Welcome or welcome back, join me now on the everlasting quest to quantify and put an equation to something that is objectively unquantifiable and un-equationable. Here’s where we are now:

PLamp: RMS lamp power (W)
VLamp: RMS lamp voltage (V)
ILamp: RMS lamp current (A)
ZBal: Ballast impedance (Ω)
OCV: RMS open-circuit voltage (V)
C: Mystery unknown constant

These variables make up the following equation:

ZBal = (C * √(OCV^2 - VLamp^2)) / ILamp

Which can be rearranged as:

ILamp = (C * √(OCV^2 - VLamp^2)) / ZBal,
and:
C = (ZBal * ILamp) / √(OCV^2 - VLamp^2)

One may think that the value C is power factor, but inputting lamp power factor calculated by PLamp/(Vlamp*ILamp) gives completely false results. Take the following examples:

400W MV:
PF = 400 / (135 * 3.2) = 0.926
ZBal = (0.926 * √(220^2 - 135^2)) / 3.2 = 50.27 ohms
50.27 / 44.5 = 1.13 = 13% off
Says the lamp needs 50.27 ohms of impedance when the specs say 45.5 ohms

55W SOX:
PF = 56 / (109 * 0.59) = 0.87
ZBal = (0.87 * √(480^2 - 109^2)) / 0.59 = 689.3 ohms
689.3 / 775 = 0.89 = 11% off
Says the lamp needs 689.3 ohms of impedance when the specs say 775 ohms

So obviously that isn’t gonna work, it gives wrong answers. So we use the rearranged equation to look specifically for this special C constant, which is different for every lamp, however reasonably consistent within technologies. I have found the following average values of C for different technologies:

MV: 0.831
MH: 0.873 (highly variable)
HPS: 0.913
LPS: 0.973
FL: 0.915

So this definitely isn’t the power factor. It seems to trend in the opposite direction of power factor, with MV being the lowest and LPS being the highest, which definitely points away from this being power factor. I thought maybe this was just the distortion power factor excluding displacement power factor, but I don’t have any evidence for this and I have no idea.

Anyway, now with these values of C:

400W MV:
C = 0.831
ZBal = (0.831 * √(220^2 - 135^2)) / 3.2 = ohms
45 / 44.5 = 1.01 = 1% off

55W SOX:
C = 0.973
ZBal = (0.973 * √(480^2 - 109^2)) / 0.59 = 770 ohms
770 / 775 = 0.99 = 1% off

So obviously these values give good results, they are averages that were trained on the existing data from the reference circuits. But do they work for circumstances outside of the reference circuits? Well apparently yes, and LightBulbFun helped me prove it.

Standard specs for 18W SOX-E:
PLamp: 18W
VLamp: 15V
ILamp: .35A
ZBal: 829Ω
OCV: 300V

And we can use the C equation to get:
C: 0.985
which is only a little bit higher than the LPS average of 0.973, but will help us get a more accurate answer.

With these values, using the ballast impedance equation, we can extrapolate what ballast impedance we will need for different OCVs, such as 220, 230, and 240V. We get the following values:
220V: 598.1Ω
230V: 627.2Ω
240V: 656.2Ω

LightBulbFun’s experimental values are below. 240V was not able to be determined just due to practical constraints.
220V: 602Ω
230V: 632Ω
Both of these values are within 1% of my calculated values, meaning whatever the values of C means, it does not change significantly with different circuit characteristics.

This makes it possible to predict fairly accurately the necessary ballast impedances and probable lamp currents for lamps running on alternative gear. This seems to be very accurate when the reference circuit is available (so an individualized C can be calculated) but the average C values can be sufficient (within 2-5% usually).

So this is my question:
What is C? Where does it come from? It is obviously some sort of characteristic of the arc discharge type because it is fairly consistent within technologies, but it is not the total power factor. Before I was greatly doubting the ability of these equations to work outside of the standard circuits, but apparently they do so quite well (for 18W SOX-E at least).

Also if anyone else has the right equipment, we might be able to prove that these equations work with other discharge technologies.

What do you think? Rip me apart, as always lol

In you "C" are actually two completely unrelated things that cause the "C" to get lower:
One is the lamp power factor (lamp power is less than Vrms*Irms),
second are the ballast losses.

Stsrting with the second: You calculated the ballast impedance as a lossless reactance, while in reality it exhibits not completely negligible resistive component, affecting the angles in the voltage triangle (one arm is lamp voltage, second ballast impedance drop, the 3rd is the OCV). When resistance is added, you get lower ballast voltage drop, hence the lower ballast impedance to maintain the current. As your equations neglect the resistance, it moves into the "mystery correction" coefficient.

The second thing is the lamp power factor. Power factor is nothing else than a correlation factor between two waveforms (voltage vs current). To be perfectly unity, both have to match in shape and phase. With just pure sinewaves (that is what classical AC electrical calculation teach) it reduces to the famous cos(phi), as because both the same shape sinewaves, the only mismatch is the phase. With any other shape you have no otyer choice than really run the integrals:
Preal = 1/T * integ[0->T](v(t) * i(t) * dt)
Xrms = sqrt(1/T * integ[0->T](X(t)^2 * dt)
And do the calculations:
PF = Preal / (Vrms * Irms)
 
It depends mainly on the shape of the current waveform. The thing is, practically all lamps have the waveform near rectangular, some with an initial spike. It depends on the exact plasma dynamics, but at technical 50Hz most lamps are pretty close to rectangular.
 Combining that idealized square wave with a sinewave current (for the start neglect the current distortion), you get 2/pi/sqrt(2) = 0.9.
The reignition spikes after zero cross make it even lower, as they contribute to the rms (because of the squaring), but do not add almost any real power, because at that time the current is very low (it just crossed zero when it chanes its polarity). Exact shape depends on the exact plasma dynamics, so it is dependent on the chemistry and to some extend also the lamp construction (the starting probe in MV facilitates reignition earlier at lower voltage, so yielding less reignition overshoot than it would be with a lamp without or with disconnected/bypassed probe).
But in reality the current is also affected by the saturation spiking, where these spikes contribute moreto the rms (because of the squaring in the rms calculus), not that much with the real power (where the arc keeps the voltage constant, so real power scales linear with current in the real power formula). The quantity depends on the exact ballast design, how much it is designed to saturate.
So in total the total power factor is a bit less than 0.9, something between 0.8..0.9 I would consider plausible. To get exact fifure, you need to take oscilloscope pictures and process the waveforms (in Excell, Calc, Matlab,...).

@Medved
So if I am understanding this correctly, two ballasts with identical total impedance will run a lamp at different currents depending on how much of that impedance is resistance vs how much is reactance? That is very interesting considering the reference circuits for these lamps just specify "ohms" and don't say whether that is impedance or resistance.

Is there any practical way to take ballast resistance into account besides actually measuring it for every ballast? If we know total impedance, OCV, current, lamp voltage, lamp power, and lamp power factor, can we calculate it?

@Medved
So if I am understanding this correctly, two ballasts with identical total impedance will run a lamp at different currents depending on how much of that impedance is resistance vs how much is reactance? That is very interesting considering the reference circuits for these lamps just specify "ohms" and don't say whether that is impedance or resistance.

Is there any practical way to take ballast resistance into account besides actually measuring it for every ballast? If we know total impedance, OCV, current, lamp voltage, lamp power, and lamp power factor, can we calculate it?

actually reference circuits do specify a Power Factor for the ballast itself, for ANSI and JIS setups (apart from SOX lamps where its 0.06) its 0.075 for all lamps from 4W T5 to 1000W mercury lamp


with IEC setups the PF will vary depending on the ballast/lamp from 0.12 for a small FL lamp to 0.04 for a big 1000W Discharge lamp (and for each lamp data sheet the IEC standards do give a PF)




https://www.doc88.com/p-39229364420492.html


https://www.doc88.com/p-68461902278270.html

I was adressing the power factor of the discharge lamp itself, not the power factor on the mains input.