I have ask before about changes of inductance vs. AC excitation in ferrites before and didn't get a satisfactory answer. We can use a particular potcore at 5ua or 10amps. but how much does the inductance change over this range? What current does the manufacturer use to make his curves. I'm thinking about a transformer at the input of a radio receiver with a signal measured in microvolts. Say we built a 50 ohm to 200 ohm transformer using the manufacturers specs, and made the primary inductance 4 times 50 ohms at our design freq.. It would work fine at 100 ma to many amps. But at 5ua is there enough inductance in the primary to make it function as a proper impedance transformer? Mike
Depends on the material. It'll change with temperature too.
They sweep the current.
Losses may be too high. Try winding one for the job.
And get a copy of Epcos's Ferrite Magnetic Designer tool.
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Most manufacturers make 'copies' of the popular ferrites. There are often direct cross-references beteen say Epcos, Ferroxcube/Philips/Yageo and TDK for example.
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The cheap Chinese manufacturers tend not to have the latest low-loss, high temp parts though but that's a PSU issue.
Epcos's data also advises different ferrites for different applications.
Graham
-- due to the hugely increased level of spam please make the obvious adjustment to my email address
Get hold of a data book on ferrite cores. Most of them include flux density versus field strength curves for the various ferrites used in their cores.
EPCOS at least makes the data availlable on the web - here's the link for their N27 material
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-- Bill Sloman, Nijmegen
Ok Bill, I'm over my head on this because I don't understand all the terms. If we look at page 3 of the link you gave, first graph, you will see the B vs u(o) curve. At 25mT it is already down 34% from the peak and I would bet a dollar that the curve drops faster the lower you go with the B field. My guess (only a guess) is that if I wound my theoretical transformer (described above) and had my 0.1 microamp driving it, it would be no where near 25mT. Somehow this B vs u(o) curve will relate to permeability which is then related to (A sub L). I would then use (A sub L) to calculate the turns on the theoretical 50 ohm primary (200 ohms inductance). It just seems that at very low currents the permeability will drop so low I will not have my
200 ohms inductance and the transformer will not work as designed. Since I have never seen this discussed and people build working radios everyday, I'm probably all wrong, but I'm not sure where. Thanks, Mike
Look at page 4, which shows the field versus the magnetising current - it looks more or less like a straight line at zero field, which suggests that permeability doesn't drop all that fast below 25mT. Presumably what is happening is that is that the "sticky" component of the alignment of the magnetic dipoles responsible for the hysterisis is falling out of the permeability at low fields - as you can see from the curves on page 4 it goes away at high fields.
mehow
A sub
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Every specific core pair has magnetic path length, specified in the data sheet, and the exciting field (in Amperes per metre) is just the current looped around that core divided by the magneitc path length in metres. In your case this is just your 10uA times the number of turns divided by the magnetic path length of your core.
I
Wrong.
Your error lies in assuming that the permeability drops to zero at zero magnetising current. It doesn't, and in fact in this case probably continues to fall off roughly linearly to about 60% of the peak permeability.
If you really want a more stable inductance, gap your core until the inductance is around 10% of the ungapped figure - which usually takes a layer or two of 60 micron transformer tape between the core halves - and wind a new core with three times the number of turns (the square root of ten times more turns, to be precise but it is difficult to gap the core this precisely, which is why the manufacturers do it for you by grinding down the centre leg of a gapped core set).
Since 90% of the magnetic path is now in the gap, current dependence goes down by a factor of ten. Bigger gaps provide even more stable inductances, if you can afford the extra series resistance and inter- winding capacitance.
Look at page 4, which shows the field versus the magnetising current - it looks more or less like a straight line at zero field, which suggests that permeability doesn't drop all that fast below 25mT. Presumably what is happening is that is that the "sticky" component of the alignment of the magnetic dipoles responsible for the hysterisis is falling out of the permeability at low fields - as you can see from the curves on page 4 it goes away at high fields.
Every specific core pair has magnetic path length, specified in the data sheet, and the exciting field (in Amperes per metre) is just the current looped around that core divided by the magneitc path length in metres. In your case this is just your 10uA times the number of turns divided by the magnetic path length of your core.
Wrong.
Your error lies in assuming that the permeability drops to zero at zero magnetising current. It doesn't, and in fact in this case probably continues to fall off roughly linearly to about 60% of the peak permeability.
If you really want a more stable inductance, gap your core until the inductance is around 10% of the ungapped figure - which usually takes a layer or two of 60 micron transformer tape between the core halves - and wind a new core with three times the number of turns (the square root of ten times more turns, to be precise but it is difficult to gap the core this precisely, which is why the manufacturers do it for you by grinding down the centre leg of a gapped core set).
Since 90% of the magnetic path is now in the gap, current dependence goes down by a factor of ten. Bigger gaps provide even more stable inductances, if you can afford the extra series resistance and inter- winding capacitance.
-- Bill Sloman, Nijmegen
Thanks, Bill I did some quick calculations of B/H on graph number 4 to find u(o)u(r). It varies from 1.7, peaks at 4 and then drops to 0.4, this is a 10 to 1 range but the variance is not at the low end where I would have expected. Again I don't know quite how B/H relates to A sub L but I think it figures into it. If I get some time I'll get numbers together for a transformer I made for a Flag antenna. I'm curious about the B in my transformer compared to the values shown on the graph. Mike
Thanks Eeyore, finally someone that KNOWS this stuff. I have a 50 ohm to 200 ohm transformer, 4 turns to 8 turns on a FT37 toroid of material 75. I have 2 microvolts driving the 50 ohm primary at 1Mhz. The effective volume of the core .163 cm^3 What is the B field? You will KNOW what other data you need if any (I'd have to guess) so let me KNOW and I'll look it up for you. It sure will be interesting to see how far off my guess was.
1Mhz, Xj=200 Rdc=0.025
I'll look at it, when you calculate the B field quantity, we'll see if the Epcos program goes to that low level. Thanks again for your help, Mike
Hey Graham, I reposted a graph from Fred Bartoli on alt.binaries.schematics.electronic, it shows inductance dropping with decreasing drive voltage. It doesn't go low enough though. Mike
-- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at
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Causing what ? This is my third attempt at finding out if my postulation has any merit at all. I've been told forget it, just build the transformer. It is just a curiosity for me, I have used the same core at 300 watts power levels and at microwatt levels. Would it have been a better transformer at lower frequencies if I had added more turns. I don't know and I have know way to measure microvolts Note his graph only goes down to 2Vrms, I'd like to see a graph down to 2 microvolts. Hey Jim, I'm a little surprised at your response, is Phil rubbing off on you? Have a drink of fine wine on me :-) Mike
I was just pointing out that the effect is not unexpected. Inductance is proportional to B/H. ...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
Help save the environment!
Please dispose of socialism properly!
Inductance is defined by V =3D L. dI/dt, which is to say that it is proportional to the rate of change of the magnetic flux threading the coil (B) divided by the rate of change of the current exciting the coil (H), which isn't the same thing.
It can be dependent on the absolute value of the current through the coil if there is a ferro-magnetic core in the magnetic circuit whose permeability changes with B, but proportionality would be unexpected and very unusual.
Nowadays, that's the sign of a creative engineer. Someone went off the boundaries of the data sheet to do the design. Before VVC diodes (voltage-variable-capacitance) there were Increductors (I think that's the right spelling) intended for this kind of thing, with specifications and everything.
So did Marconi Instruments in the 1950s - VHF sweep generator using a "magnetic reactor"
--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)
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