How to measure inductance indirectly?

I have several inductors I'd like to know the values of. I have no way to measure L directly, ie, with a meter.

I've got a scope and signal generator. Is it as simple as hooking a known value of C in series with the L and sweeping the frequency range looking for a peak in V (accross the inductor) and a drop in I?

Thanks,

Reply to
DaveC
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A lot simpler than that.

Just feed any frequency across the inductor. Read the voltage across it and the current through it (like across a 1 ohm resistor or a .1 ohm resistor)

The voltage divided by the current is the reactance of the inductor at that frequency and then just use the formula for reactance. Xl= 2piFL And viola with just your signal gen and a resistor and a voltmeter you can measure the L.

Try it at a couple different freqs and it should come out pretty close.

You have to use a high enough freq to swamp out the resistance though. But a freq with in the range of your voltmeter.

. . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

Reply to
BUSHBADEE

It can be, but your results might be better if you add a series resistance before the tank circuit (and possibly a terminating resistance to the transmission line from the signal generator).

If your tank circuit is placed in parallel with a low resistance (like 50 ohms from your signal generator, for example), the effective Q of the circuit will be reduced, and it will be more difficult to distinguish the amplitude peak when you sweep the frequency. On the other hand, increasing the series resistance before the tank will reduce the peak amplitude that you see at resonance, but the frequency peak will be sharper. Infinite resistance will give the sharpest peak, but also zero voltage - somewhere between infinite resistance and zero resistance is the 'best' measurement point.

-- Mike --

Reply to
Mike

On Tue, 2 Dec 2003 22:15:33 -0800, BUSHBADEE wrote (in message ):

RMS voltage? RMS current?

XL = V inductor / I inductor

so

L = X / (2*pi*F) right?

Can you suggest a few good frequencies to measure at? I've got an old Fluke handheld meter (don't recall the model #...)

The inductance value stays (relatively) constant across a range of frequencies? If I determine L at 1 Khz, for example, will that inductor have the same value at 20 Khz?

Thanks,

Reply to
DaveC

/|\ peak in the feed-I for a series-tuned LC.

More or less, yes. Although I go for a parallel L-C and high value resistive connection to the generator. Using different values of tuning-C then allows an easy guesstimate of the inductor's own stray-C.

Reply to
Tony Williams

The inductance is an inherrent property of the coil. There are other factors, Q, Resistance and distributed capacity. That is why you check it at a couple different frequencies. . . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

Reply to
BUSHBADEE

Not really- it depends on what the application is. If for example the inductors are for a switching power supply application, then you would want to measure the current response to a step voltage at various levels of DC bias- inductance is inferred from rise time L/R. If the inductors are to be used for RF or some other linear application where Q might be of interest then a simple technique is to use the Hay bridge at the application frequency:

Please view in a fixed-width font such as Courier.

Hay Bridge

+-----------+-----------------------------------+ | | | | \ _ \ | R1 / /| R3 / | \ \ | // / | | _ | | | /| | | C1 === +------------------------------+ --- /| | +-----+ | / \ | )|( | | | ~ | sig | )|( | | \ / gen | )|( === | --- | | | | | | +----+ | sd | +------+ | | +-|>|-+-/\/\-+-> VOM | | | \ | | ) \ | R2 / | === Lx ) Rx / | \ | detector | ) \ | / | | ) / | | | | | | | | | | | | +-----------+------+------------+---------------+------+ | | --- ///

Adjust R1 C1 for null at operating w=2pif where:

(R1-j/(wC1))(Rx*jwLx)=R2R3(Rx+jwLx)

Lx => Rx* -- = R2R3Rx or Lx=C1*R2*R3 C1

and

R2*R3 => R1RxwLx=R2R3wLx or Rx= ----- R1

Rx Q= --- wLx

Reply to
Fred Bloggs

Very nice. Tell us about the requirements on the transformer.

Thanks, - Win

whill_at_picovolt-dot-com

Reply to
Winfield Hill

The transformer can be a MiniCiruits RF type- out to several 100MHz on the high end and maybe down to 10KHz on the low end- it is the detector and its minimum detectable signal level that might require more work than I've shown.

Reply to
Fred Bloggs

The simplest way, is to get a meter that is designed to read inductance. A good inductance meter is fairly expensive. Leader has a very good laboratory type at a reasonable cost, compared to many of the others.

The more complex way is to make a resonant circuit with the inductor, and use your generator and scope, with some math to figure it out, as per the suggestions you received.

I've used a lot of Leader equipment, and found it to be very good.

Site for inductance meters

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Reply to
Jerry G.

On Wed, 3 Dec 2003 9:25:54 -0800, Jerry G. wrote (in message ):

I looked around and found Wavetek's Meterman series LCR55 which retails for about US$180. Cheapest solution for a meter.

Thanks,

Reply to
DaveC

Alternatively, get a grid dip meter! Made for the job...

Reply to
Paul Burridge

I made one of these from a kit and it has worked quite well for me.

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Reply to
John Popelish

Various folks have mentioned this one as a cheap meter that does a fairly good job:

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Its $100 in kit form, or $130 assembled and tested. That's USD.

Reply to
Robert Monsen

The *inductance* will stay pretty constant over a range of frequencies, provided you're not driving the core (if it has one) into saturation. But measure the DC resistance also, and compare with the reactance. If they are the same order of magnitude, then you may need to do some more math to figure out the true inductance.

If you put a resistor in series (say, 100 ohms), then 'tune' the frequency so the voltage across the inductor is about the same (albeit, not in phase with) as the resistor, they you should have a pretty good voltage/current reading to do the calculations.

You may also need to apply some 'gut' feeling. After all, if its a big multi-turn iron core, or little air-core inductor, your technique may have to be adjusted.

daestrom

Reply to
daestrom

On Tue, 2 Dec 2003 22:15:33 -0800, BUSHBADEE wrote (in message ):

My question is how to calculate reactance:

Or:

Your two formulae confused me. Which one should I use for calculating XL? Does it matter? Are they equivalent?

I hooked a choke in series with a 1-ohm 10 watt resistor. I took some measurements (VL, VR).

I used XL = 2*pi*F*L and reformed the formula to: L = X / (2*pi*f) -- is this right?

None of the known L values I measured worked out in calculation.

What am I doing wrong?

Thanks,

Reply to
DaveC

If you are mesuring smps type inductors (or, for that matter, any form of power inductor/transformer) then V=LdI/dt is the simplest way. All you need is a scope and a couple of bits. Charge up a large cap to some voltage V, then short your inductor across the cap while measuring current (a hall effect DCCT is useful here, but so is a resistor. beware peak pulse power and resistor ESL). If the energy stored in the cap 0.5CV^2 >> energy in L at current of interest 0.5LI^2, then V remains pretty much constant, and from the slope of the current waveform, L=V/(dI/dt). If you dont have a storage/digital scope, repetitively pulse the choke using a big FET. I typically charge the caps with a current limited PSU & a series R. This approach costs very little, takes a few seconds and gives a lot of information - saturation current, and inductance-vs-current - iron powder & step-gap chokes have very non-linear inductance, and the inductance at very low current can be order(s) of magnitude higher than that at rated current, leading to rubbish results from most inductance meters. Its easy to test a choke at 5,000A this way too.

Terry

Reply to
Terry Given

Also for SMPS or RFI filters one major thing to look out for is the saturation current (core saturation) versus the core temperature. At elevated temperatures, 100 degrees or so, the saturation current may be significantly reduced.....

Cheers

Klaus

Reply to
Klaus Vestergaard Kragelund

Is this a classic Grid Dipper?

Cheers

Reply to
Martin Riddle

can anyone be more clear on how to get the current drop (I) ? and for the voltage, is it just simply measure the highest peak value ? ( I only have a pulse generator)

thank you..

eric

Reply to
Eric

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