Sensing a 24 V AC signal

Question

I have a sprinkler controller with no separate output for a pump relay. I need a 16-input logic OR circuit that can trigger the relay when one or more of its 16 channels is live with 24 - 30 V AC. One side is common to all inputs. Due to space limitations, I need a minimum component count.

What is the simplest way to convert the 24 V AC signal to a suitable logic level with sufficiently low ripple so that the relay remains on and does not chatter? Do I need opto-isolators, transistors, diodes, or logic gates to achieve this?

Answer 1

use a current transformer to sense the solenoid current you can then drive a triac to turn on the contactor for the pump.

^{simulate this circuit – Schematic created using CircuitLab}

Here "R" is a terminal with a continuous 24VAC supply, and "C" is the common return.

or if you don't like current transformers here's an all semiconductor solution

^{simulate this circuit}

2N3055 was chosen for Q1 qnd Q2 because it has a 4A max base current a separate DC supply provides power to operate the pump contactor.

Answer 2

What you describe are sixteen 24..30 Vac voltages that share a common Vac return.

Using power diodes if pump relay has no switching supply

Each of sixteen can go through a power diode to regulator supply. That's smoothed by C1. When power is removed, C1 is mostly discharged by the relay coil, then further by the regulator, then down to 0 V by R1. The linear regulator gives an accurate voltage for driving your pump relay's coil or enable.

If the pump relay is a mechanical coil relay, it can drive the coil with the regulator 24 V output. Use another regulator for a different coil voltage but be mindful of the regulator power dissipation. A 40 mA coil current at 24 V would be reasonable and dissipate approx. 700 mW in the regulator at 30 Vac (42 V across C1), 320 mW at 24 Vac.

If the pump relay is an SSR, its control input will draw considerably less current, around 15 mA. This would dissipate approx. 260 mW in the regulator at 30 Vac, 120 mW at 24 Vac.

^{simulate this circuit – Schematic created using CircuitLab}

Using opto-isolators if pump relay has a switching supply

The opto-isolator circuit uses more parts and draws more current in total from the inputs.

^{simulate this circuit}

Answer 3

I have done it with 6 diodes, a small resistor and an optocoupler. I would make two groups of 3 diodes in series (>|-->|-->|) then connect them so one set conducts during the positive cycle and the other conduct through the negative cycle. Then use an AC optocoupler (they have back to back LEDs) with a resistor in series connected to the two strings of diodes that were connected in parallel. This string goes in series with the common for all valves and will turn the optocoupler on whenever any solenoid turns on. Be sure to size the diodes properly. The three diodes in series should give you about 2.1V which should be enough to turn on the optocoupler. If not enough add another one in each direction.

You need to read the data sheet to determine how much current you need to cause it to saturate. You can also read the transfer percentage as that will help size the circuits gain etc. I generally do not do that, I try for about 25-50% of the LEDs rated current. I do the same thing with DC, generally with a transistor but an opto coupler will work just as well it just needs a little more voltage.