It sounds like a misunderstanding: if a solenoid is required for continuous operation, all one has to do is order one that is so rated. What's the application?

The application is in a fire suppression system. The selenoid is supplied 120volts from a panel once a fire is detected. The codes require supervision of the circuit to ensure that continuity exists in the electrical line to the selenoid. We typically do this with a 24 volt selenoid but we have a growing number of customers that want to use 120 volt. This has been a very difficult problem to solve - I am a Chem Eng by trade and have EE's on my staff and so far nobody has an answer.

What's wrong with this, then?

• Wire a relay in series with it, and look at the normally open contact of that relay. If the contact opens then the continuity to the solenoid has been broken, so use that feature to operate an alarm.

Craig

How do you ensure continuity of the 24 volt ckt or is this just required for a 120 volt ckt?

It is an 8 pole unit with the selenoid coil between poles 7 & 8. There is a switching contact between poles 6 & 7 that is NC. There is an in-line supervisory device between poles 4 & 5 (I think that is an End of line resistor). Accessory power (battery back-up) is provided in poles 1-3. The input power is 750 mA at 24 VDC. This is really taxing my electrical know-how and is the best explanation I can give.

Hi Craig,

Reading your description I am wondering if you are really talking about a Double Pole, Double Throw relay that looks like this:

This relay has 8 points of contact with the coil between terminals 7 & 8. And it seems to match your description so if this looks right, then we will be on the same page.

So, until we agree that we are talking about the same thing, perhaps I can address another point. And that is the assumption that the fire is not causing the 120 VAC power to be interrupted (or maybe it could ... thereby needing a supervisory circuit). If it must be powered on 120 VAC then it seems that you need an uninterpretable power supply (UPS) like the type sold for personal computers or something larger.

The UPS may also need to provide power to the alarm devices, which when added up may draw more than your 750 mA in the 24 VDC circuit. Perhaps you should check to see if your input power includes power to alarm devices when activated. And it would be helpful if we knew about the voltage at which those devices were intended to operate (i.e. 24 VDC or 120 VAC)

Thanks for the response. The eight sets of contacts are installed in a small cabinet. The cabinet receives input from a control panel. The control panel receives a signal from a detection circuit (usually a spot thermal detector) or linear wire. The panel then sends an electrical signal to the releasing circuit (or the eight sets of contacts). As indicated in a previous message the eight contacts are set-up as follows:

24Vdc supervisory

Terminal 1 - Open until power is received and then closed by switching contact from T2 - this powers up auxilliary alarming devices and fuel shutoff valve, blower motors, door closers, etc.

Terminal 2 - NC switching contact with Terminal 3

Terminal 3 - NC switching contact with Terminal 2

Terminal 4 - + from the panel

Terminal 5 - Connected to a switching contact with terminal 6, & if supervised a jumper with in-line resistor is installed between terminals 4 & 5

Terminal 6 - connected to terminal 5 with switching contact

Terminal 7 - jumpered to terminal 6 and connected to terminal 8 with a Voltage Transient suppressor installed, terminal 7 also connected to the solenoid

Terminal 8 - connected to terminal 7 as indicated above and the solenoid

Note - this is how the 24Vdc is supervised. The newest customer requirement is a little different in that we need to supervise a 120VAC selenoid. It is my understanding that the 120 V selenoid cannot take a trickle voltage as this will burn up the coil. So far nobody has a solution for this problem.

I've never heard that a solenoid could not take a trickle current. I agree that you cannot mix 120 VAC and 24 VDC circuitry casually.

You may try checking the following information;

http://en.wikipedia.org/wiki/Fire_alarm_control_panel

I would suggest that you read it thoroughly. You solution is not cheap because you said, "The selenoid is supplied 120volts from a panel once a fire is detected. The codes require supervision of the circuit to ensure that continuity exists in the electrical line to the selenoid."

From what you have just said, the circuit/coil is de-energised until a fire is detected. Consequently, the coil/relay is NOT continuously energised (unless you have continous fires).

This problem is generally overcome in Australia by using a "failsafe" system for relays & also by using DC coils with such relays.

Here's a nice simple approach. From your power source put an LED with a current limiting resistor in series with solenoid. Scale the resistor for 5-10mA of current through the LED. Don't forget to include the impedance of the solenoid coil in you calculation. Also an additional diode in parallel but reversed to the LED. The 5-10mA of current going through the coil will likely not be sufficient to activate the solenoid. Most of your power will be dissipated by the current limiting resistor, so size it appropriately. To activate the solenoid with this circuit in place just bypass the resistors and diodes with a contact. If you need to monitor continuity more than just producing a light, either diode could be the diode of an optocoupler.

What is the purpose of the additional diode in parallel but reversed to the LED? I feel like I'm over barbed wire right now!

The reverse voltage that an LED can take is not near 120V. So by adding the other diode, the reverse current has a path not through the LED but through the other diode.

To do as RedFred suggested you will need a 2 watt, 12K ohm resistor and LED or optocoupler device between the control panel power supply and the coil (Terminal #7)of your solenoid. This assumes that Terminal 8 is the return path, or neutral of the AC power line and that some path exists whether it goes to the control panel or not.

I think you can neglect the resistance of the coil since it is likely to be much smaller than the 12K ohm resistor. When the control panel activates its emergency circuit, both ends of your supervisory circuit will be at the same potential and thereby no current will flow through the supervisory LED or optocoupler at that time. Does this make sense to you?

One caveat, the optocoupler will actually be turning ON and OFF sixty times a second. While this is not a problem for the optocoupler, your sensing circuitry may or may not appreciate this. Depending on the input of your monitoring circuitry, you may want to add a capacitor to filter this out.