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'[EE:] measuring inductance with DC present'
2006\08\08@134502 by Bob Blick

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Does anyone have a simple and fairly (+-10%) accurate method for measuring
the value of an inductor with DC present? I can add an extra winding if I
need to. The parameters are approximately 100uH, 8 amps, 20KHz.

Thanks,

Bob


2006\08\08@142007 by Martin K

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face
You can't take it out of the circuit and measure it with a function
generator, and oscilloscope?
Does inductance change drastically with frequency and DC bias? (assuming
it doesn't saturate)
--
Martin K

On Tue, August 8, 2006 5:45 pm, Bob Blick wrote:
{Quote hidden}

> -

2006\08\08@143232 by David VanHorn

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On 8/8/06, Martin K <spam_OUTmartin-distlistsTakeThisOuTspamnnytech.net> wrote:
>
> You can't take it out of the circuit and measure it with a function
> generator, and oscilloscope?
> Does inductance change drastically with frequency and DC bias? (assuming
> it doesn't saturate)
> --


Depends on the core material, some do.

2006\08\08@144722 by Bob Blick

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The inductance changes dramatically when DC is present, so that's why I
want to measure it with DC present. But if I have it in an inductance
bridge, or hook it to an inductance meter, the impedance of whatever I use
to put DC current into it is put in parallel so it lowers the reading.

Cheers,

Bob

{Quote hidden}

2006\08\08@153455 by Jack Smith

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Bob:

I did some similar measurements a few years ago and my approach was to
use three identical inductors in series across the DC power source. I
had a battery powered inductance meter so I could float it and measure
the center inductor (my meter had a blocking capacitor, so I didn't need
to add one, but yours may or may not need a blocking cap). I also added
a bypass across the DC power source, although this was almost certainly
unnecessary since the power supply had a bypass cap at the binding posts.

The measured inductance consists of Lx in parallel with 2Lx, so you have
to solve for Lx, not too hard, of course.

The idea of using three test inductors is that presumably they all have
similar L versus I curves and they have the same bias current since they
are in series. You could also rotate the three inductors through the
positions so that you measure each one and average the results.

Another approach would be to use air wound chokes for the power supply
isolation, as they should be essentially constant with current.

This arrangement has all sorts of stray C that I assumed was negligible,
as I was measuring inductors in the 1 or 2 Henry range, at 1000 Hz, but
your smaller inductors may require some further analysis.


Jack


Bob Blick wrote:
{Quote hidden}

2006\08\08@155044 by Steve Baldwin

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Measure the DC resistance out of circuit and then watch the rate of change
of current with a current probe and scope in circuit. Work backwards from
rate of change to time constant to inductance. It's one of those formulas with
e to the something.

Steve.


On 8 Aug 2006 at 11:47, Bob Blick wrote:

> The inductance changes dramatically when DC is present, so that's why
> I want to measure it with DC present. But if I have it in an
> inductance bridge, or hook it to an inductance meter, the impedance of
> whatever I use to put DC current into it is put in parallel so it
> lowers the reading.

==========================================
Steve Baldwin                          Electronic Product Design
TLA Microsystems Ltd             Microcontroller Specialists
PO Box 15-680, New Lynn                http://www.tla.co.nz
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=========================================


2006\08\08@185506 by Bob Blick

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Hi Jack,

Thanks, that's a good idea. I'll have to use an impedance bridge because
my henry meter doesn't work right with a cap in series.

Cheerful regards,

Bob


{Quote hidden}

2006\08\08@191535 by Jack Smith

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Bob:

Let me know how it works for you at the higher frequency. You'll have
more fun with those component values than I did at 1 KHz.

Thinking back on it now, I believe that I used an old GR RLC bridge
(GR-1650) for the inductance measurement. The details are in one of my
older notebooks.


Bob Blick wrote:
{Quote hidden}

2006\08\09@041712 by Alan B. Pearce

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> The inductance changes dramatically when DC is present, so that's why I
> want to measure it with DC present. But if I have it in an inductance
> bridge, or hook it to an inductance meter, the impedance of whatever I use
> to put DC current into it is put in parallel so it lowers the reading.

Are you sure you are not just measuring the inductance change you expect?

2006\08\09@110319 by Bob Blick

face picon face
>> The inductance changes dramatically when DC is present, so that's why I
>> want to measure it with DC present. But if I have it in an inductance
>> bridge, or hook it to an inductance meter, the impedance of whatever I
>> use
>> to put DC current into it is put in parallel so it lowers the reading.
>
> Are you sure you are not just measuring the inductance change you expect?


Hi Alan,

Just attaching a power supply and series resistor are enough to ruin any
meaningful measurement, without turning on the power supply. At the
current level I need to test (8 amps), the series resistor is low value
and influences the reading.

But methods using multiple identical inductors as suggested by others seem
promising and I'll be trying them.

Cheerful regards,

Bob


2006\08\09@113943 by Scott Dattalo

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flavicon
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Bob,

Here's a variation of Jack's suggestion. Use two power supplies, one for
AC excitation the other for DC excitation. Isolate the DC supply from
the inductor under test with an inductor you know doesn't saturate. The
exact value of this isolation inductance is not important. Similarly,
isolate the AC supply from the inductance under test with a large
capacitor (one whose impedance is way smaller than the inductor's
impedance). Finally, place a small valued resistor in series with the
inductance under test.

With this configuration, you can use an oscilloscope to measure the
phase shift between the AC excitation voltage and (say) the inductor's
voltage and back calculate the inductance. The accuracy of this
configuration is limited to how accurately you can measure the
resistor's value and how accurately you can measure the phase shift.

If the phase shift is phi degrees, then the inductance is:

    L = R/(2piF)*tan(phi)

You could also look at the relative amplitudes (e.g. the voltage across
the resistor relative to the voltage across the inductor and the
resistor), but I believe this is less accurate.

Scott

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