How to make an electronic fuse with your own hands. Texas Instruments Power Management: Protection, Monitoring, Switching Adjustable Electronic Fuse Circuitry
Fuses are disposable and require mandatory replacement if they fail due to power surges. Each of them is designed for a certain current, but in the absence of a suitable element, the one closest in value is installed. Such actions have a negative impact on the operation of the equipment and reduce its reliability. Therefore, modern circuits use current limiters, which are electronic fuses. These devices provide automatic protection and significantly increase the performance of devices.
Efficiency of current limiters
Fuses have been used in almost all circuits for a long time. They often failed and required manual replacement. In their absence, the practice was to use homemade devices in the form of various jumpers, which were very unreliable and dangerous in all respects.
These simplest elements have been replaced by electronic fuses that act as current limiters. According to their action, they are divided into two main categories. The first group restores the supply circuit after the causes of the accident have been eliminated. The operation of devices of the second group occurs only with the participation of specialists. In addition, there are passive protection devices that signal the occurrence of a dangerous situation using sound or light.
In radio-electronic devices, protection against current overloads is carried out using resistive or semiconductor current sensors connected in series in the circuit. If the voltage drops below the standard level, a protective device is activated, disconnecting the equipment from the power supply. This method of protection assumes the possibility of changing the magnitude of the current at which the protection is triggered.
Good and effective protection is ensured by a limited amount of limiting current passing through the load. The set level cannot be exceeded even if there is a short circuit in the circuit. Limiting the maximum current is carried out using special devices - stable current generators.
Electronic fuse diagrams
The presented diagrams display the most simple automatic protective measures against current overloads. The design of these devices is based on devices that have an initial current that cannot be exceeded. The required current value is set by selecting a specific transistor.
In diagram 1, an element of the KP302A brand is used, indicating a maximum current value of 30-50 mA. In order to increase this value, it is necessary to connect several transistors in parallel at once.
Circuit 2 works using conventional bipolar transistors with a minimum current transfer ratio of 80-100. The input voltage path begins in resistor R1, then passes through transistor VT1, opening it. The saturation mode of the transistor causes most of the voltage to flow to the output. If the current does not exceed the threshold value, in this case the transistor VT2 remains closed and the HL1 LED will not light up. In circuit 2, resistor R3 is a current sensor.
In the event of a voltage drop, transistor VT1 will close, thus limiting the flow of current through the load. Element VT2, on the contrary, will be open, with the LED turning on at the same time. The ratings of the elements indicated in diagram 2 correspond to a short circuit current with a voltage of 0.7 volts, a resistance of 3.6 ohms and a current of 0.2 - 0.23 amperes.
In diagram 3, the electronic fuse uses high-power field-effect transistor VT1 as a key. The protection operates at a current that depends on the ratio of resistive elements. An important role is played by the resistance value of the current sensor, which is connected in series with the field-effect transistor. After the protection has tripped, the load is reconnected by pressing the SA1 button.
Current limiters - stabilizers
Stabilizers are considered one of the most effective current limiters. For example, using the device in diagram 1, it is possible to obtain a stable voltage at the output, adjustable from 0 to 17 volts.
To protect against short circuits and overcurrent, special elements are used in the form of thyristor VS1 and a current sensor on resistor R2. When the current in the load increases, the thyristor is turned on with simultaneous shunting of the VT1 control circuit. After this, the output voltage value becomes zero. Triggering of the protection is confirmed by the LED turning on.
After eliminating the malfunction, the stabilizer is restarted by pressing the SB1 button and then unlocking the thyristor. There are current limiters equipped with protection and audible overload indicators. To control the audio frequency generator, a special key on a transistor is used.
(author Tonich dated August 6, 2013) does not have protection against overload and short-circuit current. In the depths of the Internet, a simple protection circuit was found - an electronic fuse. This device is connected between the load and the power source.
Here is the electrical diagram of the EP.
Contacts X1 and X2 connect the device to the power source. The load is connected to contacts X3, X4. The device is an electronic key made on transistors VT1 ... VT3. The electronic key is controlled by a current sensor assembled on resistors R1, R2 and potentiometer R4.
When the load current exceeds that set by potentiometer R4, the voltage drop across the emitter junction of transistor VT3 leads to its opening and, as a result, shunting of the emitter junction VT1. The voltage at the base of VT1 relative to its emitter turns out to be so small that VT1 is locked and no current flows through it. As a result, the VT1-R5 circuit is broken, and the voltage at the base of VT2 becomes below its operating threshold, the transistor VT2 is closed, and the load is de-energized. After eliminating the short circuit. (or overload) processes, starting with VT3, occur in the reverse order.
The switch threshold on transistor VT3 is set by potentiometer R4. This determines the maximum permissible current at which the ED will operate.
Powerful resistor R3 serves to limit the current through VT2. Capacitor C1 suppresses impulse noise (micro-sparking) that occurs when the slider slides along the resistive layer of the potentiometer.
Specifications:
Operating voltage - 5…30V.
The operating current adjustment range is 0.1…3.5A.
Components:
R3 - 0.5 Ohm, powerful 10 W, the remaining resistors are 0.25 W.
R1 - 470 Ohm.
R2, R6 - 1 kOhm.
R5-110 Ohm.
R4 - trimming resistor - 4.7 kOhm.
VT1-VT3 transistors BC 547B (KT 3102A)
VT2-transistor KT 805AM, KT 808AM, KT 819GM, 2N3055 should be installed on a radiator with an area of at least 100 sq.cm using thermal paste.
After assembly, I connected the electronic device to the power source. As a load I used a powerful wirewound resistor with a resistance of 3 ohms. Set the slider of potentiometer R4 to the minimum resistance and applied voltage to the ED from zero. On a voltmeter connected to a power source - 30 V, the load current and voltage are zero. Set the R4 slider to maximum resistance. At a current of 3.8A the ED worked. Since I wanted to increase the response current, I decided to reduce the resistance of resistor R3 to 0.3 Ohm. We managed to bring the operating current to 6 A. I didn’t try to set it anymore, because The KT805AM transistor is designed for a current of 5A. After the ED is triggered, re-activation is possible after 15 seconds.
An electronic fuse can also be made using a powerful field-effect transistor, but more on that in the next article.
Printed circuit board in Layout 6.0 program
Modern power switching transistors have very low drain-source resistances when on, which ensures low voltage drop when large currents pass through this structure. This circumstance allows the use of such transistors in electronic fuses.
For example, the IRL2505 transistor has a drain-source resistance, with a source-gate voltage of 10V, only 0.008 Ohms. At a current of 10A, the power P=I² R will be released on the crystal of such a transistor; P = 10 10 0.008 = 0.8 W. This suggests that at a given current the transistor can be installed without using a radiator. Although I always try to install at least small heat sinks. In many cases, this allows you to protect the transistor from thermal breakdown in emergency situations. This transistor is used in the protection circuit described in the article “”. If necessary, you can use surface-mounted radioelements and make the device in the form of a small module. The device diagram is shown in Figure 1. It was calculated for a current of up to 4A.
Electronic fuse diagram
In this circuit, a field-effect transistor with a p channel IRF4905 is used as a key, having an open resistance of 0.02 Ohm, with a gate voltage = 10V.
In principle, this value also limits the minimum supply voltage of this circuit. With a drain current of 10A, it will generate a power of 2 W, which will entail the need to install a small heat sink. The maximum gate-source voltage of this transistor is 20V, therefore, to prevent breakdown of the gate-source structure, a zener diode VD1 is introduced into the circuit, which can be used as any zener diode with a stabilization voltage of 12 volts. If the voltage at the input of the circuit is less than 20V, then the zener diode can be removed from the circuit. If you install a zener diode, you may need to adjust the value of resistor R8. R8 = (Upit - Ust)/Ist; Where Upit is the voltage at the circuit input, Ust is the stabilization voltage of the zener diode, Ist is the zener diode current. For example, Upit = 35V, Ust = 12V, Ist = 0.005A. R8 = (35-12)/0.005 = 4600 Ohm.
Current-voltage converter
Resistor R2 is used as a current sensor in the circuit, in order to reduce the power released by this resistor; its value is chosen to be only one hundredth of an Ohm. When using SMD elements, it can be composed of 10 resistors of 0.1 Ohm, size 1206, with a power of 0.25 W. The use of a current sensor with such a low resistance entailed the use of a signal amplifier from this sensor. The DA1.1 op amp of the LM358N microcircuit is used as an amplifier.
The gain of this amplifier is equal to (R3 + R4)/R1 = 100. Thus, with a current sensor having a resistance of 0.01 Ohm, the conversion coefficient of this current-voltage converter is equal to unity, i.e. One ampere of load current is equal to a voltage of 1V at output 7 DA1.1. You can adjust the Kus with resistor R3. With the indicated values of resistors R5 and R6, the maximum protection current can be set within.... Now let's count. R5 + R6 = 1 + 10 = 11kOhm. Let's find the current flowing through this divider: I = U/R = 5A/11000Ohm = 0.00045A. Hence, the maximum voltage that can be set at pin 2 of DA1 will be equal to U = I x R = 0.00045A x 10000 Ohm = 4.5 V. Thus, the maximum protection current will be approximately 4.5A.
Voltage comparator
A voltage comparator is assembled on the second op-amp, which is part of this MS. The inverting input of this comparator is supplied with a reference voltage regulated by resistor R6 from stabilizer DA2. Non-inverting input 3 of DA1.2 is supplied with amplified voltage from the current sensor. The load of the comparator is a series circuit, an optocoupler LED and a damping adjustment resistor R7. Resistor R7 sets the current passing through this circuit, about 15 mA.
Circuit operation
The scheme works as follows. For example, with a load current of 3A, a voltage of 0.01 x 3 = 0.03V will be released at the current sensor. The output of amplifier DA1.1 will have a voltage equal to 0.03V x 100 = 3V. If in this case, at input 2 of DA1.2 there is a reference voltage set by resistor R6, less than three volts, then at the output of comparator 1 a voltage will appear close to the supply voltage of the op-amp, i.e. five volts. As a result, the optocoupler LED will light up. The optocoupler thyristor will open and bridge the gate of the field-effect transistor with its source. The transistor will turn off and turn off the load. You can return the circuit to its original state with the SB1 button or by turning the power supply off and on again.
The article discusses the circuit of an electronic fuse for a high load current, up to 30 amperes. The article examined the circuit of a DC ammeter based on a module with an ACS712 chip; in this article, this module will be used as a load current sensor for an electronic fuse. The circuit diagram of the electronic fuse is shown in Figure 1.
The diagram shows a module designed for a load current of up to five amperes. On AliExpress you can also purchase modules for a current of 20 amperes and 30 amperes and use them in this circuit. But then the VT1 IRL2505 transistor should be replaced with two of the same transistors. Although other MOSFETs can be used. The supply voltage of this circuit is limited only by the maximum supply voltage of the LM7805 power stabilizer chip - 35 volts.
Circuit operation
After applying voltage to the input of the circuit, a voltage of five volts appears at the output of the supply voltage stabilizer of the DA3 microcircuit and the DA2 current sensor module. The diagram shows a microcircuit of the module of the same name, and not the module itself. The module has three outputs and capacitor C2 is located on its board. A voltage appears at output 7 of the DA2 chip (Module Output) of approximately 2.5 V. This voltage is supplied to input 2 of the comparator, implemented on the LM358N operational amplifier. Its inverting input, pin 3 of the DA3 chip, is supplied with a reference voltage from a resistive adjustable divider R3 and R4. Using resistor R3, the current threshold of the circuit is set. This voltage is set higher than the voltage from the ACS712 output. This means that at this level of voltage at the inputs of the op-amp, at its output there will be a voltage close to its supply voltage. This voltage will be applied to the LED circuit of optocoupler U1. Pin 1 DA3 — > pin 1 U1 — > pin 2 U1 — > quenching resistor R2 — > common wire. The LED of the optocoupler will light up, which will lead to the appearance of an opening voltage for transistor VT1 at its output in the region of eight volts. Transistor VT1 will open and, through the module, the input voltage of the circuit will be almost completely supplied to its output. Diode VD1 will be closed with a positive voltage at its cathode, and in this case it will not have any effect on the operation of the comparator circuit. Any low-power diode can be used as this diode.
Current sensor modules implemented on the ACS712 chip and designed for different load currents of 5, 20 and thirty amperes have different current-voltage conversion ratios. The corresponding coefficients are 185 mV/A, 100 mV/A and 66 mV/A. For a five-amp sensor indicated in the diagram, the output voltage relative to 2.5 volts, at a current of 5A, will increase by 5 x 185 = 925 mV = 0.925 V. That is, the total output voltage from the sensor will be approximately 2.5 + 0.925 = 3.425 V. I write: approximately, because different sensors have different output voltages in the absence of load current and are not exactly equal to 2.5 volts. And so on, when the voltage at the output of the sensor exceeds the set reference voltage at input 3 of the DA3 microcircuit, the comparator will operate and the voltage at its output will be almost zero. The cathode of diode VD1 will be connected to the common wire through the internal output transistor of the operational amplifier and will be shunted to the common wire and the reference voltage at the non-inverting input of the op-amp. Positive feedback occurs through an open diode. A “latch” effect occurs. The comparator can remain in this position for as long as desired. After removing the voltage from the optocoupler LED, the opening voltage at the gate of the key transistor VT1 will also disappear. The transistor will turn off and de-energize the load. To restore the functionality of the circuit, it is necessary to remove the voltage from it and then reapply it.
Key MOSFET transistors IRL2505 have a very low open-channel resistance, it is equal to 0.008 Ohms. Based on this, with a drain current equal to ten amperes, thermal power will be released on the transistor crystal equal to: P = I² R = 100 0.008 = 0.8 W. This suggests that the transistor can operate at a given current without additional heat sink. But I always advise installing at least a small heat sink in the form of an aluminum plate. This will protect the transistor from thermal breakdown in an emergency.
The main disadvantage of fuses when used to protect electronic circuits is their inertia, i.e. long response time, during which some elements of the circuit have time to fail. You can ensure automatic protection of the device and at the same time increase its performance through the use of electronic fuses.
These devices can be divided into two groups:
With self-healing of the power circuit after eliminating the causes of the accident;
There are also passive protection devices: in emergency mode, they only indicate the presence of a dangerous situation with a light or sound signal, without disconnecting the load. To protect radio-electronic devices from current overloads, resistive or semiconductor current sensors are usually used, connected in series to the load circuit. As soon as the voltage drop across the current sensor exceeds a preset level, a protective device is triggered, disconnecting the load from the power source. The advantage of this method of protection is that the magnitude of the protection operation current can be easily changed. Another method of protecting a load is to limit the current limit through it.
Even if there is a short circuit in the load circuit, the current will not be able to exceed the specified level and damage the load. To limit the maximum load current, stable current generators are used. The circuit of the simplest current limiter is shown in Fig. 1.
In fact, this is a current stabilizer on a field-effect transistor. The load current when using such a limiter cannot exceed the initial drain current of the field-effect transistor. The magnitude of this current can be set by selecting the type of transistor. For the KP302V transistor shown in the diagram, the maximum current through the load will not exceed 30...50 mA. The value of this current can be increased by connecting several transistors in parallel. The load current limiter (Fig. 2) uses bipolar transistors with a current transfer coefficient of at least 80...100.
In normal mode, transistor VT2 is open due to the flow of base current through resistor R1.
As the current increases, the voltage between the collector and emitter of VT2 increases and, when it becomes approximately 0.6 V, transistor VT1 opens and bypasses the base-emitter circuit of VT2, causing it to close. If a short circuit occurs in the load, then the short circuit current flows through the circuit: “+” of the power source - short-circuited load Rн - resistor R2 - base-emitter junction VT1 - source. Since VT2 is closed, the short circuit current is limited by resistor R2. After eliminating the short circuit, the limiter does not turn on independently. To do this, you need to disconnect and reconnect the load for a short time (short-circuit the base and emitter terminals of VT1 with each other). In this case, VT1 will close and VT2 will open, and voltage will be supplied to the load. In Fig. Figure 3b shows a diagram for protecting consumers from overvoltage in low-voltage circuits.
When the input voltage increases above the nominal voltage, the zener diode VD2 breaks through, transistor VT1 opens, VT2 closes, and the load is protected from overvoltage. As a device for protecting power supplies, you can use an electronic fuse (Fig. 4), connected between the source and the load.
When the power is turned on, the current flowing through resistor R1 and the emitter junction VT4 opens the composite transistor VT4-VT3. The remaining transistors remain closed. The load is supplied with rated voltage. When an overload occurs, the voltage drop across R2 becomes sufficient to open the dinistor analogue. Following it, transistor VT5 opens and bypasses the emitter junction VT4. As a result, transistors VT3 and VT4 close, disconnecting the load from the power source. The load current decreases sharply, but the dinistor analogue remains open. The fuse can remain in this state indefinitely. A residual current flows through the load, determined by resistance R1, i.e.
tens of times less than nominal. The voltage drop across the closed transistor VT3 turns on the HL1 "Alarm" LED. To resume operation of the device in nominal mode after eliminating the overload, you must briefly turn off the power source or disconnect the load. The fuse is assembled on a printed circuit board, the drawing of which is shown in Fig. 6.
With the component ratings indicated in the diagram, the fuse has the following characteristics:
Rated supply voltage - 12V;
Rated load current - 1 A;
Operation current - 1.2 A;
Residual load voltage - 1.2 V;
The voltage drop across the fuse is 0.75 V.
The control voltage to the thyristor is supplied through resistor R2 from the current sensor, which is played by resistor R1 of very low resistance (0.1 Ohm). This type of thyristor turns on when the voltage on the control electrode (relative to the cathode) is 0.5...0.6 V. In the initial state, a current of approximately 8...15 mA flows through transistor VT3, which remains stable when the output voltage of the power supply changes. This current flows through the HL2 LED, which signals the operation of the device, into the base circuit of the transistor VT2. Since the static current transfer coefficient of VT2 is several thousand, it opens and is capable of passing a current of several amperes into the load. In this case, the voltage drop across the transistor does not exceed 1 V. The load current creates a voltage drop across resistor R1, which is the opening voltage for the thyristor. In this case, it is possible to connect a constant resistor in series with it or in parallel. If, when the fuse trips, a residual current still flows through transistor VT2 (the transistor does not close), it is recommended to use the HL2 LED with a higher operating voltage or connect the KD102B, KD103B, KD105B, KD522B diode in series with it. If the power supply has a voltage stabilizer, the fuse should be connected in front of it, and not at the output of the unit. A voltage stabilizer with built-in protection (Fig. 8) allows you to obtain an output voltage that is adjustable within the range from 0 to 17 V.
The electronic fuse (Fig. 7) contains a powerful transistor VT2, which is connected to the negative power wire, two current stabilizers on field-effect transistors (adjustable on VT1 and unregulated on VT3) and a threshold element - thyristor VS1.
A voltage stabilizer is assembled on transistors VT1 and VT2 according to a traditional circuit, however, in parallel with the zener diode VD1, a relay cascade is connected on transistors VT3...VT5 with a current sensor on the resistor Rx.
When the load current increases, this cascade is triggered and shunts the zener diode. The voltage at the output of the stabilizer drops to an insignificant value. To unlock the protection circuit, just briefly press the SB1 button. To increase the stabilization coefficient, instead of the Zener diode VD1, you can turn on an integrated voltage stabilizer (three-terminal). Electronic fuses can be made using a powerful field-effect transistor as a key (Fig. 10).
The fuse is connected between the power source (switch) and the load. It operates at voltages from 5 to 20 V and load currents up to 40 A. Field-effect transistor VT1 simultaneously functions as an electronic switch and a current sensor. A voltage comparator is built on the DA1 chip, and a reference voltage source (2.5 V) is built on the DA2 chip. To start the device, use the SB1 button, when briefly closed, the supply voltage through the diode VD2 and resistor R4 is supplied to the gate of the transistor, it opens and connects the load to the power source. The device can use any op-amp (DA1) that is operable at zero voltage at both inputs under single-supply conditions. In particular, domestic analogs of the LM358 microcircuit are suitable - KR1040UD1A, K1464UD1R in the DIP-8 package and K1464UD1T in the SO-8 package. DA2 - any chip from the TL431 series. Trimmer resistor - SPZ-19a, SPZ-28 or similar imported ones. Fixed resistors - MLT, S2-33, R1-4, R1-12. Capacitor C1 - K10-17V. Button SB1 - any small-sized one with self-return. When using surface-mount parts: DA1 - LM358AM, DA2 - TL431CD (Fig. 12a), resistors P1-12, etc., the device is placed on a printed circuit board made of single-sided foil fiberglass with dimensions of 20x25 mm (Fig. 12.b).
The protection operation current is determined by the ratio of the resistive elements and depends, first of all, on the resistance value of the current sensor Rs, connected in series with the field-effect transistor VT1. The circuit diagram of a device based on a field-effect transistor of the IRL series is shown in Fig. 11.
Setting up the device comes down to setting the operating current using trimming resistor R1 (Fig. 11). The interval of change of this current can be set by selecting resistance R2. In power supplies that can briefly withstand current overload (output short circuit), passive protection devices are used.
In emergency mode, they notify about it with a light or sound alarm, without turning off the load on their own. Figure 13 shows the diagram of the LED indicator (VD2).
The diode rectifier VD1...VD4 is powered by a transformer, the secondary winding of which is designed for the voltage and current required for the operation of the voltage stabilizer. The signaling device is an audio frequency generator HA1 with an acoustic emitter (dynamic head) BA1 connected to it. The operation of the generator is controlled by a key on transistor VT1. When the stabilizer operates, the load current passes through the current sensor R1, creating a voltage drop across it. While the current is small (with resistance R1 indicated on the diagram - less than 0.3 A), transistor VT1 is closed. As the current increases, the voltage across the resistor increases. When it reaches 0.7 V, VT1 opens and the rectified voltage is supplied to the alarm device. AC electronic equipment protection circuits are typically more complex and less common. This is due to the fact that the reliability of the operation of semiconductor devices at increased voltages of the network level is less, since an accidental surge in the network voltage, for example, during transient processes, can easily break through the transition of even the highest voltage semiconductor device.
A semiconductor fuse (Fig. 16) is capable of protecting the connected electronic circuit (Rн) from overcurrent.
The load current can be smoothly adjusted with potentiometer R2 ranging from a few milliamps to 8 A. The maximum load current, if necessary, can be significantly increased by installing transistor VT1 on the radiator, equipping it with a fan and increasing the number of field-effect transistors connected in parallel. The mains load current limiter is shown in Fig. 18.
Its power characteristics are determined only by the type of field-effect transistor used. The basis of the circuit is the current source on VT2, VT3, R3 and R4. Resistor R3 ensures the opening of field-effect transistor VT3, R4 is current-setting. The device shown in Fig. 19 is designed to quickly disconnect energy consumers from the network if the current in the circuit exceeds the permissible value.
Compared to fuses and electromechanical fuses, electronic fuses have a significantly higher operating speed. In addition, this device can be easily and accurately configured to operate at any current in the range of 0.1...10 A. The protection device is powered directly from the network using a transformerless circuit using elements R7...R9, SZ, C4, VD3.. .VD5. Load switching is performed by an electronic switch - triac VS1. To open it, short pulses are sent to the control electrode through transformer T2. These pulses are generated by a self-oscillator on a unijunction transistor VT1. To open the triac, a current through the control electrode of up to 100 mA is required. This current is provided in pulse mode. Capacitor C2 is charged from the power source through resistor R2. As soon as the voltage across it reaches the opening threshold of transistor VT1, capacitor C2 is discharged through the circuit emitter-base transition 1 VT1 - winding 1 T2. This process is repeated with a frequency determined by the ratings of R2 and C2 (approximately 1.5... 2 kHz). Since the pulse repetition rate of the self-oscillator is much higher than the network voltage (50 Hz), the triac opens almost at the beginning of each half-cycle of the network voltage. The current sensor in the load circuit is the current transformer T1. When load current flows n also passes through the primary winding T1. In the secondary winding (3-4), an increased voltage is released, proportional to the load current. This voltage is rectified by the diode bridge VD1 and supplied through resistor R5 to the control electrode of thyristor VS2. If this voltage reaches the operating threshold of VS2, it opens and short-circuits C2 through the diode VD2, so that the self-oscillator stops working. When the pulses driving VS1 disappear, the load is switched off. At the same time, the HL1 indicator lights up. The sensitivity of the circuit's response can be smoothly adjusted using resistor R3. Capacitor C1 prevents the protection from triggering during short-term interference in the network. The circuit can remain in the off state for a long time, and to return it to its original state, you must press the SB1 button. And using the SB2 button, if necessary, the load can be turned off manually. Current transformer T1 is homemade. For winding, it is convenient to use the frame and magnetic circuit from any transformer used in old domestic telephones. A magnetic core made of iron or ferrite M2000NM of standard size W5x5 is suitable. Winding 3-4 is made with PEL wire Ø 0.08 mm and contains 3000...3400 turns.
If there are no such pulses, then it may be necessary to select resistance R2. The operation of thyristor VS2 when pressing the SB2 button should be detected. If the HL1 LED does not light after the button is released, reduce the resistance of R4 to increase the current required to keep VS2 open. You can check the operation of the device as a whole by connecting a lamp and a dial voltmeter to the XS1 sockets. First of all, you need to make sure that the triac VS1 opens completely (by measuring the voltage across the lamp). If this is not the case, you need to swap the terminals of any of the windings of transformer T2. The electronic fuse circuit can be simplified by removing the current transformer T1 and replacing its winding 1-2 with a resistor with a resistance of 0.2...0.3 Ohm and a diode.
The resistance of this resistor is adjusted to the required protection current. But in this case, the protection circuit will operate on one half-wave of the mains voltage, which will reduce its performance when the load is turned off. When using the circuit, it should be taken into account that some consumers, for example, lamps, switching power supplies, electric motors, etc., produce an inrush current at the moment of switching on . In this case, the threshold for triggering the protection must be increased or, which is much better, measures must be taken to reduce this throw.
