Well it works!
What, you doubted me??
However, I have to keep the power onto the relay continuously, and, as soon as I disconnect, the solenoid shuts off as intended.
Exactly. Thats why my original diagram showed two discrete switches. One for "ON" and one for "OFF". Using a single SPDT switch means it's always either ON or OFF, with no "STAY AS YOU ARE" option.
I have been thinking that 12 volts produces a very fast reaction and I wonder therefore if 9 volts would produce a slower reaction enabling a pulse effect to keep the water balance on whilst the relay is off (dead).
There is no "dead" using a double-throw switch or relay.
Again, using the timer I suggested (which has a changeover relay for its output) would indeed keep its internal relay activated during the "ON" cycle (for 1 to 8 minutes per your earlier post?), but the solenoid itself would only get a short burst of power.
Gardena must have had a reason for using a 9 volt battery, and a solenoid coil of this exact type.
small, cheap, easy to get.
I don't have any 9 volt relays but I can re-jig the solenoid power via the relay to a 9 volt battery and just use 12 volts for activating the relay. Will give it a try
Don't expect a different outcome, because it won't be.
Let me re-explain the theory of operation of my original circuit.
A capacitor, on DC, will charge up to the supply voltage. The amount of current it draws, and how long it draws it for, is a function of the resistance of the circuit, and the capacitance of the capacitor itself. When you first apply power to the circuit and operate the "on" switch, peak current will flow through the coil and capacitor. The capacitor will charge up at a rate determined by the resistance of the circuit and the voltage. The charging current will rapidly reduce as the capacitor aproaches full charge, end eventually will be virtually zero (just leakage current).
Thus, your solenoid has had a brief pulse of current even though the power could remain turned on for hours.
When you disconnect the power, the capacitor will hold the charge - it has no discharge path. The charge would (in theory) hold forever - except of course, capacitors have some internal leakage and the charge will eventually dissipate. (I've had 100,000uF capacitors stored for years that took a chunk out of a screwdriver with the charge still in them after that time!)
Finally, when you close the "off" switch, the charge in the capacitor has a discharge path back through the coil - and will again produce an initial peak current which tapers off as the charge in the capacitor is discharged, ending up with the cap discharged and there being no current through the coil - in effect, "reset" ready for the next cycle.
If you use a changeover (double pole) relay, it is only ever in one of two states - charging the capacitor ("on"), or discharging the capacitor ("off").
