Electronics > Beginners
PFC Math
TheDood:
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"
T3sl4co1l:
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
TheDood:
--- Quote from: T3sl4co1l on January 02, 2020, 01:45:59 am ---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
--- End quote ---
Thanks 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?
dietert1:
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
T3sl4co1l:
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
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