Re: Time to saturate a coil.
You are first asking a question about the behaviour of coils when they saturate, but then you give an unrelated examples with the solenoid opening time.
I will try to answer them both.
When a coil saturates it simply behaves like a resistor with the resistance equal to the DC resistance of the coil. So inductance is "gone", you are left with the parasitic resistance.
The time "to saturate" a coil can be calculated from Faraday's law:
ΔB=U*t/(N*Ae), where ΔB is the flux change, U the applied voltage, N the number of turns and Ae the effective area of the core. For ΔB large enough such that the peak flux density, Bpk is exceeded, the core saturates. Call this flux excursion ΔBmax=Bpk-Br, where Br is the remanent flux density. Bpk and Br are material-dependent. So the time becomes:
tsat=ΔBmax*N*Ae/U
The question about the solenoid can be answered by doing some calculations, but not so easily. You have to understand that the opening time is a function of the current passing through the solenoid, but also a mechanical time constant given by the inertia of moving mass and constant of the spring. This part is the hardest to estimate and I am not going to touch it.
Back to the solenoid. There is a minimum current for the solenoid to operate. So until the current in the solenoid reaches this threshold value, nothing will move. The time necessary for the current to reach this threshold can be calculated from the equation: i=Ip*(1-e^(-t/τ)), where Ip is the peak current in the solenoid, that is Ip=U/R, and τ=L/R. U is the voltage applied to the solenoid, R its resistance and L its inductance.
From this is follows that the time required for the current to reach the threshold is:
t=-L/R*loge(1-It/Ip), where It is the threshold current.
So for 12V operation Ip=12/R, for 8V Ip=8/R. The rest remains the same. Using these values you can calculate the time.
Add to this the time required for the mechanical part and you get the opening time.