Connect with us

Help - Need to slow down relay pickup

Discussion in 'Electronic Basics' started by Lemon Joke Kid, Sep 23, 2003.

Scroll to continue with content
  1. The circuit is something like this:

    + ----||--------(P)------(S)----- -


    This is an electromechanical annunciator. There are actually 40 P
    coils tied to the S coil. It operates as follows: Field contacts
    close, energizing their corresponding P coil and the S coil. The P
    coil picks up, dropping a flag to indicate which field contact picked
    up.

    The S coil drops a flag giving local indication of station trouble.
    It also closes a set of contacts to give remote station trouble
    indication.

    Under normal conditions it works fine, because the field contacts
    generally close and stay closed for most alarms. The problem is when
    we have very brief, intermittent contact closures either due to a
    mechanical problem with the field device, a device hanging very close
    to set point, or possibly grounded wiring between the annunciator and
    the field device.

    The problem is the flags on the P coils don't seem to drop as fast as
    the flags on the S coil. The P coils need to be energized longer than
    the S coil to get the flag to drop. The result is a flag down on the
    S coil indicating station trouble, but no P coil flag to indicate
    which point is giving the alarm.

    A simple fix is to give a slight delay to the S coil to ensure that
    any contact closure that is of long enough duration to get S to drop
    will also be long enough to allow P to drop. I thought I could
    accomplish this with a simple RC network across the S coil.

    I experimented with a few different capacitor values and a resistance
    decade box. I did get a slight amount of delay, but much less than I
    expected. Changing values didn't make much difference. I only need
    probably 1/4 sec. delay, 1/2 sec. at the very most and I need to do
    this as simply as possible.

    Any thoughts?
     
  2. JeffM

    JeffM Guest

    Bigger capacitor OR a darlington driver with a capacitor on the input
    (effect = beta x capacitance).
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-