IP with a smooth change in polarity. Power supply with variable polarity - power supplies (laboratory) - power supplies Power supply with variable polarity
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The figure shows the circuit of a two-channel audio power amplifier based on the LA4450 IC. The output power of the amplifier at a supply voltage of 26.4V (recommended) is 12W (per channel) into an 8 Ohm load and 20W (per channel) into a 4 Ohm load. The LA4450 IC has thermal protection, overvoltage and impulse noise protection. Main characteristics Maximum voltage...
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The figure shows a circuit of a switching power supply with an output voltage of 12V and a power of 15W, based on the integrated AC/DC converter TOP201YAI. This circuit uses a pulse transformer with an additional winding 4-5 and a rectifier at D3 to power the optocoupler transistor, which provides feedback control. The switching power supply uses a transformer to...
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This device is designed to automatically maintain voltage on the soldering iron heater. As is known, high-quality soldering with POS-61 solder is possible only in a narrow temperature range. As is known, changing the supply voltage from 180 to 250 V leads to a change in the temperature of the soldering iron tip by 38%, this device will reduce this change to 4%. Device...
- 21.09.2014
I use this device to protect against current overloads of electrical appliances operating from a 220V network. The device has relay load control and can therefore be used in conjunction with any type of electronic equipment. The circuit consists of a current sensor (optocoupler U1) and a switch on VT1 whose load is a relay. When current passes through R1 to ...
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, a respected user 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. 
Often, bipolar power supplies have a constant output voltage. The desire to construct an regulated one from an unregulated bipolar power supply at low cost usually does not lead to anything good, since this leads to an imbalance of the output voltages (in amplitude) of opposite polarities. To implement this option, it is necessary to significantly “weight” the scheme.
There is also an option when an electronic unit is added to a unipolar power supply, which generates a negative voltage from a positive one. But this version of a bipolar source also has an imbalance of opposite voltages and does not allow use in power supplies with continuously variable output voltage.
This article provides another original version bipolar power from unipolar having the right to exist. This is a prefix - built on an operational amplifier LM358, to a conventional unipolar power supply, which allows you to obtain a full bipolar output voltage.
Any power supply with a voltage of 7...30 volts can act as an input voltage source, and the output voltage will be 3...14.5 volts.
During operation, this divider does not distort the output parameters of a unipolar power supply. This divider attachment can withstand a load of up to 10 amperes without distorting the voltage, both in the positive and negative channels. For example, if a load with a current consumption of 9 amperes is connected in the negative circuit of a bipolar power source, and 0.2 amperes in the positive circuit, then the difference between the negative and positive voltage will be less than 0.01 volts.
It should be noted that only the presence of a regulator in a unipolar power supply can ensure a change in the output in a bipolar one, otherwise adjustment will be impossible.
Description of the attachment-divider of unipolar voltage into bipolar
(DA1) measures the potential difference between the common wire and the midpoint of the voltage divider assembled at resistances R1, R2, R3. When this difference changes, the LM358 op-amp leads to stabilization of the output voltage, decreasing it or increasing it.
When input voltage is applied to the circuit, capacitors C1 and C2 are charged at half the supply voltage. With a balanced load, these voltages will be the output voltage of a bipolar power supply.
Now let's analyze the situation when an unbalanced load is connected to the output of a bipolar power supply, for example, the load resistance in the positive circuit is significantly lower than the load resistance connected to the negative circuit.
Since a load is connected in parallel to capacitor C1 (diode VD1 and a small load resistance), capacitor C2 will be charged both through capacitor C1 and through the above-designated circuit (diode VD1 and a small load resistance).
For this reason, capacitor C2 will be charged with a higher voltage than capacitor C1, and this will lead to the fact that the negative voltage will be higher than the positive one. On the common wire, the voltage will increase relative to the midpoint of the voltage divider R1, R2, R3, where the voltage is 50% of the input.
This contributes to the emergence of a negative voltage at the output of the op-amp LM358 relative to the common wire. As a result, transistors VT2 and VT4 open and, similarly to the electrical circuit “diode VD1, small load resistance” in the positive electrical circuit, bypasses capacitance C2 in the negative circuit, which leads to a balance of the currents of both circuits (positive and negative)
Likewise, transistors VT1, VT3 will open if there is a load imbalance towards negative voltage.