on the plates of the capacitor, therefore a current will flow in a closed electrical circuit, called the capacitive discharge current i bit.

Current flows from the positive plate of the capacitor through a resistance to the negative plate. This corresponds to the transition of excess electrons from the negative plate to the positive plate, where they are missing. The process of row frames occurs until the potentials of both plates are equal, that is, the potential difference between them becomes equal to zero: Uc=0.

In Fig. 4, a shows a graph of the decrease in voltage on the capacitor during discharge from the value Uc o = 100 V to zero, and the voltage decreases first quickly and then more slowly.

In Fig. Figure 4b shows a graph of changes in the discharge current. The strength of the discharge current depends on the resistance value R and according to Ohm's law i discharge = Uc / R

Rice. 4. Graphs of voltage and current during capacitor discharge

At the initial moment, when the voltage on the capacitor plates is greatest, the strength of the discharge current is also greatest, and with a decrease in Uc during the discharge process, the discharge current also decreases. When Uc=0, the discharge current stops.

The duration of the discharge depends on:

1) from the capacitance of capacitor C

2) on the value of resistance R by which the capacitor is discharged.

The higher the resistance R, the slower the discharge will occur. This is explained by the fact that with high resistance, the strength of the discharge current is small and the amount of charge on the capacitor plates decreases slowly.

This can be shown on graphs of the discharge current of the same capacitor, having a capacity of 10 μF and charged to a voltage of 100 V, at two different resistance values ​​(Fig. 5): curve 1 - at R = 40 Ohm, i discharge = Uc o/ R = 100/40 = 2.5 A and curve 2 - at 20 Ohm i sig = 100/20 = 5 A.

Rice. 5. Graphs of discharge currents at different resistances

Discharge also occurs more slowly when the capacitor capacity is large. This happens because with a larger capacitance, there is a larger amount of electricity on the capacitor plates ( more charge) and it will take a longer period of time for the charge to drain. This is clearly shown by the graphs of discharge currents for two capacitors of equal capacity, charged to the same voltage of 100 V and discharged to a resistance R = 40 Ohms (Fig. 6: curve 1 - for a capacitor with a capacity of 10 μF and curve 2 - for a capacitor with a capacity of 20 mkf).

Rice. 6. Graphs of discharge currents at different capacities

From the processes considered, we can conclude that in a circuit with a capacitor, current flows only at the moments of charge and discharge, when the voltage on the plates changes.

This is explained by the fact that when the voltage changes, the amount of charge on the plates changes, and this requires the movement of charges along the circuit, i.e., an electric current must pass through the circuit. A charged capacitor does not allow direct current to pass through, since the dielectric between its plates opens the circuit.

Capacitor energy

During the charging process, the capacitor accumulates energy, receiving it from the generator. When a capacitor is discharged, all the energy of the electric field is converted into thermal energy, that is, it goes to heating the resistance through which the capacitor is discharged. The greater the capacitance of the capacitor and the voltage on its plates, the greater the energy of the electric field of the capacitor. The amount of energy possessed by a capacitor with a capacitance C, charged to a voltage U, is equal to: W = W c = CU 2 /2

Example. Capacitor C = 10 μF is charged to a voltage U = 500 V. Determine the energy that will be released into the heat at the resistance through which the capacitor is discharged.

Solution. During the discharge, all the energy stored by the capacitor will turn into heat. Therefore, W = W c = CU 2 /2 = (10 x 10 -6 x 500)/2 = 1.25 J.

When designing amplifier power supplies Often problems arise that have nothing to do with the amplifier itself, or that are a consequence of the used element base. So in power supplies transistor amplifiers With high power, the problem often arises of implementing a smooth switching on of the power supply, that is, ensuring a slow charge of electrolytic capacitors in the smoothing filter, which can have a very significant capacity and, without taking appropriate measures, will simply damage the rectifier diodes at the moment of switching on.

In power supplies for tube amplifiers of any power, it is necessary to provide a feed delay high anode voltage before warming up the lamps, in order to avoid premature depletion of the cathode and, as a result, a significant reduction in the lamp life. Of course, when using a kenotron rectifier, this problem is solved by itself. But if you use a conventional bridge rectifier with an LC filter, you cannot do without an additional device.

Both of the above problems can be solved by a simple device that can be easily built into both a transistor and a tube amplifier.

Device diagram.

The schematic diagram of the soft start device is shown in the figure:

Click to enlarge

The alternating voltage on the secondary winding of transformer TP1 is rectified by the diode bridge Br1 and stabilized by the integrated stabilizer VR1. Resistor R1 ensures smooth charging of capacitor C3. When the voltage across it reaches a threshold value, transistor T1 will open, causing relay Rel1 to operate. Resistor R2 ensures the discharge of capacitor C3 when the device is turned off.

Inclusion options.

The Rel1 relay contact group is connected depending on the type of amplifier and the organization of the power supply.

For example, to ensure smooth charging of capacitors in the power supply transistor power amplifier, the presented device can be used to bypass the ballast resistor after charging the capacitors in order to eliminate power losses on it. Possible variant inclusions are shown in the diagram:

The values ​​of the fuse and ballast resistor are not indicated, since they are selected based on the power of the amplifier and the capacitance of the smoothing filter capacitors.

In a tube amplifier, the presented device will help organize a feed delay high anode voltage before the lamps warm up, which can significantly extend their service life. A possible inclusion option is shown in the figure:

The delay circuit here is turned on simultaneously with the filament transformer. After the lamps have warmed up, relay Rel1 will turn on, resulting in mains voltage will be fed to the anode transformer.

If your amplifier uses one transformer to power both the lamp filament circuits and the anode voltage, then the relay contact group should be moved to the secondary winding circuit anode voltage.

Elements of the switch-on delay circuit (soft start):

• Fuse: 220V 100mA,
• Transformer: any low-power with an output voltage of 12-14V,
• Diode bridge: any small-sized one with parameters 35V/1A and higher,
• Capacitors: C1 - 1000uF 35V, C2 - 100nF 63V, C3 - 100uF 25V,
• Resistors: R1 - 220 kOhm, R2 - 120 kOhm,
• Transistor: IRF510,
• Integral stabilizer: 7809, LM7809, L7809, MC7809 (7812),
• Relay: with an operating winding voltage of 9V (12V for 7812) and a contact group of the appropriate power.

Due to the low current consumption, the stabilizer chip and field-effect transistor can be mounted without radiators.

However, someone may have the idea to abandon the extra, albeit small-sized, transformer and power the delay circuit from the filament voltage. Considering that the standard value of the filament voltage is ~6.3V, you will have to replace the L7809 stabilizer with an L7805 and use a relay with a winding operating voltage of 5V. Such relays usually consume significant current, in which case the microcircuit and transistor will have to be equipped with small radiators.

When using a relay with a 12V winding (somehow more common) a microcircuit integral stabilizer should be replaced with 7812 (L7812, LM7812, MC7812).

With the values ​​of resistor R1 and capacitor C3 indicated in the diagram delay time inclusions are of the order 20 seconds. To increase the time interval, it is necessary to increase the capacitance of capacitor C3.

The article was prepared based on materials from the magazine "Audio Express"

Free translation by the Editor-in-Chief of RadioGazeta.