testing for 24vac

I am currently writing software for an X10 capable sprinkler system. The system has the ability to control 8 zones. Each zone has a corresponding 24 vac output that connects to the manifold. When the voltage is present the zone turns on.

I would like to be able to test that he sprinkler is turning on the zones that I request. I was thinking that I could use 8 diodes representing the 8 outputs. When voltage is detected on one of the outputs the corresponding diode lights up.

Would someone please explain how I could accomplish this. I am open to other ideas as well that may be simpler (or maybe more elegant ?segmented LED?)

-george economos

BTW - Please keep in mind that I am a software engineer with limited electronics experience ;-)

Thanks for the reply. It seems what you have suggested would be a way for me to place your device between the controller and the manifold. Am I correct? (I hope otherwise I am more lost than I thought)

My device will only be used during the software development phase. I don't actually have to pass the current on to the manifold. ( In this way I don't actually have to look out the window and see if the sprinklers are turning on and off during testing)

I was really hoping to supply each of the 8 24vac-outputs to it's own LED? I'm not just sure how. For instance can I connect 24vac directly to a LED? To you wire the commons together?

This is a real NEWBIE question!!

thanks again for your patience, george economos

You can connect LEDs to your 24 vac outputs with series diodes and series current limiting resistors and they will light up when the output is on.

So I would connect them in this order?

That should work, normally I think I would probably put the resistor in after the 24vac and before the first diode (there is nothing wrong with two diodes being directly in series. I don't have a "magic" reason for doing so, but I like to have a current limiting resistor between my source and my load for safety.

The reason I suggested using an opto isolator (instead of a plain diode) is that you have the potential issue, pardon the pun, where you can't be sure if there is a voltage difference between the sprinker power and your power system. Using an opto isolator will give you a significant amount of electrical isolation, letting the two circuits operate independantly.

The other pieces of information you need to know, is how to read a datasheet and how to apply ohms law. The diode (opto) and your LED will all have a voltage drop assocated with them typically on the order of 1 to 2 volts, at a given nominal operating current. As a rule of thumb for diodes and LEDs, I use 10ma, but you will need to see what the spec for your chosen parts call for. Once you have these figures, you subtract the voltage drops from 24 (from the 24Vac) and then divide this voltage by the desired current., which then gives you the nominal current limiting resistor value. For example, if each diode has a 2v drop, and you want 10ma you have 24 - 4 = 20 / 10ma = 2k ohm resistor.

Okay I am almost ready! One more question:

Does it matter if I use a 1/4W or 1/2 W resisitor?

thanks, george economos

The resistor will be dropping 22V. Current flow will be whatever you choose, e.g., 10mA. Since the diode is blocking half of the AC cycle, power dissipation is halved.

So, power dissipation will be P = EI/2 = 22V * 10mA / 2 = 0.11W. You'll be fine with either a 1/4W or 1/2W resistor, assuming you choose 10mA of current (that is, a 2.2k resistor).

Note that if you hook it up the way that Jasen suggests, with the diode in parallel with the LED but in the opposite direction, then current is flowing in both half-waves, so the power dissipation in the resistor doubles compared to the above calculation. The LED is still only turned on half the time, though, so its brightness does not change. For that reason, I think that Noway2's suggested circuit is preferable to Jasen's.

// --------[2.2K]-+-->|--+ LED | | 24VAC +--|

You don't need optoisolators. At each output, I'm guessing you have two screw terminals or something, that you connect the two wires that go to the solenoid valve, right?

If so, then take one ea. 2.2K resistor, one ea. LED, and one ea.

where the two 'o's go in parallel with the solenoid coil. If all of the solenoids share a common return line, then you'll have to connect one end of each of these eight networks to it.

Have Fun! Rich

No! Don't put them in series. Put the diode in antiparallel, to limit the reverse voltage on the LED: K +-diode-+ 24 vac -> resistor --+ +---24 V. ret. +--LED--+ K

K = "Kathode" ;-) (either way will work, as long as they're opposite.)

Cheers! Rich

Hey guys thank for all the help so far.

Assuming I run the diode and reistor in SERIES, can I use the a

Hey another question what type of specifications should the LEDs have? I assume voltage would have to be around that after the resitor does it' s thing.

Again thank you for all the patience. I find this all very interesting. Great group.

thanks, george economos

No, because when the diode and LED in series are reverse-biased, there's no telling if the reverse leakage of the diode will allow the LED's reverse breakdown to be exceeded. There's no facility for them to "share" the reverse voltage, and if they did, the LED would pop.

Use the diode in series if you need to, but put another diode in antiparallel with the LED, just to protect it.

Good Luck! Rich

The 1N4000 series rectifiers (standard for this type of application) will limit reverse leakage current to nanoamperes. Nothing will pop.

Success! I managed to create a test circuit and hook it up to the sprinkler controller. I ended up wiring the LED and diode in parallel in opposite directions as suggested. I am able to turn the LED on and off using X10 commands from my pc. I would like to thank everyone for their help on this.

george economos

yes both types can handle the peak voltages (approx 32V) and currents (approx 15mA) involved

bright enough to be visible if you're working in sunlight high intensity LEDs could be a good idea, otherwise any LED will do.

Bye. Jasen