A mosfet current source has a lot of capacitance. We often put an inductor, or a compound inductor, in the drain of a mosfet current source, to keep it constant-current down into the picoseconds.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
I'd tend to put RF filters right at the load instead of at the other end of a leaky coax cable.
A 100uH inductor is not going to be much of a current source for picoseconds.
How do you measure the impedance of the current source at picosecond time frames? It can't be much higher than 377 ohms (free space).
Don't you find mosfets noisy? How about a nice bipolar or jfet?
t.)
recommends winding half-way around the toroid, then reversing the direction of rotation of the toroid and winding all the way around the toroid - init ially over the top of the first layer - before reversing again and putting the last layer of winding on top of the last half of the second lot.
There's no vertical turn and equal numbers of bigger - outside and smaller
- inside - turns going each way.
Ayrton-Perry seems to be winding all the way around the toroid and then rev ersing the direction of rotation of the toroid and winding all the way back - still no vertical turn but the turns going out are smaller than the turn s coming back.
You need a lot of magnetic field to saturate a ferrite - a good proportion of a Tesla - and one rarely see that on the average printed circuit board. It would be hard to saturate the centre pillar of a gapped pot-core or RM c ore.
The permeability of a manganese-nickel ferrite is about a couple of thousan d times higher than free space, so the outer ring of ferrite does tend to p ick most of the field around. The gaps to put the wires through don't signi fy in this environment. Even heavily gapped cores tend to limit the gap to something that will mean that the flux through the gap will be several perc ent of that through the outer ring, so they aren't nearly as good as a prop erly astatic wound toroid would be, if you could buy one.
-- Bill Sloman, Sydney
That drawing leaves a lot to be desired; best guess is that the winding is on the center part with the outer section being the ferrite shield which also helps adding inductance. Would try one out, using the infamous "sniffer" loop to see what the fields are (top and bottom) at the circular "cracks"; might need to add mu metal end plates..
The problem with the normal 2-layer ones is the hude capacitance. Putting the layers in parallel gets rid of that, but you do have to wind it all the way round to get the leads to come out in the right place.
Cheers
Phil Hobbs
Thanks, John. Where do you buy them? I only found stock in the lands of the Vikings.
-- Regards, Joerg http://www.analogconsultants.com/
Almost nothing. If you hold them side by side then a little. Pot cores a majorly better shielded magnetically than those "shielded" power inductors. If you look closely you can see the ratehr large gap that is filled with tinted epoxy.
Well, Murphy sez ... :-)
-- Regards, Joerg http://www.analogconsultants.com/
It's nice that Digikey has finally started carrying a reasonable selection of cores and bobbins. Just making a little RM6 flyback 500V step-up the other day-- works a treat.
The thing is, this geometry puts the N pole at the center, S pole at the periphery; so, there's no dipole moment. There's also no quadrupole moment. The only long-distance field has to fall off as R**-5 or faster.
My search coil wasn't well-coupled or very close. Your similar inductor held adjacent at the gap is both. Given a little free space around the unit, it'll self-shield effectively. I'd be more worried about a minor flaw (crack, inclusion) in a toroid core, causing a net dipole moment.
the layers in parallel gets rid of that, but you do have to wind it all th e way round to get the leads to come out in the right place.
Bank-winding a toroid to get low inter-winding capacitance and making the w inding astatic might be tricky.
The parallel capacitance of a single-layer winding is typically of the orde r of 1pF. Messing with the order of the turns to make the winding astatic c an probably be proven to make things worse - and in fact the the capacitanc e between the last turn on the toroid and the adjacent first turn would pro bably blow that 1pF out of the water for a single layer full winding on a t oroid.
In fact single layer windings are characteristic of RF transformers - for s tuff below 100kHz the extra capacitance of multilayer windings is something one can usually live with (though the Baxandall class-D oscillator exists because it isn't always easy).
-- Bill Sloman, Sydney
On the other hand, the boundary condition -- that the current at the terminals must be going the same direction, and therefore the magnetic field continues through the loop -- extends the first resonant mode from
1/4 wave of the simple solenoid, to 1 wave. So it's much higher. This won't be very useful for inductors (where the high impedance of the parallel resonant mode is desirable, as opposed to the low impedance of this series resonant mode), but it is pertinent to current transformers, which due to the permeable core, exhibit a resonance at fairly low frequencies -- ranging from 10s of MHz for single layers in the 100 turn range, down to 100s of kHz for 300+ turns on poor cores.To get maximal impedance from a toroidal inductor near resonance, the currents at either end must be very unequal (ideally, one zero, the other maximal), which implies not a full winding, but one ending 1/4, 1/2 or 3/4 of the way short. But that would ruin the boundary condition, so you actually *need* that extra 1/4 wave stub to complete the toroid. You would then have, actually, a complete toroid, with zero impedance across its terminals -- a shorted loop -- and a tap at any of the 1/4 wave positions for the desired impedance transformation.
The rub of it is, you very specifically aren't using a TEM toroid mode anymore, you're using the TE/M1 mode, with a quadrupole field instead.
Indeed, to get a compact field, at resonance... what would you have to do? You necessarily must have nonuniform current, which means magnetic field, and electric field, will be as well. You have your choice of arranging those connections for any kind of symmetry you like, with the result that you can achieve null stray field at infinity... but I think it necessarily must be a nonzero order polar field at its heart (with the solenoid being a dipole case, the toroid being a quadrupole, the one-half-twisted toroid being an octupole, and so on).
The reason has to be part of this: in an inductor, we assume speed of light is infinite, and electric and magnetic fields are separated (lumped entirely inside one inductor and one capacitor, and each well shielded); when the circuit element itself becomes resonant, that assumption is broken, and we must have room for both fields. Therefore, an inductor pushed to its maximum useful frequency (parallel resonance) necessarily must exhibit external fields -- no B=0 is possible outside the long solenoid or uniform toroid at AC when c < infty.
Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com
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