3
\$\begingroup\$

I need to momentarily activate an approx. 50mA @ 24V DC load, using a 90V, 30hz AC signal line. Whatever method is used, it needs to have the smallest current draw from the signal line possible.

I am not sure if it is possible to turn on a transistor or solid state relay with AC, and I cannot seem to find much on the web.

\$\endgroup\$
10
  • 3
    \$\begingroup\$ how long is momentarily? \$\endgroup\$
    – Gustavo Litovsky
    Commented Jan 3, 2013 at 17:00
  • 1
    \$\begingroup\$ Is this a telephone ringing indicator? There are many ICs designed to do precisely this. \$\endgroup\$
    – Dave Tweed
    Commented Jan 3, 2013 at 17:09
  • 1
    \$\begingroup\$ Sorry for the letdown, Gustavo, I realize this isn't terribly interesting as an academic exercise. However, as mundane as it is I would still love any help I can get. As to the other replies, yes it is a ringing indicator of sorts. I work in a manufacturing plant and we have a central telephone system, with 3 or more universal telephone alarms in the back half of the building. The reason for my question is that every time the phone rings, the load of the 3 alarms is too great for the central system and it overloads, thus I am looking to remove the load that line sees. @Dave, which IC's? \$\endgroup\$
    – Scott
    Commented Jan 3, 2013 at 17:50
  • 1
    \$\begingroup\$ It's been a long time since I needed a standalone ring detector; usually I'm working with SLIC chips that have it built in. One example is the TI TCM1520A. \$\endgroup\$
    – Dave Tweed
    Commented Jan 3, 2013 at 19:03
  • 1
    \$\begingroup\$ @BrianCarlton: An REN of 1 represents a 7K load, or about 13mA @ 90V. The OP probably wants to stay below REN=0.3, or about 4mA. \$\endgroup\$
    – Dave Tweed
    Commented Jan 3, 2013 at 19:09

1 Answer 1

Reset to default
1
\$\begingroup\$

I see from the comments so far that this circuit requirement is for a telephone ring detector circuit. There is a plethora of information on the web for telephone ring detector circuits. I suggest that you use a circuit that incorporates an opto-coupler to isolate the telephone line from what ever circuit that is used to drive the down wind ringers. The reason for this is that if you directly couple a phone line to circuits that are locally powered you can end up getting a lot of line frequency feedback into the phone line which is not a good thing. Here is one circuit that looks like it could be a good starting place for you.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ This is a bad answer for several reasons. First, it loads the line way more than necessary. Second, the output specification was 24VDC @ 50mA. Finally, this circuit turns the output OFF when the phone rings! \$\endgroup\$
    – Dave Tweed
    Commented Jan 3, 2013 at 18:50
  • \$\begingroup\$ I take back the comment about excessive load -- it's probably OK in that respect, but the other two comments stand. \$\endgroup\$
    – Dave Tweed
    Commented Jan 3, 2013 at 19:09
  • \$\begingroup\$ I will give this a try if I cannot locate any single chip solutions. Just don't want this getting overly complicated :) \$\endgroup\$
    – Scott
    Commented Jan 3, 2013 at 20:11
  • \$\begingroup\$ @Tweed - I presented this circuit as a starting point for Scott and stated so in my amswer. Your comment about the state of the relay only applies to what is happening to the current in the relay coil. The output can be connected to N.O. or to N.C. contacts to achieve the desired switching behavior. Your flippant comment about the load on the relay is just that. Since presented as a starting point is should be obvious that the relay can be used to drive almost any load. Nuff said. \$\endgroup\$
    – Michael Karas
    Commented Jan 4, 2013 at 2:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.