As you know, I am not "bogged down" by specific makes and models...I solve problems by going back to first principles.
RUNNING
For most discharge lamps to run in a stable manner with a choke in series, the
magnitude of the voltage drop across the lamp and the
magnitude of the voltage drop across the choke have to be approximately equal, (although the drop across the choke can be more than the drop across the lamp, but it is a bit inefficient). Note I say
magnitude because these are scalar quantities...but on adding them, "directional"
vector addition has to be used...i.e Pythgoras' Thorem.
The SOX 18 is rated at 350mA @ 57v arc drop at this current.
Let's set the choke drop to 50v at the same 350mA, (as it is in series with the lamp and the same current flows through both.)
Now we vector add these two to get the supply voltage we will need to keep it all going.
So, the lamp voltage drop is one side of as right-angled triangle and the choke drop is the other side at 90*, the hypotenuse is the "mains" or supply voltage that will be required. So, applying Pythgoras' Thorem to this triangle of voltages:-
(57) Squared + (50) squared = (supply) squared
3249 + 2500 = 5749
Root 5749 = 75.8v
So a supply of 75.8v will run the setup stably, but we need to know the value of the choke in mH that will drop 50v @ 350mA @ 60Hz.
Choke drops 50v @ 350mA....Using Ohm's Law we can find the Inductive Reactance...
V = I x R (Choke V = Lamp Current x Choke Reactance)
50 = 0.35 x Xl
Xl = 143 iOhms
Now use the inductive reactance equation to find the inductance of the choke required at 50Hz.
Xl = 2pi f L
143 = 3.14159 x 2 x 60 x L
143 = 6.2918 x 60 x L
143 = 377.5 x L
143/377.5 = L
L = 379mH
If you place a SOX 18 in series with a 379mH choke and run it off 75.8 volts R.M.S. @ 60Hz...once started, it should run up normally.
Now you can furnish the 76 volts using a
buck/boost transformer arrangment, you need to buck 44 volts off 120v.
Alternatively, you can recalculate the choke value that would be needed in series at the full 120v supply and then no transformer will be required. This will increase the drop across the choke to more than the drop across the lamp so it will be a bit inefficient, but it will work O.K.
Again, back to the triangle, one side is still 57v long (same lamp), but the hypotenuse has now been increased to 120v long (from 76v)...new higher supply voltage...how does this effect the third side, the drop across the choke? Again, Pythgoras' Thorem:-
(120) Squared = (57) Squared + (Unknown Choke Drop) Squared
14400 = 3249 + (Unknown Choke Drop) Squared
11151 = (Unknown Choke Drop) Squared
Root (11151) = Unknown Choke Drop
105.6v = Unknown Choke Drop
So use Ohm's Law again to find the new reactance of this choke...it will be a little more than twice that of the choke above (which was 143 iOhms).
V = I x R
106v = 0.35 x Xl
Xl = 301.7 iOhms
Now for the inductance:-
Xl - 2 x pi x f x L
302 = 377 x L (f = 60)
L = 801mH
So, you have two choices now, 76v supply with 380mH in series...OR:-
120v supply with 800mH in series
Or any other combination you might care to calculate using the same techniques as above. (E,G. running a SOX-18 off 240v @ 50Hz if you came here to visit me!).
STARTING
The 380mH in series off 76v will mean the open circuit voltage across the lamp will be 76v and the E.M.F. generated from the 380mH choke when the lamp is shorted and the short is then removed will be smaller than an 800mH choke will generate.
The chokes must be able to carry that current that will flow through them of they are placed directly across their respective supplies...So, Using Ohm's Law again:-
76v across 143 iOhms ------> 531mA
120v across 302 iOhms ------> 397mA
You can measure the inductances of fluoro chokes by three methods:-
1, Measure it in its original circuit with its original T12 or T8 lamp, i.e, its drop and the current flowing through it and calculate its inductive reactance using Ohm's Law and Inductance from that. This would be the easiest method as it only needs a multimeter.
2, Use an L,R,C, impedance bridge to measure that choke passively out of circuit. Next easiest method, requires access to an L,R,C, bridge.
3. Use the resonant method. Place known capacitor, (1uF say) in parallel with choke to be measured, place a 1K resistor in series with this parallel LC network. Feed the whole network from an audio signal generator and sweep from 10Hz to 1KHz. Measure voltage across choke with oscilloscope and note where peak occurs. Take frequency of peak and value of capacitor and plug back into parallel resonance equation to solve for unknown L.
As for those specific starters you mention, you will have to acquire them, measure their properties in circuit, i.e, how their impedance changes with applied voltage from a high source impedance, draw their circuits and analyse those circuits to extract their mode of operation...to "get back inside the head" of the bloke who designed them and work out what his intentions were and whether or not they match your intentions.
Any starter would need to be activated by the open circuit voltage..i.e. 76 or 120 for the two cases above and would need to stop acting below about 65v.
Measuring the voltage drop of the lamp right after startup (when it was still on the neon) would be helpful in designing a starter circuit that would not keep re-triggering. At starting the arc drop is slightly greater, by about 18% and the arc current is slightly less. As the lamp comes up the arc voltage rises a bit, by another 9%, then falls to the operating value of 57v. So, at the highest point the arc drop is about 73v....very close to the supply in the first case gear scenario and difficult for a starter to distinguish between the lamp in this state of half-run up (73v) and an extinguished lamp (76v). So, based on this, the 800mH/120v supply option is more advantageous from an automated starting perspective and the starter would be required to act in the 80-120v range but cease to act below 80v.