Contact US

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

L/R ratio for switch contacts

L/R ratio for switch contacts

L/R ratio for switch contacts

I am reviewing vendor data on the inductive and resistive ratings of pressure switch contacts for dc application.  The table gives an L/R ratio of .026, time to reach 63% of current peak value.  How is the L/R ratio used in applying such contacts?

RE: L/R ratio for switch contacts

Here are some thoughts (nothing concrete).

Your switch will (presumably) interrupt a load/supply circuit which can be modeled as a series inductor and resistance.

Higher inductance gives worse duty during switch interuption due to higher inductive kick (v=Ldi/dt).

Higher resistance within your circuit tends to dampen the inductive kick somewhat through energy dissipation.  Low resistance within your circuit is the worst case for circuit interuption since the circuit does not dampen it's own inductive kick at all.

Combine the above effects to say that the worst case for switch interuption is high L/R.

The units of L/R are seconds.  If you were to apply a step voltage V to your LR circuit then the current would rise as I=V/R(1-exp(-t/[L/R]).  At time t=L/R the current will be at I(t=L/R)=V/R(1-exp(-1])=v/r*68%=68% of its initial value.  Thus the rise time is simply a measure of the L/R ratio of your circuit.

RE: L/R ratio for switch contacts

Suggestion: Beside the dynamical definition of self-inductance in the above posting, i.e. v = L di/dt there is also the energetical definition of the self-inductance L (in Henries), i.e. W = (1/2)(L x I**2) where
W (in Joules or Wattseconds) is the energy of the magnetic field with inductance L (in Henries) and current I (in Amperes). Therefore, any mitigation of arcs across the DC contacts via diodes or Metal Oxide Varistors (MOVs) must consider the energy dissipation and be rated for it.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close