If a switch mode power supply uses a bridge rectifier and capacitor to create a DC voltage that is then mode into square waves and the switching action done, why is it any harder on a power source than a linear power supply … If the switch mode power supply worked as many triac dimmers do and only passed part of the AC waveform on, the problem is obvious.
What makes SMPSs tougher on power supplies is that, like triac dimmers, they do in fact pass only part of the AC waveform, but unlike triac dimmers, they also draw current in harmonically rich pulses that are many times larger in amplitude than what a comparable size (wattage) linear load on a triac dimmer would draw. To understand why this is the case, let’s look at how SMPSs operate in more detail.
The SMPSs we typically encounter in motion picture lighting convert AC power to DC, and then use some type of switch to convert the DC back to AC of a higher frequency (20-50k Hz in the case of flicker-free fluorescents) or different waveform (the square wave of flicker-free HMI ballasts.) Regardless of the end result, all SMPSs use a diode-capacitor front end to convert the supply AC to DC.
As can be seen in these power point slides from a workshop I developed for the members of New England Studio Mechanics Union (IATSE Local 481) on power quality, the diode-capacitor section converts the AC power to DC power by first feeding the AC input current through a bridge rectifier, which inverts the negative half of the AC sine wave and makes it positive. The rectified current then passes into a conditioning capacitor that removes the 60 Hz rise and fall and flattens out the voltage - making it essentially DC. The DC is then fed from the conditioning capacitor to the Switch-mode converter, which in the case of fluorescent ballast, is a high frequency inverter that utilizes a pair of MOSFETs to generate the high frequency (20-50kHZ) AC power.
To obtain a low ripple on the DC output of its’ front-end diode-capacitor section, the smoothing capacitor of a SMPS is typically quite large. Since the smoothing capacitor of a SMPS can only charge when the input voltage is greater than its stored voltage, as illustrated above the instantaneous line voltage is below the voltage on the capacitor most of the time. Consequently, as illustrated below, 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 rectifiers conduct current only for a small portion of each line half-cycle and only during the ascending portion of the supply voltage waveform - which pulls the current out of phase with the voltage. And since, during this very brief charging period, the capacitor must also charge fully as can be seen in this illustration above, large pulses of current are drawn for short durations. The end result is that the current drawn from the mains is a series of narrow pulses that peak before the voltage peaks and whose amplitude is 5-10 times higher than the resulting DC value. As such, SMPSs are a very inefficient power supply.
The inefficiency of SMPSs is apparent if we contrast the 300W incandescent Fresnel light in the slide above to a 4’-4 Bank Kino Flo. The 4’-4 Bank Kino Flo uses four 75W T-12 tubes for a total wattage of 300W (same as the 300W Fresnel. As the slide below indicates the current drawn by the 300 Fresnel on a power meter is only 2.44 amps and it’s nice sinusoidal waveform has a crest factor of 1.3. If we look at the current drawn by the 4’-4 Bank Kino we see it draws 4.12 amps and draws a distorted waveform with a much higher crest factor of 2.5. Because it draws current in short bursts to refresh its’ smoothing capacitor, it must draw more current (4.12A) for the same wattage (300) as the 300 Fresnel (2.44A.)
One measure of the inefficiency of SMPSs is “power factor.” To explain power factor, let’s take a second look at our 300W incandescent and 300W Fluorescent Lights. The significant difference in the amplitude of the current drawn by them can be attributed to the 4’-4 Bank’s relatively ineffective means of using power. Because it must charge its smoothing capacitor in short bursts, it draws significantly more current for roughly the same wattage as the 300 Fresnel. 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 the 4’-4 Bank: there is what we call “real power” or that which generates work (measured in kilowatts or kW) and the reactive power or that which does not contribute to the work (measured in kilovoltamperes or kVA.) The sum of the real power and reactive power is called the “apparent power” (also measured in kVA.)
Power factor is the ratio between real power (in kilowatts) and apparent power (in kilovoltampere). The favorite analogy film electricians like to use to explain these terms is that if apparent power is a glass of beer, reactive power is the foam that prevents you from filling it up all the way, so that you are left with less beer or real power. In other words, the thirst-quenching portion of your beer is represented by kW in this illustration. The foam is represented by kVAR. The total contents of your mug, kVA, is this summation of kW (the beer) and kVAR (the foam). In our beer mug analogy, power factor (pf) is then the ratio of Beer (real power) to the entire volume of the mug (beer plus foam or apparent power.) In the case of the 4’-4 Bank Kino, its’ apparent power of 490W (kVA = 4.12A x 119V= 490.28W) is the actual power expended by the fixture, the load that the generator and cable must support.
To determine that this is in fact the case with our respective 300W loads, let’s now take power factor readings for each. In the case of our two 300W light sources, the 300 Fresnel would considered to be more “efficient” because its kW (.29) and kVA (.29) are equal and so the power factor is 1.0 (the more efficient a source the closer the apparent power will be to the real power and the closer the power factor will be to 1.0.) Where the 4’-4 Bank Kino takes 4.12 Amps at 119 Volts it has an apparent power of roughly 480W, to generate 310 Watts of light (KW), our 4’-4 Bank Kino ballast wastes roughly 35% of the power that it uses in charging its’ smoothing capacitor and has a pf of .65.
How do we account for this wasted power? Where does it go? To answer that question let’s look at the harmonic currents generated by each load. When compared to a linear load like a 300W Fresnel, that generates virtually on harmonic distortion of the current waveform, a non-linear load like the 4’-4 bank Kino generates considerable harmonic components that distort the waveform. The harmonic components of the current drawn by the 4’- 4 Bank Kino produce no useful work and therefore are reactive in nature, but none-the-less are part of the load that the generator and cable must support making them tougher on power supplies than the 300W Fresnel light. These power quality issues have been vexing film electricians for years, to learn more about how we have learned to remediate the adverse effects of harmonics, read a white I have written on the use of portable generators in motion picture production available at http://www.screenlightandgrip.com/html/emailnewsletter_generators.html
Guy Holt, Gaffer
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