# Time of current rise in a coil

Discussion in 'General Electronics Discussion' started by seymourfroggs, Jun 2, 2016.

1. ### seymourfroggs

13
0
Sep 29, 2014
Hello

In a circuit where a short burst of current through a coil opens a valve, I need to estimate the time to reach a certain current level.

The point is that there are two resistances, but I think only that of the coil matters. This is key to my question.

We simply have 15 v applied across the circuit. There is 5.6Ω in series with the coil, whose resistance is 0.15Ω. The coil's inductance is 0.15 mH.

So tau is L/R and my question is is tau 10^-3 from 0.15/0.0015 ? Or is it (0.15 + 5.6)/0.0015?

I need to find the time to reach 5 A in the coil but I can do it if I know the value of tau.

Thank you!

Seymour

2. ### duke37

5,364
772
Jan 9, 2011
If I interpet what you have correctly, the current will only rise to 2.6A.
Any resistance in series with the coil needs to be added to the coil resistance.
To get 5A the total resistance will need to be less than 3Ω.

seymourfroggs likes this.
3. ### Harald KappModeratorModerator

11,997
2,812
Nov 17, 2011
tau = L/R -> I(t) = I0*(1-exp(-t/tau))
At t=tau the current has reached 63% of the max value I0.
At t=3+tau the current has reached 95% of the max value I0.
R=Rseries+Rcoil, of course.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
2,841
Jan 21, 2010
Do you mean 3 * tau?

5. ### Harald KappModeratorModerator

11,997
2,812
Nov 17, 2011
Of course, thanks for spotting the typo.

Harald

6. ### seymourfroggs

13
0
Sep 29, 2014
Thanks, all.
I made a mistake - of course with that R the current would only be 2.6 A. The separate R should have been 2.8 Ω. (Two 5.6 Ω in parallel for a particular reason).
But you have made it clear that the downstream R is included as I expected but wasn't sure.

7. ### seymourfroggs

13
0
Sep 29, 2014
Just for completeness, (I must have been nodding off!),
The tau value (s) would be L/R which is 0.00015/2.95 which is
0.05 ms,
so we get to 5 A in ~150μs - 200μs.
This fits with what I observe and is satisfactory.
Thanks, guys.