- Transformer winding direction - 4 Updates
- Old Ampex reel to reel audio recorders - 1 Update
- Diode in series with the mains - 1 Update
- Car alternator failure -- twice! - 9 Updates
| whit3rd <whit3rd@gmail.com>: Feb 16 09:14AM -0800 On Sunday, February 15, 2015 at 9:08:03 PM UTC-8, mike wrote: > stressed if I turned it on at zero voltage when the primary current was > zero. And from the unpowered state, the voltage and current can't > be anything but zero. So, this is a concern that the SCR's dI/dt rating may be exceeded if peak voltage is present at the turnon time (the specific microsecond of time, because SCRs turn on faster than an AC period). That's a valid concern, with two solutions: (1) use an IGBT instead of SCR (yeah, I know, it's a big deal), or (2) use a transformer with enough stray inductance that the worst-case current risetime is tolerable. One hopes that the transformer is appropriate to this kind of frequent-switching use, and fulfills requirement (2). You really HAVE to hope, the microsecond-scale inductance of the transformer cannot be easily measured (the core isn't fully magnetized that quickly, so 60 Hz measurement tells an inappropriate value). Flux coupling to the secondary will also be poor for that short time. A 'typical' SCR (TYN640, in stock at DigiKey, needs dI/dt under 50A per microsecond...) doesn't need much inductance to keep its turnon conditions satisfied. > And that, if I could arrange the resting place on the B-H curve > from the previous pulse such that the first half-cycle wouldn't > saturate the core... Again, this depends on the transformer design and material. Remember Phil's comment that a 240V transformer on 120V excitation wouldn't saturate- because a 240V transformer has an oversize core for 120V excitation. The key concept is that the core remanent field also might be zero, and a first half-cycle starting at zero magnetization is more stressful than that same half-cycle starting at the inverse remanent field (which is what subsequent cycles of AC excitation provides). Starting at peak V is a safe bet if the remanent field is negligible, like when the gadget has been powered down for a while. |
| mike <ham789@netzero.net>: Feb 16 03:35PM -0800 On 2/16/2015 4:52 AM, Chris Jones wrote: > current transducer and DSO, (and a vastly over-sized triac or even > better a pair of big SCRs, just in case!). It would be nice to see the > plots, and it is one way to end an argument. This has been much more of a discussion than the "argument" you typically see in these newsgroups. I appreciate your use of logic instead of the usual name-calling. A discussion doesn't need to end with a loser. I had it in my head that remenence was the primary source of the inrush problem and all you'd need to do is prevent the first peak from driving the core into the saturation. If you left the core in the optimal state given your intent to turn it on at zero voltage crossing, you'd be done. I'll certainly examine it more carefully next time I use a transformer. I dug out the documentation for my welder. Turns out that I couldn't delay the turn-on if I wanted. All the solid state relays I had were/are zero voltage turn on units. Best I can do is control the slope of the AC input at turn on and turn off. The relay is rated at 600A non-repetitive for the duration of a typical weld. I was more concerned about drawing 40 amps out of a 15 amp breaker thru a daisy chain of connections thru outlets along the way from the breaker to the test bench in the back room. I never popped the breaker, but I'm sure I wasn't doing it any favors. I took it out to the garage and hooked it up to the dryer plug when I wanted to use it. Along the way, I picked up a 125 joule CD welder off ebay for $20 shipped. Solved all the battery tab welding problems without stressing the electrical system. The MOT is in the pile of superseded projects. > microcontrollers or even 555 timers are cheap enough that as Phil > mentioned, you can figure out the time of the voltage peak from the zero > crossing. I used a GAL20V8 as the controller. Wouldn't have been hard to delay the control, but the relay would have waited for zero volts anyway. > right time in a microwave oven. The current that your transformer draws > in steady state might be so high (due to the secondary current) that > saturation doesn't make it all that much worse, that was my assumption. I'm already running the transformer way past its rated steady state load capability. but given the choice (or > given a bigger transformer than your circuit breaker likes), you might > as well make it optimum if that is just a matter of inserting a small > delay. Yep, I learned something. I'll take a look at that next time I do something with a transformer. > If instead you turn it off for an integer number of mains cycles that > adds up to a few seconds, the core flux will not be the same when you go > to switch it back on again. Would be interesting to learn the trajectory of the magnetic characteristics when you turn off the input with the output (almost) shorted. > Then give it to me and I will hold the terminals a week later. > I think you will notice the difference. The state of the flux in the > core matters. I think that's a red herring. I can't argue the result, but don't think it's relevant. Short the output and repeat the experiment. > Note that in the case of a transformer, it is possible that some value > of secondary current could result in the primary current being zero (or > any other chosen value) at the zero-crossing of the mains voltage. If the secondary is loaded heavily with a resistor, why wouldn't the primary voltage be pretty much in phase with the primary current? The leakage inductance should be a small part of the equation? That > is not relevant to my point, which relates to the flux density in the > core, which won't be affected much by the secondary current if a > low-impedance supply is driving the primary winding. OK, but the (almost) shorted output should have a major effect on what happens when you stop supplying primary current and where the core ends up on it's B-H curve. I appreciate the principle. But, how much does it matter? If you have an unloaded transformer and see a 40 amp input surge, you're likely to be concerned. If your load causes a 40 amp input current and you do what you can to minimize the surge current, how much does it really matter? Unfortunately, I have no easy way to measure it given my SSR characteristics. > (insufficient) current to heat them, but this wasn't really satisfactory > because getting the force and contact resistance just right was not > reliable. That was my problem. Slight changes in contact pressure made huge differences in the energy contributing to the weld. Here's what I did to improve MOT welds in order of decreasing benefit. Control the turn on/off phase of the input voltage and control the number of integral cycles of 60Hz. When your weld takes six cycles, being off by one is significant. Spring load the contacts for pressure repeatability. Cutting a slot lengthwise down the battery tab forces more of the current thru the weld points decreasing sensitivity to contact resistance somewhat. With all that, I still had trouble constructing a whole battery pack without blowing at least one weld. The same fixture with a CD welder gives almost 100% good welds. http://i.imgur.com/yd8c0rf.jpg http://i.imgur.com/er1BqSb.jpg http://i.imgur.com/OeQZdWH.jpg > If I made a transformer big enough to weld thick workpieces with proper > contact pressure, it might cause excessive drop in my mains supply, > and/or trip breakers. It's all about time. For battery tabs, you need very localized heating and a very short weld time. I have a commercial spot welder. You just push the button and wait several seconds for the metal to turn red. Works great for angle iron, but would kill a battery instantly. > think at least 2 banks of boostcaps in series will be desirable. Due to > the capacitors holding more energy than the total that you would want > for one weld, it would be necessary to find a way to switch them off, Just lower the initial voltage on the caps. E = C*V^2/2. You could also disconnect some of the caps. Or maybe not fire all the FETs...although that may be easier said than done. I had great difficulty measuring output current because the huge magnetic fields generated coupled into everything. > and it would also be very useful to be able to adjust the current by PWM > during the weld. See the link below Therefore a lot of MOSFETs would be required. It seems > about what would happen in the event of one failed MOSFET, and I would > like to think of a way to mitigate that. Perhaps the bondwire or package > pin would be an adequate fuse. Each FET has to be able to handle all the current it can deliver with the output shorted at the weld. Limit the charge current to the caps and a short won't cause much problem. I wonder if you couldn't run a wire from each FET thru a common toroid before the junction point to help equalize the peak currents?? As for failures, anything that would "blow" under short conditions would seriously undermine your ability to deliver current to the "short" at the weld. > Chris The advantage of CD is that you get a defined energy pulse. The energy delivered to the weld point is relatively insensitive to the resistance of the contact. Far less so than with a MOT. I've watched a lot of youtube videos of successful welds using racks of low voltage caps and arrays of FETs. My Unitek uses about 400V discharged into a pulse transformer. That V^2 factor for energy storage adds up rapidly. The magic is in the pulse transformer. It's smaller than my fist and claims to put 7KA into .001 ohms. Would be interesting to learn how to build one of those. From this: http://i.imgur.com/ZeZerGx.png it appears that they might pass current backwards thru the transformer to "reset" it. Tripped over this overview today: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=12&ved=0CFcQFjAL&url=http%3A%2F%2Fwww.amadamiyachi.com%2Fservlet%2Fservlet.FileDownload%3FretURL%3D%252Fapex%252Feducationalresources_fundamentals%26file%3D01530000000Jybm&ei=-lXiVL3lBMi5oQTb4YDYDQ&usg=AFQjCNFc_ush2PjqSdItUaNVEJsM3Wmh7g&sig2=ZJclqkBCdA3AOHUL_l1eew&bvm=bv.85970519,d.cGU&cad=rja I've given up trying to weld thin aluminum. With wire feed and Argon, I can lay a bead on THICK aluminum. I experimented briefly with TIG. Better, but would take practice that exceeds my attention span and my Argon budget. Welco 52 rod with a torch makes a very strong bond on stuff that can take the heat. There's very small margin between a successful bond and a puddle on the floor. I learned the hard way that most stuff that looks like aluminum is really potmetal and has the same melting temperature as the Welco 52. |
| Phil Allison <pallison49@gmail.com>: Feb 16 04:35PM -0800 whit3rd wrote: > Remember Phil's > comment that a 240V transformer on 120V excitation wouldn't saturate- > because a 240V transformer has an oversize core for 120V excitation. ** Switching at a zero crossing creates a short term, low frequency component in the AC wave - a component approximately equal to the same AC voltage at half frequency. Most AC supply transformers operate right on their low frequency limit - but if you only apply half voltage then that limit is now at half the original frequency - so the switch on condition is tolerated. IOW, a given saturation condition is proportional to V / f. I often see this in action when testing US model amplifiers meant for a 120V at 60Hz supply. A step-down device only adjusts the voltage leaving the frequency at 50Hz. The resulting magnetising current is the SAME as if one had applied 144VC to the transformer instead of 120V so Imag is way higher. Some transformers are OK with this and others run very hot PLUS it has a significant effect on the VA rating of a suitable step down transformer. ... Phil |
| Chris Jones <lugnut808@spam.yahoo.com>: Feb 18 12:03AM +1100 On 17/02/2015 10:35, mike wrote: > delay the turn-on if I wanted. All the solid state relays I had > were/are zero voltage turn on units. Best I can do is control the slope > of the AC input at turn on and turn off. Ok. >> core matters. > I think that's a red herring. I can't argue the result, but don't think > it's relevant. Short the output and repeat the experiment. My point was just that the state of the field in the core would be different just a nanosecond after you turn off the primary current vs a week later in your examplesl, hence the consequences of switching it back on would be different. > primary voltage > be pretty much in phase with the primary current? The leakage inductance > should be a small part of the equation? It quite possibly would, but I was just saying that the primary current by itself isn't a good way to determine what the flux in the core is. Anyway I am pretty sure you get it. > OK, but the (almost) shorted output should have a major effect > on what happens when you stop supplying primary current and where > the core ends up on it's B-H curve. I guess it might in that the SSR will turn off when the primary current is zero which might be a different time instant, and therefore a diferent point on the voltage waveform from what it would have been without the secondary current. Still, some cores might go to nearly zero flux density, regardless of when they are turned off, if they don't have much remanence. > input current and you do what you can to minimize the surge current, > how much does it really matter? Unfortunately, I have no easy way > to measure it given my SSR characteristics. Ok. >> the capacitors holding more energy than the total that you would want >> for one weld, it would be necessary to find a way to switch them off, > Just lower the initial voltage on the caps. E = C*V^2/2. To get the energy in these capacitors low enough for one weld, the voltage would be tiny and I think it may not deliver enough current (but would deliver it for too long) in that case. Also I would like the possibility of current waveforms other than exponentially decaying. > You could also disconnect some of the caps. I was hoping to stay within the current rating of the caps which is 1900A each, so I would want several in parallel to get enough welding current. The caps are capable of developing nearly 10kA each according to the datasheet but that supposedly damages them and they are somewhat expensive. > Or maybe not fire > all the FETs...although that may be easier said than done. Again, I was intending to stay within the current rating of the FETs so I would want all of them on to share the current as evenly as I can, even if I have to do something else to reduce the current. I could reduce the current by adding some cable resistance if there is not already enough. I would still want to have the FETs there as a way of switching it off at the end of the weld. Maybe PWMing it to regulate the current would be worthwhile, but it probably means twice as many FETs. Still I might need catch diodes if I didn't put the second lot of FETs so maybe that is no extra problem. > I had great difficulty measuring output current because the huge > magnetic fields generated coupled into > everything. Yes, that sounds difficult. I guess you could use that as the way of measuring the current, e.g. a Rogowski coil or whatever it is called. > Each FET has to be able to handle all the current it can deliver with > the output shorted at the weld. Limit the charge current to the caps > and a short won't cause much problem. I think it could be a problem... Hehe these caps are pretty amazing, one of them will easily melt a nail well after the charging current is disconnected. Then another nail, and another. 3000 Farads is more like a small capacity battery (but with less resistance than a battery). > I wonder if you couldn't run a wire from each FET thru a common toroid > before the junction point to help equalize the peak currents?? I was more thinking putting the wire from each FET stage through its own toroid, and sensing the currents in each wire and regulating it with individual PWM for each stage. That sounds like too many parts so maybe just having separate longish wires from each FET stage would be enough sharing resistance even with common PWM for all stages. Anyway I won't build this for ages so I will have forgotten by then. > As for failures, anything that would "blow" under short conditions would > seriously undermine your ability to deliver current to the "short" > at the weld. Well I was not intending to exceed the ratings of the FETs, so if there are 100 FETs, each with wires that fuse at say 200A, but normally each conducts 100A, then if one FET fails short and starts carrying a few kA it will be good if its wires fuse cleanly rather than keeping conducting and starting a fire. As long as I don't exceed each FET's ratings it would not be a problem for each FET to have a fuse action at some higher current. I am reminded of the photos of inside a Tesla's battery pack that seems to have a lot of cells in parallel, each cell with an individual thin fuse wire in series with it. > http://i.imgur.com/ZeZerGx.png > it appears that they might pass current backwards thru the transformer > to "reset" it. Interesting, but I hope I don't need to build one, partly becuase it restricts pulse length etc. |
| Phil Allison <pallison49@gmail.com>: Feb 16 08:00PM -0800 Nutcase Kook wrote: > I use a variac with an AC ammeter in line. ** On the output side? > series with the mains, the one-sided "switching" 50/60 times a second > causes horrendous magnetising current problems that would overheat the > variac I'm sure, used on the same mains ring main. ** You are exaggerating a bit. > Replace that diode > with a preset triac for balanced "switching" ? **Could be a gotcha there - how does the fan get its power? In my old Black & Decker (Made in England) 1600 watt gun, the DC fan runs via one diode from a tap on the main heater winding. The half power diode has the same wiring polarity so fan operation is not stopped at half setting. Fit a triac and the fan would lose half its power, so the air would be just as hot with either setting. > What effect , if any , would such diodes in say hot air guns, falsely? > register on a moving disc mains kWh meter ** Suspect half wave loads make them under read a bit. Why not try it out while counting turns on the big wheel .... Phil |
| Tauno Voipio <tauno.voipio@notused.fi.invalid>: Feb 16 06:48PM +0200 On 16.2.15 16:01, Sparky wrote: > Until next time. > What could cause these symptoms? Twice? > Thanks! Bad battery which does not let the voltage to rise, or too much load, but not enough to blow the fuse. A winding can be fried with a slight overload and limited cooling. Do you happen to have quadruple 300 W audio channels on board? -- -TV |
| Jon Elson <jmelson@wustl.edu>: Feb 16 05:27PM -0600 Sparky wrote: > If a diode shorts, there also wouldn¹t be any increase, just a > decrease in DC output voltage. No, ONE a shorted diode would cause huge currents in the winding connected to that phase. Two shorted diodes on the same phase would cause huge DC currents and blow the fusible link. Jon |
| Martin Riddle <martin_rid@verizon.net>: Feb 16 06:32PM -0500 >Until next time. >What could cause these symptoms? Twice? >Thanks! Does the alternator use Avalanche Diodes? A bad battery wiring connection will cause a diode to short. I'd check the terminals and crimps. BTW I had this problem with the double GM battery lugs, they used a lead spacer that shrunk when it thermal cycled. \Cheers |
| "Tom Del Rosso" <td_03@verizon.net.invalid>: Feb 16 08:51PM -0500 Sparky wrote: > regulator from the local parts supplier. > Now (7 or 8 months later) the replacement has failed in the same > mode. Išm suspicious. Also confused: If you asked in rec.autos.tech you'd get answers with less knowledge of theory but with more knowledge of common failure modes. -- |
| jurb6006@gmail.com: Feb 16 06:52PM -0800 Alternators are shit today. In the old days you could replace one, take the cablesoff a good but completely dead battery and start it with a another and then connect the battery to charge. Sometimes it would slow down the engine because of the load. A dead but good battery is almost a short circuit. Those old alternators could handle it but not the new ones. It even should says in the instructions to make sure the battery is charged before starting that car after replacing the (shit) alternator. (let's not even go into a few Chevy engines where you have to damnear remove the engine to change it, and they are a high failure part) You would not get the same thing likely with two alternators in a row unless it is a common failure. The problem MIGHT be an overload but it would be one that is only there when it is running, otherwise any short that heavy would kill the battery and it would not start the next morning. Problem is they don't put a proper ammeter in cars anymore. I don't even want to go into telling you how to hook one up. The starter motor has to be isolated and on some cars this is damnear impossible. You don't want to read the starter amps on the ammeter, it is too high and you would need a proper scale and that would be too high to see if this is the problem. The only real way to do it these days is at the alternator itself, that will not indicate discharge like in the old days but it can tell you how much current is being pulled. The fattest red wire on it, take it off and put the ammeter there. The alternator has a rating, if the constant drain without the headlights and AC on is near that rating there is an overload. Not a dea short so an ohmmmeter will not tell you where it is. It could be almost anywhere, one of the coils or cokl drivers, fuel pump, or some modification like those road lights or blue diamond headlights with the ballasts or who knows. It could also be that the engine is running too hot, concievably. That is not the most likely though. It is also possible that the alternator is simply underbuilt for the car. that is actually pretty likely, it would not be the first time I saw it happen. |
| Sparky <see@thesig.net>: Feb 16 07:14PM -0800 > Did you change the battery any time after the alternator was changed or is > it the same one still? Replaced with new from NAPA auto parts, not the cheapest, but a quality one. |
| Sparky <see@thesig.net>: Feb 16 07:16PM -0800 > Do you happen to have quadruple 300 W audio channels on board? No, mostly stock old Toyota. Only additional load is Xenon lamps. These pull 10A total which isn't much for lights... Maybe I'll upgrade the alternator to a later-model Toyota 70A or such. |
| Sparky <see@thesig.net>: Feb 16 07:20PM -0800 > If you asked in rec.autos.tech you'd get answers with less knowledge of > theory but with more knowledge of common failure modes. THANK YOU! Hadn't crossed my mind (being the eletrickery kind of guy that I am...) Best day to you! |
| Sparky <see@thesig.net>: Feb 16 07:31PM -0800 > connected to that phase. Two shorted diodes on the same > phase would cause huge DC currents and blow the fusible link. > Jon Ah. Then might be one shorted diode... Should probably upgrade alternator to a higher-amp one that can handle the car's load more easily. |
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