The devices require a power supply unit (PSU), which has adjustable output voltage and the ability to regulate the level of overcurrent protection over a wide range. When the protection is triggered, the load (connected device) should automatically turn off.
An Internet search yielded several suitable power supply circuits. I settled on one of them. The circuit is easy to manufacture and set up, consists of accessible parts, and fulfills the stated requirements.
The power supply proposed for production is based on the LM358 operational amplifier and has the following characteristics:
Input voltage, V - 24...29
Output stabilized voltage, V - 1...20 (27)
Protection operation current, A - 0.03...2.0
Photo 2. Power supply circuit
Description of the power supply
The adjustable voltage stabilizer is assembled on the DA1.1 operational amplifier. The amplifier input (pin 3) receives a reference voltage from the motor of the variable resistor R2, the stability of which is ensured by the zener diode VD1, and the inverting input (pin 2) receives the voltage from the emitter of the transistor VT1 through the voltage divider R10R7. Using variable resistor R2, you can change the output voltage of the power supply.
The overcurrent protection unit is made on the DA1.2 operational amplifier; it compares the voltages at the op-amp inputs. Input 5 through resistor R14 receives voltage from the load current sensor - resistor R13. The inverting input (pin 6) receives a reference voltage, the stability of which is ensured by diode VD2 with a stabilization voltage of about 0.6 V.
As long as the voltage drop created by the load current across resistor R13 is less than the exemplary value, the voltage at the output (pin 7) of op-amp DA1.2 is close to zero. If the load current exceeds the permissible set level, the voltage at the current sensor will increase and the voltage at the output of op-amp DA1.2 will increase almost to the supply voltage. At the same time, the HL1 LED will turn on, signaling an excess, and the VT2 transistor will open, shunting the VD1 zener diode with resistor R12. As a result, transistor VT1 will close, the output voltage of the power supply will decrease to almost zero and the load will turn off. To turn on the load you need to press the SA1 button. The protection level is adjusted using variable resistor R5.
PSU manufacturing
1. The basis of the power supply and its output characteristics are determined by the current source - the transformer used. In my case, a toroidal transformer from a washing machine was used. The transformer has two output windings for 8V and 15V. By connecting both windings in series and adding a rectifier bridge using medium-power diodes KD202M available at hand, I obtained a constant voltage source of 23V, 2A for the power supply.
Photo 3. Transformer and rectifier bridge.
2. Another defining part of the power supply is the device body. In this case, a children's slide projector hanging around in the garage found use. By removing the excess and processing the holes in the front part for installing an indicating microammeter, a blank power supply housing was obtained.
Photo 4. PSU body blank
3. The electronic circuit is mounted on a universal mounting plate measuring 45 x 65 mm. The layout of the parts on the board depends on the sizes of the components found on the farm. Instead of resistors R6 (setting the operating current) and R10 (limiting the maximum output voltage), trimming resistors with a value increased by 1.5 times are installed on the board. After setting up the power supply, they can be replaced with permanent ones.
Photo 5. Circuit board
4. Assembling the board and remote elements of the electronic circuit in full for testing, setting and adjusting the output parameters.
Photo 6. Power supply control unit
5. Fabrication and adjustment of a shunt and additional resistance for using a microammeter as an ammeter or power supply voltmeter. Additional resistance consists of permanent and trimming resistors connected in series (pictured above). The shunt (pictured below) is included in the main current circuit and consists of a wire with low resistance. The wire size is determined by the maximum output current. When measuring current, the device is connected in parallel to the shunt.
Photo 7. Microammeter, shunt and additional resistance
Adjustment of the length of the shunt and the value of additional resistance is carried out with the appropriate connection to the device with control for compliance using a multimeter. The device is switched to Ammeter/Voltmeter mode using a toggle switch in accordance with the diagram:
Photo 8. Control mode switching diagram
6. Marking and processing of the front panel of the power supply unit, installation of remote parts. In this version, the front panel includes a microammeter (toggle switch for switching the A/V control mode to the right of the device), output terminals, voltage and current regulators, and operating mode indicators. To reduce losses and due to frequent use, a separate stabilized 5 V output is additionally provided. Why is the voltage from the 8V transformer winding supplied to the second rectifier bridge and a typical 7805 circuit with built-in protection.
Photo 9. Front panel
7. PSU assembly. All power supply elements are installed in the housing. In this embodiment, the radiator of the control transistor VT1 is an aluminum plate 5 mm thick, fixed in the upper part of the housing cover, which serves as an additional radiator. The transistor is fixed to the radiator through an electrically insulating gasket.
Many homemade units have the disadvantage of lacking protection against power reverse polarity. Even an experienced person can inadvertently confuse the polarity of the power supply. And there is a high probability that after this the charger will become unusable.
This article will discuss 3 options for reverse polarity protection, which work flawlessly and do not require any adjustment.
Option 1
This protection is the simplest and differs from similar ones in that it does not use any transistors or microcircuits. Relays, diode isolation - that’s all its components.
The scheme works as follows. The minus in the circuit is common, so the positive circuit will be considered.
If there is no battery connected to the input, the relay is in the open state. When the battery is connected, the plus is supplied through the diode VD2 to the relay winding, as a result of which the relay contact closes and the main charging current flows to the battery.
At the same time, the green LED indicator lights up, indicating that the connection is correct.
And if you now remove the battery, then there will be voltage at the output of the circuit, since the current from the charger will continue to flow through the VD2 diode to the relay winding.
If the connection polarity is reversed, the VD2 diode will be locked and no power will be supplied to the relay winding. The relay will not work.
In this case, the red LED will light up, which is intentionally connected incorrectly. It will indicate that the polarity of the battery connection is incorrect.
Diode VD1 protects the circuit from self-induction that occurs when the relay is turned off.
If such protection is introduced into , it’s worth taking a 12 V relay. The permissible current of the relay depends only on the power . On average, it is worth using a 15-20 A relay.
This scheme still has no analogues in many respects. It simultaneously protects against power reversal and short circuit.
The operating principle of this scheme is as follows. During normal operation, the plus from the power source through the LED and resistor R9 opens the field-effect transistor, and the minus through the open junction of the “field switch” goes to the output of the circuit to the battery.
When a polarity reversal or short circuit occurs, the current in the circuit increases sharply, resulting in a voltage drop across the “field switch” and across the shunt. This voltage drop is enough to trigger the low-power transistor VT2. Opening, the latter closes the field-effect transistor, closing the gate to ground. At the same time, the LED lights up, since power for it is provided by the open junction of transistor VT2.
Due to its high response speed, this circuit is guaranteed to protect for any problem at the output.
The circuit is very reliable in operation and can remain in a protected state indefinitely.
This is a particularly simple circuit, which can hardly even be called a circuit, since it uses only 2 components. This is a powerful diode and fuse. This option is quite viable and is even used on an industrial scale.
Power from the charger is supplied to the battery through the fuse. The fuse is selected based on the maximum charging current. For example, if the current is 10 A, then a 12-15 A fuse is needed.
The diode is connected in parallel and is closed during normal operation. But if the polarity is reversed, the diode will open and a short circuit will occur.
And the fuse is the weak link in this circuit, which will burn out at the same moment. After this you will have to change it.
The diode should be selected according to the datasheet based on the fact that its maximum short-term current was several times greater than the fuse combustion current.
This scheme does not provide 100% protection, since there have been cases when the charger burned out faster than the fuse.
Bottom line
From an efficiency point of view, the first scheme is better than the others. But from the point of view of versatility and speed of response, the best option is scheme 2. Well, the third option is often used on an industrial scale. This type of protection can be seen, for example, on any car radio.
All circuits, except the last one, have a self-healing function, that is, operation will be restored as soon as the short circuit is removed or the polarity of the battery connection is changed.
Attached files:
How to make a simple Power Bank with your own hands: diagram of a homemade power bank
This is a small universal short-circuit protection unit that is intended for use in networks. It is specially designed to fit into most power supplies without redesigning their circuitry. The circuit, despite the presence of a microcircuit, is very easy to understand. Save it to your computer to see it in a better size.
To solder the circuit you will need:
- 1 - TL082 dual op-amp
- 2 - 1n4148 diode
- 1 - tip122 NPN transistor
- 1 - BC558 PNP transistor BC557, BC556
- 1 - resistor 2700 ohm
- 1 - 1000 ohm resistor
- 1 - 10 kohm resistor
- 1 - resistor 22 kom
- 1 - potentiometer 10 kohm
- 1 - capacitor 470 uF
- 1 - capacitor 1 µF
- 1 - normally closed switch
- 1 - relay model T74 "G5LA-14"
Connecting the circuit to the power supply
Here, a low value resistor is connected in series with the output of the power supply. Once current starts flowing through it, there will be a small voltage drop and we will use this voltage drop to determine whether the power is the result of an overload or a short circuit. This circuit is based on an operational amplifier (op-amp) included as a comparator.
- If the voltage at the non-inverting output is higher than at the inverting output, then the output is set to a “high” level.
- If the voltage at the non-inverting output is lower than at the inverting output, then the output is set to a “low” level.
True, this has nothing to do with the logical 5-volt level of conventional microcircuits. When the op amp is "high", its output will be very close to the positive potential of the supply voltage, so if the supply is +12V, the "high" will be close to +12V. When the op amp is "low", its output will be almost at minus supply voltage, therefore, close to 0 V.
When using op amps as comparators, we typically have an input signal and a reference voltage to compare that input signal to. So we have a resistor with a variable voltage that is determined according to the current that flows through it and the reference voltage. This resistor is the most important part of the circuit. It is connected in series with the output power. You need to select a resistor that has a voltage drop of approximately 0.5~0.7 volts when there is an overload of current passing through it. Overload current occurs when the protection circuit operates and closes the power output to prevent damage to it.
You can select a resistor using Ohm's law. The first thing to determine is the power supply overcurrent. To do this, you need to know the maximum permissible current of the power supply.
Let's say your power supply can output 3 amps (the voltage of the power supply does not matter). So, we got P = 0.6 V / 3 A. P = 0.2 Ohm. The next thing you should do is calculate the power dissipation across this resistor using the formula: P=V*I. If we use our last example, we get: P = 0.6 V * 3 A. P = 1.8 W - 3 or 5 W resistor will be more than enough.
To make the circuit work, you will need to apply voltage to it, which can be from 9 to 15 V. To calibrate, apply voltage to the inverting input of the op-amp and turn the potentiometer. This tension will increase or decrease depending on which way you turn it. The value needs to be adjusted according to the input stage gain of 0.6 Volts (something around 2.2 to 3 Volts if your amplifier stage is like mine). This procedure takes some time, and the best method for calibration is the scientific poke method. You may need to set the potentiometer to a higher voltage so that the protection does not trip during load peaks. Download the project file.
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 loadAbsolutely 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 necessary, 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
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 the sake of clarity, I've 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
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
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.
Only registered users can participate in the survey. , Please.
The integrated circuit (IC) KR142EN12A is an adjustable voltage stabilizer of the compensation type in the KT-28-2 package, which allows you to power devices with a current of up to 1.5 A in the voltage range of 1.2...37 V. This integrated stabilizer has thermally stable protection according to current and output short circuit protection.
Based on the KR142EN12A IC, you can build an adjustable power supply, the circuit of which (without a transformer and diode bridge) is shown in Fig.2. The rectified input voltage is supplied from the diode bridge to capacitor C1. Transistor VT2 and chip DA1 should be located on the radiator.
Heat sink flange DA1 is electrically connected to pin 2, so if DAT and transistor VD2 are located on the same heatsink, then they need to be isolated from each other.
In the author's version, DA1 is installed on a separate small radiator, which is not galvanically connected to the radiator and transistor VT2. The power dissipated by a chip with a heat sink should not exceed 10 W. Resistors R3 and R5 form a voltage divider included in the measuring element of the stabilizer. A stabilized negative voltage of -5 V is supplied to capacitor C2 and resistor R2 (used to select the thermally stable point VD1). In the original version, the voltage is supplied from the KTs407A diode bridge and the 79L05 stabilizer, powered from a separate winding of the power transformer.
For guard from closing the output circuit of the stabilizer, it is enough to connect an electrolytic capacitor with a capacity of at least 10 μF in parallel with resistor R3, and shunt resistor R5 with a KD521A diode. The location of the parts is not critical, but for good temperature stability it is necessary to use the appropriate types of resistors. They should be located as far as possible from heat sources. The overall stability of the output voltage consists of many factors and usually does not exceed 0.25% after warming up.
After switching on and warming up the device, the minimum output voltage of 0 V is set with resistor Rao6. Resistors R2 ( Fig.2) and resistor Rno6 ( Fig.3) must be multi-turn trimmers from the SP5 series.
Possibilities the current of the KR142EN12A microcircuit is limited to 1.5 A. Currently, there are microcircuits on sale with similar parameters, but designed for a higher current in the load, for example LM350 - for a current of 3 A, LM338 - for a current of 5 A. Recently on sale imported microcircuits from the LOW DROP series (SD, DV, LT1083/1084/1085) appeared. These microcircuits can operate at a reduced voltage between input and output (up to 1... 1.3 V) and provide a stabilized output voltage in the range of 1.25...30 V at a load current of 7.5/5/3 A, respectively . The closest domestic analogue in terms of parameters, type KR142EN22, has a maximum stabilization current of 7.5 A. At the maximum output current, the stabilization mode is guaranteed by the manufacturer at an input-output voltage of at least 1.5 V. The microcircuits also have built-in protection against excess current in the load of the permissible value and thermal protection against overheating of the case. These stabilizers provide output voltage instability of 0.05%/V, output voltage instability when the output current changes from 10 mA to a maximum value of no worse than 0.1%/V. On Fig.4 shows a power supply circuit for a home laboratory, which allows you to do without transistors VT1 and VT2, shown in Fig.2.
Instead of the DA1 KR142EN12A microcircuit, the KR142EN22A microcircuit was used. This is an adjustable stabilizer with a low voltage drop, which allows you to obtain a current of up to 7.5 A in the load. For example, the input voltage supplied to the microcircuit is Uin = 39 V, output voltage at the load Uout = 30 V, current at the load louf = 5 A, then the maximum power dissipated by the microcircuit at the load is 45 W. Electrolytic capacitor C7 is used to reduce output impedance at high frequencies, and also reduces noise voltage and improves ripple smoothing. If this capacitor is tantalum, then its nominal capacity must be at least 22 μF, if aluminum - at least 150 μF. If necessary, the capacitance of capacitor C7 can be increased. If the electrolytic capacitor C7 is located at a distance of more than 155 mm and is connected to the power supply with a wire with a cross-section of less than 1 mm, then an additional electrolytic capacitor with a capacity of at least 10 μF is installed on the board parallel to the capacitor C7, closer to the microcircuit itself. The capacitance of filter capacitor C1 can be determined approximately at the rate of 2000 μF per 1 A of output current (at a voltage of at least 50 V). 
To reduce the temperature drift of the output voltage, resistor R8 must be either wire-wound or metal-foil with an error of no worse than 1%. Resistor R7 is the same type as R8. If the KS113A zener diode is not available, you can use the unit shown in Fig.3. The author is quite satisfied with the protection circuit solution given in, as it works flawlessly and has been tested in practice. You can use any power supply protection circuit solutions, for example those proposed in. In the author’s version, when relay K1 is triggered, contacts K 1.1 are closed, short-circuiting resistor R7, and the voltage at the output of the power supply becomes equal to 0 V. The printed circuit board of the power supply and the location of the elements are shown in Fig. 5, the appearance of the power supply is shown in Fig.6.
