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If the blocking voltage is low enough to allow you to use Schottky diodes, use them. If you need higher blocking voltage, i.e. silicon diodes, you need to add resistors in each branch to avoid thermal …

Yes, that's the basic effect of the ESR (equivalent series resistance) and the capacitance of the cell. If you use any free circuit simulator and play around with reasonable values of R and C, you wi …

C10? A capacitor. Possibly X-type. Most likely part of an EMC filter. We will need better images of where it sits in the circuits and remnants of marking on the component itself to tell for sure.

240 < 250 so no. If it’s a one-off supervised test, I wouldn’t worry, but it it goes into a product you intend to sell, it’s at the very least a liability and possibly violation of CE, UL or similar.

Instantaneous increase in current with lower input voltage is only possible with power electronics with regulation. For example a TV will adjust the duty cycle of the SMPS to keep all DC rail voltage …

Here is a really quick and dirty simulation for you. Several downsides: no filters what so ever and would not transfer the same amount of power. Just the top switch waveform difference.

If you can accept inverted logic into your PLC, this would do it: simulate this circuit – Schematic created using CircuitLab If you can't have inverted logic, use two: simulate this circuit You …

That's difficult to answer in general since it comes down to how it's used, mainly soft versus hard switching but for a simplified hard switching topology, you want to limit the switching losses to so …

With SOT223 1117 having 110 K/W junction to ambient thermal resistance and you are pushing 1.6 W of dissipation, you are hitting the thermal protection which is killing your output. Your solution wi …

You need a minimum load for them to regulate. Add a resistor to give 100 uA of load and the situation should improve.

It will adjust automatically, hence voltage regulator. If you take a look under the hood, there is a feedback mechanism which measures the output voltage, compares with an (internal) voltage reference …

A starting point would be 20 % magnetizing current of your full rated load, in this case 2 VA. At 230 V, that’s 8.6 mA. Z = 2*Pi*L*f. Solve for L. If the transformer would have been 2 kW, about 5 % wo …

The spark is caused by your battery charging the output capacitor on your power supply, which is zero volt when you start and have very low impedance so the current consumption from the battery is hug …

Yes, it’s possible. You need to provide base drive above your supply voltage in order to not make an emitter follower and at this point you might as well use a MOSFET instead. If your supply voltage i …

No, you will seriously unbalance them when you charge and discharge your pack. Find another solution, using either all three in series with BMS or all three in parallel.