Halogen lamp transformers for tube amps. Is power factor an issue? - Page 1
Forum section: electronics / beginners

author: dazz
author_rank: Frequent Contributor
timestamp: April 20, 2022, 07:41:30 pm
content: Hi there everyone, I'm back with one of my (probably) stupid questions.As the title suggests, it's about using these 50W (50VA?) trannies for tube amps as power transformers.I have measured the resistance and inductance of both windings and then run a simulation to calculate the theoretical power factor of a linear power supply, using two of these back to back trannies, to get the 12V for the filaments in between the trannies, and isolated high voltage for the plates as well. The result is a measly 32%, much lower than the expected 60-70%, apparently due to the low-ish inductance of the windings. If I increase the inductance of the primary winding 10 fold, the PF shoots up to almost 80% in the simulation.Does that sound about right? I tried to find a typical value for the inductance in commercial power transformers to no avail. Is 2.4H for the primary low? I believe these things are used to drive halogen lamps at 12V AC, in which case the power factor wouldn't be an issue, correct? I'm guessing this configuration is OK for a single ended amp, but nothing much more power hungry than that.And the final question, is there anything I can do to improve the power factor and squeeze a few extra watts from a power supply like that? (assuming that's the issue, of course)Thanks!

author: TimFox
author_rank: Super Contributor
timestamp: April 20, 2022, 08:04:37 pm
content: For a magnetizing inductance of 2.4 H, at 50 Hz that is a reactive impedance of +j750 ohms, which will pull a magnetizing current of 0.3 A rms at 240 V rms.I assume you measured the primary inductance with the secondary open-circuit to obtain the magnetizing inductance for your model, and similarly for the secondary inductance.With your ideal coupling co-efficientk= 1, the 2.4 H primary and 8.5 mH secondary calculate to a voltage ratioN= (2400/8.5)1/2= 16.8:1, which agrees approximately with the 240 V to 12 V rating, but indicates that the coupling co-efficient is a bit lower (as expected for a small transformer). You can measure the leakage inductance, and thereby the co-efficientk, by measuring the primary inductance with the secondary shorted to improve the model.

author: dazz
author_rank: Frequent Contributor
timestamp: April 20, 2022, 08:42:29 pm
content: Quote from: TimFox on April 20, 2022, 08:04:37 pmOh, my. So much to unpack there. I need to do some reading on the topic apparently. Thanks Tim.Yes, I measured the inductance of both windings open circuit. Great to know I can estimate the coupling coefficient measuring inductance with the secondary shorted, I'll do that in a minute. I didn't think K was all that relevant because it didn't seem to affect the PF result, and the sim runs much faster with K=1ETA: oh, you calculated the impedance using the inductor formula XL=2 x pi x f x L, and then the magnetizing current with ohms law, I see

author: dazz
author_rank: Frequent Contributor
timestamp: April 20, 2022, 08:57:08 pm
content: The primary reads just 115mH with the secondary shorted out. Not sure what to make of that, seems like a massive difference.

author: TimFox
author_rank: Super Contributor
timestamp: April 20, 2022, 09:25:30 pm
content: That measurement is the "leakage inductance" (referred to the primary).The coupling co-efficientkmust be < 1 for a physical transformer.To calculatek, use 1 -k2=LSC/LOC(approximately),where "OC" means with the other winding open-circuited (your first measurement) and "SC" means with the other winding short-circuited (your second measurement).This value ofkcan be used in your Spice model of the transformer.A perfect transformer, whereLSC= 0, hask= 1.Physically, the leakage inductance indicates how much of the flux from the primary winding misses linking the secondary winding and goes elsewhere.Lightweight transformers usually have too little iron to achieve very low leakage inductance (cost and weight considerations).See textbooks for other equivalent circuits which show the magnetizing and leakage inductances as parasitic components in parallel and series (respectively) with an ideal transformer's primary winding.The magnetizing inductance limits performance at low frequencies, and the leakage inductance limits performance at high frequencies.

author: dazz
author_rank: Frequent Contributor
timestamp: April 20, 2022, 09:47:52 pm
content: The formula says k=0.97575 in this case. Pretty terrible magnetic coupling there, isn't it?So am I right that you're saying that's the issue here, and not the low power factor due to low inductances in the windings, as I initially speculated?

author: TimFox
author_rank: Super Contributor
timestamp: April 20, 2022, 10:07:43 pm
content: That's not so bad for a lightweight transformer.The PF depends on the parasitic inductances and the relevant resistances.You can use thatkin the Spice model.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 20, 2022, 11:31:14 pm
content: A note, mind that SPICE just cranks the numbers on whatever it is that it's given. If your simulation starts from zero, you're getting all the inrush current as part of that calculation. The PF calculation should probably be done midway along, or with setting an initial voltage on the capacitor.Just glancing at the screenshots here, maybe this has been done, not sure. So, just ideas.Note that k=1 is highly unrealistic for mains transformers, and the source itself should be something like 0.1 ohm plus 100uH. This will noticeably increase the PF due to harmonic current draw -- and also reduce the output regulation (try different load currents!).Which is particularly poor for two transformers cascaded, rather than just one with everything -- so keep that in mind. A class A, or shallow class AB, amplifier will be fine with this. (Or for deeper AB, maybe intentionally used, to get more compression / gradual onset of distortion, say for guitar amps where that's a feature rather than a bug!)Tim

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 07:44:38 am
content: Quote from: TimFox on April 20, 2022, 09:25:30 pmOK, so I can see in the sim what you're saying about leakage inductance and how it's effects are far more evident at higher frequencies. I increased the signal freq to 1K and the voltage dropped to 170V. Yikes!Fortunately this is a power supply operating at 50Hz, so even though the effect of the higher leaking inductance / lower coupling coefficient (which I understand are two sides of the same coin, right?) is definitely noticeable (especially because there's two of these transformers back to back, hence core losses are doubled), the low magnetizing inductance of the primary at 2.4H could play a bigger role here at limiting performance of this setup at just 50Hz. Did I get it right so far?What would be a typical primary inductance in a commercial power transformer as in a Hammond or the likes?Thanks so much for your input, I think I'm learning a lot here.

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 08:07:57 am
content: Quote from: T3sl4co1l on April 20, 2022, 11:31:14 pmYes, thanks. I start running the PF measurement half way through the simulation at a steady state. Note the MEAS directive starts at 400ms and ends at 800ms.Quote from: T3sl4co1l on April 20, 2022, 11:31:14 pmYes, I noticed the power factor increases as the load increases, but I'm not sure I understand what you're saying there. I'll try adding a .1 ohm plus 100uH in series with the voltage source and see what that does.A friend has tried it in single ended amps and even a push-pull with 2 EL84. It seems to work fine for SE, but it sags quite a bit in that PP config. I seem to remember plate voltage at the power stage dropped to some 210-215V or something like that. I'm guessing the power output in that case will be nowhere near the 15-18W it could do with a proper mains tranny.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 21, 2022, 08:26:17 am
content: Oh BTW, measure that  inductance at level. Steel's initial permeability is pretty ass, it needs some voltage to get going. If you measured at like 1V that'll give you a lowball. Really, might as well just measure the current with a meter on the primary, unloaded secondary, and assume the current flow is inductive (which, shouldn't be too bad an assumption). If you have a variac, use that and stop it down a bit (say 100-180V) so it's definitely out of saturation (since that will also increase the RMS well above the fundamental or average figure you're looking for).Quote from: dazz on April 21, 2022, 08:07:57 amAh, got it.QuoteKind of both ways: 0.32 is suspiciously low, you should be seeing more like 0.5-0.7 for a typical cap-input filter like this. Or, probably not a whole lot more than that once leakage inductance is in, but, not sure. Anyway, you should get something different with the, what is that anyway, 2.4H at 240V 50Hz is 0.32A out of 0.55Arms, so yeah that's dragging down your PF a lot. Which is just plain old phase shift PF (or DPF if you like), not harmonics (which is predominantly what a cap-input filter causes in terms of PF).QuoteOh yeah, no way those will do that. Figure a 50W resistive = 50VA transformer is good for---well, I mean, PF ~ 0.5 means what it says on the tin, 25W of real output. Which at typical class A efficiency is hardly 6W of real output, maybe 10 if you don't mind the compression that class AB brings.A somewhat bigger transformer, without the loss due to cascaded windings, will do the job -- but obviously, won't be as handy either. So, other than that... use two or three in parallel, or see if you can find a similarly larger pair?Tim

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 10:00:20 am
content: Quote from: T3sl4co1l on April 21, 2022, 08:26:17 amNot sure what you mean there, sorry mate. Keep in mind I'm a complete noob. I'm meassuring inductances with an el-cheapo LCR meter. It's cheap, but it seems to work OK. At least it seems to be in the ballpark when measuring known inductors and guitar pickups. The meter works at 3V, 150Hz, 44% duty cycle. You mean I should put a much larger sine wave through the primary, then measure the primary current, and calculate the inductance of the primary with that somehow? My function generator is pretty aweful, but worth a try... maybe? It won't do anywhere near 100V RMS, but still better than 3V.Quote from: T3sl4co1l on April 21, 2022, 08:26:17 amI need to do some proper reading on this, cause I don't understand how you know that "2.4H at 240V 50Hz is 0.32A out of 0.55Arms" is bad for power factor. I think I get the part where you say that will screw up the phase, hence dropping the PF beyond what the rectifier + filter will because of signal distorsion (harmonics), but I'm missing a good chunk of the big picture. Will do my homework. I think I need to get a better understanding of how these things relate to one another, like how leakeage inductance effects power factor and stuff like that. I'll google it all.Quote from: T3sl4co1l on April 21, 2022, 08:26:17 amI've got a dirty little secret: never assembled a tube amp, not once, haha. I'm mostly in it for the theoretical knowledge, but I'm aware nothing beats getting your hands dirty and testing these things in RL. I will prolly build a simple linear supply and see if I can build a constant current load with a mosfet or something, to put it to the test.BTW, I have a wall power meter. Problem is it doesn't measure power factor. Can I estimate it somehow comparing the wattage draw at the wall vs the load of the linear power supply?

author: jonpaul
author_rank: Super Contributor
timestamp: April 21, 2022, 10:11:33 am
content: Hello, Bonjour a tous:As designer and mfg of transformers, since 1970s, in lighting and audio, a moment please!1/ "Halogen lamp" transformers are optimized for very low cost of production, minimum copper/iron so, high core and copper losses.Result is poor efficiency .85% not unusual) the flux density is higher than normal transformer, running closer to saturation.The transformer voltage regulation with load is very poor, perhaps 30% drop no load to full load. Most commercial transformers for PSU mains use are 90% ...98% efficient at full load.2/ Modern trend since 20 years is "electronic transformers" that are switching supplies at high freq. similar appearance but completely unsuitable for audio PSU use.3/ PF of conventional FW rectifier capacitor input DC supplies is known for a very long time ~ 0.7The performance of a transformer/rectifier/capacitor DC PS is derived in the classic 1930 RCE 1946 paper below..4/ Measurement of mains transformers is quite different that the discussion, which seems fixed on a linear ideal equ ckt.Due to the nonlinear core and high Pcu, Pfe, such modeling is very inaccurate.The classic test of a power transformer:a. Open ckt sec, measure pri mag current and power losses vs mains voltage.This is the magnetizing current adn Pfe = core losses. Steinenmetz in 1920 t GE, researched and developed the curves and math.b. Short ckt sec, carefull and slowly increase input FROM Zero V, observe losses at rated sec current.That is the Pcu = copper or ohmic losses. as Pfe is negligible.In general the idea seems unfeasible as efficient transformers are readily available .Bon Chance,JonREFS: Rectifiers:O.H. SCHADE 1943 Proceedings of the IRE ( Volume: 31, Issue: 7, July 1943)https://ieeexplore.ieee.org/document/1694774/authors#authorsCore losses:Charles Proteus Steinmetz : On the Law of Hysteresis ...AIEE 1892https://en.wikipedia.org/wiki/Charles_Proteus_Steinmetz

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 10:24:56 am
content: Thank you, Jon Paul.I think I'm going to need a primer on transformer theory, because most of what you said there went straight over my headThe only thing I kind of knew already is that the PF of a linear power supply with a rectifier + filter cap is typically circa .7No idea what "equ ckt" "Pcu" or "Pfe" mean. Off to google again!The idea behind using these halogen transformers is that they're phased out now, so there's a crap load of them available essentially for free. Seems like a good way to save some e-waste from hitting the landfills, as long as it works decently, of course.

author: jonpaul
author_rank: Super Contributor
timestamp: April 21, 2022, 11:45:26 am
content: Rebonjour Cher Dazz, mille mercis pour votre gentil notesAm a veteran magnetics, power electronics and high voltage design and consultant....Bravo for your initiative and curiosity.Equ ckt = équivalent circuit1/ Transformers losses are mostly in the wire resistance ,I,sq,R, called Pcu= copper power and core losses Pfe= power in iron ,due to hysterysis.2/A pair of halogen transformers in series will double the already high losses.3/ linear models,and,simulation are difficult and inaccurate, due to,non-linéaire and saturation of cores.A simple resistance and short circuit /open circuit test can,determine the losses. Indeed FW rectification PF is about 0.7Suggest that you read papers and text books on power transformers theory and practice before proceeding.Bon courageJonPS....the high magnetic fields of,these,halogen transformer will radiate mains fréquence harmonics that can be heard in the AudioMost fine power amplifier transformers are toroids or have magnetic shields,and shading bands.

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 12:03:32 pm
content: Quote from: jonpaul on April 21, 2022, 11:45:26 amMerci beaucoup, monsieurI have a bunch of electronics books here yet to read. I'll see if I can find the relevant theory.BTW, I found this article in the LTSpice wiki, and apparently it's capable of simulating non-linearities in trannies to some extent.https://ltwiki.org/?title=Transformers#How_about_transformer_saturation_effects.3F_Can_LTspice_model_those.3F

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 12:08:45 pm
content: Do you guys reckon this constant current load will do the trick as a load for the power supply for my tests?I have some beefy heatsinks here for the mosfet

author: jonpaul
author_rank: Super Contributor
timestamp: April 21, 2022, 02:17:52 pm
content: Load tests are much simpler, use power resistor, rehostat, heater etc.No need for FETS!jon

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 21, 2022, 03:19:33 pm
content: Quote from: dazz on April 21, 2022, 10:00:20 amAh, fair...Suffice it to say: magnetic materials are nonlinear to begin with, and steel particularly so. That is to say, the inductance varies with signal level. And it happens to be lower at low levels (the initial bit of the curve, hence, initial permeability), gets going nicely in the middle bit, then slows down again (saturation).Meanwhile, this is nonlinear, so, generating harmonics as well; though mostly around the peaks, due to saturation (which can be much deeper than the drop at low signal levels).The fact that you can start with one frequency (a sine wave), and get other frequencies out, is an important fact; there's absolutely no way to do this that consists of merely scaling, adding, shifting, etc. sine waves together. Some other process is needed -- and, necessarily so, such a process breaks the rules of superposition, so our job of figuring out what's going on, gets harder.More concretely, a function f is linear if it obeys f(a + b) = f(a) + f(b). A nonlinear function, when we put in a = A sin(t) say, will give us something other than (or, more likely, in addition to) a' = A' sin(t) + B out.Well, perhaps that's still not beginner-friendly enough, but by example: say you have two circuits, one measuring 3A and the other 4A; when the currents are summed together, do you read (a) -1A (b) 7A (c) 5A (d) anything inbetween? For DC, in a linear circuit, we can only get (a) or (b), but a nonlinear circuit might show something quite different out the other side. (Also, an AC circuit will read anything inbetween (well, from 0, not from -1, since we read magnitude; we can't tell direction of AC without further trickery), if the currents are coherent (same frequency and waveform, more or less); the special case of (c) being when they are phased 90 degrees apart. Extra credit: what else is special about the numbers 3, 4, 5 and 90°?)Oh nice, here's a real waveform, hooray for citizen science:https://www.physicsforums.com/threads/power-absorbed-and-b-vs-h-plot-for-microwave-oven-transformer.989092/MOTs are notoriously stingy, for similar reasons as mentioned above; they use undersized core and winding and don't much care about the crap efficiency because they're only used intermittently. At low voltages, or at the low currents between peaks, a stable slope is recognizable, but approaching the peaks, the current shoots way up -- and much faster than proportionally, see the curves are spaced by 10V each (arithmetic progression) yet the current goes up in a geometric progression (roughly as V^2, if I had to guess?).Anyway, it's the stable slopes that you're after. Ideally, that waveform would be a smooth sine, and you simply take X_L = |V| / |I| (assumption: insignificant losses, inductance dominant), then L = X_L / (2 pi F).The other thing you can do is (duh!) measure from the secondary. Still might not be enough from the LCR, but maybe a better hope of getting there with a signal generator, say (5 or 10V is a sizable fraction of the 12V rating).QuoteYes, it'll all be a bit of a whorl without some background... basically, there are more ways to "hide" an ampere when it's moving up and down.Inductance draws current, without drawing power; this reactive current is registered on the meter, same as anything else (because the meter doesn't know what the current is in relation to, it's just some current), but it's out of phase with the voltage, always chasing each others' tails as it were, and this has the effect of averaging zero power draw.Likewise, the harmonic currents from the rectifier (or core saturation) cause wiggles at multiples of the line frequency, but the line voltage has little to no wiggles of its own, so the real power draw at those frequencies is close to 0V * (x)A, again averaging towards zero -- but this again reads as (RMS) line current, so it's reactive power spent without real power (work done).In the extreme, you can have resonant circuits which draw hundreds of times more reactive power than real power transmitted; usually for special purposes like close-in signal filtering, or wireless power transmission.The basic definition of PF is, the ratio of real to apparent power. Real power P is P = average(V*I), for instantaneous V and I, averaged over time. Apparent power S is exactly what it says on the tin: it's apparent from meter readings, S = |V| * |I| (where | | denotes RMS value).QuoteIf it tells you V and I RMS, you can do the above still.And yep, get your hands, well, dirty perhaps, just not shocked, better to avoid that. Tube amps are particularly easy and forgiving, aside from the usual rigors of loop compensation if you'll be exploring negative feedback.Which, compensation is another topic that's rather complicated; not so much so that it's intractable -- in fact analytical solutions do exist -- at least, to certain limited problems -- linear systems, as it happens (linearity is quite an important property in analysis!) -- but they require advanced calculus to work with directly, so one can't be blamed for finding them complicated, at least to complete depth. Fortunately, one does not need to understand the grand scheme, to be able to swap around a few resistors and capacitors, and test for best signal response. I think a hand-waving explanation is quite reasonable, and simply testing values (starting from a good guess) gets you there quite quickly in practice.But I digress.Tim

author: jmelson
author_rank: Super Contributor
timestamp: April 21, 2022, 03:47:36 pm
content: Are you SURE these are really plain mains-frequency transformers? They may be some kind of switching power supply.Jon

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 09:41:44 pm
content: Quote from: jmelson on April 21, 2022, 03:47:36 pmYes, 100% positive. I believe it's one of the old transformers used for halogen lamps back when they worked with AC. Defo no SMPS

author: dazz
author_rank: Frequent Contributor
timestamp: April 21, 2022, 10:28:52 pm
content: Quote from: T3sl4co1l on April 21, 2022, 03:19:33 pmThanks so much once again. I'll see if I can get some measurements of the primary inductance at higher voltages.As for the rest, it'll take me a while to process. Much appreciated, as always.

author: dazz
author_rank: Frequent Contributor
timestamp: April 22, 2022, 09:59:26 pm
content: Quote from: T3sl4co1l on April 21, 2022, 03:19:33 pmSo I'm ready to do some measurements at higher voltages. As for measuring from the secondary, would you mind elaborating a little bit there please? I already measured the secondary with the primary shorted, the result is 420uH, which is consistent with the previous result for K: SQRT(1-(0.42/8.6)) = 0.975, but that's not telling me anything about the primary inductance, is it?Should I measure the voltage and the current at the primary with the LCR in the secondary injecting those 3V AC to calculate the primary inductance, and then maybe repeat the process with the function generator at higher voltages? Please excuse my ignorance

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 23, 2022, 12:59:10 am
content: Oh just do inductance measurement on the secondary, primary open.Point being, if all you have is a couple volts, you're going to get different measurements primary and secondary, relative to the ratio, because it's such a small fraction of the primary nominal voltage, but a modest fraction of the secondary.Tim

author: dazz
author_rank: Frequent Contributor
timestamp: April 23, 2022, 06:22:02 pm
content: I ended up using mains voltage to measure the primary inductance. Got 147mA AC @ 235V AC. If I got the math right, that's XL=1600 ohms, and L=5H. Double the inductance vs measuring at 3V. Interesting.I'm guessing the secondary inductance will increase by the same ratio, right?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 23, 2022, 07:55:53 pm
content: Sounds about right. Yeah, I mean, measure Vsec at the same time, it's gonna be the same ratio right?Tim

author: dazz
author_rank: Frequent Contributor
timestamp: April 23, 2022, 08:05:13 pm
content: OK, next I will rectify and filter and will load it to see how much power I can get out of it.I have a wall power meter, but it doesn't measure power factor. Can I estimate it by comparing the power draw at the load vs the power draw at the wall, or will all readings be strictly real power?

author: dazz
author_rank: Frequent Contributor
timestamp: April 24, 2022, 02:40:22 am
content: Full bridge rectifier plus 100uF filter cap are in... and I have a whopping 17W of no load power draw at the wall. Seems like a lot, doesn't it? No load DC voltage is 310V, and it drops to 278V with a 10K ohms load. That's actually better than the simulation. There the voltage drops to 259V DC with the same 10K load.