LC filter design with predefined capacitors

At ~100kHz can you use an active filter?

George H.

Yes, I can, and have already done it with some success (using the glorious LTC1562). The question is out of sheer curiosity, an offspring of my recent experiments with transductors. I can easily achieve 4:1 L tuning range (at I_max=10mA, which I consider the largest practical current in this application), so I started to wonder if an IDAC-tunable LC bandpass filter is feasible. I'm experimenting in Spice, but the results are disappointing: BW depends on Q (surprise, surprise...), which is proportional to sqrt(L/C), which in turn cannot be kept constant for a fixed value of C. So a purely L-dependent compensation scheme is needed.

Best regards, Piotr

No. Not without series/parallel combinations. Or, what you can get, will not be a Bessel type (despite Bessel being the lowest Q of common types).

You can fix one component kind in an RC active filter, because impedance doesn't matter. This is a consequence of the zero Zout, inf Zin asssumption of opamp circuits -- one which you need to maintain, hence needing GBW so-and-so above Fc. In an LC filter, Zin must match Zout, so a similar condition is not possible.

Now... if you're really serious about this "full custom" business... you can use _transformers_, and any ratios and capacitors you like. ;-)

Mind that a transformer is three degrees of freedom (Lp, Ls, k), so this makes it considerably more complicated. Example: use LL instead of (or with) series L to tune series stages, or use it for zeroes so your bandpass is sharper on the skirts. (Note that, even if you make Lp, Ls very large with high-mu core material, LL is still a very relevant value.)

Coupled inductors are also great for building coupled resonator BPFs, but that's a synthesis method really only useful at higher Q (BW < 30% say), so not as applicable here.

Tim

Sadly, this confirms my simulation results.

Interesting, didn't think of that. Just wanted to build a current-tunable LC bandpass for fun and use transductors in the resonant sections.

Can you make them electrically tunable?

Best regards, Piotr

Transductors, as saturated inductors, are not good for resonont circuits or filters.

I would think so, at least more reasonably so than the other case. The input and output couplings (or impedances) need to be variable, though, and probably the intermediate couplings too.

And then there's the matter of tracking, but I'm assuming you're okay with that.

Can you make a tunable variac? ;-)

For simple variable Q or coupling, check RDH4. Doing that with passive components, I don't know.

In general, for an arbitrarily complex network of moderately variable inductors (including couplings), and a few capacitors, I think you can get a reasonable range. Say, an octave or two in Fc and BW. But you need not like the solution. :^)

Tim

Why? The one I've just made has tuning range of 2.2mH--450uH (@0--10mA/100kHz) and AC coupling of 18mVpp (tunable winding) when there's 3Vpp sine/100kHz at the magnetizing winding. I'd say it's at least decent.

Best regards, Piotr

I know nothing of transductors, I know of no one using them for low noise front end 'audio' designs. Fun, but probably noisy is my guess.

George H.

Oh make one into a flux gate magnetometer, maybe.

GH

I wonder how Barkhausen noise factors into it. Parametric with signal? With change in control level, certainly?

Tim

I think that's because they are pretty hard to make, large and burn a lot of current, not because of their inferior performance. By their very nature they should be a lot less sensitive to the parasitic modulation caused by large signal levels. You can also make one to match your specific needs, while the high C varactors like BB112 a) quickly become unobtainium; b) don't have many values to choose from, which constrains your L and c) limit the front-end's dynamic range.

I've heard exactly the opposite, but it's all anecdata. An honest measurement would be very welcome.

Best regards, Piotr

The change in effective inductance of a transductor relies on non-linear behavior of the core. It also distorts the payload signal, and that's not a good idea in a filter or resonator.

There is a device with linear adjustable inductance: the variometer. It has two coupled windings, and one winding can be turned with respect to the other, thus varying the coupling and the effective inductance of the series connection of the windings. For images, google for 'variometer inductor'.

Will there be any significant contribution compared to a regular ferrite-based inductor? Barkhausen noise is caused by domain boundary jumps, but there should be none for a stable magnetizing field.

But only during the change/sweep and cease when H is stable? SWAG, but why shouldn't it be that way?

Best regards, Piotr

Exactly the same applies to varactors. I'd even say they are more prone to this kind of distortion because of their voltage control mode. Transductors need significant current to change their properties considerably and it is always easier to get spurious voltage than spurious current due to the absence of low Z paths.

There even self-winding coils, (the Coil and Capstan section):

but that's an overkill.

Best regards, Piotr

They use magnetically tunable circuits at the other end, though:

So I'll give it a chance and breadboard the complete circuit.

Best regards, Piotr

Sure build one and let us know. It's pretty easy to measure noise to ~10%- 20% with a bunch of gain and some filters. (At least in the audio range.)

George H.

'waves hands over head', "I know nothing.". But given it's magnetic there will be temperature effects too.

George H.

Clearly observable.

Best regards, Piotr

Has to do with changing magnetic domains. So, a very small change might not produce any noise (a 'pop' as a domain leaves a pinning site), but large scale changes (which should include signal, especially if we're talking 10s or 100s mT signal flux here.

I don't know what magnitude it is, whether it's something that can only be avoided down in the (nT?) range where Johnson noise dominates anyway, or if it's small enough not to mind.

Oh, huh, it should be the noise resistance corresponding to hysteresis loss, shouldn't it?

Tim