# Protect devices from improper power polarity

• Tutorial

When designing industrial devices with high reliability requirements, I have repeatedly faced the problem of protecting the device from the wrong polarity of the power supply. Even experienced installers sometimes manage to confuse plus with minus. Probably even more acute such problems are during the experiments of novice electronics engineers. In this article, we will consider the simplest solutions to the problem - both traditional and rarely used in practice protection methods.

The simplest solution that begs right away is the inclusion of 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 on the open diode.

Here is a typical CVC for direct connection of a diode. At a current of 2 amperes, 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, such a drop plays a lesser role, but there is another unpleasant factor. In circuits with high current consumption, a very significant power will be dissipated on the diode. 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 significantly warm up!
However, if you are ready to part with a little more money, then you can use the Schottky diode, which has a lower voltage drop.

Here is a typical CVC for a Schottky diode. We 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 to the device in the reverse connection, which should burn out if the power supply is mixed up and lead to a short circuit. In this case, your device will most likely suffer a minimum of damage, but the power source may fail, not to mention 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.
However, there is another somewhat more expensive, but very simple and devoid of the above disadvantages 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 has fallen dramatically. Perhaps the fact that they are rarely used to protect critical circuits from the wrong polarity of power supply can be largely explained by the inertia of thinking. Consider the following scheme:

When power is applied, voltage to the load passes through a protective diode. The fall on it is large enough - in our case, about a volt. However, as a result, a voltage exceeding the cutoff voltage is generated between the gate and the 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 an open transistor.

Let's move on to the specifics. For example, for the FQP47Z06 transistor, the typical channel resistance will be 0.026 Ohms! It is easy to calculate that the power dissipated by the transistor in this case will be only 25 milliwatts, and the voltage drop will be close to zero!
When changing the polarity of the power source, the current in the circuit will not flow. Of the drawbacks of the circuit, perhaps only the fact that such transistors have not too much breakdown voltage between the gate and the source, but having slightly complicated the circuit, it can be used to protect higher-voltage circuits.

I think it’s not difficult for readers to figure out how this scheme works.

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