and made several measurements. I assumed internal resistance r of my wavetek to be 50 Ohm. At first, I forgot about that r and got bad results (clearly increasing L as function of R).
The results show a relatively clear pattern, that my welder's inductance is between 2 and 5 mH. That's not bad.
Table follows Computing Inductance of reactor V1 6.5 v voltage on wavetek's terminals V2 v Voltage across resistor R Ohm Resistance Frequency, f 5000 Hz Wavetek's frequency
Internal resistance 50 Ohm R V2 x Inductance, H Inductance,mH
The web site says, "Measure the wire resistance of the inductor under measurement; call it r.", so r isn't the internal resistance of your Wavetek; it's the resistance of wire that makes up the inductor. Since you don't have a way to measure the wire resistance taking into account proximity and skin effect, you'll have to settle for the DC resistance as measured by an ohmmeter.
The web site also says, "Adjust the generator to a frequency of 2 to
10 KHz, sine wave, and 1.00 volt rms. The voltmeter must be connected across points A and B, that is, reading the voltage from the generator." And then " 3. Move the voltmeter test leads across points C and B and take down the reading; call it x."
So if you adjust the voltage at the Wavetek's terminals to 1 volt like Dr. Rico says to do, why are you saying above that V1 is the voltage on wavetek's terminals and has a value of 6.5 volts?
Furthermore, since x is supposed to be the voltage across the resistor according to the web site, and you say above that V2 is the voltage across the resistor, how can they be different, as shown in the table? I don't think you're correctly following Dr. Rico's procecdure.
If you'll describe the physical details of your inductor (dimensions, number of turns, wire size, etc.), there are formulas for estimating the inductance and distributed capacitance.
On Sat, 15 Oct 2005 06:12:28 GMT, Ignoramus3498 wroth:
Your inductor also has a parasitic capacitance associated with it. Each turn is a conductor acting like a capacitor plate with respect to adjacent turns.
The method you used ignored that capacitance.
A "better" method to estimate the inductance would be to purposly add an external capacitor across the inductor's terminals and make the R very large (10K to 100K, it isn't at all critical) so that your generator/R combination acts like a current source. Then sweep the generator's frequency looking for a resonant peak in the voltage across the L/C combination. Make the added capacitor's value much larger than the inductor's own parasitic capacitance. About .1 uF should do it.
Once you have that *single* frequency measurement, the inductance should "fall out" of a simple calculation.
I have to wonder, just what would I get from not ignoring that capacitance, in terms of accuracy of measurement, given that my voltmeter is not the pinnacle of accuracy. For all I know, my inductance is between 2 mH and 5 mH. Would I get dramatically different result from considering the parasitic capacitance?
On Sat, 15 Oct 2005 15:51:08 GMT, Ignoramus24693 wroth:
Yes. Using the method I suggested, the voltmeter could be very inaccurate and you would still be able to make a good measurement since it only has to read a relative peak and the calculated inductance would be as accurate as the tolerance of the capacitor, usually 5%, and the accuracy of the signal generator readout.
"Between 2 mH and 5 mH" is approaching 100% uncertainty.
On Sat, 15 Oct 2005 19:37:37 +0200, "Fred Bartoli" wroth:
Fine. And if the inductance is 2 mH? Does Ignoramus have to keep adding and subtracting capacitors until the resonant frequency comes out to exactly 5 KHz?
But you *do* understand that r in Dr. Rico's formula is not the internal resistance of the Wavetek, right? If you set r to the Wavetek's internal resistance, then the formula won't give the correct inductance. You have to set it to the wire resistance, and for your inductor, zero is probably appropriate.
If you were using V1 = 1 volt, as Dr. Rico says, the formula V2/V1 *would* be dimensionally correct.
If you set r (which is *not* the Wavetek's internal resistance) in the formula to zero and redo the calculations, you should get:
I'm assuming you already got these results, because as you say earlier, "At first, I forgot about that r and got bad results (clearly increasing L as function of R)."
You *should* get increasing L with increasing R if you use Dr. Rico's formula with your measurements. The internal resistance of the Wavetek doesn't have any effect on your measurements, because you are measuring V1 at the terminals of the Wavetek, and therefore the drop across the internal resistance comes
*before* your measurement of V1, If it were possible to measure the *ideal, hypothetical* internal voltage source of the Wavetek, then it would matter, but you can't do that anyway, so set r to zero in the formula (assuming it's really effectively zero, as you say).
It looks like something else is wrong. Make sure that you are using the
*sine* wave output of the Wavetek. Double check your setup.
Do you have an oscilloscope so you can look at the waveshapes out of the Wavetek, and across the resistor R?
How about making your measurements at a different frequency, maybe 20 kHz?
Is your voltmeter a true RMS responding meter? And when measuring AC, does it respond to AC + DC?
Does this inductor have an iron core?
What will be the frequency of the welding current passing through the inductor?
Too bad you didn't tell us that this was what you were doing, but left us to guess that might be it. Your table didn't list the V1 values; I suppose we could infer them from the x values, but it's good practice to show all the measured values you use. What I'm trying to say here is that if you reference a web page that describes a technical procedure, and you modify that procedure then you should say so, because some change you made may have contributed to your bad results. And when you ask for help in determining what is wrong, we need to know *exactly* what you have done. Don't leave it to your readers to guess; spell it out exactly.
It's also *not* good practice to show 15 digit results from a computation that has only 3 digit numbers as input.
I assumed (falsely, it would seem) that you had built the inductor, and would therefore know its dimensions.
If it's a *saturable* inductor, then when DC passes through it, it will saturate, and then the inductance will decrease and AC *will* pass, the amount depending on what the residual inductance is after it saturates.
reference a
No, no. It is possible to calculate what V1 was from your given V2 and calculated x. I'm just offering advice for the future when you ask for help. Tell us *all* the details.
I think I have seen your postings here before about building an H-bridge. Are you planning to use this inductor with your H-bridge setup? Is that why you want to know the inductance, or is it just curiosity?
I have a suggestion (it involves getting access to the inductor again). It is often the case when measuring inductors with an iron core that are designed to handle high power, that inductance measurements with a small excitation (such as your Wavetek) are substantially in error.
If you have a variac, or a small transformer (such as the filament transformers that can be bought at Radio Shack), apply a small 60 Hz voltage to the inductor and measure the current. I'm guessing that the inductance should be in the range of a few millihenries, so remembering that 2*Pi*f is 377 for f=60 Hz, if you apply 1 volt AC to a 10 millihenry inductor (for example), the current would be V/(377*L) = 1/377*.01 = .265 amps. You could measure that with any old ammeter. You could apply voltages up to 10 volts, or whatever it takes to get a few amps and plug the numbers (applied voltage and resultant current) in the formula: L = V/(377*I). You usually get better numbers with high power iron core inductors if you make the measurement with several amps through the inductor. I wouldn't be surprised if you see some variation in the inductance versus current with an iron core inductor, but the inductance values you get with this procedure will be more like what is happening under actual operating conditions (assuming the inductor isn't saturated).
If you use a variac, remember that they aren't isolated from the line. If you have a small transformer, 120 volts in (drive this with the variac) and perhaps
12 or so volts out, you get the safety factor of isolation.
a écrit dans le message de news: snipped-for-privacy@4ax.com...
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That was not the point. It was to show that a realistic parasitic capacitance has negligible impact on the measured value due to the low measurement frequency. For 2mH the case is even more favourable and he can safely ignore the parasitics influence in his measurement (for the required accuracy).
It will be even better if he lowers the frequency so that the inductance and his generator have the same impedance value.
Because x is V2/V1. I do not like to use dimensionally incorrect formulas.
Further on the page, he explains derivation of the formula and x is clearly V2/V1. That's what I used. I reasoned that my voltmeter would be more accurate at higher voltages. His page really has all details, his use of reference voltge of 1v.
Inductor is hard to access and so turns are hard to count. I could barely attach test leads.
This is a good point, in one of my early electronics classes we measured an iron core inductor at low power. About eleven groups (two people each) measured the inductor at about .15 uh. Everyone in the class was happy, except me, the inductor was clearly marked 5H. It took a couple of weeks for me to learn about B/H curves and how our measurement was low on the graph where the slope was very slight. Those were a very exciting two weeks for me!
Saturable? Does the manual say that somewhere? It seems to me that this inductor saturating would have a detrimental effect on current regulation and arc stability.
Well mostly DC anyhow, it is powered by a 6 phase SCR rectifier. Low current setting procedure adjusts to 20A, high current 200A and max "arc force boost" another 200A for 400A max current through this inductor (at some unknown duty cycle, the manual leaves much to be desired). Yes, this is the energy storage inductor in a 360 Hz buck mode 400A current regulating power supply with no output capacitor that i planning to drive his H-bridge with :-).
Well I suspect whoever said it is wrong, even if it was me :-). Any inductor will saturate at some current but I expect they designed it not to saturate even at 400 amps by providing an appropriate amount of iron and an air gap.
Which part are you questioning? The current settings are on PDF page
43 (manual page 31), paras 3,4 and 5. This would show up on the Volt/Ampere curve for the machine if the bums provided one, go look at the curve for the default Cybertig (the one you get from their web page with no serial number) to see the shape of the curves, which you could easily redraw shifted a bit to match the current settings from the adjustment procedure for your machine.
They do not explain the operation of the power supply, but it is evident from the schematic.
The main interesting issue is still the inductor energy. I think you are underestimating the inductance and recommend that you take the advice of those who suggested a higher current measurement, on the order of 20 amps to get you up to the bottom end of design current.
Then calculate your snubber component values, enter it into LTSpice (free is a very good price) and simulate it (with your inductor). You do not have nearly enough parts to be conducting your design experiments with real parts, and simulation changes are a lot quicker than breadboard changes. I am sure that you will find LTSpice easy to use, the only hard part is understanding its limitations.
When you have the H-bridge with snubber simulation working you can then use it to look at the effects of your snubber on welding current.
A working simulation does not guarantee a working circuit but at least it will give you some chance, especially if you can get someone here to check out your simulation before you build it.
How well do you understand the main power circuit of your welder? The
3-pahse to 6-phase transformer is a bit tricky but I think you should take time to figure it out.
And pardon me for insulting your welds, but I was welding high pressure steam piping for a living well before my estimate of when you were born, and my standards are such that even I can't meet them anymore, what with being 30+ years out of regular daily practice!
I got it. That you found it was quite insightful. Thank you.
Yes, I also think so.
That would, obviously, be 20 amps AC, right?
I am thinking of making this circuit, not quite for 20 amps but still:
/--- Inductor ----B A | AC line/\\/\\/\\ | | | \\_____UUUUUUUU___| Room heater load
I would do that with the welder fully unplugged and be mindful that it may be energized (even though I would not expect that to be the case).
If I measured the AC line voltage, current and voltage across A and B, would I be able to get the inductance of the inductor?
I do have the ability to parallel room heaters to get about 25 or so amps, although that feels like something I would rather avoid.
Never heard of it, but I will check it out.
I just bought some varistors for experimenting and will place one across the welding terminals, to see if it would blow up when an arc is stopped. I am aware that this may be a much softer turn off condition than turnoff of the H bridge. On the other hand, I do not expect the Hbridge to turn off at full current, except for microseconds, but it would regularly experience arc interruptions.
That's indeed very interesting. Note also that I will use an RCD snubber.
Very interesting.
I looked at the schematic and I think that I have the basic idea of how it works, but it may be lacking. I will be happy to investigate whatever issues that you bring up.
What I am really curious about, tangentially, is whether this three phase machine can be converted to 220V single phase by disconnecting the three transformers, and adding some caps to compensate for extra ripple. That would be just awesome if I could avoid the use of my phase converter.
Oh, I was not at all insulted... My total welding experience is under
That should work, although I had in mind a low voltage output of a transformer for supplying the ~20A AC. But you might as well use the parts you have. You could also consider the "splat" test Terry suggested; I think he meant discharging a big capacitor through your inductor while watching current and voltage on your scope, in hopes of reaching Isat at (I think) something over 400 amps.
Convienient that you have a current shunt already connected to your inductor.
You had better plan for that unexpected full current hard turnoff, since you cannot make it impossible. Consider drive circuit failure.
Just to be sure you have it I will post a book explanation of the
6-phase rectifier to ABSE. Note that it refers to your SCRs as mercury ignitrons, since it was written before the EU lunatics in Brussels outlawed mercury and they started making them from silicon :-).
No chance, look at the circuit and inductor in the default single phase Cybertig manual. Note the addition of a freewheeling diode which allows welding current to flow during the zero crossings of line voltage, and the larger inductor to carry the current through the no drive available period. Your 6-phase half wave rectifier provides a much more constant drive, and I have found that 3-phase machines are easier to weld with than single phase machines, possibly because of less ripple or possibly because of lower inductance, not sure. (The easiest stick welding supply I have used had no deliberate inductance in the welding circuit, with current control by a big resistor bank!) Plus there are other issues too mumerous to mention regarding the control circuits.
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