Measured INDUCTANCE of my welding reactor

[Ignoramus3498]

I am not yet completely sure of my numbers.

I used this schematic:

http://et.nmsu.edu/~etti/fall96/electronics/induct/induct.html

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
100 3.75 0.576923076923077 0.0027662025946487 2.77
200 4.82 0.741538461538462 0.00322321281768636 3.22
300 5.29 0.813846153846154 0.00368392086575143 3.68
400 5.57 0.856923076923077 0.00395077803262996 3.95
500 5.75 0.884615384615385 0.00414877825161414 4.15
600 5.87 0.903076923076923 0.00438069122639741 4.38
1000 6.13 0.943076923076923 0.00470874461427637 4.71
2000 6.33 0.973846153846154 0.00393088999695529 3.93

i

 

[[email protected]]

I am not yet completely sure of my numbers.

I used this schematic:

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.

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.

Jim

 

[Ignoramus24693]

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?

i

 

[Fred Bartoli]

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.

The measurment frequency is 5kHz. How much does it take of a capacitance to
resonate a 5mH inductance at 5kHz?

A: 200nF.

It's more than unlikely for the parasitics to be significant wrt this 200nF
value.

 

[[email protected]]

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?

i

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.

j

 

[[email protected]]

The measurment frequency is 5kHz. How much does it take of a capacitance to
resonate a 5mH inductance at 5kHz?

A: 200nF.

It's more than unlikely for the parasitics to be significant wrt this 200nF
value.

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?

Jim

 

[Fred Bartoli]

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?

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.

 

[The Phantom]

I am not yet completely sure of my numbers.

I used this schematic:

http://et.nmsu.edu/~etti/fall96/electronics/induct/induct.html

and made several measurements. I assumed internal resistance r of my
wavetek to be 50 Ohm.

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.

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

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?

R Ohm Resistance
Frequency, f 5000 Hz Wavetek's frequency

Internal resistance 50 Ohm

R V2 x Inductance, H Inductance,mH
100 3.75 0.576923076923077 0.0027662025946487 2.77
200 4.82 0.741538461538462 0.00322321281768636 3.22
300 5.29 0.813846153846154 0.00368392086575143 3.68
400 5.57 0.856923076923077 0.00395077803262996 3.95
500 5.75 0.884615384615385 0.00414877825161414 4.15
600 5.87 0.903076923076923 0.00438069122639741 4.38
1000 6.13 0.943076923076923 0.00470874461427637 4.71
2000 6.33 0.973846153846154 0.00393088999695529 3.93

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.

 

[Ignoramus24693]

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.

it's effectively zero.

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?

Because x is V2/V1. I do not like to use dimensionally incorrect
formulas.

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.

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.

i

 

[The Phantom]

it's effectively zero.

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.

Because x is V2/V1. I do not like to use dimensionally incorrect
formulas.

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:

Frequency, f 5000 Hz Wavetek's frequency

Internal resistance 50 Ohm

R V2 x Inductance,mH
100 3.75 0.576923076923077 4.51
200 4.82 0.741538461538462 5.76
300 5.29 0.813846153846154 6.82
400 5.57 0.856923076923077 7.66
500 5.75 0.884615384615385 8.39
600 5.87 0.903076923076923 9.08
1000 6.13 0.943076923076923 11.2
2000 6.33 0.973846153846154 14.9

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?

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.

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.

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.

I assumed (falsely, it would seem) that you had built the inductor, and would
therefore know its dimensions.

 

[Ignoramus24693]

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.

I see. Gives me food for thought...

If you were using V1 = 1 volt, as Dr. Rico says, the formula V2/V1 *would* be
dimensionally correct.

I just did not like it, although mathematically it is alright.

If you set r (which is *not* the Wavetek's internal resistance) in the formula
to zero and redo the calculations, you should get:

Frequency, f 5000 Hz Wavetek's frequency

Internal resistance 50 Ohm

R V2 x Inductance,mH
100 3.75 0.576923076923077 4.51
200 4.82 0.741538461538462 5.76
300 5.29 0.813846153846154 6.82
400 5.57 0.856923076923077 7.66
500 5.75 0.884615384615385 8.39
600 5.87 0.903076923076923 9.08
1000 6.13 0.943076923076923 11.2
2000 6.33 0.973846153846154 14.9

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)."
yep.

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.

I was indeed using sine wave. I made sure of that.

Do you have an oscilloscope so you can look at the waveshapes out of the
Wavetek, and across the resistor R?

Sure, I have two scopes, but I am positive that it was sinewave.

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?

Not sure, good question.

Does this inductor have an iron core?

Yes, indeed. It is a saturable inductor from a TIG welder:

http://igor.chudov.com/projects/Welding/00-Hobart-CyberTig-Welder/

What will be the frequency of the welding current passing through the
inductor?

Only DC will pass through inductor. Past that, but before the high
frequency arc starter, there will be an H bridge based inverter.

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.

Fair enough. I could redo it, however, it is a PITA and takes about an
hour to unscrew the side panel, get there, it is all dusty and yucky.

It's also *not* good practice to show 15 digit results from a computation that
has only 3 digit numbers as input.

I just had hard times adjusting precision in that Gnumeric, but I
agree.

I assumed (falsely, it would seem) that you had built the inductor, and would
therefore know its dimensions.

The dimensions, I could guess more easily than the # of turns.

i

 

[The Phantom]

I see. Gives me food for thought...


I just did not like it, although mathematically it is alright.


I was indeed using sine wave. I made sure of that.


Sure, I have two scopes, but I am positive that it was sinewave.


Not sure, good question.


Yes, indeed. It is a saturable inductor from a TIG welder:

http://igor.chudov.com/projects/Welding/00-Hobart-CyberTig-Welder/


Only DC will pass through inductor.

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.

Past that, but before the high
frequency arc starter, there will be an H bridge based inverter.


Fair enough. I could redo it, however, it is a PITA and takes about an
hour to unscrew the side panel, get there, it is all dusty and yucky.

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.

 

[amdxjunk]

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.

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!

 

[Glen Walpert]

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 .

 

[Don Foreman]

I am not yet completely sure of my numbers.

I used this schematic:

http://et.nmsu.edu/~etti/fall96/electronics/induct/induct.html

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

If your inductor has an iron core (likely), it will behave quite
differently at 5KHz than at 60 Hz due to differences in eddy current
and hysteresis losses. It will also have different values at higher
levels of excitation.

A better way to measure it might be to use a small current sense
resistor and note phase shift with excitation similar in magnitude
and frequency where it would be used.

 

[Ignoramus29878]

If your inductor has an iron core (likely), it will behave quite
Correct

differently at 5KHz than at 60 Hz due to differences in eddy current
and hysteresis losses. It will also have different values at higher
levels of excitation.

A better way to measure it might be to use a small current sense
resistor and note phase shift with excitation similar in magnitude
and frequency where it would be used.

I was going to wire a space heater in series with the inductor, and
measure the voltage across the inductor.

i
--

 

[Ignoramus29878]

Glen, I made another attempt to measure inductance. It now came to be
1.85 mH.

I wired a space heater in series with the inductor, and measured
relevant values.

Another attempt at inductance

Vac 124.4
Voltage across inductor 8.69
Current 12.5

Formula: V=I*2*pi*f*L

or L=V/(I*2*pi*f)
Inductance= 0.0018450 Henry
1.8450 mH

i

 

[Winfield Hill]

Ignoramus29878 wrote...

Glen, I made another attempt to measure inductance.
It now came to be 1.85mH. I wired a space heater in
series with the inductor, and measured relevant values.

Vac 124.4
Voltage across inductor 8.69
Current 12.5

Formula: V = I*2*pi*f*L
or L = V/(I*2*pi*f)
Inductance = 0.0018450 Henry
1.8450 mH

That's cute, a space heater, all warmed up and glowing. :>)

So let's see, an inductor's stored energy is E = 0.5 L I^2,
which is = 37J for you at 200A. That's nothing to sneeze at!
If a node capacitance is say 2nF (a MG200Q2YS40 has less than
1000pF of capacitance above 40V), the flyback spike voltage
will be V = I sqrt (L/C) = 200 sqrt (1.85mH/2nF) = 192kV.
Ouch!! Obviously this means the open-circuit flyback spike
will be high enough to cause breakdown someplace, hopefully
at the welding tip, etc., but if not, where will it be???

I assume it won't be at the IGBT -- it had better not be!!
Just a quick analysis:

OK, let's say the IGBT breaks down at 1200V. This would
mean we're expecting the IGBT to handle a peak power of
1.2kV times 200A = 240kW. Ahem... The inductor's current
decay rate with a 1.2kV drop will be dI/dt = V/L = 0.65A/us,
which means it'll take about 310us to ramp down to zero. So,
we have a 240kW peak-power pulse lasting 300us. <cough!>
Looking at the MG200Q2YS40's Transient Thermal Resistance
curve (page 4), we see the thermal resistance is 0.01 C/W
at 1ms. We can extrapolate, using the square-root rule, and
estimate the 30us region has a thermal resistance of about
0.0018 C/W. Doing the crude back-of-the-envelope calculation,
this corresponds to a junction temperature rise of 444C for
240kW. A full calculation would show a higher value. We're
in the order-of-magnitude "ballpark" of safe breakdown, but
sadly, without some other protection to prevent breakdown,
the IGBT will probably be toast. Perhaps a 5-10x larger
part could safely handle such a large breakdown...

 

[mike]

I am not yet completely sure of my numbers.

I used this schematic:

http://et.nmsu.edu/~etti/fall96/electronics/induct/induct.html

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
100 3.75 0.576923076923077 0.0027662025946487 2.77
200 4.82 0.741538461538462 0.00322321281768636 3.22
300 5.29 0.813846153846154 0.00368392086575143 3.68
400 5.57 0.856923076923077 0.00395077803262996 3.95
500 5.75 0.884615384615385 0.00414877825161414 4.15
600 5.87 0.903076923076923 0.00438069122639741 4.38
1000 6.13 0.943076923076923 0.00470874461427637 4.71
2000 6.33 0.973846153846154 0.00393088999695529 3.93

i

I've been trying to make sense out of this thread.
There was one tiny mention of "saturable reactor". The only reason one
would care, since most cored inductors are saturable to some extent or
other, is if one intended to operate in saturation.

One way to describe saturation is that the inductance is VERY dependent
on the current. A low current measurement may be interesting, but does
not accurately describe the small signal operating condition for a
higher fixed current.
We haven't even started talking about the time variant (large signal)
behavior. Or the effects of hysteresis.

So, even if you could determine the inductance to 15 decimal
places, what could you do with the number?

In a welder, don't you care most about what happens at welding current?
mike

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[Ignoramus15297]

Ignoramus29878 wrote...

That's cute, a space heater, all warmed up and glowing. :>)

So let's see, an inductor's stored energy is E = 0.5 L I^2,
which is = 37J for you at 200A.

Makes sense, assuming strict linearity of this inductor.

That's nothing to sneeze at! If a node capacitance is say 2nF (a
MG200Q2YS40 has less than 1000pF of capacitance above 40V), the
flyback spike voltage will be V = I sqrt (L/C) = 200 sqrt
(1.85mH/2nF) = 192kV. Ouch!! Obviously this means the open-circuit
flyback spike will be high enough to cause breakdown someplace,
hopefully at the welding tip, etc., but if not, where will it be???

I assume it won't be at the IGBT -- it had better not be!!
Just a quick analysis:

OK, let's say the IGBT breaks down at 1200V. This would
mean we're expecting the IGBT to handle a peak power of
1.2kV times 200A = 240kW. Ahem... The inductor's current
decay rate with a 1.2kV drop will be dI/dt = V/L = 0.65A/us,
which means it'll take about 310us to ramp down to zero. So,
we have a 240kW peak-power pulse lasting 300us. <cough!>
Looking at the MG200Q2YS40's Transient Thermal Resistance
curve (page 4), we see the thermal resistance is 0.01 C/W
at 1ms. We can extrapolate, using the square-root rule, and
estimate the 30us region has a thermal resistance of about
0.0018 C/W. Doing the crude back-of-the-envelope calculation,
this corresponds to a junction temperature rise of 444C for
240kW. A full calculation would show a higher value. We're
in the order-of-magnitude "ballpark" of safe breakdown, but
sadly, without some other protection to prevent breakdown,
the IGBT will probably be toast. Perhaps a 5-10x larger
part could safely handle such a large breakdown...

Winfield, is 37J enough to raise the temperature of the junction so
much?

In any case... The power to be handled at full shutdown (not temporary
shutdown lasting a fraction of a microsecond), is obviously large. In
a welding situation, it amounts to turning the bridge off, hard, at
full welding current. Imagine what would need to happen: I am welding
at high current, and someone or something would switch off the bridge.

Could that happen? Sure. If the power supply of my circuit dies, or
the power goes out, or my wife does something she'd never do (touch
the welding machine), all that could happen this.

But this would not be an event that would happen repeatedly. On the
repeated basis, we have fast switching events occurring very quickly
happen several dozen or hundreds of times per second. But these do not
last long.

Surely, I want to have a snubber circuit. I decided to have an RCD
snubber circuit, based on Terry's suggestion. In addition, I would add
a bunch of varistors to handle these unusual shutdowns.

i