# how do u test an inductor?

Discussion in 'Beginner Electronics' started by Louie, Aug 14, 2003.

1. ### LouieGuest

can someone tell me how to test/measure an inductor on a power supply
circuit board. can it be done in circuit with a VOM?

TIA
louie

2. ### Stepan NovotillGuest

It should measure pretty close to zero ohms.
That won't tell you if it has shorted turns though.
Stepan

3. ### LouieGuest

well i do have an analogue scope and i'd rather not fork out for an ESR
mater. i'll try what you describe. it should be plenty accurate for my
purposes.

thanks Terry, that's what i needed.
Louie

4. ### tim kettringGuest

Whats the good reason , closing tolerance by a factor of 10 ? In
effect a precision 1 ohm resistor ?

tim

5. ### Terry GivenGuest

Good question: see below...
No. A "grunty" 1Ohm resistor, with very low inductance is the reason.
Lots of people forget to calculate the Peak Pulse Power seen by a resistor,
and this can be crucial. Consider for example a 5k6 10W wirewound resistor
connected directly across 230Vac 50Hz mains: We all know that the power
dissipation = (230V)^2/5600 Ohms = 9.45W. But if we tried it, using this
part, it wouldnt live very long. The peak pulse power is of course
Vpeak^2/R, and given that Vpeak = sqrt(2)*Vrms in this case the resistor
power dissipation is actually an 18.9W peak sine wave, at 100Hz.

A classic (and fatal) examples of this mistake are FET gate drive
resistors - 10R, 12V gate waveform, the peak pulse power = 14.4W. I have
seen plenty of people using 0805 or 1206 smt resistors for this job, and
they work fine, for a while....

the other good one crops up in EMI filters - people often stick small
resistors in series with the filter caps, to damp out resonances, and
effectively convert HF energy into heat. BUT they forget that the resistor
will see Vpeak^2/R peak power. I did this myself in an AC motor controller,
using 2W Philips PR02 1R resistors, running from 400V three phase supply.
During development/testing we noticed a bright flash, and upon close
inspection discovered the resistors were totally snotted. A little thought
showed why: sqrt(2)*400Vac = 566V peak, so the poor old resistors saw a
worst-case 320kW peak pulse power. The datasheet (philips are good for this)
has a pulse overload curve, and the part could handle about 100W for a short
while (100us), so 320kW rooted it completely. The solution there was to use
Carbon Composition resistors, with an appropriately wicked P.P.Power rating.

Ultimately it all boils down to thermal mass. Most resistors are a
non-conducting drum, with a thin coating of resistive material, often cut
into a spiral. For a very short length of time, the resistive material will
absorb energy "adiabatically" i.e. no heat will flow into the surroundings.
If you know the energy of the pulse, and the mass m & specific heat capacity
of the material c (Joules/(kg*Kelvin)), it is easy to calculate change in
temperature dT = E*m/c. This is true for ANY thermal problem. With a FET, if
dT + Tambient > 200 degrees C (473 Kelvin) then the silicon goes "intrinsic"
i.e. turns into a conductor, and the FET self-destructs.

A carbon composition resistor is made from a solid lump of suitably
resistive material, so there is LOTS of mass compared to a normal resistor,
so it can absorb a bigger belt. Companies like IRC and Philips make
resistors designed specifically for this job, eg IRC's CHP series, Philips
MMA/MMB series etc.

Be careful with totally enclosed wire-wound resistors. Good brands like
Vitrohm have the wound element totally enclosed in thermally conductive
ceramic goop, so there is no air and hotspots dont form - you can beat the
hell out of this type of resistor, and it will cope. Cheap, crappy resistors
usually have a groove cut in the ceramic body, into which the resistive
element is dropped. They then pour ceramic goo over the top, usually
resulting in a lot of air around the "back" of the element. These will often
emit a nice orange flash when you thump them with a big pulse - thats the
wire heating up red hot, as air is a lousy conductor of heat. These will
crap out, sooner or later.

so back to the original question:
If Vsupply = 12V and R=1 Ohm, then Ppeak = 144W, which will fry an ordinary
resistor. I fwe make it from 10 * 10R in parallel, then each one only sees
14.4W peak, so will probably not die. And dont forget, when the Inductor
under test saturates (it will unless its air cored) the inductance will
plummet to the air-cored value, so the current will rise very rapidly,
ultimately limited by the current sense resistor.

I actually did this a few years back, trying to measure an inductor in a
physics lab. I got odd results, which got weirder every time i re-tried the
experiment. eventually i realised what was going on, and used a lot of 22R
resistors to do the job......only to discover that the HP56000 series scope
i was using had way to slow a single-shot-sample rate, and all i got was 3
dots for my whole waveform.....*sigh*.......out with the 30 year old
analogue scope....

6. ### Terry GivenGuest

Some very good points. The unusual devices require unusual measuring
yep.

Two basic techniques: The splat test in reverse - apply a current source
through the inductor in parallel with a cap, then disconnect it - use a good
cap, and the damping is due entirely to the inductor losses. You really need
to use a FET to disconnect the current source, to get a nice waveform. Very
hard to do at 1,000A. Then fit an exponential decay envelope to the ringing
waveform and voila. Horribly hard to get better than about 20% accurate
measurements.

The best technique is of course calorimetry. Dump the inductor into a
container of suitable liquid, of known specific heat capacity c, mass m and
temperature. Make sure the container is in an isothermal environment (i.e. a
chilly bin made of polystyrene). Run it for a suitably long time - 4-5
thermal time constants are required; this can be hours for a decent sized
choke. Then stir the liquid and measure its temperature. from dT, m & c,
know energy E, and if time t is also recorded, can calculate average power
loss. this can be done quite accurately, but in real terms 10% is good and
5% is outstanding for thermal measurements. Careful you dont set the
container on fire though - estimate the loss and thermal time constant first
(measuring the therm. tc is easy) then use that to figure out how much
liquid is required. De-ionised water is good, but then so is cooking oil.
(note: dont use a flammable liquid. You can easily calibrate your setup -
bung a power resistor in there, and stick a known amount of power thru it
for a known period of time, measure the temp rise, and voila. Shit, if you
want, just keep fucking with the power supplied to the resistor until the 2
temperature rises are the same......)

Its not too difficult to get a very good idea of what the losses are going
to be in an inductor - just make sure you take skin & proximity effect into
account for the copper loss, and core temperature for core loss. But what
you get is not always what you asked for - we had a problem with some 150uH
1,500A chokes for a 10kHz UPS that had astonishingly high core loss -
between 3 and 20x what we calculated. It took us a while to figure out why -
and it was the core manufacturers fault. The core was a tape-wound "C"
core - an oval, but with 2 cuts = 4 x 1 inch air gaps, to minimise fringing
flux. The material was basically metglass, 0.001" thick. When Honeywell cut
the cores, in a fit of inspired stupidity they neglected to lap & etch the
cut faces - result was metal smearing across the faces, therefore effective
tape thickness ranged from 0.001" to 1" hence eddy current losses were all
over the show. We fixed it with hydroflouric acid, and they were VERY
apologetic when we pointed out what they had done.

I also once re-designed a transformer for a 1500W ups PFC front end - boost
converter + dc-dc converter down to 24Vdc (battery level). The idiot who
designed it (ME, MIT - ha!) used 12 layers of 0.6mm Cu foil at 100kHz. skin
depth at 100kHz is 66mm/sqrt(1e5) = 0.208mm, so the foil was about 3 skin
depths thick. At 12 layers, split-wound (1/2P, S, 1/2P) that is 6 effective
layers, the ac-dc resistance ratio Fr = 150. So his AC resistance was about
150 times the DC resistance, which kind of explained all the fires (and i do
mean fires). I reduced the copper thickness to 0.1mm, so the dc resistance
went up by a factor of 6, BUT Fr dropped from 150 to 1.2, so the total loss
went down by a factor of 150/(6*1.2) = 21. And no-one at worked believed me
when i said that would solve the problem, but it did.

see "soft ferrites" by e.c. snelling OR switchmode power supply handbook, K.
Billings, OR the philips app notes by Jongsma for the relevant
nomographs/formulae. I once messed around with calculating it analytically,
but its easier to read it from a table......I'm lazy.

7. ### JeffMGuest

apply a current source through the inductor in parallel with a cap,
I was waiting for someone to say that.
What no one has yet said is that a common failure mode for an inductor
is a shorted turn (think "step-down transformer, then think "autotransformer".
When this happens, the Q (quality factor) goes to hell and the thing won't ring.

My Russian physicist boss calls these "bulk resistors"
and while they can absorb a lot of abuse, they're getting harder to find.
I did find a distributor who bought up a bunch of Allen-Bradley parts
after they quit making them, but the guy makes me pay dearly for them.