Protection of the power supply from short-circuit. Short circuit protection circuit Diode bridge protection from short circuit

Beginner radio amateurs, who are the majority, choose simpler circuits to assemble a regulated power supply. I also decided to make such a circuit, since it’s unlikely that I’ll be able to get expensive parts and set up a complex power supply.

The most basic thing for any design is the body. Here I was lucky to get a non-working ATX power supply from the computer, where the future power supply will be placed.

I left the connectors on the back for the 220V network, and screwed a regular socket in place of the cooler, since there are always not enough of them for the mass of my electronic devices. In short, it won’t be superfluous.

The printed circuit board for the power supply is simple and easy to make even for beginners. As a last resort, you can cut out the tracks with a cutter rather than etching. For maximum current protection - and this must be included in an amateur radio power supply, I chose an electronic fuse circuit with an overload indication on an LED.

The front panel of the power supply is made of plastic, PCB or even plywood - whoever is rich in what. Dial indicators will be attached to it - a voltmeter and an ammeter (as it later became clear that this is much better and more convenient than a digital indication), a voltage regulator and buttons for turning on and switching protection modes. I chose 0.1 and 1A, but you can calculate the current protection resistor to any value.

There will also be two terminals on the front panel of the power supply for connecting the power supply output wires.

It turns out that this is something similar to a power supply. We choose a transformer such that it fits into the housing. So if you go to buy it at a radio market, first measure the dimensions of the box.

We cover the body with self-adhesive film or paint it with varnish.

The green LED will light up when the power supply is turned on, and the red LED will indicate that the overcurrent protection has tripped.

Here it is written how to calculate the shunt for dial indicators. And in order to put new values ​​of volts and amperes on the scale, you will have to open their cases and carefully stick pieces of paper with new values ​​on top of the old ones.

That's all. An excellent simple power supply made from scrap materials is completely ready. Working with it for several months showed its high reliability and ease of use. Material provided by in_sane.

Discuss the article SIMPLE POWER SUPPLY WITH PROTECTION

This circuit is a simple transistor power supply equipped with short circuit (short circuit) protection. Its diagram is shown in the figure.

Main parameters:

• Output voltage - 0..12V;
• The maximum output current is 400 mA.

The scheme works as follows. The input voltage of the 220V network is converted by a transformer to 16-17V, then rectified by diodes VD1-VD4. Filtering of rectified voltage ripples is carried out by capacitor C1. Next, the rectified voltage is supplied to the zener diode VD6, which stabilizes the voltage at its terminals to 12V. The remainder of the voltage is extinguished by resistor R2. Next, the voltage is adjusted by variable resistor R3 to the required level within 0-12V. This is followed by a current amplifier on transistors VT2 and VT3, which amplifies the current to a level of 400 mA. The load of the current amplifier is resistor R5. Capacitor C2 additionally filters output voltage ripple.

This is how protection works. In the absence of a short circuit at the output, the voltage at the terminals of VT1 is close to zero and the transistor is closed. The R1-VD5 circuit provides a bias at its base at a level of 0.4-0.7 V (voltage drop across the open p-n junction of the diode). This bias is enough to open the transistor at a certain collector-emitter voltage level. As soon as a short circuit occurs at the output, the collector-emitter voltage becomes different from zero and equal to the voltage at the output of the unit. Transistor VT1 opens, and the resistance of its collector junction becomes close to zero, and, therefore, at the zener diode. Thus, zero input voltage is supplied to the current amplifier; very little current will flow through transistors VT2, VT3, and they will not fail. The protection is turned off immediately when the short circuit is eliminated.

Details

The transformer can be any with a core cross-sectional area of ​​4 cm 2 or more. The primary winding contains 2200 turns of PEV-0.18 wire, the secondary winding contains 150-170 turns of PEV-0.45 wire. A ready-made frame scan transformer from old tube TVs of the TVK110L2 series or similar will also work. Diodes VD1-VD4 can be D302-D305, D229Zh-D229L or any with a current of at least 1 A and a reverse voltage of at least 55 V. Transistors VT1, VT2 can be any low-frequency low-power ones, for example, MP39-MP42. You can also use more modern silicon transistors, for example, KT361, KT203, KT209, KT503, KT3107 and others. As VT3 - germanium P213-P215 or more modern silicon high-power low-frequency KT814, KT816, KT818 and others. When replacing VT1, it may turn out that short-circuit protection does not work. Then you should connect another diode (or two, if necessary) in series with VD5. If VT1 is made of silicon, then it is better to use silicon diodes, for example, KD209(A-B).

In conclusion, it is worth noting that instead of the p-n-p transistors indicated in the diagram, n-p-n transistors with similar parameters can be used (not instead of any of VT1-VT3, but instead of all of them). Then you will need to change the polarity of the diodes, zener diode, capacitors, and diode bridge. At the output, accordingly, the polarity of the voltage will be different.

List of radioelements

Designation	Type	Denomination	Quantity	Note	Shop	My notepad
VT1, VT2	Bipolar transistor	MP42B	2	MP39-MP42, KT361, KT203, KT209, KT503, KT3107		To notepad
VT3	Bipolar transistor	P213B	1	P213-P215, KT814, KT816, KT818		To notepad
VD1-VD4	Diode	D242B	4	D302-D305, D229Zh-D229L		To notepad
VD5	Diode	KD226B	1			To notepad
VD6	Zener diode	D814D	1			To notepad
C1		2000 µF, 25 V	1			To notepad
C2	Electrolytic capacitor	500 µF. 25 V	1			To notepad
R1	Resistor	10 kOhm	1			To notepad
R2	Resistor	360 Ohm	1			To notepad
R3	Variable resistor	4.7 kOhm	1			To notepad
R4, R5	Resistor

Today my article will be of an exclusively theoretical nature, or rather, it will not contain “hardware” as in previous articles, but do not be upset - it has not become less useful. The fact is that the problem of protecting electronic components directly affects the reliability of devices, their service life, and therefore your important competitive advantage - the ability to provide a long-term product warranty. The implementation of protection concerns not only my favorite power electronics, but also any device in principle, so even if you are designing IoT crafts and you have a modest 100 mA, you still need to understand how to ensure trouble-free operation of your device.

Current protection or short circuit (short circuit) protection is probably the most common type of protection because neglect in this matter causes devastating consequences in the literal sense. As an example, I suggest looking at a voltage stabilizer that was sad because of a short circuit:

The diagnosis here is simple - an error occurred in the stabilizer and ultra-high currents began to flow in the circuit; the protection should have turned off the device, but something went wrong. After reading the article, it seems to me that you yourself will be able to guess what the problem could be.

As for the load itself... If you have an electronic device the size of a matchbox, there are no such currents, then do not think that you cannot become as sad as the stabilizer. Surely you don’t want to burn bundles of $10-$1000 chips? If so, then I invite you to familiarize yourself with the principles and methods of dealing with short circuits!

Purpose of the article

I am targeting my article at people for whom electronics is a hobby and novice developers, so everything will be told “at a glance” for a more meaningful understanding of what is happening. For those who want an academic touch, go and read any university textbook on electrical engineering + the “classics” of Horowitz, Hill “The Art of Circuit Design”.

Separately, I would like to say that all solutions will be hardware-based, that is, without microcontrollers and other perversions. In recent years, it has become quite fashionable to program where it is necessary and where it is not necessary. I often observe current “protection”, which is implemented by simply measuring the ADC voltage with some arduino or microcontroller, and then the devices still fail. I strongly advise you not to do the same! I will talk about this problem in more detail later.

A little about short circuit currents

In order to start coming up with methods of protection, you must first understand what we are fighting against. What is a “short circuit”? Ohm’s favorite law will help us here; consider the ideal case:

Just? Actually, this circuit is the equivalent circuit of almost any electronic device, that is, there is an energy source that supplies it to the load, and it heats up and does or does not do something else.

Let’s agree that the power of the source allows the voltage to be constant, that is, “not to sag” under any load. During normal operation, the current acting in the circuit will be equal to:

Now imagine that Uncle Vasya dropped a wrench on the wires going to the light bulb and our load decreased 100 times, that is, instead of R it became 0.01*R and with the help of simple calculations we get a current 100 times greater. If the light bulb consumed 5A, then now the current from the load will be about 500A, which is quite enough to melt Uncle Vasya’s key. Now a small conclusion...

Short circuit- a significant decrease in load resistance, which leads to a significant increase in current in the circuit.

It is worth understanding that short-circuit currents are usually hundreds and thousands of times greater than the rated current, and even a short period of time is enough for the device to fail. Here, many will probably remember electromechanical protection devices (“automatic devices” and others), but everything here is very prosaic... Usually a household socket is protected by a circuit breaker with a rated current of 16A, that is, shutdown will occur at 6-7 times the current, which is already about 100A. The laptop power supply has a power of about 100 W, that is, the current is less than 1A. Even if a short circuit occurs, the machine will not notice it for a long time and will turn off the load only when everything has already burned out. This is more fire protection than equipment protection.

Now let's look at another frequently encountered case - through current. I will show it using the example of a dc/dc converter with a synchronous buck topology; all MPPT controllers, many LED drivers and powerful DC/DC converters on boards are built exactly on it. Let's look at the converter circuit:

The diagram shows two options for overcurrent: green way for a “classic” short circuit, when there is a decrease in load resistance (“snot” between roads after soldering, for example) and orange path. When can current flow through the orange path? I think many people know that the open channel resistance of a field-effect transistor is very small; in modern low-voltage transistors it is 1-10 mOhm. Now let’s imagine that PWM with a high level came to the keys at the same time, that is, both keys opened, for the “VCCIN - GND” source this is equivalent to connecting a load with a resistance of about 2-20 mOhm! Let's apply the great and mighty Ohm's law and get a current value of more than 250A even with a 5V power supply! Although don’t worry, there won’t be such a current - the components and conductors on the printed circuit board will burn out earlier and break the circuit.

This error very often occurs in the power system and especially in power electronics. It can occur for various reasons, for example, due to control errors or long-term transient processes. In the latter case, even the “dead time” in your converter will not help.

I think the problem is clear and familiar to many of you, now it’s clear what needs to be dealt with and all that remains is to figure out HOW. This is what the next story will be about.

Operating principle of current protection

Here you need to apply ordinary logic and see the cause-and-effect relationship:
1) The main problem is the large current in the circuit;
2) How to understand what current value? -> Measure it;
3) Measured and obtained the value -> Compare it with the specified acceptable value;
4) If the value is exceeded -> Disconnect the load from the current source.

Measure the current -> Find out whether the permissible current has been exceeded -> Disconnect the load

Absolutely any protection, not only current, is built this way. Depending on the physical quantity on which the protection is built, various technical problems and methods for solving them will arise on the way to implementation, but the essence is unchanged.

Now I propose to go through the entire security chain in order and solve all the technical problems that arise. Good protection is protection that is planned in advance and it works. This means we can’t do without modeling, I’ll use the popular and free one MultiSIM Blue, which is actively promoted by Mouser. You can download it there - link. I will also say in advance that within the framework of this article I will not delve into the circuitry and fill your head with unnecessary things at this stage, just know that everything will be a little more complicated in real hardware.

Current measurement

This is the first point in our chain and probably the easiest to understand. There are several ways to measure current in a circuit, and each has its own advantages and disadvantages; which one to use specifically in your task is up to you to decide. I will tell you, based on my experience, about these very advantages and disadvantages. Some of them are “generally accepted”, and some are my worldviews; please note that I’m not even trying to pretend to be some kind of truth.

1) Current shunt. The basis of the fundamentals “works” on the same great and powerful Ohm’s law. The simplest, cheapest, fastest and generally the best method, but with a number of disadvantages:

A) No galvanic isolation. You will have to implement it separately, for example, using a high-speed optocoupler. This is not difficult to implement, but it requires additional space on the board, decoupled dc/dc and other components that cost money and add overall dimensions. Although galvanic isolation is not always needed, of course.

B) At high currents, global warming accelerates. As I wrote earlier, it all “works” on Ohm’s law, which means it heats up and warms the atmosphere. This leads to a decrease in efficiency and the need to cool the shunt. There is a way to minimize this disadvantage - to reduce the shunt resistance. Unfortunately, it cannot be reduced indefinitely and at all I wouldn't recommend reducing it to less than 1 mOhm, if you still have little experience, because the need arises to combat interference and the requirements for the design stage of the printed circuit board increase.

In my devices I like to use these shunts PA2512FKF7W0R002E:

Current measurement occurs by measuring the voltage drop across the shunt, for example, when a current of 30A flows across the shunt there will be a drop:

That is, when we get a drop of 60 mV on the shunt, this will mean that we have reached the limit and if the drop increases further, then we will need to turn off our device or load. Now let's calculate how much heat will be released on our shunt:

Not a little, right? This point must be taken into account, because The maximum power of my shunt is 2 W and it cannot be exceeded, and you should also not solder the shunts with low-melting solder - it can come off, I’ve seen that too.

• Use shunts when you have high voltage and not very high currents
• Monitor the amount of heat generated by the shunt
• Use shunts where you need maximum performance
• Use shunts only from special materials: constantan, manganin and the like

2) Hall effect current sensors. Here I will allow myself my own classification, which fully reflects the essence of various solutions for this effect, namely: cheap And expensive.

A) Cheap, for example, ACS712 and the like. Among the advantages, I can note the ease of use and the presence of galvanic isolation, but that’s where the advantages end. The main disadvantage is the extremely unstable behavior under the influence of RF interference. Any dc/dc or powerful reactive load is interference, that is, in 90% of cases these sensors are useless, because they “go crazy” and rather show the weather on Mars. But it’s not for nothing that they are made?

Are they galvanically isolated and can measure high currents? Yes. Don't like interference? Yes too. Where to put them? That's right, into a low-responsibility monitoring system and for measuring current consumption from batteries. I have them in inverters for solar power plants and wind power plants for a qualitative assessment of the current consumption from the battery, which allows you to extend the life cycle of the batteries. These sensors look like this:

B) Expensive. They have all the advantages of cheap ones, but do not have their disadvantages. An example of such a sensor LEM LTS 15-NP:

What we have as a result:
1) High performance;
2) Galvanic isolation;
3) Ease of use;
4) Large measured currents regardless of voltage;
5) High measurement accuracy;
6) Even “evil” EMPs do not interfere with work; affect accuracy.

But what is the downside then? Those who opened the link above clearly saw it - this is the price. $18, Karl! And even for a series of 1000+ pieces, the price will not fall below $10, and the actual purchase will be $12-13. You can’t install this in a power supply unit for a couple of bucks, but I would like it... Summarize:

A) This is the best solution in principle for measuring current, but expensive;
b) Use these sensors in harsh operating conditions;
c) Use these sensors in critical components;
d) Use them if your device costs a lot of money, for example, a 5-10 kW UPS, where it will definitely justify itself, because the price of the device will be several thousand dollars.

3) Current transformer. Standard solution in many devices. There are two minuses - they do not work with direct current and have nonlinear characteristics. Pros - cheap, reliable and you can measure enormous currents. It is on current transformers that automation and protection systems are built in RU-0.4, 6, 10, 35 kV enterprises, and there thousands of amperes are quite normal.

To be honest, I try not to use them, because I don’t like them, but I still use them in various control cabinets and other AC systems, because They cost a couple of dollars and provide galvanic isolation, not $15-20 like LEMs, and they perform their task perfectly in a 50 Hz network. They usually look like this, but they also appear on all sorts of EFD cores:

Perhaps we can finish with current measurement methods. I talked about the main ones, but of course not all. To expand your own horizons and knowledge, I advise you to at least google and look at various sensors on the same digikey.

Measured Voltage Drop Gain

Further construction of the protection system will be based on the shunt as a current sensor. Let's build a system with the previously announced current value of 30A. At the shunt we get a drop of 60 mV and here 2 technical problems arise:

A) It is inconvenient to measure and compare a signal with an amplitude of 60 mV. ADCs usually have a measurement range of 3.3V, that is, with 12 bits of capacity we get a quantization step:

This means that for the range of 0-60 mV, which corresponds to 0-30A, we will get a small number of steps:

We find that the measurement depth will be only:

It is worth understanding that this is an idealized figure and in reality they will be many times worse, because... The ADC itself has an error, especially around zero. Of course, we will not use an ADC for protection, but we will have to measure the current from the same shunt to build a control system. Here the task was to clearly explain, but this is also relevant for comparators, which in the area of ​​ground potential (0V usually) operate very unstable, even rail-to-rail.

B) If we want to drag a signal with an amplitude of 60 mV across the board, then after 5-10 cm there will be nothing left of it due to interference, and at the moment of short circuit we definitely won’t have to count on it, because EMR will further increase. Of course, you can hang the protection circuit directly on the leg of the shunt, but we will not get rid of the first problem.

To solve these problems we need an operational amplifier (op-amp). I won’t talk about how it works - the topic is easily googled, but we’ll talk about the critical parameters and choice of op-amp. First, let's define the scheme. I said that there won’t be any special graces here, so let’s cover the op-amp with negative feedback (NFB) and get an amplifier with a known gain. I will model this action in MultiSIM (the picture is clickable):

You can download the file for the simulation at home - .

The voltage source V2 acts as our shunt, or rather, it simulates the voltage drop across it. For clarity, I have chosen a drop-off value of 100 mV, now we need to boost the signal to move it to a more convenient voltage, usually between 1/2 and 2/3 V ref. This will allow you to get a large number of quantization steps in the current range + leave a margin for measurements to assess how bad everything is and calculate the current rise time, this is important in complex reactive load control systems. The gain in this case is equal to:

This way we have the opportunity to amplify our signal to the required level. Now let's look at what parameters you should pay attention to:

• The op amp must be rail-to-rail to adequately handle signals near ground potential (GND)
• It is worth choosing an op-amp with a high slew rate of the output signal. For my favorite OPA376, this parameter is 2V/µs, which allows you to achieve the maximum output value of the op-amp equal to VCC 3.3V in just 2 µs. This speed is quite enough to save any converter or load with frequencies up to 200 kHz. These parameters should be understood and turned on when choosing an op-amp, otherwise there is a chance to put an op-amp for $10 where an amplifier for $1 would suffice
• The bandwidth selected by the op-amp must be at least 10 times greater than the maximum load switching frequency. Again, look for the “golden mean” in the price/performance ratio, everything is good in moderation

In most of my projects I use an op-amp from Texas Instruments - OPA376, its performance characteristics are enough to implement protection in most tasks and the price tag of $1 is quite good. If you need cheaper, then look at solutions from ST, and if even cheaper, then at Microchip and Micrel. For religious reasons, I only use TI and Linear, because I like it and sleep more peacefully.

Adding realism to the security system

Let's now add a shunt, load, power source and other attributes in the simulator that will bring our model closer to reality. The resulting result looks like this (clickable image):

You can download the simulation file for MultiSIM - .

Here we already see our shunt R1 with a resistance of the same 2 mOhm, I chose a power source of 310V (rectified network) and the load for it is a 10.2 Ohm resistor, which again, according to Ohm’s law, gives us a current:

As you can see, the previously calculated 60 mV drops on the shunt and we amplify it with the gain:

At the output we receive an amplified signal with an amplitude of 3.1V. Agree, you can feed it to the ADC, to the comparator and drag it across the board 20-40 mm without any fears or deterioration in stability. We will continue to work with this signal.

Comparing Signals Using a Comparator

Comparator- this is a circuit that accepts 2 signals as input, and if the signal amplitude at the direct input (+) is greater than at the inverse input (-), then a log appears at the output. 1 (VCC). Otherwise log. 0 (GND).

Formally, any op-amp can be turned on as a comparator, but such a solution in terms of performance characteristics will be inferior to the comparator in terms of speed and price/result ratio. In our case, the higher the performance, the higher the likelihood that the protection will have time to work and save the device. I like to use a comparator, again from Texas Instruments - LMV7271. What you should pay attention to:

• The response delay is, in fact, the main speed limiter. For the comparator mentioned above, this time is about 880 ns, which is quite fast and in many tasks is somewhat redundant at a price of $2, and you can choose a more optimal comparator
• Again, I advise you to use a rail-to-rail comparator, otherwise the output will not be 5V, but less. The simulator will help you verify this; choose something that is not rail-to-rail and experiment. The signal from the comparator is usually fed to the driver failure input (SD) and it would be nice to have a stable TTL signal there
• Choose a comparator with a push-pull output rather than an open-drain and others. This is convenient and we have predicted performance characteristics for the output

Now let's add a comparator to our project in the simulator and look at its operation in the mode when the protection has not worked and the current does not exceed the emergency one (clickable image):

You can download the file for simulation in MultiSIM - .

What do we need... If the current exceeds 30A, it is necessary that there is a log at the output of the comparator. 0 (GND), this signal will feed the SD or EN input of the driver and turn it off. In the normal state, the output should be a log. 1 (5V TTL) and turn on the power switch driver (for example, the “folk” IR2110 and less ancient ones).

Let's return to our logic:
1) We measured the current on the shunt and got 56.4 mV;
2) We amplified our signal with a factor of 50.78 and got 2.88V at the op-amp output;
3) We apply a reference signal with which we will compare to the direct input of the comparator. We set it using a divider on R2 and set it to 3.1V - this corresponds to a current of approximately 30A. This resistor adjusts the protection threshold!
4) Now we apply the signal from the op-amp output to the inverse and compare the two signals: 3.1V > 2.88V. At the direct input (+) the voltage is higher than at the inverse input (-), which means the current is not exceeded and the output is log. 1 - the drivers are working, but our LED1 is not lit.

Now we increase the current to a value of >30A (twist R8 and reduce the resistance) and look at the result (clickable image):

Let's review the points from our “logic”:
1) We measured the current on the shunt and got 68.9 mV;
2) We amplified our signal with a factor of 50.78 and got 3.4V at the op-amp output;
4) Now we apply the signal from the op-amp output to the inverse and compare the two signals: 3.1V< 3.4В. На прямом входу (+) напряжение НИЖЕ, чем на инверсном входе (-), значит ток превышен и на выходе лог. 0 - драйвера НЕ работают, а наш светодиод LED1 горит.

Why hardware?

The answer to this question is simple - any programmable solution on an MK, with an external ADC, etc., can simply “freeze” and even if you are a fairly competent software writer and have turned on a watchdog timer and other anti-freeze protections - while it is all being processed, your device will burn out.

Hardware protection allows you to implement a system with performance within a few microseconds, and if the budget allows, then within 100-200 ns, which is generally enough for any task. Also, hardware protection will not be able to freeze and will save the device, even if for some reason your control microcontroller or DSP is frozen. The protection will turn off the driver, your control circuit will quietly restart, test the hardware and either report an error, for example, in Modbus, or start if all is well.

It is worth noting here that specialized controllers for building power converters have special inputs that allow you to disable the generation of a PWM signal in hardware. For example, the beloved STM32 has a BKIN input for this.

Separately, it is worth saying about such a thing as CPLD. In essence, this is a set of high-speed logic and its reliability is comparable to a hardware solution. It would be quite common sense to put a small CPLD on the board and implement hardware protection, deadtime and other amenities in it, if we are talking about dc/dc or some kind of control cabinets. CPLD makes this solution very flexible and convenient.

Epilogue

That's probably all. I hope you enjoyed reading this article and it will give you some new knowledge or refresh old ones. Always try to think in advance which modules in your device should be implemented in hardware and which in software. Often the hardware implementation is orders of magnitude simpler than the software implementation, and this leads to savings in development time and, accordingly, its cost.

The format of an article without hardware is new to me and I would like to ask you to express your opinion in the survey.

The simplest short circuit protection is relevant for both experienced and novice radio amateurs, since no one is immune from errors. This article provides a simple but very original diagram that will help you protect your device from unwanted failure. The self-resetting fuse de-energizes the circuit, and the LEDs signal an emergency, quickly, reliably and simply.

Short circuit protection circuit:

The circuit shown in Figure 1 is a very easy-to-set up protection for an amateur radio power supply or any other circuit.

Figure No. 1 – Short circuit protection circuit.

Operation of short circuit protection circuit:

The scheme is very simple and understandable. Since the current flows along the path of least resistance while the fuse FU1 is intact, the output load Rн (Figure No. 2) is connected and current flows through it. In this case, the VD4 LED is constantly lit (preferably green).

Figure No. 2 - Operation of the circuit with a full fuse

If the load current exceeds the maximum current permissible for the fuse, it trips, thereby breaking (bypassing) the load circuit, Figure No. 3. In this case, LED VD3 lights up (red) and VD4 goes out. In this case, neither your load nor the circuit suffers (of course, provided that the fuse trips in a timely manner).

Figure No. 3 – The fuse has tripped

Diodes VD1, VD5 and zener diode VD2 protect LEDs from reverse currents. Resistors R1, R2 limit the current in the protection circuit. For fuse FU1, I recommend using a self-resetting fuse. And you select the values ​​of all elements of the circuit depending on your needs.

Almost everyone has experienced a short circuit in their life. But most often it happened like this: flash, clap and that’s it. This happened only because there was short circuit protection.

Short circuit protection device

The device may be electronic, electromechanical, or a simple fuse. Electronic devices are mainly used in complex electronic devices, and we will not consider them in this article. Let's focus on fuses and electromechanical devices. Fuses were first used to protect household electrical circuits. We are used to seeing them in the form of “plugs” in the electrical panel.

There were several types, but all the protection boiled down to the fact that inside this “plug” there was a thin copper wire that burned out when a short circuit occurred. It was necessary to run to the store, buy a fuse, or store at home a supply of fuses that might not be needed soon. It was inconvenient. And automatic switches were born, which at first also looked like “traffic jams”.

It was a simple electromechanical circuit breaker. They were produced for different currents, but the maximum value was 16 amperes. Soon higher values ​​were required, and technical progress made it possible to produce machines as we now see them in most electrical panels of our homes.

How does a machine gun protect us?

It has two types of protection. One type is based on induction, the second on heating. A short circuit is characterized by a large current that flows through the short-circuited circuit. The machine is designed in such a way that current flows through a bimetallic plate and an inductor. So, when a large current flows through the machine, a strong magnetic flux arises in the coil, which sets the machine’s release mechanism in motion. Well, the bimetallic plate is designed to carry the rated current. When current flows through wires, it always causes heat. But we often don’t notice this, because the heat has time to dissipate and it seems to us that the wires are not heating up. A bimetallic strip consists of two metals with different properties. When heated, both metals deform (expand), but as one metal expands more than the other, the plate begins to bend. The plate is selected in such a way that when the nominal value of the machine is exceeded, due to bending, it activates the release mechanism. Thus, it turns out that one protection (inductive) works on short-circuit currents, and the second on currents flowing for a long time through the cable. Since short circuit currents are rapid in nature and flow in the network for a short period of time, the bimetallic plate does not have time to heat up to such an extent as to deform and turn off the circuit breaker.

Short circuit protection circuit

In fact, there is nothing complicated in this scheme. It is installed in the circuit, which disconnects either the phase wire or the entire circuit at once. But there are nuances. Let's look at them in more detail.

1. You cannot install separate machines in the phase circuit and the zero circuit. For one simple reason. If suddenly, due to a short circuit, the zero circuit breaker turns off, then the entire electrical network will be energized, because the phase circuit breaker will remain on.
   
2. You cannot install a wire with a smaller cross-section than the machine allows. Very often, in apartments with old wiring, in order to increase power, more powerful circuit breakers are installed... Alas, this is the most common cause of short circuits. This is what happens in such cases. Suppose, for clarity, there is a copper wire with a cross-section of 1.5 sq. mm, which is capable of withstanding a current of up to 16 A. A 25A machine is placed on it. We connect a load to this network, say 4.5 kW, and a current of 20.5 amperes will flow through the wire. The wire will start to get very hot, but the machine will not turn off the network. As you remember, the machine has two types of protection. The short circuit protection does not work yet because there is no short circuit, and the rated current protection will operate at a value greater than 25 amps. So it turns out that the wire gets very hot, the insulation begins to melt, but the machine does not work. In the end, an insulation breakdown occurs and a short circuit appears and the machine finally trips. But what do you get? The line can no longer be used and must be replaced. This is not difficult if the wires are laid openly. But what if they are hidden in the wall? New repairs are guaranteed to you.
3. If the aluminum wiring is more than 15 years old, and the copper wiring is more than 25 years old, and you are going to make repairs, definitely replace it with new wiring. Despite the investment it will save you money. Imagine that you have already made a repair, and there is a bad contact in some junction box? This is if we talk about copper wire (in which, as a rule, only the insulation ages or the joints oxidize or weaken over time, then begin to heat up, which leads to the destruction of the twist even faster). If we talk about aluminum wire, then everything is even worse. Aluminum is a very ductile metal. With temperature fluctuations, the compression and expansion of the wire is quite significant. And if there was a microcrack in the wire (manufacturing defect, technological defect), then over time it increases, and when it becomes quite large, which means the wire in this place is thinner, then when current flows, this area begins to heat up and cool down, which only speeds up the process . Therefore, even if it seems to you that everything is fine with the wiring: “It worked before!”, it’s better to change it anyway.
4. Junction boxes. There are articles about this, but I will briefly go through them here. NEVER DO SCROLLS!!! Even if you make them well, it's a twist. Metal tends to shrink and expand under the influence of temperature, and the twist weakens. Avoid using screw terminals for the same reason. Screw terminals can be used in open wiring. Then at least you can periodically look into the boxes and check the condition of the wiring. Screw clamps of the “PPE” type or terminal connections of the “WAGO” type are best suited for this purpose; screw clamps of the “Nut” type are best suited for power wiring (such clamps have two plates that are held together with four screws, in the middle there is another plate, i.e. using such clamps you can connect copper and aluminum wires). Leave a reserve of at least 15 cm of stripped wire. This serves two purposes: if the twist contact is poor, the wire has time to dissipate heat, and you have the opportunity to redo the twist if something happens. Try to place the wires in such a way that there is no overlap between the phase and neutral wires with the ground wire. The wires can cross, but not lie on top of each other. Try to place the twists so that the phase wire is on one side, and the neutral and ground wires are on the other.

   

5. Do not connect copper and aluminum wires directly. Either use WAGO terminal blocks or Walnut clamps. This is especially true for wires intended for connecting electric stoves. Usually, when they make repairs and move a stove socket, they extend the cable. Very often these are aluminum wires that are extended with copper.
6. A little special. Do not skimp on switches and sockets (especially for electric stoves). The fact is that nowadays it’s quite difficult to find good sockets for electric stoves (I’m talking about small towns), so it’s best to either use the “Nut” U739M clamps or find a good socket.
7. When tightening the terminals on sockets, do it more tightly, but do not break the thread; if this happens, it is better to change the socket immediately, do not rely on “maybe”.
8. When laying a new electrical route, use the following standards: 10-15 cm from corners, ceilings, walls (along the floor), jambs, window frames, floor (along the wall). This will protect you when installing, for example, suspended ceilings or baseboards, which are secured using dowels for which you need to punch a hole. If the wire is located in the corner between the floor and the wall, it is very easy to get caught in the wire. All wires must be positioned strictly horizontally or vertically. This will make it easier for you to understand where you can make a new hole if you suddenly need to hang a shelf or a picture or a TV.
9. Do not daisy chain (from one to another) more than 4 sockets. In the kitchen, I generally do not recommend connecting more than two, especially where you plan to use an oven, kettle, dishwasher and microwave in one place.
10. It is best to lay a separate line for the oven or connect it to the line from which the hob is powered (because very often they consume about 3 kW.) Not every outlet can withstand such a load, and even if another powerful consumer is connected to it ( for example, a kettle), you risk getting a short circuit due to the strong heating of the connection in the socket by the cable.
11. Avoid using extension cords to power high-power electrical appliances, such as oil heaters, or use extension cords from reputable manufacturers rather than Chinese "no name" brands. Read carefully what power a given extension cord can handle, and do not use it if it has less power than you need to power. When using an extension cord, try to avoid stranded wire. If the wire just lies there, it has time to dissipate heat. If the wire is twisted, the heat does not have time to dissipate and the wire begins to heat up noticeably, which can also lead to a short circuit.
12. Do not connect several powerful consumers to one outlet (through a tee or an extension cord with several outlets). A load of 3.5 kW can be connected to a good outlet, and up to 2 kW to a not-so-good outlet. In houses with aluminum wiring, no more than 2 kW in any socket, and even better, do not include more than 2 kW in a group of sockets powered by one circuit breaker.
13. Before installing a heater in each room, make sure that the rooms are powered from different machines. As they say: “And sometimes a stick can shoot,” the same is with machine guns: “And sometimes a machine gun can fail to work,” and the consequences of this are quite cruel. Therefore, protect yourself and your loved ones.
14. Handle heating devices carefully, making sure that the wire does not come into contact with the heating elements.

Short circuit circuit breaker

Why did I make this a separate point? It's simple. It is the machine that provides short circuit protection. If you install, then you must install an automatic machine next, or install it immediately (this is a two-in-one device: an RCD and an automatic machine). Such a device turns off the network in case of a short circuit, and when the rated current value is exceeded, and when there is a leakage current, when, for example, you are under voltage and electric current begins to flow through you. Let me remind you again: the RCD DOES NOT PROTECT FROM SHORT CIRCUIT, the RCD protects you from electric shock. Of course, it may be that the RCD will turn off the network in the event of a short circuit, but it is not intended for this. The operation of an RCD during a short circuit is completely random. And all the wiring may burn out, everything may be in flames, but the RCD will not turn off the network.

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