then you are not following the article...

look at the current equation:


at initial time t=0, exponential part becomes e^0 which is 1.
then we obtain that initial current is iL(0)=(E/R)*(1-1)=(E/R)*0 = 0. and since current is zero, impedance at that point in time is infinite, not R

at some large t value exponential part becomes zero (negative exponent),
and iL(large enough)=(E/R)*(1-0)=(E/R)*1=E/R so at this time, impedance is really just R

tjhe same is also shown by graphs... at t-0, current is zero (infinite impedance).

Yes it is correct my solenoids work at 12V, but I also have an MCU that works at 3.3VDC and my doubt is that the solenoids when the switching happens, the input capacitor may discharge and create issues for the MCU.

i think i read that solenoids draw some 0.6A or so. and there are several of them (5-6?). so we are likely talking about 12VDC source that needs to be capable of some 4A or so of current. also even if people are racing through those doors, i would say 5 seconds is the minimum time you may need to bridge, possibly 10x that.
voltage regulators have some overhead, 1-3V. this is called voltage drop. assuming generous 3V for voltage drop, it means 5V regulator input should never see less than 8V. if you get voltage drop from 12V to 8V under load for ANY amount of time - problem is the PSU. adding capacitor that would prevent brownouts that last for 50 seconds or so for loads that draw some 4A would be impractical and power up time would be ridiculous,

capacitors placed around regulators are used to ensure stability / prevent oscillations of the regulator.

MCUs tend to draw very little current, so this circuit is much easier to back by some capacitor.

 

I think by magnetic flux delay we mean the behavior of the inductor described here:

Inductor Charging and Discharging in RL Circuit Analysis Equations

I see no initial current spike in that analysis.

 

Especially in a circuit where every conductor resistance forms an undesired coupling. Switcher power supplies are a prime example, but certainly not the only application where unintended coupling due to "fairly small" wiring impedance caused problems. Sometimes that small bit of disturbance can be ignored, but some times it can't be ignored.


And it may be that I was thinking of voltage, but said current. It is not that I am 100% infallible, you know. I never claimed to be, either. First error this month, so I am doing OK.

 

I was thinking about various configurations including the possibility of isolating control from power. I thought of these four possibilities:

1. 1) Power the microprocessor with an isolated DC DC converter and opto-isolate the logic output going to the MOSFET with an opto-isolator (but this option has the cons of having a very high quiescent current because of the possible built-in flyback converter and very expensive).
2. 2) Connect the pin of the microprocessor to the input of the optoisolator and the output to the MOSFET albeit sharing grounds.
3. 3) Connect the pin of the microprocessor to the input of the optoisolator and the output to the MOSFET albeit sharing the grounds, but with the possibility of splitting the 12V and GND of the power supply feeding the solenoids with the 12V power supply feeding the logic. Is this a good idea?
4. 4) Do not use any kind of isolation and connect directly via resistors to the gate of the MOSFET.

Can anyone with experience help me?