Whats the time needed to saturate a coil?

[future]

Time to saturate a coil.

Hi,

I am looking for theories about coils and how they behave when it saturates.

Ex: A selenoid takes 10mS to open when 12v is applied, how long it will take to open if we apply 8v?

Thank you.

 

[flatulent]

Re: Time to saturate a coil.

You have a nonlinear system. As the core of the solenoid changes position, the inductance changes. Your best bet would be to measure a physical sample of the solenoid or consult the data sheet.

 

[future]

Re: Time to saturate a coil.

And what about if it was a linear system?

Like a relay or something with a fixed core.

Just thinking about the relation between different drive voltages.

 

[Borber]

Re: Time to saturate a coil.

Ideal inductor has no ohmic resistance, it is 0. When you connect voltage accross it the current will start to flow and will increase to infinity. Real inductor has allso resistance which as consequence limits maximum current. Current in such inductor raises exponentionaly to I=V/Rcoil. Relay coils have inductance and series resistance. Current at which they activate depends on number of turns because some certain magnetics force has to be accheived for this purpouse. This force depends on mechanical construction of relay. When current rises the product I*number of turns reaches the value where relay activates. Voltage connected on relay must be high enough so the current can be high enough. Current rise slope depends on resistance of coil and connected voltage so it produces activation delay. When you connect smaller voltage on relay you can expect greater delay.

 

[VVV]

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.