PFC Math - Page 1
Forum section: electronics / beginners

author: TheDood
author_rank: Regular Contributor
timestamp: December 21, 2019, 01:00:58 pm
content: Energy in Capacitor:J = (1/2)·(F·V^2)P = W = V·AW = J/sI'm thinking if I were to divide each half sine wave into X amount of equi-power sections, let's say 14 sections per half sine wave, that I could then sense the V in the storage cap at every zero cross pulse to determine the nessecary power needed to be flowed per half sine wave, or the necessary duty cycle % for every section. With the wave divided up into equi-power sections or sections of equal area the frequency will be dynamic with it having greater time periods near the 0 points of the curve than in the middle.Its been awhile since calc days, but I'm trying to integrate the sin function with respect to time to get total relative power. From this total area under the curve I'm trying to divide it into X amount of sections, lets say 14, and then solve for (t) setting integrals equal to 1/14 areas of the total. Then after solving for (t), plug (t) back into the instantaneous equation to find V(t) for each interval. This way I can create equal wattage intervals and sense V to trigger interval duty cycles.I'm not sure my integration or setup of the sin function representing a 120VACrms 60Hz AC is correct? I'm also not sure if I'm solving in the correct units?? Can someone look over my procedure and give me any confirmation or corrections?Example:1000uF 400V Storage Cap = 80JZero cross comes and the cap reads 300V, or 45J. I've set the PFC to carry out in (14) individual sections per half sine wave.35J/0.00833s = 4201W4201W ÷ 14 sections = 300.07W/sectionEach period has a predefined constant average V, so I'd divide 300W by each intervals average V to find the required avgI to provide 300W, then I'd compare this Current nessecary to the possible current flow per interval avgV due to the total ESR of the entire cap charging cct, to then arrive at a % duty cycle calc for each interval. Because the time intervals are dynamic or vary in time duration from interval to internal, once a % "On" duty cycle calc was found for 1 section it should be able to apply to all sections?

author: unitedatoms
author_rank: Frequent Contributor
timestamp: December 21, 2019, 01:25:57 pm
content: The power factor is best at non reactive load for mains frequency. Not best for constant power over time. Consider controlling amount of power per interval proportional to what equivalent resistor absorbs for given interval's rms voltage.

author: TheDood
author_rank: Regular Contributor
timestamp: December 21, 2019, 03:03:43 pm
content: Quote from: unitedatoms on December 21, 2019, 01:25:57 pmHmmm @unitedatoms, Im not sure Im following.Also the integration or math I posted wouldn't be power, but rather V*s? For power Id have to find the peak power (pk V * pk I) and use that as my Amplitude in the sine function, I think. I think you're saying to use a resistor to sense when current has flowed sufficiently per interval rather than sensing line V and managing via anticipated/calculated/expected V rather than resistor feedback? Still use the same dynamic frequency, ie longer intervals near zero points ect but multiply the intervals rms V by the calculated I measured across R? So Id conduct until a certain % drop in V occurred? Thanks for the reply.

author: rstofer
author_rank: Super Contributor
timestamp: December 21, 2019, 03:48:51 pm
content: The integral of sin(x) is -cos(x) plus, perhaps, a constant of integration. If you want the area under some portion, say 0 to PI just solve for -cos(PI) - (-cos(0))

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2019, 09:32:58 pm
content: This is probably an alright, detailed explanation:https://www.st.com/content/ccc/resource/technical/document/application_note/72/e5/be/2c/74/20/45/bb/CD00195944.pdf/files/CD00195944.pdf/jcr:content/translations/en.CD00195944.pdfQuote from: TheDood on December 21, 2019, 01:00:58 pmYou're missing:- Definition of power factor- A measure of distortion- A load (you're just charging a cap??)- Variation in supply voltage and load currentAll of these are solved by traditional approaches like the above, so it should be quite fruitful to study and understand them.You hinted elsewhere that you might implement a control in Arduino or something. Judging by your current state of knowledge, this is doomed to fail, in a wide variety of hilarious ways -- the practical result being a lot of magic smoke, and possibly injury.Consider this: I'm an expert in this subject, I know everything from the finest layout parasitics to the highest control schemes (well, some of them anyway; I get the jist of others but have yet to implement them). And despite all of this, I have yet to implement a fully digital, direct microcontroller-based, control for any kind of power supply. (I did do an FPGA-based one, but that's still easier than software.)This is very much on the Dunning-Kruger spectrum: you will, at the very first, feel ignorant of the subject; next, you will gain confidence as you gain familiarity with immediate material, but remain ignorant of all the other materials they connect with. This is where one "knows enough to be dangerous", and gets overconfident. You literally don't know what you don't know. If you continue, you will hopefully discover those connections, expanding your perspective massively,In short: create a lesson plan for yourself. Break it down to the component parts, and understand them. Explode the edge cases, where a given subject hits its limitations, and where it connects to others.A rough outline for this project might be:- Basic circuits, KVL, KCL, analysis- AC circuits, steady-state analysis (phasors, complex numbers), transient analysis+ Likely prerequisites: calculus, differential equations- Switching power supplies: types, topologies, operation and waveforms, intro controls- Controls: feedback, pole-zero analysis, stability, compensation; cascaded and nested loops- Digital controls: DSP theory, pole-zero equivalence (Z and F transforms), PID controller, implementation+ Prerequisites: a programming language, on a hardware platform (e.g. ASM or C on AVR); familiarity with the platform (read and understand the datasheets, operation of the CPU and peripherals); hard real-time computing; interrupt-driven IO, and atomic access, race conditions and so on- And some E&M to understand layout parasitics wouldn't be a bad idea either.With intensive and directed study, expect to spend a month or two on each of these items. And that's if you learn quickly; the average case where these would be covered in college courses several months long will be more typical, and even longer still if poorly directed.So, expect your whole project to take at least half a year, more likely several. Budget accordingly, in terms of time, materials, money, whatever. Don't be afraid to pick up some college level courses to cover the more opaque or heavy topics. If you have more budget, you could get a tutor to provide additional direction.Or if the end product is really your ultimate goal, you can ask others to do the grunt work -- probably at a similar budget to the accelerated plan above. I would expect a project like this to take the better part of 2-6 months (or you can always go higher)by professionals, depending on how much optimization is necessary to push efficiency and such.Without this, just plodding along -- you can certainly arrive at something that works, but you won't have any of the tools necessary to understand reliability or efficiency, and how to improve them. Let alone real-world problems like manufacturability and production, electromagnetic compatibility (EMC) and safety (UL/ETL).Case in point: the big project I was working on through highschool was an induction heater. This is essentially a resonant power supply without a rectified DC output; instead it heats metal directly by magnetic field. I started with small scale tests, blew up a few transistors, honed my knowledge of circuitry and theory, and eventually (some years later), demonstrated a supply at 5kW, with frequency, phase and current controls, and overcurrent and desat fault protection. (To this day, most of the home-grown designs out there on the internet, are open loop and unprotected; they're a minefield to operate, not practical to do real work with. It's easy to see why a, say, $3000 Chinese power supply is such a good deal, when you just want to make some metal hot.)Tim

author: TheDood
author_rank: Regular Contributor
timestamp: December 22, 2019, 12:04:05 am
content: Quote from: T3sl4co1l on December 21, 2019, 09:32:58 pmThanks Tim,That's quite a daunting list of topics lol. I'm not sure I know all the technical names or ideas attached to their labels but I think I'm comprehending the issue of PF in a basic way.Its my interpretation that PF is implemented due to the stress on the mains supply lines when everyone is pulling their entire load energy at peak. Normally a cap charges up but it wont until it's V is exceeded, thus charging only happens at peak I/V without PFC. If everyone were pulling their load energy at peak, theoretically, the power transmission lines would have to be bulkier than needed if power were consumed in a more peak like way vs a more constant average draw. You have to have thick gauge to deal with peak current vs smaller gauge to deal with current that's been averaged over the cycle. This is why I'm assuming capacitive droppers are only allowed under 75W. It flies in the face of the 120VAC power transmission principle. Instead of running DC we run AC so current doesn't need to be as high and lines don't need to be as big and power can be transmitted further by stepping up V ect, but when using a capacitive dropper to supply 75W of power you're pulling much more current per line (lets say 75W at 40V, or1.875A) compared to (75W/120V, or 0.625A).Its my understanding that PFC is implemented as a way to comply with the way the grid is setup and designed. That its an ideology based around a collective effort to maintain grid efficiency?Power is what we care about. Joules/s. Components can transform Voltage (J/C) and Current (C/s) but J/s is what we're after in terms of Pin to Pout. A load will draw energy from a cap at the rate equivalent to the load energy demands. The load switching side will convert V & I pulled from cap to the correct V & I needed by load but the energy will be the same. This is what an inductor + switch + cap does. Takes low V and current and transforms it into higher V lower current? The coil is charged magnetically and once the voltage is removed the field collapses and the energy that went into creating the magnetic field rushes to find a pathway to travel through, ie the capacitor.In order to supply the Load side with power, Id think you'd want to charge the storage cap to its peak and hold it there as best you could. This means attempting to fully charge the storage cap at every interval of AC supplied. If I'm aware of the exact power needed to charge the cap per AC cycle, then I should be able to determine the duty cycle of the 14 sections (or whatever #) needed in order to correctly distribute the required power over the entire AC wave instead of just topping it off at peak. One thing among many that I still need to look into (lol I'm sure it's in your list you've supplied) is the relationship between current flow, and coil discharge V. But if I'm able to learn the journey of the J's at input to the J's at output then I should be able to adjust the PFC sectional duty cycle to equate the 2 over time and achieve corrected PF?I didn't know how practical an arduino would be. If it would be capable of computing the necessary shut off times per interval fast enough. I'm not sure it will work, you're right, it may fail hilariously lol and then Ill be forced to use a typical controller IC. I figured at under 20kHz that an arduino would be able to keep up but idk, there's lots of stuff I need to research up on still. Forums are my tutors and I appreciate your pointers and hintsI've taken calc i, ii, iii, & diff eqs but it's been awhile since my undergrad engineering days lol I've a tiny bit of a CS background but bolstering that as well. When you speak about the confidence vs naivety I definitely am a victim of that but it seems its a natural outcome when making new personal discoveries and good to be aware of though its a more or less uncontrolled reality surrounding learning in general imo.After a quick view of the link posted it looks like my interval spacing is backwards, lol, but generally speaking if I'm able to know the power left to fill a cap, and the amplitude and Hz of the incoming waveform, I should be able to average the necessary power needed to "top off" cap in the X # of intervals that I divide the AC wave into? That there's a maximum amount of power and a maximum possible peak current given by the ESR of the coil shorting/charging cct, and that this maximum can be used to determine the duty cycle % per interval? Multiplying max I by max V will give a sine function with an amplitude of peak power and that this equation is the 1 that I want to integrate & divide into 14 equal parts (or whatever number is deemed most efficient per given target PF and switching losses) and solve for (t) at each progressive area (1/14A, 2/14A, 3/14A, etc..), and then solve for V using the (t) values found from the Psin function and plugging into the Vsin function?A lot of the Q's are rhetorical, and you've given me a link to look into so Ill pruse over your comment and see what I can link together. Thanks.

author: unitedatoms
author_rank: Frequent Contributor
timestamp: December 22, 2019, 01:19:56 am
content: My point is, that perfect Power Factor is not the same thing as constant power across 14 portions of single half wave.Perfect power factor is when each of 14 intervals energy is equal to energy delivered to equivalent resistor.For example for 300W and 120V RMS, the resistor is 48 Ohm. That is perfect power factor.That means the energy for interval 1 = Integral[0..1/1680] of ((Sqrt(2)*120*Sin(2*3.1415926*60*x))^2)/48, where x is time in seconds over interval from 0 seconds to 1/(60*2*14) seconds.Equals 0.00593.. Joules.For second interval x will be for different span. Equals 0.0401.. Joules. Etc.Total over whole 2 half waves will be 5 joules, meaning at 60 Hz it is 300 Joules per second, or watts.For capacitor with zero volts of initial charge the energy, instead of resistor, come with method of limiting the end energy to this values: 0.00593.. Joules for first interval, 0.0401.. for next, etc.The useful calculator can be this one:https://www.mathcalculator.org/integral-calculator/

author: TheDood
author_rank: Regular Contributor
timestamp: December 22, 2019, 09:30:38 am
content: Quote from: unitedatoms on December 22, 2019, 01:19:56 amThanks unitedatoms,P = (V^2)/RSo essentially you're saying to not change the interval time spacing, but keep them all equal, t1=t2=...tn, that's why your Joules/interval increase per time step? I've seen some sin waves divided up with varying intervals of t, but they show bigger intervals in the middle of the wave vs smaller intervals towards the zero points, I can't seem to reconcile why this is, and not the inverse with smaller intervals in the middle vs longer intervals near zero points? (Btw seeing the 48 on the bottom, and the top squared, was confusing for a second till I tracked down another expression for P lol)I'm thinking I have to determine the efficiency of coil charge and collapse. I've been assuming that the energy transferred back into the line from the collapsing mag field is equal to the energy that was expelled through it during charging, ie if a coil were charged at 7V for a duration of 10ms, given the coils ESR there'd be an equivalent amount of J's that can be calculated during that event and that once the cct is broken or V is removed and the field collapses, that the calculated J's found during the charging time interval is equal to the delta J calculated in a capacitor pre collapse and post collapse. Perhaps my assumption that (aside from cap ESR losses) the energy required to charge the mag field will be transferred at 100% efficiency, and equal the energy added to the capacitor, is incorrect?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2019, 11:42:30 am
content: A sufficient definition for unity power factor, is to draw current proportional to the instantaneous voltage.That is why PFC controllers almost always use a multiplier function.The general definition of PF, is the ratio between apparent and real power.Apparent power is Vrms * Irms, and real power is avg(V*I).When V and I are out of phase, or contain harmonics in different proportions, the RMS of each will be higher than the real power draw would suggest, and PF will be less than 1.This is not a very prescriptive formula -- you can measure or calculate the RMS and average of various waveform pairs, but you can't so easily solve backwards for the one pairing that gives unity PF. (Eh, actually you probably can, something about the sin^2 x <--> (1 - sin(2x))/2 identity should be involved.)These limits keep the mains waveforms sinusoidal and in phase. Passive rectifiers create harmonics, which increase losses in transmission equipment (especially transformers). Phase shifts draw current (causing losses) without delivering meterable* power. Both reduce the capacity and efficiency of the system.*To residential customers, usually; industrial customers may have PF based metering though.Harmonics are also undesirable on a 3-phase system, as multiples of 3 add constructively, which can draw nasty neutral currents, or develop N-GND voltages.So overall, due in part to how we have designed our power transmission systems (i.e., historical reasons), but also in part because motors are still a huge part of power draw (polyphase is self-starting, does not deliver torque ripple, and 3-phase saves a bit on wiring capacity compared to 1 or 2 phase), and there isn't much real competition for plain old transformers -- these are why high PF is desirable.You certainly can't draw "equal power sections" because near zero crossing, the current would have to go towards infinity. A current draw of csc(t) has an insanely low power factor.Even if one were to draw constant current (as a choke-input filter does, for L --> inf), that only gives a modest power factor, because the square wave current has harmonics which the sine wave voltage lacks. Those harmonics account for 43% of the square wave so the power factor would be around 0.57.So we can't do that either. As it happens, we try to emulate a resistor (which has PF = 1 by definition). Our load might not be a constant resistance, so we have to change the resistance in real time. And obviously, we can't change it very quickly, otherwise we'd be dicking up that nice smooth sine wave current we meant to be drawing. So the loop bandwidth is intentionally very low (a few Hz), and the output filter cap relatively large.And so, very quickly you can see, it becomes very difficult indeed to talk about these systems, without also understanding control loops, how to set current draw in a switching circuit, and so on.A riddle for you: in a PFC system, if the output voltage, and load current, is ~constant, what will the PFC's output current waveform (into the filter cap and load) look like? What is special about it (or not very special, for that matter)? You should only need to use the facts given in this post, I think(?), though that's probably not much of a hint I'll admit...And from that waveform, what can you deduce about the minimum required filter capacitor, given some ripple (peak-to-peak voltage), mean output voltage, and load current?Tim

author: Siwastaja
author_rank: Super Contributor
timestamp: December 22, 2019, 12:59:23 pm
content: I don't understand the idea of calculating equi-power time-divided sections, or doing any time sectioning at all. For PFC, the formula indeed is:I(t) = V(t) * k,where k is a multiplier to reach your desired average output current. It can be based on a slow feedback of the output voltage, for example, for the simplest case.The algorithm should be trivial; the devil is in the engineering details like understanding component ratings, designing the layout, etc., as usual.The PFC thing becomes interesting when you are tasked with increasing a power factor of a certain load while minimizing component count and maximizing efficiency. Then it may be something else than the classical extra boost PFC stage. The idea is still the same: you try to draw a current which is directly proportional to the instantenous voltage.I'm guessing that you have a basic misunderstanding that a constant current during the sine wave period is optimal for the distribution network. This is not the case.

author: TheDood
author_rank: Regular Contributor
timestamp: December 22, 2019, 05:33:25 pm
content: Quote from: Siwastaja on December 22, 2019, 12:59:23 pmQuote from: T3sl4co1l on December 22, 2019, 11:42:30 amWhoo doggy lol you guys are awesome! Thanks, I can see its going to take me a bit of a paradigm shift (I think). This guy is what I started watching today and will see if I cant gain a bit of understanding by the next time I come back.Siwastaja,I think Im following what you are saying, but still a bit confused. Its my understanding that current is C/s, and voltage is J/C. Lets imagine that the time interval from 0 to 1/2Pi is 1s, and the amount of possible Joules transferred is 10 (based on V and I of wave). So 10W. If I were to divide the wave into sections of 1J each, Id need to have a greater time interval near the 0 point than in the middle/end. This is because the V is not not very large (J/C), and so more time would be needed to flow enough coulombs to accommodate 1J. This demonstrates what I think you were saying about current flow being proportional to V, Id need more time at lower V's to flow the same amount of energy? This is why I was thinking about dividing the wave into sections of varying time intervals. If I know that my load (and cap before the load) draws 5J/s, then Id need to adjust duty cycle to 50% in each interval. If I didn't adjust the duty cycle my current draw would spike till the storage cap was full and then Id have an imbalanced current draw, or Id have more current draw in one or 2 sections compared to that of the others sections of the wave, and this would then result in poor PF? If I know what the cap needs in terms of J's to be fully charged then I can adjust the %duty cycle of each interval so that the transferred power resembles the sin wave and not only certain sections that flow spikes.Tim,Ahh, yes, not equal power, but equal Joules, but not intervals of equal J/s. Perhaps this is not the way forward though. You and Siwastaja have both have been hitting on the proportionality of Current to V, this is where Im getting confused because isnt the force wanting to flow I proportional to V all the time? Im assuming we are drawing bits of current and then boosting it onto a cap in bursts. That this way of burst charging a cap negates the leading phase shift compared to if the cap were allowed to charge without interruption? Also this type of sectioning the wave off into bits of current allows for the mitigation of only peak times transferring power, ie the power is transferred throughout the entire wave. Harnessing the property of an inductor to step up V and deliver I when mains V is too low for load to flow? V is set by mains, so Power is adjusted by current regulation, and current is C/s and dependent upon V so ... idk, lol wouldn't the time the switch was on and allowing current to flow match that of the V present during the on duration, ie the current flows more at higher V and less at lower V, so wouldn't the proportionality be maintained, just the amount of C flowed, and the sine function of the V will act upon current regardless the interval spacing, so really only current or coulombs is being regulated, but still in proportion to instantaneous V?I think I am misunderstanding. Does the equation posted, suggest that we impede current by an opposing V? K is some multiple needed to reduce delta V to an amount needed to flow the current given by the load R? I feel like I'm over complicating this lolPer your riddle, Tim, Im guessing the input current waveform and input V waveform match in proportionality and that the constant current on Load side is achieved by a separate process down the line. That while the I wave and the V wave are both sin waves in proportion to each other, that the amplitude of the I wave is in proportionality to the Load current draw or Load constant current? As far as filter cap sizing, im guessing youd need a large one, like you said because C = (I*t)/ripple, and if your ripple is smaller than 1 then your C needed will be large?Im going to go and learn some more and Ill be back with new questions and hopefully some new knowledge stemming from researching your guys' insight/commentary.Riddles for the time being lol..Richard's father has 3 sons, "Snap, Crackle, and ___?"The more you take the more you leave behind, what am I?What is brown, has a head and tail but not legs?

author: unitedatoms
author_rank: Frequent Contributor
timestamp: December 22, 2019, 06:15:16 pm
content: Power curve will be double frequency sin(t) in shape and always positive, not negative, unless the load is reactive.

author: dietert1
author_rank: Super Contributor
timestamp: December 23, 2019, 09:14:42 am
content: Assuming the usual configuration with a bridge rectifier followed by a step-up DC->DC charging a large capacitor at above peak input voltage, you have two aims:a) Measure and control the input current to be proportional to input DC Voltage, Iin(t) = k * Uin(t). In order to do this you need measurements of Uin(t) (voltage divider) and for the input current Iin(t) (sense resistor). Input current is controlled by changing the duty cycle of the step-up converterb) Regulate the capacitor voltage at a certain value. This is done by adapting k: increase with load current out of capacitor. There needs to be a voltage divider for output voltage, too.The controller needs a multiplier and some filters to determine average input current (average over one period of step-up) and keep k roughly constant over one period of mains input. You can find all the details in the data sheet or app notes of any "single chip PFC controller". Nowadays there are digital controllers, too.Regards, Dieter

author: TheDood
author_rank: Regular Contributor
timestamp: December 23, 2019, 03:37:32 pm
content: Thanks Dieter,Instead of thinking of it as in reference to storage capacitor level, think of it as in reference to instantaneous load demands?So I'd sense my load I allowed to flow from storage cap, then find the corresponding mains peak I by setting mains Irms = to equivalent load I (you would have to convert load I to equivalent load I, like you would a transformer, power being conserved?) and then the ratio of (peak V to peak I) multiplied by the instantaneous mains V, can be used to determine time ON per period of sin wave division (sensed and ON till k·V(t)mains was met then off)?And the sensing would have to be done through a filter so to avoid large V fluctuations in the storage capacitor? Ie if large ripple were picked up by load current sensor it would give incorrect k constant which also needs reseting every 0->Pi sin cycle?When I'm picking inductors and the switching Hz, do I want the [5·(time constant)] duration to be equal to the minimum load current desired? So that the coil would be fully energized at even the smallest duty cycle%? Or do I want to use the time constant as a regulator, ie setting [5·(time constant)] equal to the entire period?

author: dietert1
author_rank: Super Contributor
timestamp: December 23, 2019, 05:06:56 pm
content: If you measure the instantaneous load current as well, you can work it into the controller to improve load regulation, but i am not sure it's necessary. Anyway the capacitor needs to be a buffer for some mains periods at high loads, otherwise there can't be PFC. PFC is relevant at loads above 75 or 100 W. The whole scheme is relatively simple because the boost converter works at a much higher frequency, so energy stored in the inductor is always much smaller compared to energy in the capacitor.Usually there are more converters with better voltage regulation operating from the capacitor. Those regulators should be able to work correctly at reduced capacitor voltage like 300 instead of 400 V.The inductor is determined by maximum load current and something like half the capacitor voltage as input. It's the most critical situation and then you want a nice triangle modulation without current breaks. With low load some integrated PFC controllers seem to implement burst mode or tricks like that. But that's more about radiated noise and standby efficiency. Again i recommend to study some data sheets. They explain how to dimension the power parts. For low loss the inductor needs to run hot, so it should not sit next to the capacitor. And the boost controller needs a second, very sturdy capacitor on its input side (between bridge and inductor) to buffer high frequency current peaks.Regards, Dieter

author: TheDood
author_rank: Regular Contributor
timestamp: January 01, 2020, 08:03:55 pm
content: Trying to determine the amount of coulombs flowed during inductor discharge compared to the energy stored during charging.If I charge an inductor for 1 time constant, in 3 time constants it will have discharged ~95% of the stored energy, but how many coulombs will have flowed? Can I equate coulombs flowed during charge up to coulombs flowed during discharge?If I put a non linear load after and in series with a discharging inductor, the voltage across the [inductor + nonlinear load] during discharge should equal the Vf of the nonlinear load at initial inductor discharge (steady state current flow), and then the voltage across the [inductor + nonlinear load] would drop at the nonlinear rate of the load Vf per load current draw (and the load current draw would drop at the inductor discharge rate di/dt?). Is the rate of change in current in a discharging inductor the same function as the rate of energy discharge in an inductor? If I'm putting a load between the inductor and the return line, will the inductor essentially create the required V to satisfy the di/dt of the discharging inductor?I'm starting to morph my concentration more to coulombs than joules but trying to comprehend both. At first glance it seems like you have to burn much more wattage during the inductor charge duration than the amount of energy actually stored or transferred during discharge. In order to store the amount of energy needed to power a cct your current flowing through the switching inductor has to be considerably more than the intended amount of current through the load (which the inductor is storing/dissipating power to). I'm probably missing something..Attached is the spreadsheet I'm working on (disregard cap & inductor suggestion section, dark gray). Anything orange is an input box, anything green I just thought as arbitrarily significant. The switching Hz is populated automatically based on a charging duration of 1 Tau. Perhaps I should create charging times greater than 3 Tau, or the time needed to discharge ~95%, but I thought that the charge function during the first time constant was more linear and would allow for a more linear power manipulation under varying duty cycle %'s compared to charging for longer durations than 1 Tau. L/Ω inputs which result in calculating switching Hz greater than 3500 pulses per AC sine cycle won't populate graphs because I only dragged formulas down 3500 cells... Also, the Joules vs time graph, is that right? I thought it'd look like a sin curve but with greater amplitude, was not expecting the graph I got..EDIT:Ok, I think I got a little further, I have to determine what the f(n) of R will be of any non linear load in series with inductor, given the forward current vs forward voltage curve of the non linear load. This would effect discharge times and the rate that Coulombs flowed because Tau would be smaller initially and then larger as R dropped, and then back to large as I(L) dropped??Also Joules per Time Graph sub-labeled "(V/I)/s per s" is supposed to be sub-labeled as "(V/I)*s per s"

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 02, 2020, 01:45:59 am
content: Why coulombs? Inductors are not charge invariant. An inductance charged to nonzero current then shorted, will maintain that current forever, delivering I*t charge across a given point in its path. Infinite as t-->inf.Capacitance is charge invariant: around a cycle, delta Q = 0. Analogously, inductance is flux invariant: around a cycle, delta Phi = 0 (Phi = V*t).This is why, when we draw inductor switching waveforms, we draw rectangles of equal area, and thus determine peak current (or current change), duty cycle and voltage ratio.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 02, 2020, 02:15:10 pm
content: Quote from: T3sl4co1l on January 02, 2020, 01:45:59 amThanks Tim,Lol always giving me new stuff! Thanks, I'll have to look up charge invariance and phi, but heres my response in the time between..Coulombs because coulombs seem to be the limiting reactant in the process of powering non linear components. Ie, after a certain V is met, power consumption is more closely correlated with I than V.Example:Energy stored in an inductor during charge up = 100J.If that 100 J is realized as 1C at 100V, compared say to 50C at 2V, that makes a difference? If my nonlinear load has a Vf of 1.5V, it seems the 100J discharge depicting more coulombs flowed is the scenario I'd find more favorable to my particular needs. This is why I'm starting to look more at coulombs than Joules, though I'm just not sure..The energy stored in an inductor is 0.5L×(I^2), and I've been trying to equate that to coulombs flowed because that is what I'm trying to control for. I can't flow coulombs when load Vf is greater then AC V(t) so that's why we use the boost inductor (instead of pure buck topology) but I'm trying to calculate inductor size, switching Hz, and duty cycle per my desired coulombs flowed on the other side. Sure they'll need a certain energy per electron to flow, but I think if I update Tau with the R(I of NL load) that calculating J is less than necessary as it's the qty of coulombs ( does qty flowed during charge up = qty flowed during discharge?) and the rate at which they're discharged from inductor that I'm concerned about (because I'm assuming inductor V will adjust to whatever is needed in order to satisfy the documented rate of change of I of an inductor during discharge) The question I'm trying to answer now is, 'is the total qty of coulombs flowed during the charge up stage the same qty to be expected at discharge regardless the load placed in series?' Does the inductor create a force, equivalent to satisfy the documented di/dt during inductor discharge, regardless the Ω placed in its path, or is the documented di/dt of a discharging inductor, actually due to a documented dJ/dt of a discharging inductor, and that the common di/dt graph (refer to attachment) looks the way it does due to R being constant?The rate at which the coulombs are discharged from the inductor as well as the qty being discharged from the inductor, compared to the V ripple desired for load, will determine my capacitor size (small capacitance increases cap V lots with a little qty of coulombs added to plates, while large capacitances increase cap V very little with a little qty coulombs added to plates), but the AC Hz will be the thing to calculate capacitor size for Iripple from (store enough excess during amplitude to cover lulls of zero points)? Preliminarily, it seems capacitor has to be huuuge to stave off current ripple over 120Hz input, but still mainly working on following and understanding the path of the coulombs from mains to inductor to elsewhere atm.EDIT:After looking at the graph a little more, (coulombs in) does not equal (coulombs out), but is the ratio consistent between in/out regardless the type of load placed in series at discharge? Can I calculate switching Hz, and duty cycle based off the ratio of in/out and my desired load I?

author: dietert1
author_rank: Super Contributor
timestamp: January 02, 2020, 03:17:59 pm
content: When you take the basic situation of 200 V input voltage and 400 V output voltage, then the "charging" voltage of the inductor is 200 V and the "discharging" voltage also (400-200). For one booster cycle those voltages are nearly constant. So is dI/dt, except it's negative during discharge. Input current will have a nice symmetric sawtooth modulation. It will flow from the input during "charge" (increasing) and "discharge" (decreasing). Effective input current will be twice the effective output current into the cap.If you succeed to model this situation, you will also succeed to model other input voltages. For example at 300 V input voltage "charging" the inductor will be three times faster than "discharging" it. So the symmetric sawtooth becomes more like a triangle. This time the effective booster input current is about 4/3 of its output current.This example demonstrates how output current of the PFC booster cannot be proportional to input voltage. If you want PFC, you need to control the input current. The booster output current will have a strong 100/120 Hz modulation and the buffer cap does the rest.Regards, Dieter

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 02, 2020, 04:09:11 pm
content: You still don't get anything for power factor, from looking at coulombs or joules. Only thing PF cares about is current averaged over the filter time constant, and that current being proportional to the line voltage.Easy way to control inductor current to track an average: a hysteretic controller. Switch on when current is below threshold, and off when above. The duty cycle is just whatever turns up, you don't need to know or care what it is. Frequency either.You do of course care about minimum and maximum times, on and off. Frequency can't go too high (at high voltage and light load) because of switching losses. Frequency can't go too low (at low voltage) because of fixed filter cutoff. On and off times are limited by the gate driver and other logic. So you should probably have some logic to account for that, to limit pulse widths and/or frequency to keep things reasonable. Which of course will screw up the current, you're no longer switching at the expected points; which leads to further logic, like adjusting the hysteresis band, or implementing pulse skipping, or you might go with average current mode control instead, etc.There's BCM (boundary control mode) PFC, where one threshold is varied, the inductor peak current (switch-off point), and switch-on is timed when inductor current falls to zero. The average of a zero-based triangle wave is peak/2, regardless of the duty cycle. Easy. Frequency isn't too wild, though still gets too fast at low currents, for which some holdoff is needed, and subsequent adjustment of the setpoint.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 02, 2020, 06:53:07 pm
content: Quote from: dietert1 on January 02, 2020, 03:17:59 pmThanks Dieter,So if I had a 15Vinput & 31Vload, and it took 1s to charge up an inductor to the point that it was flowing 1A, that that amount of energy stored, is equivalent to enough Joules, such that during discharge, my load would flow 1A down to 0A, and deteriorate its instantaneous current flow at the same rate that the current increased in the inductor during charge up? So 0-1s the inductor charged up to reach a final current of 1A, then discharge begins, and current through load starts at 1A and drops to 0A from time period 1s-2s?If 1Vinput instead of 15Vinput were used (still 31V load), it would mean that charge up time would be 1/30th of discharge time?I'm thinking that in order to draw current in a sin wave, as well as keep current constant through my load, that I'm back to a variable frequency switching solution (but all duty cycle pulses same time duration, because L is constant and I'd want to charge L to same % of max with every pulse), varying duty cycle to match desired current. Creating more pulses near zero points while less near peaks. Splitting the waveform up into an arbitrary # of equi-coulomb segments and then determining Hz per segment, or the amount of pulses needed per segment to equate to the average max coloumbs desired to flow. Then use the inductor to squirt the amount of coloumbs I wanted per segment of sin wave onto a capacitor that powers the load? The current waveform would match the proportionality of the voltage waveform due to the constant R during inductor charge durations, but the amount of coulombs transferred or flowed to load would be constant because in times of greater Vinput more coulombs would want to flow so a larger off time would be desired per period, compared to zero points where more pulses would be desired, and less time off per periods. In each pulse, the current rate pulled from mains would match the sine wave proportionality, but the change in pulses per time would create a constant string of coloumbs for my load?I'd set the R in the inductor charging loop to match Load R (as best as possible so that input I wasn't distorted so much greater than load I if/when R is small), then I'd size the inductor by setting the max pulse duration of the variable switching frequency to 5 Tau (?, maybe 1Tau?) and solving for L, or something there abouts. The smallest variable frequency will also play into the L sizing, and then perhaps the inductor charging R will have to be less than load R because we're trying to accumulate enough coulombs during peak to cover the off portion of switching period, so depending on discharge times maybe half as big?If not counting the time off and only connecting the dots of pulse current per V, the input waveform would look like its consuming more I than it is, when being billed for meter costs, what would I be billed for? Having a greater "connect the dots" current compared to actual current is what correcting PFC is all about? So I would be back to bad power factor if the "connect the dots" waveform showed higher Irms than actual I?

author: TheDood
author_rank: Regular Contributor
timestamp: January 02, 2020, 07:04:07 pm
content: Quote from: T3sl4co1l on January 02, 2020, 04:09:11 pmThanks Tim,Oh man, didn't see this until now, but I'm going to have to take a break to digest this info for a second (at work). Ill look up the terms BCM, invariance, hysteretic controller ect, and get back. Quickly though, if my "connect the dots" current is greater than my actual current draw, my PF is crap again, or is it a-ok?EDIT:Also, if Vpk is 170V or 170× greater than 1V, Ipk is also 170× greater at Vpk than I(v) when V = 1? So maybe 170× more pulses at 1V than 170V? Or time period at 170V is 170× longer than time period when V = 1?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 03, 2020, 02:08:55 am
content: Quote from: TheDood on January 02, 2020, 07:04:07 pmReality doesn't care if you think you can connect dots or not; all that matters is it averages out right.QuoteSo, pulse density modulation? Assuming each pulse can charge the inductor to a given peak current? Yeah, that works, you'll find it takes a far higher frequency than is practical though.Tim