Capacitive Reactance, Inductive Reactance, Power Factor, Switching Hz - Page 1
Forum section: electronics / beginners

author: TheDood
author_rank: Regular Contributor
timestamp: December 11, 2019, 05:55:51 am
content: Capacitive reactance = 1/(2π·Hz·F)Inductive reactance = 2π·Hz·HIf you've a series C L cct, and both the capactive reactance and the inductive reactance were equal, does your ideal cct have a PF of 1?If you've a capacitive dropper cct, and you're bypassing the inductor at a certain Hz, can you effectively manipulate your inductive reactance to match the capacitive reactance by changing your Hz at which you bypass? The series X1 cap is at a constant 60Hz from the AC side, but the coil Hz can be manipulated by the frequency of your bypass period?2.2μF X1 cap @ 60Hz = 1205.72Ω0.1H L @ X = 1205.72ΩX = 1,918.96HzIf you bypassed @ ~1919Hz, would you correct your PF?Thanks

author: TheDood
author_rank: Regular Contributor
timestamp: December 11, 2019, 08:35:14 am
content: Nevermind.

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 11, 2019, 10:58:19 am
content: Quote from: TheDood on December 11, 2019, 05:55:51 amNo, PF would be zero if the components are ideal (no resistive losses) because the lagging PF of the inductor is exactly cancelled by the leading PF of the capacitor at resonance (ie - when XCand XLare equal).

author: TheDood
author_rank: Regular Contributor
timestamp: December 11, 2019, 12:37:25 pm
content: Quote from: MagicSmoker on December 11, 2019, 10:58:19 amThanks MagicSmoker,Is that what we want? No lead, no lag? If a capacitive dropper creates leading, an appropriately sized coil would cancel and correct PF to what is desired? Im confused on what is wanted because I thought .95 was the threshold for compliance?Can a UC385, or an MC33262 be used to correct PF of a capacitive dropper?https://www.onsemi.com/pub/Collateral/AND8179-D.PDFhttps://www.ti.com/lit/ds/symlink/uc3854.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 11, 2019, 01:47:38 pm
content: Well, yes and no, depends where you measure. Assuming there's some resistance in that circuit (because if not, the circuit is meaningless!), the PF as seen by the source will go to 1.If there's no resistance, then at resonance, reactive power goes to infinity while real power is still zero, so you have a \$\frac{0}{\infty}\$ form. Since it's symmetrical around resonance, the limits match and the result is zero (as noted above).You cannot synthesize a reactance from behind a FWB though, and you'll also have a hard time doing so in front of the FWB by using lossy devices (transistors can only switch or dissipate power, they cannot produce it). Note the definition of reactance is that it alternately consumes and returns power to the circuit!This does mean you can synthesize a reactance by using, say, a nice compact capacitor, a switching converter, and a controller to set the voltage and current accordingly (within some limited frequency and energy range, since the capacitance stores energy inversely to the desired inductance; this is quite doable at a fixed mains frequency). But that's a rather worse solution than what would satisfy your underlying problem.(Such solutions are actually useful, though -- when absolutely required. Such a device can be used to effectively multiply a capacitor, allowing one to implement a filter that would otherwise require, say, a lot of bulky electrolytics. The Google Little Box contest winner (2014) used such an approach. Similarly, industrial equipment can be power-factor-corrected by bolting on an inverter and controlling its current draw inversely of the attached load, thus using its supply (DC link) capacitor as PFC storage.)Tim

author: TheDood
author_rank: Regular Contributor
timestamp: December 11, 2019, 03:57:58 pm
content: Thanks Tim,Haha thanks for the reply, worse in what way? Im trying to avoid a transformer and/or a voltage source. Its my assumption that a transformer is only about 96% efficient and my supply would only decrease from there, on top of being bulky and costly. With an intended load of LEDs Id rather have a current source and preferably one without using a transformer simply for mains isolation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 11, 2019, 10:54:29 pm
content: You seem to be looking to power LEDs, with good power factor, good efficiency, and preferably dimming.Standard approach -- simply look up a controller chip that does all of that. Follow the application circuit. Most appnotes are rich with information, so there is much to learn; though what's provided is typically poorly supported and narrowly focused, of course. (Or, notoriously wrong, as many appnotes are wont to do. Well...)Or even simpler still, buy the ready-made LED module or lamp; you certainly won't beat it on cost, though you may not learn much from it either.It will involve a switching circuit (possibly an integrated switch, possibly integrated driver with external MOSFET), inductor, EMI filter, and an isolation transformer is optional (the driver modules are isolated, but enclosed lamps don't need to be).I'm not sure what you intend to gain from eliminating the remaining 10 or 5 or 3% or whatever of heat loss; the LEDs themselves are, whatever they are, 20-50% efficient (i.e. in terms of raw power, not just luminous efficacy), so losing a few points on the converter is essentially meaningless. It equates to a few more minutes of run time off a battery that stores, say, one or a few hours of energy. Even a 90% efficient LED isn't worth optimizing the converter much beyond say 95%.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 12, 2019, 12:07:38 am
content: Alternate approach: the "valley fill" rectifier delivers reasonably good PF and is often used in consumer-grade lighting (LED and CFL). A random web page article about it (and other forms of PFC):https://www.powerelectronicstalks.com/2018/09/types-of-power-factor-correction.html

author: TheDood
author_rank: Regular Contributor
timestamp: December 13, 2019, 09:01:10 am
content: Thanks guys,One waveform attached was while being switched at 240Hz and implementing the charge pump smoothing resistor passive valley fill. I'm simulating 10kHz now to see if I can more closely match the waves together but the sims take forever at higher Hz. The AC input is at 60Hz.When I switch at greater multiples of 60Hz, higher than 240Hz, the nice smooth current shown spikes erratically (many times at ~10s, but not sure if it's due to PWM on-pulse time) and regardless the multiple used (480hz, 600hz, ect). It seems that the 3cap or 4cap valley fill with R smoothing but without charge pump is greater effected than the 2 cap charge pump + smoothing resistor valley fill topology.When I don't use a multiple of 120Hz, it seems to stabilize more or less throughout the sim. I tried 1kHz using the VFCP&SR (valley fill charge pump and smoothing resistor) and had good stable current up to 45s when I cut the sim short to try 10kHz. After the 10kHz is stimulated I'll post the results.2 waveforms attached come from 2 different VF topologies. 1 is the VFCP&SR (240Hz), and the other is from a VF4Cap&SR (600Hz).**Also, I'm not sure if I'm looking at the correct nodes to determine PF, I've been selecting the AC source V and AC source I, but can't remember if it's the top side AC source V or the bottom side AC source V. That's where I should be looking? If I select AC source V before X1 cap I get the full AC wave form stretching to 170V, but after the X1 cap it's only ~40V peak. Also, the -AC waveform seems to only be effected by the load in any SIM I run (so if load V needed 40V, -VAC peaks are reduced to ~-80V, while +VAC is maintained at 170V peak?), is this typical and due to the ground node placement? Can I/should I try to change things around?

author: TheDood
author_rank: Regular Contributor
timestamp: December 13, 2019, 12:04:35 pm
content: The Valley Fill PFC really drops the load current compared to without implementing it. Ive tried adding a voltage doubler but I think the LED load is regulating the doubler because I cant squeeze any more current out of the cct with the doubler added. The only way that I can see of maintaining the current flow I had before implementing the VFPFC, is by increasing the VAC input, but then Im using a transformer (Im trying to get around) and I believe Id be doubling my current draw pre step-up transformer compared to no transformer?

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 13, 2019, 06:07:46 pm
content: What the heck are V2 and the related components around it supposed to do? Err... I just noticed you have this across the AC side of the mains, not the DC side, so even more perplexing. Just delete all the stuff and try running the simulation again.

author: TheDood
author_rank: Regular Contributor
timestamp: December 13, 2019, 08:33:43 pm
content: Quote from: MagicSmoker on December 13, 2019, 06:07:46 pmV2 + FETS are a way to rob some of the constant current produced by the X1 cap. It's the dimming portion of the cct. Maybe it won't work? Just trying to reduce V drop when the current is bypassed and I figured before the bridge was more efficient than after the bridge?How glitchy is LTspice? I'm pretty sure I've been getting incorrect results or sims. I close it all out, reopen and re-run, and then I get different results. Very frustrating. I thought I was just not remebering correctly the mods I've made in between sims, but I'm pretty sure its been acting up and giving me shit data after awhile?

author: TheDood
author_rank: Regular Contributor
timestamp: December 13, 2019, 08:42:31 pm
content: I've been working on a buck converter if the capacitive dropper is not going to cut it. Ill start a new thread and post it later. I figured if you poured rectified AC into a cap that you could then run a buck converter topology after it to power a load requiring lower V? The sims have been all over the place with op amp outputs greater than the V supplying it, with LEDs at 21V+ but only 6mA current flow, ect ect, taking a break.. Maybe a computer reboot will help..

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 13, 2019, 10:14:11 pm
content: LTSpice isn't glitchy at all for me. Sometimes it has convergence problems when the ideal diode is used (the default one, unfortunately) but putting a few pF of capacitance in parallel usually fixes that. I also generally run the alternate solver (select Control Panel under the Simulate tab to change) because it gives more accurate results at the expense of taking longer to run.Of course, it also helps if your circuit makes sense and is solvable by SPICE...

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 14, 2019, 02:06:55 pm
content: Quote from: TheDood on December 13, 2019, 12:04:35 pmIncrease the value of C1 - it is acting as a "lossless" current limiter due its capacitive reactance. This reactance, in Ohms, is found using the classic formula: 1/(2*Pi*f*C) where f is in Hz and C is in Farads.12.2uF gives a reactance of approximately 217 ohms which will limit current into a short circuit of 0.55A when supplied by 120Vrms (remember to use the RMS value of the AC voltage here). Note that the higher the voltage drop of the LED string the lower the current the capacitor will let through since it is the difference in voltage between the (rectified) source and LED string that is dropped across the capacitor.Making this kind of circuit dimmable with a conventional triac-based dimmer is nightmarish. There's probably some specialized ICs already on the market that enable this functionality, but I leave it to you to do that research.In the meantime, I attached an LTSpice simulation of a rectifier + valley fill PFC + LED load circuit that simulates just fine and rather quickly, even with the alternate solver. I did not take any care in selecting components to leave you with something to do, but it gives you a good - and working - starting point.

author: TheDood
author_rank: Regular Contributor
timestamp: December 15, 2019, 05:06:41 am
content: Quote from: MagicSmoker on December 14, 2019, 02:06:55 pmThat's awesome, thanks, I appreciate it, I'll look it over and see what I can see, you've been very helpful.