I continue to publish publications in the Q&A format for electrical. Real questions and professional answers. " You can familiarize yourself with part 1 and part 2 of the answers to the questions by clicking on the links that are at the bottom of this publication. In this publication you will find 4 questions and answers to them.
Question N1. Alexey asked. The question itself is literally:
Hello. Previously, I had a three-phase hob, but the recently purchased Gorenje ECT 650 CP hob turned out to be powered by two phases. What to do now, go to hand over the hob back to the store, or is there a way to connect?
I gave Alexey the following answer to this question:
Good day. It is unnecessary to change the hob in the store. It can be powered from both a single-phase electrical circuit and a three-phase one (in the latter case, two out of three phase conductors are involved). The connection options should be described in detail in the user manual and also shown on the nameplate near the terminal block.
An example of a hob label is shown below.
As you can see, in your case it is enough to connect the neutral conductor to the "N" terminal, the protective conductor to the "PE", and any two of the three phase conductors to "L1" and "L2".
Question N2. Fedor asked. I quote the question itself:
The advantages of a cordless screwdriver are negated when it is necessary to work and the tool is charging. Tell me, is it possible to create a power supply so that the screwdriver is powered from a 220 volt network?
I gave the following answer:
Hello. There are quite a few such schemes on thematic sites, we will give one of them as an example.
Legend:
- U1 - Toshiba 60W electronic transformer. It is usually used for 12 volt halogen lamps.
- T1 is a transformer that acts as an additional galvanic isolation. The output voltage can be adjusted with the primary winding, which makes it possible to connect various models of screwdrivers. A ferrite ring НМ2000 is used as a core, with 28x16x9.
The windings must be placed opposite each other. The primary winding is wound with a double wire Ø 0.8 mm (16 turns). Secondary, 12 turns with the same wire. - VD1 - Schottky diode SBL2040CT, in this case it plays the role of a full-wave rectifier.
- R1 is a 1.5 kΩ resistor.
- HL1 - LED indicator AL307BM.
- R2-R7 - 100 Ohm resistors. Needed to start an electronic transformer.
- C1 - electrolytic capacitor 470 μF x 25 V.
- C2 - 0.1 uF capacitor.
Question N3. Fedor asked. I quote the question itself:
Good day. Recently I disassembled the hood in order to clean it from dust. The hood has been operating for more than five years, the warranty has expired long ago. I switched it on, I hear the fans are working, I thought, then everything is fine. But my wife noticed that one of the fans works for blowing, that is, in reverse mode. Tell me how to fix it?
I gave the following answer:
Hello. The problem you described indicates that one of the engines (the one that works on injection) is not connected correctly. You should disassemble the hood and reverse the polarity of the power supply to the motor running in freelance mode. Do not confuse it with what is functioning normally, otherwise both will work in reverse.
Question N4. The question was asked by Stepan. The question itself is as follows:
Hello! Recently the battery charging for the radio at 1A failed. An ownerless phone charger is available with the same voltage, but with a current of 2 A. I plan to re-solder the cables on the chargers, but I'm afraid that the higher current may damage the battery on the radio. What can you suggest on this issue?
Stepan received the following answer:
Good day! From time to time you can hear the statement that an increase in the charge current leads to a rapid deterioration of the battery. This is, to put it mildly, wrong. Almost all devices, the batteries of which are charged by external devices, have a controller for limiting the maximum current. In your case, the difference between 1 and 2 amperes is insignificant, the maximum in which the difference can be noticeable is an increase in the rate of battery charging.
Previous answers to questions:
- Part 1
- Part 2