Inductor in saturation ??

Hi,

How do I tell if my inductor is saturated? I am generating an EM field
using an RLC circuit. When I measure my field on a receiver circuit it measures
that the generated field is steadily dropping (somewhat exponentially) until it
stabilizes out several minutes later... it can take > 6 minutes.

Can anyone provide insight into this.

In particular:
1. Can you describe this using the BH curve?
For example, if it is going into saturation, then it should start to draw more
current. However, from my measurements it appears as though the current is
slowly decreasing over time. I.e. I put a 1 ohm in series to measure the
voltage (i.e. get current) and this voltage just kept dropping until it reached
a somewhat steady point.
2. How would this be traced out on the curve, is the curve settling so that
Bmax = |Bmin|, and it is just moving around forever >5min... or is it a issue
with core heating, or is it some other thing...
3. Does it fit that I see some memory effect. I.e. if I turn the EM field back
on after 5 minutes of being turned off it hasn't "recovered" to start at the
value that it initially started to "drop" from when it has been sitting for a
while.
4. Is this just the ferrite material (if this is a saturation problem). Because
I know I have to drop my current to fix it then - however, would it be possible
to keep the same current, and thus the same "stronger" EM field if I just had a
different ferrite that was less permeable...

Thanks a ton!

 

The winding of the inductor is probably heating up. This has two
effects; the resistance goes up so reducing the current. Also, if the
ferrite gets hot, its permeability probably reduces, thus reducing the
induction produced by a given current.

Well, you could but I don't see any point.

Core heating may well be involved.

Yes, *thermal* memory.

Maybe, but you also need to stop the current varying. You need a
current-source drive.

 

First, permeability of a ferrite core changes due to temperature, and
may increase or decrease over temperture.
I would not attribute what you see to temperature as the first
"explaination".
If the average current in the inductor for the RLC circuit is not
zero, then the average DC current can bias the core "upwards" from zero,
and then it can possible become more saturated over time.
That magnetic field can reside when powered down (remenance) and will
be less than when previously running.
If one does an X-Y plot of inductor current vs inductor voltage, one
would see the approach of an exponentially increasing current as the
core approaches "saturation", which is not sharp.
Now one can add a gap to reduce the effective permeability, but then
one needs more turns for the same inductance, but there still will be a
(higher) drive level that will produce the same results.
Only air as a "core" material would give almost total relief of a
non-saturable nature.
The first solution would be to eliminate the DC component in the
inductor's current; capacitively couple it!
Another way is to add a second winding and pass an opposing DC current
- but prevent AC fromflowing thru this circuit (seperate inductor or
parallel LC filter or MOSFET current source or combination).
Or use the gap method.

 

It sounds as if your inductor is heating up - the time constant of the
order of minutes seems most unlikely to have anything to do with
saturation.

It also sounds as if the inductive impedance of your inductor (which
would decrease quite fast with increasing core temperature) is quite a
lot smaller than its resistance, which you'd expect to increase with
temperature at a rate of about 0.3% per degree Kelvin (which is not a
lot).

This does tend to suggest that your inductor is well and truly
saturated, and not working as an inductor at all, but without lots
more detail about what you are doing (frequency, inductance,
resistance, current level), this is strictly guesswork.

 

Hi,
There shouldn't be any DC current as it is capacitivly coupled (RLC ). But I
guess it still could be an average current thing if maybe my oscillator isn't
going fully -5 and 5??
And the reason I would like it expalined with the BH curve is I think that will
give a better idea why this could be happening. I don't feel any heat when I
touch the coil at all. So as long as the heat change is small that you guys are
claiming could be the problem.

More details are I have +/-5 volts switching on lines A, B. (i.e. A=5, B=-5,
then A=-5, B=5). f=10kHz, square wave.
Then I have this circuit: A - C - L - (R) - B.
Current draw on the power supply is around 80mA, the signal I obtained seems
to show a peak currenet draw of 140mA-ish. Coil has fine wire... I forget # of
wrapps - lots - have to double check my notes and get back to you - 4.8mH, a
resistance of 50ohms if measured with a meter.

R is in () because I just put in a 1 ohm resistor trying to obtain voltage
measurements and thus get current. However, this is a big part of my question -
HOW do I get these measurements.
Someone mentioned plot Coil Voltage VS Coil Current. I can try this but it
seems hard to get the coil voltage. RMS meter or Scope - my scope seems to
sometime not read the correct signal because you need to put the gnd clip on a
node that is switching too. However, I have obtained the voltage as one would
expect.
I mean the voltage is distorted in the shape that I have seen in books listing
how the drive current will look. So that makes sense that the voltage has the
same shape. (Sort of a sin wave with the peaks more skinny). This is suppose to
be the shape of the drive current signal according to one book I read and is
due to the BH curve - as it isn't linear.

I would most like then details of how to check if it is saturating... or how to
attribute it to a thermal problem that some others mentioned.

A doctor in passing mentioned to look at the BH curve and that he thought I was
going back and forth and that eventaully settling somewhere towards the
middle... however, the books I am reading describe this all different - some
say moving the whole curve up or down until it is symetric (how long does that
take?) Others mention jacking it up, like someone mentioned here where you
drive into saturation more... so I'm still confused what and how to check...


1. Do the results I mentioned tell you guys anything?
I took voltage measurements over the 1ohm and they droped over 6 minutes from
100mvRMS to 72 mv... and at 6 minutes was stable or at least moving so slow
that I stopped taking measurements around here...
NOTE I am confused on the measurements a bit as this gives a peak rms current
of like 62mA for my resistor value. However, with the powersupply it seems more
like it should be higher... plus the voltage signal I was able to get off the
scope a few times puts it higher at 140mA peak/98marms, but as i mentioned
having to put the gnd clip on one side of the sense resistor that has a
changing voltage can give me a wacky signal sometimes with the scope so I
wasn't sure on the scopes signal - except that it somtimes matches the
"expected" dirve current signal from the book which was reasurring.

- So I know that voltage drop (and thus current drop) is a problem, but what
does it mean? (Also like I asked before, the BH curve material claims if it was
saturating it should be drawing more current... and it does say the L won't
support the voltage as it saturates. So why is current dropping if it is
saturating. Or has it saturated hard and is somehow settling out of this or
something.... ie. the more current draw occurts very quick at the start...)

For example, I drove a different coil with a signal generator using same signal
and it didn't change the voltage over the sense resistor - however, obviously a
signal generator didn't have much drive current. So that is my reasoning to
decrease the current in my system - HOWEVER, that will make my field quite a
bit weaker. Which is just a whole different issue.

So basically I should try and get these values for an X-Y plot?

I have to check the L voltage but if it is dropping like the R (and thus
current) I mentioned, this XY plot should be growing exponentially.
So that would confirm it is going into Saturation... over a long period of
time?
However, you think it is due to asymetrical drive current, not the coil?

 

As long as the inductor is capacitively coupled and there is no DC
current via a resistor or FET, then the average bias is zero.
Does not matter what the waveform is (within reasonable limits), that
will be true.
Even if the drive is so large as to saturate the core, it will tend to
be symmetrical as the DC average is still zero.

However, what you report sounds like that there is a DC bias from some
source, so check the circuit carefully; even put an analog meter in
series with the inductor (a DVM could miss sampling "critical" parts and
might not read average, or the aliasing may give useless widely varying
values).

 

One amp goes to a C, then there is an L, then there is an R, then there
is the other Opamp. Not sure what you mean by FET. I didn't mention having a
FET in the circuit. I have those two lines being driven opposite +/-5V by high
current output opamps, then the descirbed series RLC. Where R right now is
just my own 1.6ohm sense resistor for debugging. NOTE MY CIRCUIT IS IN
RESONANCE.it is being driven at its resonant frequency incase that helps you.

Could it take this long for the BH curve to settle? I read somewhere that at
the start Bmax and -Bmin usually aren't equal as we are starting from H=0,B=0,
so basically the whole curve moves down slowly to make the BH curve symetrical
(i.e. Bmax = Bmin). Could this take minutes?
Can this have anthing to do with the coils heating up that I have heard
about but haven't heard explained. AND most importatn as I noted - I don't FEEL
ANY heat with my finger. However, I can't touch the core, but the coils seem
fine.

Tomorrow I plan to try a larger R value to to limit the drive current. Today
I confirmed that the voltage over the L does drop from 100Vpp to 74Vpp over
about 4 minutes. (Exponentially, so in the 70x-ish V range its slow droop). So
I am quite convinced that the current is too high. This supports the
description that I have heard that when it saturates the L can't support the vo
ltage any longer. However, it still doesn't support that I also heard it should
draw MORE current beucase the voltage drops at the same time over my resistor
so current is going down as this is occuring not up.
I don't see how DC could get through the capacitor. But I will see if I can
check for this somehow, I don't have an analog meter for this though.

 

L=5mH, r=1.69ohm, C=35pF, resonant frequency its being driven at 12kHz, current
drops from 60mArms to 44-ish rms in 4 minutes at this same time coil voltage
drop 100Vpp to 74Vpp. Circuit as described in other posts.

 

Does the inductor get warm?

SioL

 

I think the L is so small as to be unnoticeable. This isnt being run
as an inductor in the usual sense. Perhaps the OP wants to tell us
what theyre doing and why.


Regards, NT

 

Don't forget that will reduce the Q of your series-resonant circuit.

What convinces you of that?

The current is going down. OK, how hot are the op-amps getting? Maybe
they are running into thermal shutdown. Maybe your circuit is drifting
off resonance: that would affect the current considerably.

The effects you observe don't seem to be explainable at all by
'saturation' This is an *instantaneous* thing (well, it takes a quarter-
cycle of the frequency, so is very quick). The settling time for the B-H
loop to become bipolar is just a few cycles at most, not minutes.

Indeed. You wrote in another article:

QUOTE
L=5mH, r=1.69ohm, C=35pF, resonant frequency its being driven at 12kHz,
current drops from 60mArms to 44-ish rms in 4 minutes at this same time
coil voltage drop 100Vpp to 74Vpp. Circuit as described in other posts.
ENDQUOTE

This isn't resonant at 12 kHz. Your capacitor needs to be 35.2
NANOfarads to do that. With 35 pF, almost anything could be happening,
because the self-capacitance of your inductor may be more than that.

Why not get one? They are not costly. But, whether you have a 35 pF or a
35 nF capacitor, the type of capacitor indicates that it isn't leaking
DC unless you have zapped it with excessive voltage. Don't forget that
in a series resonant circuit, the voltages across the L and C can be
VERY much higher than the input signal voltage.

 

John brings up a good point. You may be concentrating on the wrong
component. Generating RF using an L-C combination as the antenna is
tricky. It only works well at or near resonance (a few tens of Hz at
12 kHz, if that's your frequency). The capacitor may have a large AC
voltage across it.

In a recent project, I designed a transmitter with 1400V p-p across
the capacitor (and coil). Wima makes a special cap for this. Depending
upon the frequency and voltage, you may have to derate the cap.

L-C antennas have radiation resistance. If they didn't, they wouldn't
work as antennas. The radiation resistance and losses can be modeled
as a series R. The value will depend upon core material and geometry,
diameter, frequency, winding resistance and number of turns. Ball
park: 1 to 16 ohms + winding resistance. Are you driving the antenna
at resonance? It seems far more likely that your power decrease (over
time) is due to changing resonance point. Re-adjust the frequency
after 6 minutes to confirm this.

If it isn't resonance drift or capacitor heating, check the drive
circuit for losses and thermal effects. I would be surprised if you
find any significant effects due to inductor saturation.

Frank Raffaeli
http://www.aomwireless.com/

 

yes the frequency is solid - obviously just a typo.

 

No

 

What do you mean? The frequency comes from your input signal, not from
the LCR circuit. How did you make up 35.2 nF?

 

L-C antennas have radiation resistance. If they didn't, they wouldn't

yes, the circuit has been tuned for resonance.

This seems quite unlikely, I have been incontact withsome who has experience
with this design too and they have a new coil for me to try that apparently
works at upto 170mA, so it is still sounding like it is coil saturation.

The resonance shouldn't "drift" as we aren't relying on an oscillator but using
a driving cct. at resonance. I have yet to see one of our antennas "detune"
that amount of voltage by itself... nothing feels hot, you wouldn't expect to
see the same thing when you turn it on for 30ms and off for 2 minutes, then
30ms etc., that look like core/saturation memory...

But thanks for the ideas, I'll let you guys know what happens.

What I'm surprized about though is no one has a "test" for saturation. I.e. do
this and you should know... nor explained my measured results which are suppose
to be what I've read to check. Anyways, I should know by tomorrow since we are
testing with limited current today.

 

Yes I already know this. I meant yes that was quick typing typo. If I had put
in a 35pF the circuit would NOT even be working.
Like I said - that was a typo 35pf was suppose to be 35nF. PLUS one wouldn't
even go off the numbers you are calculating 35.2nF because if you have done
this you will see you still have to adjust your C to tune the RLC for its
resonant frequency.
I drive the RLC circuit at it resonant frequency. The RLC cct been tuned for
the given freq. and the drive circuit is at this frequency.

 

how much signal is lost when the L or C changes enough to move the
resonance point 70 Hz?

It's not coil saturation, but it (or the cap) may be warming up just
enough to de-tune itself. The other posters were correct when they
told you that saturation effects occur immediately, not over 6 minutes
time. All cores are a non-linear to some degree. Here is your test to
see how linear your core is and whether you are tuned at resonance:

Apply a signal from a generator at lower power (100mV or so). Put
about 50 ohms in series. When the generator voltage and the antenna
current are exactly in phase, the circuit is at resonance. Increase to
full power and check the phase relationship again. A saturating
inductor will tune "high". If it's just a few percent or less, it will
make a fair antenna.

How does the driving circuit stay in tune with the drifting L-C
resonance point? ;-)

Thanks,

Frank Raffaeli
http://www.aomwireless.com/

 

You are correct, DC will not go thru a (good) capacitor.
If the signal drive remains constant from start (off a long time) to
"steady state" in what you say is minutes (or so), then the observed
current level should be relatively constant.
Now if it really warms up or gets hot, the average permeability could
change and the amount of change INCREASE or DECREASE or BOTH depends on
the ferrite and its permeability VS temperature characteristic.
Generally, saturation will not cause heating, it is the drive
frequency creating a rapid enough of flux change to then make for core
losses.
See if you can measure the low-level self-resonance of the inductor;
your drive frequency should be well below that. and the ferrite should
have a low loss at the drive frequency.

 

Well, if it is erlatively high Q, then a few Hz can make all the
difference.
And a drive at a frequency causing core losses *or* a saturating drive
will change the effective losses --> different resonant frequency due to
different R.