About protection electrical diagrams from incorrect power polarity using field effect transistor, I remembered that I have had an unresolved problem for a long time automatic shutdown battery from charger when the latter is de-energized. And I became curious whether it was possible to apply a similar approach in another case, where, from time immemorial, a diode was also used as a shut-off element.

This article is a typical bicycle building guide, because... talks about the development of a circuit whose functionality has long been implemented in millions of finished devices. Therefore, the request does not apply to this material, as something completely utilitarian. Rather, it is simply the story of how an electronic device is born: from recognition of a need to a working prototype through all the obstacles.

What is this all for?

When redundant low voltage power supply direct current The easiest way to turn on a lead-acid battery is as a buffer, simply in parallel with the mains source, as was done in cars before they had complex “brains”. Although the battery does not operate in the most optimal mode, it is always charged and does not require any power switching when the mains voltage at the power supply input is turned off or on. Below we will talk in more detail about some of the problems of such inclusion and an attempt to solve them.

Background

Just 20 years ago, such an issue was not on the agenda. The reason for this was the circuitry of a typical mains power supply (or charger), which prevented the battery from discharging to its output circuits when the mains voltage was turned off. Let's look at the simplest block circuit with half-wave rectification:

It is quite obvious that the same diode that rectifies the alternating voltage of the mains winding will also prevent the battery from discharging onto the secondary winding of the transformer when the mains supply voltage is turned off. The full-wave bridge rectifier circuit, although somewhat less obvious, has exactly the same properties. And even the use of a parametric voltage stabilizer with a current amplifier (such as the widespread 7812 microcircuit and its analogues) does not change the situation:

Indeed, if you look at the simplified circuit of such a stabilizer, it becomes clear that the emitter junction of the output transistor plays the role of the same shut-off diode, which closes when the voltage at the rectifier output is lost, and keeps the battery charge intact.

However, in recent years everything has changed. Transformer power supplies with parametric stabilization have been replaced by more compact and cheaper switching AC/DC voltage converters, which have a much higher efficiency and power/weight ratio. But with all the advantages, these power supplies have one drawback: their output circuits have a much more complex circuit design, which usually does not provide any protection against current flow back from the secondary circuit. As a result, when using such a source in a system of the form “BP -> buffer battery -> load”, when the mains voltage is turned off, the battery begins to intensively discharge to the output circuits of the power supply.

The simplest way (diode)

The simplest solution is to use a Schottky barrier diode connected to the positive wire connecting the power supply and the battery:

However, the main problems of such a solution have already been voiced in the article mentioned above. In addition, this approach may be unacceptable due to the fact that a 12-volt lead-acid battery requires a voltage of at least 13.6 volts to operate in buffer mode. And almost half a volt falling across the diode can make this voltage simply unattainable in combination with the existing power supply (exactly my case).

All this forces us to look for alternative ways of automatic switching, which should have the following properties:

1. Low forward voltage drop when on.
2. The ability to withstand, without significant heating, the direct current consumed from the power supply by the load and the buffer battery when switched on.
3. High reverse voltage drop and low off-state self-consumption.
4. Normally off state, so that when a charged battery is connected to an initially de-energized system, it does not begin to discharge.
5. Automatic transition to the on state when mains voltage is applied, regardless of the presence and charge level of the battery.
6. The fastest possible automatic transition to the off state in the event of a power failure.

If the diode were an ideal device, then it would fulfill all these conditions without any problems, but harsh reality casts doubt on points 1 and 2.

Naive solution (DC relay)

When analyzing the requirements, anyone who is even a little “in the know” will come up with the idea of ​​​​using an electromagnetic relay for this purpose, which is capable of physically closing contacts using a magnetic field created by the control current in the winding. And he’ll probably even scribble something like this on a napkin:

In this circuit, the normally open relay contacts close only when current flows through the winding connected to the output of the power supply. However, if you go through the list of requirements, it turns out that this circuit does not correspond to point 6. After all, if the relay contacts were once closed, a loss of mains voltage will not lead to their opening for the reason that the winding (and with it the entire output circuit of the power supply) remains connected to the battery through the same contacts! There is a typical case of positive feedback when the control circuit has direct communication with the executive, and as a result the system acquires the properties of a bistable trigger.

Thus, such a naive approach is not a solution to the problem. Moreover, if you analyze the current situation logically, you can easily come to the conclusion that in the interval “BP -> buffer battery”, under ideal conditions, no other solution other than a valve conducting current in one direction simply can be. Indeed, if we do not use any external control signal, then no matter what we do at this point in the circuit, any of our switching elements, once turned on, will make the electricity created by the battery indistinguishable from the electricity created by the power supply.

Roundabout (AC relay)

After realizing all the problems of the previous point, a “rummaging” person usually comes up with a new idea of ​​using the power supply itself as a one-way conductive valve. Why not? After all, if the power supply is not a reversible device, and the battery voltage supplied to its output does not create AC voltage 220 volts (as happens in 100% of cases in real circuits), then this difference can be used as a control signal for the switching element:

Bingo! All requirements are met and the only thing needed for this is a relay capable of closing contacts when mains voltage is applied to it. This may be a special AC relay designed for mains voltage. Or a regular relay with its own mini-power supply (any transformer-free step-down circuit with a simple rectifier is enough here).

We could have celebrated the victory, but I didn’t like this decision. Firstly, you need to connect something directly to the network, which is not good from a security point of view. Secondly, the fact that this relay must switch significant currents, probably up to tens of amperes, and this makes the entire design not as trivial and compact as it might seem initially. And thirdly, what about such a convenient field-effect transistor?

First solution (FET + battery voltage meter)

The search for a more elegant solution to the problem led me to the realization of the fact that a battery operating in buffer mode at a voltage of about 13.8 volts, without external “recharge”, quickly loses its original voltage even in the absence of load. If it starts to discharge on the power supply, then in the first minute it loses at least 0.1 volts, which is more than enough for reliable fixation by a simple comparator. In general, the idea is this: the gate of a commutating field-effect transistor is controlled by a comparator. One of the comparator inputs is connected to a stable voltage source. The second input is connected to the power supply voltage divider. Moreover, the division coefficient is selected so that the voltage at the output of the divider when the power supply is turned on is approximately 0.1..0.2 volts higher than the voltage of the stabilized source. As a result, when the power supply is turned on, the voltage from the divider will always prevail, but when the network is de-energized, as the battery voltage drops, it will decrease in proportion to this drop. After some time, the voltage at the output of the divider will be less than the voltage of the stabilizer and the comparator will break the circuit using a field-effect transistor.

An approximate diagram of such a device:

As you can see, the direct input of the comparator is connected to a source of stable voltage. The voltage of this source, in principle, is not important, the main thing is that it is within the permissible input voltages of the comparator, but it is convenient when it is approximately half the battery voltage, that is, about 6 volts. The inverse input of the comparator is connected to the power supply voltage divider, and the output is connected to the gate of the switching transistor. When the voltage at the inverse input exceeds that at the forward input, the output of the comparator connects the gate of the field-effect transistor to ground, causing the transistor to turn on and complete the circuit. After de-energizing the network, after some time the battery voltage drops, along with it the voltage at the inverse input of the comparator drops, and when it is below the level at the direct input, the comparator “tears” the transistor gate from the ground and thereby breaks the circuit. Subsequently, when the power supply “comes to life” again, the voltage at the inverse input will instantly rise to a normal level and the transistor will open again.

For practical implementation For this circuit, I used the LM393 chip I had. This is a very cheap (less than ten cents at retail), but at the same time economical and has quite good characteristics, a dual comparator. It allows power supply voltages up to 36 volts, has a transmission coefficient of at least 50 V/mV, and its inputs have a fairly high impedance. The first of the commercially available high-power P-channel MOSFETs, the FDD6685, was taken as a switching transistor. After several experiments, the following practical switch circuit was derived:

In it, the abstract source of stable voltage is replaced with a very real parametric stabilizer consisting of resistor R2 and zener diode D1, and the divider is made on the basis of trimming resistor R1, which allows you to adjust the division coefficient to the desired value. Since the comparator inputs have a very significant impedance, the value of the damping resistance in the stabilizer can be more than a hundred kOhms, which allows minimizing the leakage current, and therefore the total consumption of the device. The value of the trimming resistor is not critical at all and can be selected in the range from ten to several hundred kOhms without any consequences for the performance of the circuit. Due to the fact that the output circuit of the comparator LM393 is built according to an open collector circuit, a load resistor R3 with a resistance of several hundred kOhms is also required for its functional completion.

Adjusting the device comes down to setting the position of the trimmer resistor slider to a position where the voltage on leg 2 of the microcircuit exceeds that on leg 3 by approximately 0.1..0.2 volts. To set up, it is better not to use a multimeter in high-impedance circuits, but simply by setting the resistor slider to the lower position (according to the diagram), connect the power supply (we do not connect the battery yet), and, measuring the voltage at pin 1 of the microcircuit, move the resistor contact upward. As soon as the voltage drops sharply to zero, pre-setting can be considered completed.

You should not strive to turn off with a minimum voltage difference, because this will inevitably lead to incorrect operation of the circuit. In real conditions, on the contrary, you have to deliberately lower the sensitivity. The fact is that when the load is turned on, the voltage at the input of the circuit inevitably drops due to non-ideal stabilization in the power supply and the finite resistance of the connecting wires. This can lead to the fact that an overly sensitive device will consider such a drawdown to be a disconnection of the power supply and break the circuit. As a result, the power supply will be connected only when there is no load, and the battery will have to work the rest of the time. True, when the battery is slightly discharged, the internal diode of the field-effect transistor will open and current from the power supply will begin to flow into the circuit through it. But this will lead to overheating of the transistor and to the fact that the battery will operate in a long-term undercharge mode. In general, the final calibration must be carried out under a real load, monitoring the voltage at pin 1 of the microcircuit and ultimately leaving a small margin for reliability.

Significant disadvantages of this scheme are the relative complexity of calibration and the need to tolerate potential losses of battery energy in order to ensure correct operation.

The last drawback haunted me and after some reflection led me to the idea of ​​measuring not the battery voltage, but directly the direction of the current in the circuit.

Second solution (field effect transistor + current direction meter)

To measure the direction of the current, some clever sensor could be used. For example, a Hall sensor that registers the magnetic field vector around a conductor and allows you to determine not only the direction, but also the strength of the current without breaking the circuit. However, due to the lack of such a sensor (and experience with such devices), it was decided to try to measure the sign of the voltage drop on the field-effect transistor channel. Of course, in the open state, the channel resistance is measured in hundredths of an ohm (this is what the whole idea is for), but, nevertheless, it is quite finite and you can try to play on it. An additional argument in favor of this solution is that there is no need for fine adjustments. We will only measure the polarity of the voltage drop, and not its absolute value.

According to the most pessimistic calculations, with an open channel resistance of the FDD6685 transistor of about 14 mOhm and a differential sensitivity of the LM393 comparator from the “min” column of 50 V/mV, we will have a full voltage swing of 12 volts at the comparator output with a current through the transistor of just over 17 mA. As you can see, the value is quite real. In practice, it should be approximately an order of magnitude smaller, because the typical sensitivity of our comparator is 200 V/mV, the transistor channel resistance in real conditions, taking into account installation, is unlikely to be less than 25 mOhm, and the control voltage swing at the gate may not exceed three volt.

The abstract implementation would look something like this:

Here the comparator inputs are connected directly to the positive bus on opposite sides of the field-effect transistor. When current passes through it in different directions, the voltages at the inputs of the comparator will inevitably differ, and the sign of the difference will correspond to the direction of the current, and the magnitude will correspond to its strength.

At first glance, the circuit turns out to be extremely simple, but here a problem arises with the power supply to the comparator. It lies in the fact that we cannot power the microcircuit directly from the same circuits that it is supposed to measure. According to the datasheet, the maximum voltage at the LM393 inputs should not be higher than the supply voltage minus two volts. If this threshold is exceeded, the comparator stops noticing the difference in voltages at the direct and inverse inputs.

There are two potential solutions to this problem. The first, obvious one, is to increase the supply voltage of the comparator. The second thing that comes to mind, if you think a little, is to reduce the control voltages equally using two dividers. Here's what it might look like:

This scheme is captivating with its simplicity and conciseness, but, unfortunately, it is not feasible in the real world. The fact is that we are dealing with a voltage difference between the comparator inputs of only a few millivolts. At the same time, the spread of resistance of resistors even of the highest accuracy class is 0.1%. With a minimum acceptable division ratio of 2 to 8 and a reasonable divider impedance of 10 kOhm, the measurement error will reach 3 mV, which is several times greater than the voltage drop across the transistor at a current of 17 mA. The use of a “tuner” in one of the dividers is eliminated for the same reason, because it is not possible to select its resistance with an accuracy of more than 0.01% even when using a precision multi-turn resistor (plus do not forget about time and temperature drift). In addition, as already written above, theoretically this circuit should not need calibration at all due to its almost “digital” nature.

Based on all that has been said, in practice the only option left is to increase the supply voltage. In principle, this is not such a problem, considering that there are a huge number of specialized microcircuits that allow you to build a stepup converter for the required voltage using just a few parts. But then the complexity of the device and its consumption will almost double, which I would like to avoid.

There are several ways to build a low-power boost converter. For example, most integrated converters use the self-induction voltage of a small inductor connected in series with a “power” switch located directly on the chip. This approach is justified for relatively powerful conversion, for example, for powering an LED with a current of tens of milliamps. In our case, this is clearly redundant, because we only need to provide a current of about one milliamp. A DC voltage doubling circuit using a control switch, two capacitors, and two diodes is much more suitable for us. The principle of its operation can be understood from the diagram:

At the first moment in time, when the transistor is turned off, nothing interesting happens. The current from the power bus passes through diodes D1 and D2 to the output, as a result of which the voltage on capacitor C2 is even slightly lower than that supplied to the input. However, if the transistor opens, capacitor C1, through diode D1 and the transistor, will charge almost to the supply voltage (minus the direct drop across D1 and the transistor). Now, if we close the transistor again, it turns out that the charged capacitor C1 is connected in series with the resistor R1 and the power source. As a result, its voltage will add up to the voltage of the power source and, having suffered some losses in resistor R1 and diode D2, will charge C2 to almost double Uin. After this, the entire cycle can be started over. As a result, if the transistor switches regularly, and the energy extraction from C2 is not too great, from 12 volts you get about 20 at the cost of only five parts (not counting the key), among which there is not a single winding or dimensional element.

To implement such a doubler, in addition to the elements already listed, we need an oscillation generator and the key itself. It may seem like a lot of details, but in fact it is not, because we already have almost everything we need. I hope you haven't forgotten that LM393 contains two comparators? And what about the fact that we have only used one of them so far? After all, a comparator is also an amplifier, which means that if you embrace it with a positive feedback by alternating current, it will turn into a generator. At the same time, its output transistor will regularly open and close, perfectly performing the role of a doubler key. This is what we get when we try to implement our plan:

At first, the idea of ​​powering a generator with the voltage that it actually produces during operation may seem quite wild. However, if you take a closer look, you can see that the generator initially receives power through diodes D1 and D2, which is enough for it to start. After generation occurs, the doubler begins to operate, and the supply voltage smoothly increases to approximately 20 volts. This process takes no more than a second, after which the generator, and along with it the first comparator, receive power that significantly exceeds the operating voltage of the circuit. This gives us the opportunity to directly measure the voltage difference at the source and drain of the field-effect transistor and achieve our goal.

Here is the final diagram of our switch:

There is nothing left to explain about it, everything is described above. As you can see, the device does not contain a single adjustment element and, if assembled correctly, begins to work immediately. In addition to the already familiar active elements, only two diodes have been added, for which you can use any low-power diodes with a maximum reverse voltage of at least 25 volts and a maximum forward current of 10 mA (for example, the widely used 1N4148, which can be desoldered from an old motherboard).

This circuit was tested on a breadboard, where it proved to be fully functional. The obtained parameters fully correspond to expectations: instantaneous switching in both directions, no inadequate response when connecting a load, current consumption from the battery is only 2.1 mA.

One of the wiring options printed circuit board also attached. 300 dpi, view from the side of the parts (so you need to print in mirror image). The field effect transistor is mounted on the conductor side.

Assembled device, completely ready for installation:

I wired it the old-fashioned way, so it turned out a little crooked, but nevertheless, the device has been regularly performing its functions for several days in a circuit with a current of up to 15 amperes without any signs of overheating.

n-channel MOSFET + 7.2...15V zener diode + resistor of a couple of tens of kilo-ohms = SAFETY

The task seems to be trivial. And why would anyone ever need to protect any electronic products from power supply reverse polarity?

Alas, an insidious case has a thousand and one ways to slip a minus instead of a plus onto a device that you spent many days assembling and debugging, and now it just started working.

I will give just a few examples of potential killers of electronic breadboards, and finished products too:

• Universal power supplies with their universal plugs, which can be connected either with a plus on the internal contact or with a minus.
• Small power supplies (such boxes on the power plug) - they are all produced with a plus on the central contact, aren’t they? NO!
• Any type of connector for power supply without a hard mechanical “key”. For example, convenient and cheap computer “jumpers” with a pitch of 2.54mm. Or screw clamps.
• How do you like this scenario: the day before yesterday there were only black and blue wires at hand. Today I was sure that the "minus" is the blue wire. Chpok - that's a mistake. At first I wanted to use black and red.
• Yes, just if you have a bad day - mix up a couple of wires, or plug them in the other way around simply because you were holding the board upside down...

There will always be people (I know at least two such peppers) who, looking straight into the eyes, will firmly and categorically declare that they will never do such a stupid thing as reversing the polarity of the power source! God is their judge. Maybe after they themselves assemble and debug several original designs own development- they will become wiser. In the meantime, I won't argue. I'll just tell you what I use myself.

Life stories

I was still quite young when I had to resolder 25 out of 27 cases. Luckily, these were good old DIP microcircuits.
Since then, I almost always place a protective diode next to the power connector.

By the way, the topic of protection against incorrect power polarity is relevant not only at the prototyping stage.
Just recently I witnessed the heroic efforts of a friend to restore a giant laser cutter. The cause of the breakdown was a would-be technician who mixed up the power wires of the sensor/stabilizer for the vertical movement of the cutting head. Surprisingly, the circuit itself seems to have survived (it was, after all, protected by a diode in parallel). But everything burned out completely afterwards: amplifiers, some kind of logic, control of servos...

This is perhaps the simplest and safest option for protecting the load from power supply reverse polarity.
There is only one bad thing: the voltage drop across the diode. Depending on which diode is used, it can drop from about 0.2V (Schottky) and up to 0.7...1V - on conventional rectifier diodes with p-n junction. Such losses may be unacceptable in the case of a battery-powered or stabilized power supply. Also, at relatively high current consumption, power losses on the diode can be very undesirable.

With this type of protection there are no losses during normal operation.
Unfortunately, in the event of a polarity reversal, the power supply runs the risk of breaking. And if the power source turns out to be too strong, the diode will burn out first, and then the entire circuit it protects.
In my practice, I sometimes used this type of reverse polarity protection, especially when I was sure that the power source had overcurrent protection. However, one day I earned very clear prints on my burnt fingers when I touched the radiator of the voltage stabilizer, which was trying to fight against a thick Schottky diode.

p-channel MOSFET - a successful but expensive solution

This relatively simple solution has virtually no drawbacks: a negligible voltage/power drop across the pass-through device in normal operation, and no current in the event of a polarity reversal.
The only problem: where to get high-quality, inexpensive, high-power p-channel field-effect transistors with an insulated gate? If you know, I will be grateful for the information 😉
All other things being equal, a p-channel MOSFET in any parameter will always be approximately three times worse than its n-channel counterparts. Usually, both the price and something to choose from are worse: open-channel resistance, maximum current, input capacitance, etc. This phenomenon is explained by approximately three times less mobility of holes than electrons.

n-channel MOSFET - the best protection

It’s not at all difficult to get a powerful low-voltage n-channel CMOS transistor these days; sometimes you can even get them for free (more on that later;). So to ensure a negligible drop in open channel for any imaginable load current - a trifle.

N-channel MOSFET + 7.2...15V zener diode + resistor of a couple of tens of kilo-ohms = SAFETY

Just as in a circuit with a p-channel MOSFET, if the source is connected incorrectly, both the load and the unlucky source are out of danger.

The only “drawback” that a meticulous reader can notice in this protection scheme is that the protection is included in the so-called. "ground" wire.
This can indeed be inconvenient if a large earth star system is being built. But in this case, you just need to provide the same protection in the immediate vicinity of the power supply. If this option is not suitable, there will probably be ways to either provide such a complex system with unique power connectors with reliable mechanical keys, or install a “constant”, or at least “ground” without connectors.

Caution: static electricity!

We have all been warned many times that field-effect transistors are afraid of static discharges. This is true. Typically the gate can withstand 15...20 Volts. A little higher - and irreversible destruction of the insulator is inevitable. At the same time, there are cases when the field operator seems to still be working, but the parameters are worse, and the device can fail at any moment.
Fortunately (and unfortunately) powerful field-effect transistors have large capacitances between the gate and the rest of the crystal: from hundreds of picofarads to several nanofarads and more. Therefore, the discharge of the human body is often withstood without problems - the capacity is large enough so that the drained charge does not cause a dangerous increase in voltage. So when working with powerful field workers, it is often enough to observe minimal caution in terms of electrostatics and everything will be fine :)

I'm not alone

What I describe here is, without a doubt, a well-known practice. But if only those military industry developers had the habit of publishing their circuit designs on blogs...
Here's what I came across on the Internet:

> > I believe it is pretty well standard practice to use an N-channel
> > MOSFET in the return lead of military power supplies (28V input).
> > Drain to supply negative, source to the negative of the PSU and
> > the gate driven by a protected derivative of the positive supply.
1600 Hz, sitting on one board, is also protected:

Happy experiments!

Were you interested? Write me!

Ask, suggest: in the comments, or in a personal message. Thank you!

All the best!

Sergei Patrushin.

When designing industrial devices that are subject to increased reliability requirements, I have more than once encountered the problem of protecting the device from incorrect polarity of the power connection. Even experienced installers sometimes manage to confuse plus with minus. Probably, such problems are even more acute during the experiments of novice electronics engineers. In this article we will look at the simplest solutions to the problem - both traditional and rarely used protection methods.

The simplest solution that suggests itself right away is to connect a conventional semiconductor diode in series with the device.


Simple, cheap and cheerful, it would seem that what else is needed for happiness? However, this method has a very serious drawback - a large voltage drop across the open diode.


Here is a typical I-V characteristic for direct connection of a diode. At a current of 2 Amps, the voltage drop will be approximately 0.85 volts. In the case of low-voltage circuits of 5 volts and below, this is a very significant loss. For higher voltage ones, such a drop plays a lesser role, but there is another unpleasant factor. In chains with high current consumption, the diode will dissipate very significant power. So for the case shown in the top picture, we get:
0.85V x 2A = 1.7W.
The power dissipated by the diode is already too much for such a case and it will heat up noticeably!
However, if you are ready to part with a little more money, then you can use a Schottky diode, which has a lower drop voltage.


Here is a typical I-V characteristic for a Schottky diode. Let's calculate the power dissipation for this case.
0.55V x 2A = 1.1W
Already somewhat better. But what to do if your device consumes even more serious current?
Sometimes diodes are placed in parallel with the device in reverse connection, which should burn out if the supply voltage is mixed up and lead to a short circuit. In this case, your device will most likely suffer minimal damage, but the power supply may fail, not to mention the fact that the protective diode itself will have to be replaced, and along with it, the tracks on the board may be damaged. In short, this method is for extreme sports enthusiasts.
However, there is another slightly more expensive, but very simple and devoid of the disadvantages listed above, method of protection - using a field-effect transistor. Over the past 10 years, the parameters of these semiconductor devices improved sharply, but the price, on the contrary, fell sharply. Perhaps the fact that they are extremely rarely used to protect critical circuits from incorrect polarity of the power supply can be largely explained by the inertia of thinking. Consider the following diagram:


When power is applied, the voltage to the load passes through the protective diode. The drop on it is quite large - in our case, about a volt. However, as a result, a voltage exceeds the cutoff voltage is formed between the gate and source of the transistor and the transistor opens. The source-drain resistance decreases sharply and the current begins to flow not through the diode, but through the open transistor.


Let's move on to specifics. For example, for the FQP47З06 transistor, the typical channel resistance will be 0.026 Ohm! It is easy to calculate that the power dissipated by the transistor in our case will be only 25 milliwatts, and the voltage drop is close to zero!
When changing the polarity of the power source, no current will flow in the circuit. Among the shortcomings of the circuit, one can perhaps note that such transistors do not have a very high breakdown voltage between the gate and source, but by slightly complicating the circuit, it can be used to protect higher-voltage circuits.


I think it will not be difficult for readers to figure out for themselves how this scheme works.

After the publication of the article, the respected user Keroro in the comments provided a protection circuit based on a field-effect transistor, which is used in the iPhone 4. I hope he will not mind if I supplement my post with his find.

When designing industrial devices that are subject to increased reliability requirements, I have more than once encountered the problem of protecting the device from incorrect polarity of the power connection. Even experienced installers sometimes manage to confuse plus with minus. Probably, such problems are even more acute during the experiments of novice electronics engineers. In this article we will look at the simplest solutions to the problem - both traditional and rarely used protection methods.

The simplest solution that suggests itself right away is to connect a conventional semiconductor diode in series with the device.

Simple, cheap and cheerful, it would seem that what else is needed for happiness? However, this method has a very serious drawback - a large voltage drop across the open diode.

Here is a typical I-V characteristic for direct connection of a diode. At a current of 2 Amps, the voltage drop will be approximately 0.85 volts. In the case of low-voltage circuits of 5 volts and below, this is a very significant loss. For higher voltage ones, such a drop plays a lesser role, but there is another unpleasant factor. In circuits with high current consumption, the diode will dissipate very significant power. So for the case shown in the top picture, we get:

0.85V x 2A = 1.7W

The power dissipated by the diode is already too much for such a case and it will heat up noticeably!
However, if you are ready to part with a little more money, then you can use a Schottky diode, which has a lower drop voltage.

Here is a typical I-V characteristic for a Schottky diode. Let's calculate the power dissipation for this case.

0.55V x 2A = 1.1W

Already somewhat better. But what to do if your device consumes even more serious current?

Sometimes diodes are placed in parallel with the device in reverse connection, which should burn out if the supply voltage is mixed up and lead to a short circuit. In this case, your device will most likely suffer minimal damage, but the power supply may fail, not to mention the fact that the protective diode itself will have to be replaced, and along with it, the tracks on the board may be damaged. In short, this method is for extreme sports enthusiasts.

However, there is another slightly more expensive, but very simple and devoid of the disadvantages listed above, method of protection - using a field-effect transistor. Over the past 10 years, the parameters of these semiconductor devices have improved dramatically, but the price, on the contrary, has dropped significantly. Perhaps the fact that they are extremely rarely used to protect critical circuits from incorrect polarity of the power supply can be largely explained by the inertia of thinking. Consider the following diagram:

When power is applied, the voltage to the load passes through the protective diode. The drop on it is quite large - in our case, about a volt. However, as a result, a voltage exceeds the cutoff voltage is formed between the gate and source of the transistor and the transistor opens. The source-drain resistance decreases sharply and the current begins to flow not through the diode, but through the open transistor.

Let's move on to specifics. For example, for the FQP47З06 transistor, the typical channel resistance will be 0.026 Ohm! It is easy to calculate that the power dissipated by the transistor in our case will be only 25 milliwatts, and the voltage drop is close to zero!

When changing the polarity of the power source, no current will flow in the circuit. Among the shortcomings of the circuit, one can perhaps note that such transistors do not have a very high breakdown voltage between the gate and source, but by slightly complicating the circuit, it can be used to protect higher-voltage circuits.

I think it will not be difficult for readers to figure out for themselves how this scheme works.

After the publication of the article, the respected user Keroro in the comments provided a protection circuit based on a field-effect transistor, which is used in the iPhone 4. I hope he will not mind if I supplement my post with his find.

When designing industrial devices that are subject to increased reliability requirements, I have more than once encountered the problem of protecting the device from incorrect polarity of the power connection. Even experienced installers sometimes manage to confuse plus with minus. Probably, such problems are even more acute during the experiments of novice electronics engineers. In this article we will look at the simplest solutions to the problem - both traditional and rarely used protection methods.

The simplest solution that suggests itself right away is to connect a conventional semiconductor diode in series with the device.


Simple, cheap and cheerful, it would seem that what else is needed for happiness? However, this method has a very serious drawback - a large voltage drop across the open diode.


Here is a typical I-V characteristic for direct connection of a diode. At a current of 2 Amps, the voltage drop will be approximately 0.85 volts. In the case of low-voltage circuits of 5 volts and below, this is a very significant loss. For higher voltage ones, such a drop plays a lesser role, but there is another unpleasant factor. In circuits with high current consumption, the diode will dissipate very significant power. So for the case shown in the top picture, we get:
0.85V x 2A = 1.7W.
The power dissipated by the diode is already too much for such a case and it will heat up noticeably!
However, if you are ready to part with a little more money, then you can use a Schottky diode, which has a lower drop voltage.


Here is a typical I-V characteristic for a Schottky diode. Let's calculate the power dissipation for this case.
0.55V x 2A = 1.1W
Already somewhat better. But what to do if your device consumes even more serious current?
Sometimes diodes are placed in parallel with the device in reverse connection, which should burn out if the supply voltage is mixed up and lead to a short circuit. In this case, your device will most likely suffer minimal damage, but the power supply may fail, not to mention the fact that the protective diode itself will have to be replaced, and along with it, the tracks on the board may be damaged. In short, this method is for extreme sports enthusiasts.
However, there is another slightly more expensive, but very simple and devoid of the disadvantages listed above, method of protection - using a field-effect transistor. Over the past 10 years, the parameters of these semiconductor devices have improved dramatically, but the price, on the contrary, has dropped significantly. Perhaps the fact that they are extremely rarely used to protect critical circuits from incorrect polarity of the power supply can be largely explained by the inertia of thinking. Consider the following diagram:


When power is applied, the voltage to the load passes through the protective diode. The drop on it is quite large - in our case, about a volt. However, as a result, a voltage exceeds the cutoff voltage is formed between the gate and source of the transistor and the transistor opens. The source-drain resistance decreases sharply and the current begins to flow not through the diode, but through the open transistor.


Let's move on to specifics. For example, for the FQP47З06 transistor, the typical channel resistance will be 0.026 Ohm! It is easy to calculate that the power dissipated by the transistor in our case will be only 25 milliwatts, and the voltage drop is close to zero!
When changing the polarity of the power source, no current will flow in the circuit. Among the shortcomings of the circuit, one can perhaps note that such transistors do not have a very high breakdown voltage between the gate and source, but by slightly complicating the circuit, it can be used to protect higher-voltage circuits.


I think it will not be difficult for readers to figure out for themselves how this scheme works.

After the publication of the article, the respected user Keroro in the comments provided a protection circuit based on a field-effect transistor, which is used in the iPhone 4. I hope he will not mind if I supplement my post with his find.