@billrvolz : premature to be flippant. The answers you've been given are good. Just take some time to think about them.

To Bluejets, there is no such thing as a 'zener resistor', it's a zener diode.

I think he meant the combination of zener diode and resistor.

Do you? Then you should understand

@crutschow 's explanation.

The 470 ohm resistor drops the remaining 3V.

Only for the case when no current is drawn by the solenoid.

Have a look here:

--- Operating Point ---

V(out): 10.1772 voltage

V(n001): 12 voltage

I(D1): -0.00387824 device_current

I(R1): -0.00387824 device_current

I(V1): -0.00387824 device_current

With no load current through R1 is ~ 4 mA, output voltage is 10 V.

Your solenoid has ~47 Ω (9 V ( 0.19 A). With this load the circuit looks like this:

--- Operating Point ---

V(out): 1.09091 voltage

V(n001): 12 voltage

I(D1): -6.01091e-010 device_current

I(R2): 0.0232108 device_current

I(R1): -0.0232108 device_current

I(V1): -0.0232108 device_current

R1 and R2 form a voltage divider 1:11. your sourece voltage of 12 V is reduced to 1.1 V (V(out = 1.09091 V). The zener diode is ineffective as the voltage across it is less than 10 V. The resulting current of 0.023 mA is insufficient to operate the solenoid.

No consider

@crutschow 's proposal:

--- Operating Point ---

V(out): 8.95238 voltage

V(n001): 12 voltage

I(R2): 0.190476 device_current

I(R1): -0.190476 device_current

I(V1): -0.190476 device_current

You now have 0.19 A through the solenoid (R2) and a voltage of 9 V across R2. Perfect for your application.