I am putting together a lighting order and have a question about my power distro. I have to run three 18k's, some smaller units, as well as a 50k Softsun. I know the softsun runs off 240v, while the 18k's run 220v. Can I run these at the same time? Also while having smaller units with 120v? I have a 1400amp tow plant on order, just could use some input about running it all together, thanks.
Generator & Voltage question, running 18k's and 50k Softsun.
#1
Posted 25 November 2014  04:50 PM
#2
Posted 26 November 2014  11:43 AM
I know the softsun runs off 240v, while the 18k's run 220v. Can I run these at the same time? Also while having smaller units with 120v? I have a 1400amp tow plant on order, just could use some input about running it all together, thanks.
What vintage 18k ballasts will you be using? What are the power factors of the 18ks and softsun ballasts?
Guy Holt, Gaffer, ScreenLight & Grip, Lighting Rental & Sales in Boston
#3
Posted 27 November 2014  01:43 PM
It's not my country we work in 220/240 and for long distances and big loads we start with 380v but....
the 18ks are something around 100amps depending like Guy ask from different variables like distances of cable, line loss and types of ballast and the 50softsun cuold be around 250amps. Thinking on 3 18k and a 50softsun i think you are on a big job with a budget and the 18ks are arrimax or daylight but with beautiful new elettronic ballasts so if i'm right with 1400amps genny you will be fine. Distribute the load between legs it's another matter, if the genny has a cam lock out go with that, starting with 240 for the soft sun.
In italy we start from 380v so three legs of 220/240 each, so i need to know how your gennys work with 240 one leg or 3 legs? if it's only one put your distro behind the 50k, put stingers for 18ks that also works on 220/240 it's the same and then split another outlet with a box balancing the 110v loads.If you don't need all this only for paper 1400 could be enough.
if you have 3 legs of 240v from the genny it's easier: 50k for one 18+18k for another and 18k and minor loads in the third (always balanced between 2 legs of 110v)
i hope helps despite my english
good luck
Leonardo Brocato
Gaffer Rome Italy
#4
Posted 01 December 2014  03:41 PM
The 50K is 225 amps 3phase input 208240v just by itself
#5
Posted 10 December 2014  10:41 AM
I have a 1400amp tow plant on order, just could use some input about running it all together, thanks.
Sizing generators is a complicated matter. Lets start with the easy part. 18k HMIs will operate anywhere between 190 250V single phase (hot, hot, ground.) A 50K SoftSun will operate anywhere between 208240V three phase (hot, hot, hot, ground.) If you are going to run 120V PhasetoNeutral loads as well and don’t want to use transformers to step down to 120V, you would use a generator in three phase that would give you 120V phasetoneutral and 208V phasetophase.
Now let’s look at the load place upon the generator by the lights. Let’s take the HMIs first. Koby doesn’t specify what the smaller 120V fixtures are, but for the sake of argument, let’s assume one is a 2.5k HMI. If a 2.5k HMI were a resistive load like quartz lamps it would be a simple matter of dividing the wattage (2’500W) by the voltage (120V) to calculate the current it would draw (2’500W/120V = 20.83 Amps). Unfortunately, it is not that easy.
Schematic of an electronic HMI ballast
HMI ballasts, along with Kino & LED power supplies, fall into a class of power supplies called Switch Mode Power Supplies or SMPSs for short. As illustrated above, SMPSs have a diode/capacitor front end that first converts the input AC power to DC, before a highspeed switching device (micro processor controlled IGBTs in the case of HMIs) turns the DC current into an alternating wave again.
As illustrated in the simplified schematic above, the front end of SMPSs convert AC to DC power by first feeding the AC input current through a bridge rectifier consisting of four diodes, which inverts the negative half of the AC sine wave and makes it positive. The rectified current then passes into a bank of smoothing capacitors that remove the 60 Hz rise and fall by charging and then discharging power to the load while the rectified AC current descends to zero volts and then ascends again. By filling the gaps between the peaks of the rectified AC, the discharge from the capacitors flattens out the voltage  making it essentially DC. As such, a nonpower factor corrected (nonpfc) 2.5kW HMI ballast draws considerably more power from the generator than would a 2.5kw Incandescent lamps because it has to draw power in short bursts.
Thin Black Trace: Rectifier Bridge converts AC power to rectified sine wave. Thick Black Trace: Stored Capacitor Voltage. Red Trace: Current drawn by capacitors once input voltage is greater than voltage stored in the capacitor (thick black trace.)
The reason that the 2.5kW HMI ballast draws power in short bursts is that, as can be seen in the illustration above, the instantaneous line voltage (thin black line) is below the voltage on the capacitor most of the time (thick black line). Consequently the time the capacitor has to charge is only a very brief period of the overall cycle time while voltage ascends to its’ peak. That is because, after peaking, the half cycle from the bridge drops below the capacitor voltage; which back biases the bridge, inhibiting further current flow into the capacitor. Thus the diodes conduct only for a small portion of each line halfcycle and only during the ascending portion of the supply voltage waveform  which pulls the current out of phase with the voltage so that current leads voltage (a leading power factor.) And since, during this very brief charging period, the capacitor must also charge fully, as can be seen in this illustration in red, large pulses of current are drawn for short durations. The end result is that the current drawn from the generator is a series of narrow pulses that peak before the voltage peaks.
Because it must charge its’ smoothing capacitor in short bursts, a nonpfc 2.5kW ballast draws significantly more current (32.5A) for roughly the same wattage as an incandescent light (20.8A). Because this difference in the amount of current drawn does not contribute to an increase in work (wattage) it is called reactive power. There is then two components to the energy expended by a 2.5kW HMI: there is what we call “real power” or that which generates work (measured in watts) and the reactive power or that which does not contribute to the work (measured in kilovoltamperes reactive or kVAR.) The sum of the real power and reactive power is called the “apparent power” (also stated as kVA). In the case of a 2.5kW HMI, its’ apparent power, or the actual power expended by the fixture that the generator and cable must support is 3900VA (32.5A x 120V= 3900VA.)
The ratio of “true power” to “apparent power” is called the “Power Factor” of the light. A favorite analogy electricians like to use to explain power factor is that if apparent power is a glass of beer, power factor is the foam that prevents you from filling the glass all the way up with beer. When lights with a low power factor are used, a generator must be sized to supply the apparent power (beer plus foam), even though only the true power (beer) counts. Since a 2.5kW HMI ballast draws nearly twice the apparent power (3900VA) for is true power output of 2’500W, it has a power factor of .64 (2’500W true power/ 3900VA apparent power = .64 PF).
The next part of our load calculations are the 18kW HMIs. Since the 18kW power class is relatively new, it is safe to assume that the 18kW ballasts in this case are power factor corrected (pfc), which means that they incorporate a multistage boost converter typology to accumulate energy in their smoothing capacitors over the entire AC cycle rather than just a brief portion of it. Now that the capacitors charge throughout the AC cycle rather than just a brief portion of it, the peak current drawn by the ballast is greatly reduced. And, since the output voltage of the boost converter is set higher than the capacitors input voltage (which is why they are called boost converters), the load is forced to draw current in phase with the AC main line voltage. In this fashion, the pfc circuit realigns voltage and current (no longer a leading power factor) and induces a smoother power waveform at the distribution bus. As a result, the ballast use power more efficiently and generate less heat, thereby increasing their reliability. PFC 18K HMI ballasts will draw approximately 81A on each of two phase legs for an apparent power of 19’440 VA (162Ax120V=19’440 VA) for a near unity power factor of .93 (18’000W/19’440 VA=.93)
Where I have reached the allowable images that can be contained in a post, I will have to pick up with the final part of our load calculations, the 50kW SoftSun, in a subsequent post.
Guy Holt,
Gaffer,
ScreenLight & Grip,
#6
Posted 10 December 2014  10:44 AM
I have a 1400amp tow plant on order, just could use some input about running it all together, thanks.
The final part of our load calculations is the current drawn by the 50kW SoftSun, which like HMIs, cannot be calculated by simply dividing its’ Wattage by its’ Voltage. SoftSun Plasma lamps operate on DC, which means that their power supply (ballast) must convert the AC provided by the generator to DC. It does this in a very similar fashion to the Diode/Capacitor front end of HMI ballasts described above but without switching the DC back to an alternating waveform. In other words, one difference between an electronic HMI ballast and SoftSun ballast is that the HMI ballast goes one step further and generates an alternating current out of the DC generated by the diode/capacitors where the SoftSun ballast does not.
HMI ballasts and SoftSun ballasts differ in several other respects as well. Since SoftSun ballasts operate three phase, as illustrated above, their rectified AC voltage consists of six pulses per cycle rather than the two created by the diode bridge of a phase to neutral HMI ballast.
And, given their intended use for highspeed cinematography, the SoftSun ballasts adds smoothing capacitors across the threephased rectifier to remove the DC ripple of the six pulse rectified AC illustrated above. As such, SoftSun ballasts also draw current in short bursts of a high magnitude because of the short interval in which its’ smoothing capacitors must charge. But, unlike an HMI ballast, a SoftSun ballast draws two pulses of current for each cycle of the AC supply voltage, as illustrated below, rather than the one drawn by a SMPSs.
Six pulse power supplies draw two current peaks per cycle rather than one
The final difference between a SoftSun ballast and an HMI ballast is that, without a Switch Mode Converter final stage, the DC value of the rectified AC is a function of the AC supply voltage (0.955 times the AC amplitude, or 1.35 times the rms value.) Which means that the DC voltage to the lamp will vary depending on whether the AC supplied by the generator is 208v or 240V. So that the SoftSun’s Plasma lamp receives its’ prescribed DC value, the SoftSun Ballasts substitutes three SCRs ( thyristors) for three of its diodes. A SCR is a solidstate device like a diode but with a
third electrode, which prevents the device passing current forward until the third electrode is “triggered.”
Since the SCR will not conduct until signaled to do so on the third electrode, the “firing” can be delayed in order to reduce the value of the DC output. An electronic circuit provides a firing pulse with a variable delay, so that the waveform appears as illustrated above. The mean DC level – that is, the line where areas above and below it are equal – will be different with differing delay times. Thus the bridge can be used to give different DC output levels simply by controlling the electronic delay circuit. The ballast’s 100,000 uP capacitor bank then removes the remaining voltage ripple in the DC output. As such, a SoftSun ballast, like an HMI ballast, draws considerably more power from the generator than would a comparable Incandescent lamp. The power factor of a SoftSun Ballast is .55, which means that a 50’000W lamp has an apparent power of 90’720VA (Note: Luminys Corporation, the manufacturer of SoftSuns, will be introducing pfc power supplies that will draw 40% less power in the near future.)
Now, we are finally in a position to approximate the load a lighting package consisting of a 50K SoftSun, three 18k HMIs, and some smaller HMIs will place upon a generator. Putting aside the smaller HMIs, between the three 18Ks there is an apparent power of 58’320VA (19’440VA x 3 = 58’320VA) on top of the 90’720VA of the 50K SoftSun for a total apparent power of 149’040Van (149.4kVA.) The winding of a 1400A tow plant (MQ model EGS1400C3) is rated for an apparent power of 210kVA at unity power factor without allowance for reactive loads, which means that Koby will need a bigger generator or split his load over two generators. Why can he not run an apparent power of roughly 149VA on a generator rated for 210kVA? The reason has to do with the type of load.
The greater apparent power of lights with a poor power factor is not the only consideration when sizing a generator. When powering lights that use nonpfc SMPSs (HMIs, Kinos, CLF lamp banks, LEDs) and lights that use power supplies consisting of SCR/capacitor circuits (SoftSuns), you must also take into account the harmonic currents these nonlinear loads draw that can have a severe adverse effect on the power waveform of generators.
As is evident in the power quality meter reading of a nonpfc 2.5kW HMI above, the high peaked pulsed current (lower waveform) drawn by smoothing capacitors is a distorted waveform that does not resemble the sinusoid of AC voltage or the current drawn by an incandescent light. The same is true of the two current pulses drawn per half cycle by a SoftSun ballast. As such, the current drawn by these loads include a number of harmonic currents in addition to the 60hz fundamental (see Fast Fourier Transformation of a nonpfc 2.5kW HMI ballast below.) These harmonic currents will generate excessive heat in a generator’s winding, core, and rotor. Since generator ratings are limited by allowable temperature rise, harmonic currents act as derating factors.
But that is not all: on the soft power of a generator, harmonic currents also interact, according to Ohm’s Law, with the impedance of the generator to create voltage drop. Since smoothing capacitors consume power only at the peak of the voltage waveform, voltage drop due to system impedance occurs only at the peak of the voltage waveform – causing the “Flat Topping” we see in the top waveform of the first power quality meter above that is characteristic of capacitive loads on generators. If the voltage waveform distortion is severe, it can cause voltage regulator sensing problems and inaccurate instrument readings in a generator’s control systems.
Caterpillar’s “Limit Characteristic” graph below illustrates the effects that nonlinear loads have on generators. The fluctuations in the Engine Stability line represent the generator’s operating limits depending on whether its load has a leading or lagging power factor. As you can see by the right quadrant, the effect of the harmonics generated by leading power factor capacitive loads like those being discussed here are a significant limiting factor on generators.
A Limit Characteristic graph for a generator illustrates the effect of leading or lagging Power Factor on the generator's output.
What the Limit Characteristic graph above tells us is that, operating a capacitive load (the leading power factor quadrant right of the Unity Power Factor center line), a generator first reaches a thermal limit as a consequence of heat generation in the generator's rotor from harmonic currents. And, since increasing harmonic content leads to a tendency toward erroneous selfexcitation (SE), the generator output is inevitably affected by instability of the Excitation circuit from lower leading power factor loads. Finally, since there comes a point as the power factor of the load decreases, when harmonics inhibit the successful operation of the generator’s Automatic Voltage Regulator all together, and hence the generator’s capacity to generate any power at all, the kW output eventually drops to zero. Hence the power factor of the load the generator must operate is another derating factor.
To calculate the final power factor of the load Koby wants to put on a 1400A generator we simply divide the total true power of all the lights (104kW) by their total apparent power of 149.04kVA (19.44kVA+19.44kVA+19.44kVA+90.72kVA =149.04kVA) to come up with a pf of .71. According to the manufacturer of the alternators used in show power generators (Marathon Electric model 431RSL4007) a load with a leading pf of .71 would require that the generator be oversized by 100% in order to keep the temperature rise in its’ rotor and stator to an acceptable 80°C R/R and the subtransient reactance to 6% or less to limit voltage waveform flat topping to an acceptable level (< 5% VTHD.) So, after taking into account that the load has a leading power factor of .71, Koby will need a generator capable of producing 298.08kVA, unless of course the production were taking place in the summer in Denver, Colorado.
The final factor one must take into account when sizing generators is the environment in which it will operate. For instance, a generator’s power rating is based on its operation at sea level. Generator engine power decreases as altitude increases (thinner air), and a generator’s maximum electrical output drops accordingly. A power loss of about 3.5% per 1,000 feet of elevation gain is typical.
Another environmental factor that must be taken into account is the ambient temperature in which it will operate. The power of a generator is typically derated by about 1 to 2% for each 10°F above its nominal rating at 60°F. Combining the conditions of high altitude and high temperature may require further oversizing of a generator. For example, say this production was taking place in Denver, Colorado (elev. 5,000 ft.), rather than Los Angeles (elev. 233 ft.) during the summer when temperatures can reach 90°F. Compensating for altitude would result in a 17.5% loss (5 x 3.5%/1000ft elevation). While compensating for the 30°F rise above its’ nominal rating at 60°F would result in an additional 3% loss [(90°F  60°F) × 1% ÷ 10°F]. So in the final analysis, if we take into account the environmental factors of elevation and temperature of operating our load at 5000 ft above sea level in the summer, Koby would need a generator capable of providing 363.60kVA (298.08kVA x 1.25 = 372.60kVA) to power a lighting package consisting of a 50kW Softsun, three 18kW HMIs, with additional smaller 120V lights.
Rather than renting a 300kVA container plant, Koby would probably be better off renting two generators: a 1400A (210kVA) tow plant to power the 50kW SoftSun alone, as well as a 1200A (150 kVA) tow plant to power the three 18kW HMIs along with his smaller 120V loads.
For more details about these issues see an article I wrote for our company newsletter that explains the electrical engineering principles behind these issues and how to resolve them. The newsletter article is available at http://www.screenlig...generators.html.
Guy Holt,
Gaffer,
ScreenLight & Grip,
#7
Posted 10 December 2014  12:31 PM
Guy,
Thanks for the detailed explaination. As a former Gaffer myself I enjoy the topic. Also it clearly demonstrates why the original question could also be answered by saying,"You need to hire a qualified technician to handle this type of equipment which is not only easily damaged by improper use but also can cause phyisical injury. Thats what Best Boy Electrics are for."
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