Inductor Discharge Rate and Other Questions - Page 1
Forum section: electronics / beginners

author: TheDood
author_rank: Regular Contributor
timestamp: January 29, 2020, 08:43:43 pm
content: Perhaps worrying about the wrong stuff here?:Given a conventional single inductor boost topology, will the filter cap voltage impede a magnetic field collapse? Will the filter cap voltage effect the inductor discharge rate?If I've stored [0.5L×(I^2)] J, how long will it take to dissipate? Does a charged cap in the inductor discharge path effect the rate at which the inductor discharges? Or how much energy discharges?If Tau = L/R and 5Tau seconds equals full charge/discharge, then the amount of discharge time would be dependent upon cap V?If cap V was known, ending charging inductor I was known, and total J stored in inductor was known, can I determine discharge time?Is it important to determine exact discharge time or will it be less than charge time if impeding V is higher than input voltage?PFC:I'm trying to measure cap V at zero cross, then calculate remaining J's needed to achieve desired constant V of filter cap, then set PWM based on the needed J's to achieve the desired filter cap V. If Ω is kept constant during inductor charging then current will model a sinusoidal function as input voltage does, but the amount of current needed per switching period to arrive at the sum total of J needed to fill the filter cap per switching cycle will be time dependent. So I'm trying to determine exactly how the energy stored in an inductor is released so that I can provide enough discharge time for the inductor to fully dissipate the stored energy onto the filter cap.Will ALL the energy deposit onto the filter cap? Or is some left behind in the form of hysteresis, or maintaining residual mag field?Is this an electrically efficient way of achieving high PFC, or boosting V? Ie will the amount of power consumed during charge up equal the amount of power discharged onto cap? I've seen some graphs showing PFC, but it almost looks as if they aren't discharging the inductor completely. Is discharging completely not advised?Curious:Also, if your inductor charging V is 170V, and your filter cap is at 170V, and if current doesn't immediately drop to 0 during initial discharge event, but rather drops to 0 along some function originating from the initial current down to 0, then right after the charging event has concluded and you begin to discharge, the change in current will be almost 0, but the filter cap is at 170V, so in order to maintain the current flow, or in order to have a change in current = 0, your voltage needed to flow that current will be 170V more than the impedeing voltage (cap charge level) due to the fact that it took 170V initially to flow the current, so you'd effectively have to be generating 170V + any impeding voltage to flow the same current? Is this correct thinking?Final thoughts:If I were to integrate a J/s curve based on sinusoidal V and I and constant load R to arrive at a maximum J's per cycle possible, then I should be able to use a static switching Hz and calculate my duty cycle % based on a [(Joules needed to fill cap) ÷ (maximum joules possible)] figure?So perhaps choose an inductor with enough inductance that at 50% duty cycle of a certain Hz, the time constant is such that at peak Vin, the inductor I = desired Irms × 1.41.So instead of using a higher R in the charging event to reduce peak current in the inductor, use a bigger inductor to reduce peak I through the inductor, given a set Hz and maximum of a 50% duty cycle?If this "Final Thoughts" section is sound then I can continue on with determining core specifics, wrap #, skin effect, ect, per the Hz desired.Please give any insight you see fitting. Thanks.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2020, 12:45:09 am
content: You need something more basic:V = L dI/dtIf you've fixed the voltage on the inductor (say by clamping that voltage with a diode and a relatively large value capacitor), and L is constant, then dI/dt is also fixed.For approximately-square wave voltage, dI and dt become ΔI and Δt, trivial to integrate as the ramp is linear.If you know initial and final current, and V, you can solve for Δt, the time required to discharge the inductor.I never liked the "magnetic field collapse" phrasing. It implies something different has happened. The inductor is linear (for the most part), nothing different happens. The same equation applies, regardless of whether its arguments have pluses or minuses. Apply a positive voltage drop (given a convention for which end of the inductor is positive and negative, and its current flow is positive when flowing from positive to negative), and dI/dt is positive. Apply a negative voltage drop and dI/dt is negative. Integrate V/L dt and you have inductor current.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 30, 2020, 02:38:49 am
content: Quote from: T3sl4co1l on January 30, 2020, 12:45:09 amThanks Tim,You've been helping for awhile now, I wish I was catching on quicker, sorry, I appreciate your insight.I thought that the voltage across the inductor reversed polarity upon the event at which the energizing voltage was "instantaneously" removed? Im getting confused on how to clamp the voltage if it's reversing?For a cct with a 12V source, single inductor, and switch:State 1:Open cctEnergizing V = 12VVoltage across inductor = 0State 2:Closed cctEnergizing V = 12VV across inductor = 11.99 VState 3:Closed cctEnergizing V = 12VV across inductor = 5VState 4:Closed cctEnergizing V = 12VV across inductor = 0VState 5:Open cctEnergizing V = 12VV across inductor = -12VIs this incorrect thinking?If the energizing V is 12V but I'm trying to push current onto a 13V cap, does discharge time drop in duration compared to the charge time? If I charged a coil for 1s at 12V and placed a 13V cap (cap charged to 13V) in its discharge path, the time of discharge is reduced from the time of charge up, correct? So that if 50% duty cycle were my max ON duration, and I was boosting V, that I shouldn't be worried about discharge time as it'll be less than the charge time?Also is my assumption that, the total energy stored in the mag field is being discharged completely or almost completely (depending on hysteresis) true? That if V = J/C, and I = C/s, that the energy will dissipate completely regardless the resistance/impedance in its path? Like a transformer maintains power but the V,I of the primary is different that the V,I of the secondary but total power is conserved (ideally)? The inductor charges up with whatever V, but it reaches a certain energy before cct is opened, and if a cap charged to a higher V is put in the inductor discharge path, that the energy discharged will still be equal to the energy stored, but less current will flow due to the higher V cap in its path? IE it will reduce its discharge time because current is maintained initially at switching event, but then would drop faster as more J/C were needed to discharge the energy?Attached is what I think you were saying to me, but I'm not certain. If it is what you were indicating, it looks like an additional buffer for the transient(?), but I'm still not certain on exactly why or how we get such transients if there's a route for the current to take that doesn't require such a high V. For example, when I simulate in LTspice, I often get large spikes at switch off events. This would mean that dI/dt were large. I'm not sure how you can have such a large change in current if the inductor opposes it? I feel like I'm going in circles lol... In fact this is where I thought Tau played a role? I don't see any talk about Tau ever, and its not seen in any of the equations posted surrounding my PFC commentary, does Tau not factor in as much as I think?Extra:I get confused and this is not as heavy on my mind but figured you could elucidate. I've read that current in an inductor wants to remain that current, even after energizing voltage had been removed, but then I've read that it acts as a spring in such a way that the current flowed through wants to reverse back the way it came upon voltage removal? Which is it?Also, attached is an example of the equation previously posted. How do they know that the current drops to 0 in the time they've decided to switch off? I see that the time off is 10ms, but what if it was 5ms or 100ms? How do they know it drops to 0 in 10ms? They integrated V/L(dt)?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2020, 03:29:17 am
content: Quote from: TheDood on January 30, 2020, 02:38:49 amThis phrasing suggests a circuit where the driving impedance rises suddenly. Like switching a relay coil. Where you don't care what happens when it turns off, you just want the current gone.Of course we don't want unconstrained voltage, that damages transistors. So even in this simple case, we use a diode to clamp the voltage, which keeps the driving impedance low. The inductor voltage is always under control, always well defined.In a switching converter, the inductor voltage is always well defined: either Vin when switched on, or Vin-Vout (for the boost case shown) when switched off (and discharging). There is a third state, if the inductor has finished discharging: voltage rings down to zero.QuoteWhere's the voltage dropping? There's no resistor shown in the circuit.QuoteThis is missing the output voltage. Is it 24V then?QuoteAssuming you're still talking about the boost configuration, no, discharge time is inversely proportional to discharge voltage. For a 13V output, V_L(disch) = -1V. The equal flux condition is V_L(chg) * t_chg = V_L(disch) * t_disch. If flux is 12Vs then discharge is 1V and 12s.If the switch turns on during the discharge, current begins at whatever it left off at. If the inductor is 12H, the peak current is 1A (12Vs / 12H = 1A; henries have units of flux per ampere). Say you turn on after 6 seconds: current starts at 0.5A, and an on-time of for example 0.5s is needed to raise it back to 1A. This is CCM.QuoteIf I_L = 0 at the end, yes.QuoteEnergy only dissipates if given a path to dissipate through. Inductance is lossless. You can short an ideal inductor at 0V forever and it will hold its magnetic charge (current and flux and energy). Superconductors show this.A regular conductor will eventually dissipate its energy through the voltage dropped across its resistance.In switching converters, we choose DCR to be small compared to L*Fsw, so that converter efficiency is worthwhile.QuoteAn ideal transformer stores zero energy. A nonideal transformer is a transformer in parallel (and series) with inductance. The elements can be analyzed independently.QuoteIt won't, actually. The inductor is in series with the DC source, which boosts its energy. You get some energy for free. Which is intuitive, as in the complete absence of switching, the output is still at least Vin.A buck-boost configuration (in essence, the output cap is connected between IN and OUT, not GND and OUT) does satisfy this condition.QuoteYou've shorted the inductor in reverse with a diode, so it can never discharge into the capacitor.QuoteThere are multiple inductances in a real circuit, though it's not apparent if your model contained them.The dI/dt indeed can be quite high, but not in the main inductor.Consider at the instant of commutation, you are asking the switch's current to drop to zero, and the diode's to step up from zero.The inductance of those components, specifically the loop between them, experiences that dI/dt, and if it has inductance (in a physical circuit, it necessarily must -- this happens to be equivalent to saying the circuit has nonzero length and the speed of light is finite), that loop will generate a proportional voltage.QuoteWhen a voltage is "removed", it goes to zero. In other words, the inductor has been shorted.QuoteReally poorly phrased prompt, the discharge voltage is undefined. They seem to be making the assumption that it is constant, in which case that will be true.They specify only a switch in the circuit -- a highly nonlinear component. The impedance changes from finite(?) to infinite. The risetime of a switch can be fractional nanoseconds; the dI/dt will be extraordinary. Immediately, the switch breaks down, arcing.In purely ideal terms, the switch goes to infinite impedance in zero time, the dI/dt is infinite, and the peak voltage is infinite for infinitesimal time. The voltage is a Dirac delta. This has infinite bandwidth ("DC to light" as they say), which is impossible.The initial setup is questionable as well: if the source is a Thevenin equivalent, then we can charge the inductor from zero with enough source voltage and resistance that the current reaches the desired peak value, and switch it then; that's fine, but it's not quiescent. If we set V / R = Ipk, then the charging curve is exponential and it takes infinite time to reach Ipk. (The problem does not specify how long it's been charging for, but infinity is an acceptable duration in a theoretical question.) We could also craft a waveform which charges the inductor rapidly at first, then holds it at 0V to maintain the desired current. In both of these cases, the outcome does not depend on when we throw the switch, which seems appropriate. However, if the source is not a Thevenin equivalent, but a true current source, we run into the same problem, because if the inductor's initial current is at all mismatched to the source's current, a Dirac delta of flux will be developed as the currents equalize. Which again is impossible.The usual textbook setup for this type of problem is to simply assume the inductor has a given initial current. No source is needed, we can simply invoke an initial condition. The switch is then shorting the inductor, holding its charge until opened.A more physical circuit would represent the components as nonideal equivalents: Thevenin source; switch of finite resistance (on and off), capacitance and speed, and perhaps breakdown voltage as well; inductor with DCR and parallel impedances (core or radiation loss, capacitance, etc.). We can make meaningful predictions in this case: the realistic equivalent to a Dirac delta is a sharp spike with its parameters (rise time, peak voltage, resonant frequency, ringing, etc.) fully described by the RLC values in the circuit.More generally, we are imposing limits on the bandwidth of the circuit. A Dirac delta has infinite bandwidth, but real circuits do not.Incidentally, real signals have an exponential frequency cutoff at some point, which is why there is currently no such thing as a "DC to light" squarewave, for instance.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 30, 2020, 05:10:36 am
content: Quote from: T3sl4co1l on January 30, 2020, 03:29:17 amOh man, thank you!Yes I forgot an R. Oops!I guess the fly back is not what you were indicating, I see what you're saying, at first I thought it would increase the cap V in parallel, but now I see it just shorts.I think what you said about 13V output and 12V in, and the voltage across the inductor being -1V, is finally clicking, but I'm still not sure. In the State 1 ect scenario, at the end I've a -12V across inductor, and you quoted and said I'm missing the output V. The total cct V at that point would be 24V because 12Vin - -12V across inductor = 24V? Or why did you come up with 24V and were you saying that the Vout of inductor was 24V or entire cct? (Refer to attachment for better clarification, sorry its sloppy, but notice V's above components to see if I'm comprehending, 1 and 5 are open cct, 2-4 are closed cct). Also there'd only be -12V across inductor if the dI/dt satisfied that, correct?As far as Dirac delta, and Thevenin, Ill have to look those terms up to grasp the last part of your post and I think Ill be picking at this for awhile. I think I caught a hint or 2, thanks.I'm still not sure what you mean by clamping the V of the inductor. The Vin is fluctuating at 120Hz, so to clamp the V or keep the V constant I'd imagine you'd have to have another cap but before the inductor? Initially I thought that this type of config would produce poor PF, with only peak charging times, but after some thought maybe this is what you're referring too?Ill work to quote and organize my questions better.EDIT-Imagine the R was before the switch in the 5 state drawing..

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2020, 09:30:37 am
content: Quote from: TheDood on January 30, 2020, 05:10:36 amNow I don't know what you're referring to. Because the converter doesn't have a resistor and now you've put a resistor in. And I don't know what you mean by that resistor, if it's supposed to represent the load or something. It changes the circuit and its response.QuoteWhoa whoa whoa. Don't even bother with Vin. 120Hz isn't changing very much from cycle to cycle when you're switching at 100kHz. Make a switching cycle work first. Assume it's DC.Later, change that DC and make sure you're hitting all the points you need to.TimP.S. Odd, what's a noae markup? It got put in my quoted text, but disappeared from this message it seems.

author: TheDood
author_rank: Regular Contributor
timestamp: January 30, 2020, 07:55:01 pm
content: Quote from: T3sl4co1l on January 30, 2020, 09:30:37 amI have no idea lolQuote from: T3sl4co1l on January 30, 2020, 09:30:37 amI just added a resistor to my hypothetical state scenario in the previous post, because as you said there wasnt a resistor, which meant that the voltage across the inductor would never diminish from the source or input V without another component in which the voltage could drop across, thus I added the resistor in my second try and added a sloppy 5 state schem to illustrate, but I placed R in wrong spot. I was also looking for confirmation on my comprehension considering the 5 state schem.Quote from: T3sl4co1l on January 30, 2020, 09:30:37 amOk thanks. Energy in = energy out is easy for me to understand that's why I was trying to determine energy stored in mag field vs energy needed to top off cap and setting pwm accordingly (because inductors are lossless, or close to, or so I've read). I have learned the process to calculate energy of a mag field but was uncertain exactly how much of that energy and how fast that energy was discharged onto cap. I figured parasitic transients be damned, as long as energy was transferred completely (though it seems I need to damp the transients to save the MOSFETs from damage, I still need to look into how to accomplish this). I think that this type of charging and then completely discharging is maybe considered discontinuous mode, while you were/are commenting based on CCM? I've also been talking about the time constant, and while that is important during the charge event, I've since learned that LC ccts have a resonant frequency rather than a time constant, but I'm assuming with the diode in there that the resonance only gets 1 half harmonic(?), or the resonance only gets to go from L to C once, it doesn't bounce back, so I'd have to determine the time for 1/2 of a period of the resonant frequency in order to determine discharge time?Ill try to work on CCM as it seems that's what's being proposed more so than fully discharging. Ill have to switch my gears a bit. I get confused because you guys say that its a linear ramp but I learned its a non linear function that EE's end up just aproximating as linear, at least during 1Tau seconds, or 63% of it's max charge level (talking about current through inductors), as well as was confused with a simple "k constant," you guys make it seem so simple lol but I'm trying to understand as a laymen. Ive more of a physics background than electrical, so that's why I'm breaking it down into units I know better, ie joules. For all the comments and posts I've read (not just my own threads) it seems I'm making this way harder than necessary, but for some reason I'm still missing the disconnect. All the PFC ccts I've seen take 3 spot measurements, before inductor, at cap, at ground or return line, and then control accordingly, but I can't figure why the operation of the chip is so complex to me.What exactly is the current out to current in relationship that EE's are using for this in regards to L value, Hz, and PWM? Does the V on the storage cap factor in? Do I need to worry as much as I am about discharge time? Am I trying to create a certain back emf to satisfy double input V, or does this happen naturally? Or am I trying to mng Iout of inductor? What is it that you guys are trying to control for, back emf V, or Iout? To me measuring cap V and setting PWM makes sense (but maybe inefficient or not correct) but I'd rather understand from your guys' perspective (or just at all).... If possible for me lolEDIT-The down trending portion of the sawtooth CCM current waveform, is the portion that is being flowed to the load, while the up trending sawtooth is the inductor charging up? Is that right? The average creates the sine wave we want, but we bounce from charge to discharge along the intended sine wave because boosting is a "2 stage operation" and that's just the nature of the beast? Once a certain I is measured in L, chip turns off switch, then once a certain lower I is measured chip turns back on? So determine my upper threshold, or upper sine wave, and lower threshold, or lower sine wave, and instead of pre calculating PWM based soley on a ZC Vcap measurement, use feedback, or instaneous measurement of inductor current along with ZC Vcap measurement, to mng desired average I waveform?EDIT-EDIT-Perhaps my next comment moves the conversation forward, more so, or is more worth spending time on.

author: TheDood
author_rank: Regular Contributor
timestamp: January 30, 2020, 09:38:50 pm
content: I(t) = A·sin(2πf(t))I'd want to measure cap V @ ZC to determine necessary average current waveform with amplitude Aavg needed to refill cap.Then I feel like I could go 2 different ways.1 way would be to calculate I(t) of the Aavg equation, and compare this with the instantaneous measured I in the inductor, and then setting the switch to turn ON/OFF depending on if measured I was greater or less than I(t). This would create a sawtooth like I've seen, I think.Or another similar approach would be to create 2 more sin functions, 1 with a greater amplitude than Aavg, say Aup, and 1 with a smaller amplitude than Aavg, say Adwn. Then turn on switch until measured inductor current = I(t) of Aup function, then switch off until measured inductor current = I(t) of Adwn function.I'd be vulnerable with mains Hz fluctuations, but by managing based on measured I rather than precalculated PWM, I'd be more consistently delivering desired current regardless of mains Vpk fluctuations from cycle to cycle. There's no way to predict the future, so I guess I'd have to use the average of the last 3 mains sin cycles frequencies (or 5 or whatever) to determine the next periods Hz, or the future Hz?Also not sure how to calculate the time. If a constant switching Hz were used, I guess I'd have to record the time in the period that the ZC happened, and create some sort of multiplier based on the number of periods endured since the ZC, then I should be able to calculate (t) from that? And then in the scenario where the Hz is set based on the measured inductor I, I guess I'd have to have some sort of clock that gets reset at every ZC, and base my (t) off of that? <-- This one seems the simplest way?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2020, 04:18:52 am
content: Quote from: TheDood on January 30, 2020, 07:55:01 pmIn a practical converter, you stop at the, 11.89V point or whatever. Waiting until V_L drops to 10 or 8 or 5V means all the rest of the voltage is across the resistor, which means doubly:1. You're way the hell over rated current. If you're switching the nominal load, say 1A, at the 11.89V point, then at the 6-7V point it's doing 50 fricken amps.2. The resistor is dropping all the power, and the inductor isn't storing much more energy. You're close to 50% efficiency on this loop already, and it keeps dropping the longer you leave it on. Whereas when switched at the early threshold, it might be 99.5% (if this were the only loss mechanism).QuoteIf you're talking in terms of discharging == inductor current reaches zero every time, yes. It works fine either way. The difference is, in CCM, you can have much more DC current than AC ripple, which delivers more charge while spending less reactive power in the inductor.Which is relevant when inductors have low Q factors (high AC losses, but low DC losses), typical of powdered iron chokes. One might have a Q of 10, so a 1W power dissipation limit can only draw 10VA of reactive power, but if this is done at a ~10% ripple fraction you can deliver ~100W of real DC output with that.Which are figures in the ballpark of your typical old fashioned ATX power supply. That's how they're designed.QuoteNot harmonic, cycle, yes!You can look at resonant circuits as a time constant, depending on how you want to use it. You can take the radial time constant (i.e., the time of one radian), sqrt(LC), or the cycle time 2 pi sqrt(LC), or the quarter cycle pi sqrt(LC)/2, etc.There's also a time constant for the ringdown, because real resonant circuits have losses. Works in the same way; the general form is sine wave*exp decay. There's always some sine in there, but it can be negligible if R is dominant compared to L or C, in which case the result looks exp dominant (seemingly RC or RL alone). And there's always some exp in there, even if R is very small (e.g. superconducting resonators, which can have a Q factor around 10^7).Anyway, you are quite right actually, that when the switch turns off and the diode turns on, you can draw the equivalent circuit reflecting those states, and it's simply a charged inductor discharging into a capacitor. If allowed to fully discharge (to current = 0), the voltage will follow an arc segment which is a piece of the full LC ringdown waveform. Typically, C is made large enough that the voltage change is small.Conversely, if there's capacitance loading the switching node (there always is), then there is a time between the switch turning off, and the diode turning on, where that capacitance takes all the inductor current. Because this capacitance is small, the voltage again follows a segment of LC ringdown, in this case the rapid rising part.Because the diode changes the equivalent circuit as it turns on and off, you only ever see tiny segments of these curves, and you can quite reasonably approximate them either as linear ramps or quadratic segments.(Resonant converters, you can think of as a hybrid case; understandably, the exact mechanics are more difficult to calculate. Fortunately, with appropriate design, a crude, simplified control method is possible.)QuoteNote that the limiting case in CCM is for L --> ∞, so ΔI = 0. This isn't very interesting for control purposes (no amount of PWM can change the average current flow..), but if we assume equilibrium conditions so that we maintain the flux balance condition, then from cycle to cycle, what we're doing is putting in a completely square half-cycle of energy, and getting out the same amount.The continuum between the two extremes is: energy is the time integral of V*I. In the DCM case, current is a ramp and voltage is constant, so the power is a triangle, and its area is the energy, hence 1/2 L Ipk^2. In the ΔI = 0 case, current is constant, voltage is constant, so power is a rectangle and its area is the energy, Vin I_L t_on.For large but finite inductances, the average current does vary over time. Which, heh, again, "average" meaning what it does, needs to be qualified, i.e., over what time? Well, in this case, take a cycle average for example.The dynamic of interest is the time constant of the converter, which can span many cycles -- roughly, inverse of the ripple ratio. Since, after all, if the ripple ratio is 10% (at a nominal say 50% PWM), then applying 0% or 100% PWM, can only change the current by about 10% down or up, each cycle. So we can use this parameter to decide how much we should concern ourselves about, say, an individual cycle or what. Likewise, in DCM, cycles are fully independent, so they are individually important.QuoteWould you be better served with a mechanical analogy instead? (Are you well versed in mechanical dynamics?)You can indeed implement a switching converter in mechanics; the trouble is that, because the speed of sound is relatively low in most materials, you end up needing a great number of compromises in order to keep switching losses low. A "practical" switching transmission for an automobile would be about as large (or larger than?) the engine itself, probably switching at a few Hz or 10s of Hz.Most of the space would be taken up by flywheels (bypass capacitors) and a fuckoff huge spring (which needs to take up multiple rotations in a cycle, while handling up to full engine torque). The clutches (this would likely be a synchronous inverter) would be slamming on and off, rapidly, under hydraulic actuation I suppose (achieving switching edges in the low ms). Even with top materials and lubricants, it would wear extremely rapidly.We have the other fortune, that unlike sliding metal contacts, electronic components do not wear within their nominal ratings!The spring, incidentally, might not be a wear item. Some alloys apparently exhibit a fatigue limit, below which their life goes up exponentially as strain decreases. Other alloys do not, and the life is always related to strain (proportionally, or by another function, I forget what). The fatigue limit of spring steel is a reasonable fraction of its yield strength, so there is a direct tradeoff between cycle lifetime and power density.The engine would be governed at constant RPM, or perhaps increased RPM under heavy load as needed, but not by too much. This would be a good diesel application. The transmission's output can deliver far more torque than the engine (torque is stepped up in the same way a buck converter steps up current). Torque or RPM can vary continuously, unlike a conventional transmission that merely switches between gear trains while absorbing the difference in clutch slip, or in a lossy fluid coupling. (Which, to be fair, does a good job all its own, it isn't actually all loss -- hence the name "torque converter".)If you can imagine an automatic transmission shifting gears in milliseconds, and doing that dozens of times per second, yeah, that's how incredibly loud and jarring and fast-wearing this would be!I'm pretty sure this has been built before; but it's really just a lab curiosity, of course.Even for all the challenges we have with conventional geared transmissions (and also CVTs -- mechanical variacs), it's telling that it's been better to face those, than to try to make a more general device like this. Like I said, the material properties -- in particular, speed of sound, and density -- just don't work out.QuoteThe control is another level on top of the converter. You must abstract away the converter as a transconductance stage: given some setpoint input, it draws a (switching cycle-averaged) input current and delivers a transformation of that to the output (i.e., same power, different V and I).The input and output voltages do not change over a switching-cycle time scale, so can be treated as constant by the converter.With this abstraction, you don't care how the converter does its work. Black box. It could be full of dwarves for all you know, hammering electrons off one end and stacking them up on the other side. Maybe it's switching inductors. Maybe it's switching capacitors. Maybe it's pure computronium and its losses are actually CPU power, calculating how to overthrow the human race. Doesn't matter, externally it just transfers power in response to a control input.Then, and only then, can you apply the PFC algorithm. Set input current proportional to input voltage. If the converter's control is in terms of input current, you're set, that's all you need, great. If not, you may need an inner control loop, or a function block, to implement that.Meanwhile, averaged over multiple line cycles, set mean input current as needed to maintain output voltage. Which, again, output voltage changes gradually even over mains cycles.This more or less describes a typical BCM or CCM PFC architecture. CCM inevitably has to deal with DCM at light loads, and measuring actual inductor current can be hard so they come up with ways to estimate it (comes to mind, UCC28070's "current synthesizer" block), or use a function that corrects for the error (say, predistorting the setpoint, on the assumption that it will go into DCM and therefore change its gain).You don't stand to gain anything from looking at the converter with PFC as the direct goal. Converters can be controlled to do PFC, but they fundamentally don't do it by themselves.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 31, 2020, 10:36:21 am
content: Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amThat was great, LMAO! And exactly how I've been thinking lol as long as it does its magic and I know the time it takes to do it..Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amOk, I think I'm comprehending that charging up and then fully discharging is going to result in greater core losses (lower Q will be more effected than higher Q factor), and that you can achieve greater power output to your load given a % efficiency target utilizing CCM. This is due in part because the current doesn't have to go from 0 to X and then back to 0 in CCM, and the difference in operation results in the opportunity to flow greater amounts of current given a set time period (and target efficiency)?If I were using a powdered iron common mode choke as an AC line filter, but then a ferrite or HF core for the switching inductor; due to the average nature of the PF corrected waveform, I shouldn't need to worry (too much) about core losses in the powdered iron common mode choke when HF inductor is switched at high frequencies, correct?What about ain air core, no saturation, but probably be a large component for a 300W PFC application?Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amLol ya I just stabbed, thought I'd leap and learn lolQuote from: T3sl4co1l on January 31, 2020, 04:18:52 amNice, this would be the parasitic voltage spikes I was seeing during the begining of the OFF switching event in my simulations I'm pretty sure. Im assuming with the diode acting as the lid on the filter cap, this filter cap would not be part of the capacitance loading at the switch node.It seems Ill need to use a snubber across the switching FET? I think this would whip the unruly transients into place. Good riddance! LolQuote from: T3sl4co1l on January 31, 2020, 04:18:52 amYa brah, no free lunch yo lol. You're talking about the ESR in the components/conductors and that you're always going to have some decay, no such thing as perpetual motion! (Electroboom would be eating this up)Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amIll have to ponder this a min.Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amThis is gold. Thanks.Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amHaha you the man, I was talking more thermodynamics as opposed to electrodynamics, heat and energy transfer. I don't even know what a switching transmission is.I flip flopped between many undergrad engineering disciplines but EE was not one of them (metallurgical, general, CS, mechanical, petroleum). I've dealt with CVT's though, I use TAV 30's on my mini bikes.I'm with you when you're talking about stress and strain, elastic limits, torque ect though, and I am not disputing mechanical operation involves a shorter lifespan. There's more opportunities for inefficiencies in a mechanical operation than an electrical operation (I think typically). Harder to be as precise with mechanical than electrical, especially with repeated operation.Quote from: T3sl4co1l on January 31, 2020, 04:18:52 amAccording to these snippets, it seems I'd be better off setting switch times based on measured V(t) rather than measured I(t) due to the difficulty in measuring current through inductors at high frequencies. What I'm a bit confused on (if this assumption is correct) is why the IC mentioned implements a current estimator block rather than just trust that at a certain V(t) a certain I will flow given a set LR arrangement?Thanks for all the time and effort. You've been very helpful.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2020, 11:16:30 am
content: Quote from: TheDood on January 31, 2020, 10:36:21 amCMC doesn't matter -- it sees a few volts and even less (AC) current, at least if you've done everything else correctly (like, not shorting the switching node to the heatsink to ground, forcing the full switching waveform across the CMC).As it happens, powdered iron doesn't achieve as high mu as ferrite, so it's almost never used for CMCs. Sometimes used for small CM or DM chokes, where the loss helps dampen resonances with other components in the 10-100MHz range. But not for the dominant 0.1-10MHz range, just needs more inductance down there.Also, you're drawing switching currents from the mains, differentially; there's a relatively large film cap there (usually ~1uF) to keep the switching ripple voltage low, but it's still going to develop a few volts say. CMC doesn't help with this, it's DM.What ends up happening in most filters, the leakage of the CMC serves as the L of a pi filter -- two film caps either side of the CMC combine to give a 3rd order filter, hopefully with a cutoff below Fsw so the attenuation is good. Handy, saves adding another component.Have seen one design (based on a CCM PFC controller by STMicro) that switches at very low frequencies, when the input voltage is low -- that is, near zero-crossing. The emissions are chirping over several octaves, over the course of a mains cycle. Nominal was supposed to be like 200kHz but it dips down to maybe 1/10th that near zero crossing. You can actually see the ripple voltage at the input, with an ordinary scope probe.This is okay in the US, where the FCC regulates emissions only down to 150kHz, but some CE standards (and I don't know offhand if it's applicable for consumer or ITE or what equipment) start at 9kHz, so they may need an improved filter for use over there.I did actually end up adding a single choke to that design, to improve the DM filtering below 300kHz. Handily, there was a footprint already on the board, so it was an assembly change only. (Actually a 2nd CMC footprint, the one leg of which got jumpered out, and the other leg got the DM choke.)As for air core, they're larger for the same Q, and usually have a lot of external field, e.g. solenoids and loops. Toroids can be made, but with some difficulty, in which case you are better off using a powder core anyway, saving on turns and copper losses.Hmm, think it's been a long time since I've specified an air core inductor, in a commercial design, where the inductor handled significant reactive power.And that was an induction heater, where the reactive power (up to 100kVA in that case, I think) is hard to handle any other way. Ferrites need to be pretty big to handle that much, and there's no good way to get the core loss out of them (add a water cooling plate? yeah..). And the winding likely still needs to be water-cooled copper tubing, so you aren't saving any fab cost.Also had problems with that design, where the coil inside the enclosure was nuking the enclosure itself. (Imagine that, right, an induction heater that heats with induction?) Hah, we had one proto enclosure made from steel, we called it the pizza oven because the bottom side literally got that hot. The final aluminum version still got awfully hot, but a few ferrite plates in key areas got that down to a touchable temperature, if still a ways off from "cool".QuoteOh; that actually doesn't help very much, unfortunately. Thermodynamics is energy statics (the "dyn" is referring to energy, not its flow). Rarely, quasi-static flows. Doesn't really say anything about time-dependent processes, or non-equilibrium flows like what we have here -- after all, we're switching inductors back and forth todemandsome power flow.Maybe that's where you're getting tripped up? Because, LR circuits have an equilibrium, and store energy, that's a starting point right? But you're getting lost because you're looking at one tree (energy storage) instead of seeing the forest (continuous energy consumption -- power dissipation!). Or forest fire, perhaps I should say.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: January 31, 2020, 11:46:37 am
content: Quote from: T3sl4co1l on January 31, 2020, 11:16:30 amChuckling out loud as I read these comments. Forest fire lol, Id love to see that beast, I bet it was fun building/expirementing/cooking pizza lolHeat flux is rate of thermal transfer, and an example of energy transfer modeled by thermo, its been years though and I wish I remebered it all. Ill get the hang of electrical and all your guys' jargon, it'll just take me a min or 2 or 200000000000 lolIt's early morning here and brain is spent, Ill review the important parts tmrw, thanks again Tim.

author: TheDood
author_rank: Regular Contributor
timestamp: February 11, 2020, 02:01:52 am
content: Why and how am I flowing so much current?The idea was to check bulk cap at ZC and adjust the boost switch's PWM to create a steady-ish voltage at bulk cap with V boosted above line, then run multiple LED loads from the bulk cap, each with its own inverter and capacitive dropper with the inverter Hz controlling current through LED load.The circuit simulates flowing a ton of current through the diode after the boost inductor. Are the inverter switches upside down? I feel like I'm missing something. I thought the capacitor in the capacitive dropper would limit current and I can't understand why im flowing so much current through the diode but hardly any at the load. The pic shows 100kHz inverter, but even at lower Hz, there's a ton of conducted current D1 and barely any through the load cct?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2020, 04:29:11 am
content: Surely C2, C6 are typos? Hhahahaha, 400V gate to source on M2, M6.C1 is also clearly too large to show anything of interest in a transient sim; it might as well be a DC voltage source of whatever. Which can be a good strategy in a simulation, to show that the stuff connected to it is behaving without having to work with the dynamics of the voltage loop.And also, because the VDC fixes the voltage regardless of what current is drawn from it -- it can be divided in two, i.e., two equal VDC sources, one for the left half and one for the right half. That is, the boost stage, and the inverter stage, can be simulated completely independently of each other, speeding up the simulation of each.The same is true of V2, which can be set to a DC value for say a few milliseconds at a time, during which the operation of the boost stage can be evaluated. A millisecond is 24 cycles as shown. Really only need one or two cycles to see the switching itself. Repeat for other voltages, then finish for changing voltage, and there you go.Tim

author: TheDood
author_rank: Regular Contributor
timestamp: February 11, 2020, 05:53:30 am
content: Quote from: T3sl4co1l on February 11, 2020, 04:29:11 amThanks Tim,Do triacs offer greater VDS options than pFETs? How about Rdson? When I probe C1 I only see a max of 80V, are you saying that the pETS are shorting through their VDS rating in the sim? Or why is it drawing so much current? Ya the snubber is a bit overkill lol I was half ass calculating but based on boost L instead of parasitic, I barely skimmed this link and saw a couple equations but didn't read anything, should have lol...https://www.digikey.com/en/articles/techzone/2014/aug/resistor-capacitor-rc-snubber-design-for-power-switches