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Sensing small inductances

Phil Hobbs wrote:

Here is a 330Mhz Colpitts with split supplies. I had to add some base inductance and capacitive feedback to get it to oscillate at a low enough frequency for the 2N3904. That should not affect the question.

Where do you measure the negative resistance and how do you control it? Version 4 SHEET 1 880 708 WIRE -96 -112 -160 -112 WIRE -32 -112 -96 -112 WIRE -32 -96 -32 -112 WIRE -160 -80 -160 -112 WIRE -464 -32 -528 -32 WIRE -336 -32 -384 -32 WIRE -240 -32 -336 -32 WIRE -224 -32 -240 -32 WIRE -528 -16 -528 -32 WIRE -336 -16 -336 -32 WIRE -32 0 -32 -16 WIRE -336 64 -336 48 WIRE -160 64 -160 16 WIRE -160 64 -336 64 WIRE -128 64 -160 64 WIRE -112 64 -128 64 WIRE -336 80 -336 64 WIRE -160 80 -160 64 WIRE -336 160 -336 144 WIRE -160 176 -160 160 WIRE -160 272 -160 256 FLAG -32 0 0 FLAG -336 160 0 FLAG -128 64 Vout FLAG -240 -32 Q1B FLAG -528 -16 0 FLAG -96 -112 VCC FLAG -160 272 0 SYMBOL voltage -32 -112 R0 WINDOW 123 0 0 Left 2 WINDOW 0 33 33 Left 2 WINDOW 3 33 80 Left 2 SYMATTR InstName V2 SYMATTR Value 5V SYMATTR SpiceLine Rser=0 SYMBOL npn -224 -80 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL cap -352 -16 R0 SYMATTR InstName C1 SYMATTR Value 10pf SYMBOL cap -352 80 R0 SYMATTR InstName C2 SYMATTR Value 10pf SYMBOL ind -368 -16 M270 WINDOW 0 32 56 VTop 2 WINDOW 3 5 56 VBottom 2 WINDOW 39 -16 53 VBottom 2 SYMATTR InstName L1 SYMATTR Value 20nh SYMATTR SpiceLine Rser=1m SYMBOL voltage -160 272 R180 WINDOW 123 0 0 Left 2 WINDOW 0 33 33 Left 2 WINDOW 3 33 80 Left 2 SYMATTR InstName V1 SYMATTR Value 5V SYMATTR SpiceLine Rser=0 SYMBOL res -176 64 R0 SYMATTR InstName R1 SYMATTR Value 200 TEXT -400 -240 Left 2 !.tran 0 200n 0 50p TEXT -408 -264 Left 2 ;'Negative Resistance

Reply to
Steve Wilson
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tirsdag den 27. august 2019 kl. 21.10.58 UTC+2 skrev John Larkin:

ADC/DAC and an FPGA and you have all you need for a vector analyser

Reply to
Lasse Langwadt Christensen

What Spice doesn't include in the transistor models is the wire bond inductances.

Reply to
John Larkin

But 50 ohms is easier, and an accurate wideband 50 ohm resistor costs under 1 cent. The ADC can measure stuff and a uP or PC can do the math.

Actually, the ADC can sample at F or F/N, as long as its trigger phase can be shifted around in steps of 90 degrees. That needs a couple flipflops.

If you sample both ends of the 50 ohm resistor, you know the vector current.

That would be a cool instrument, something that would plot the vector impedance and equivalent R/L/C components vs frequency. A cheap little USB thing. I'd like to go down to 1 Hz for power magnetics.

Reply to
John Larkin

And some software!

Reply to
John Larkin

tirsdag den 27. august 2019 kl. 22.52.10 UTC+2 skrev John Larkin:

and a directional coupler, but once you have the arrays of complex data for forward and reverse it is only a few lines of formulas

and if you have open/short/load measurements on the same frequencies calibration is also easy

Reply to
Lasse Langwadt Christensen

tirsdag den 27. august 2019 kl. 22.50.59 UTC+2 skrev John Larkin:

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Reply to
Lasse Langwadt Christensen

Jpegs do work OK for line art. Better than you might expect given that they are optimised for continuous tone photographic images.

PNG is better though in terms of size if you optimise the palette.

Though with many digital cameras it needs a stop or so over exposure if the pure white isn't to come out as 18% grey.

--
Regards, 
Martin Brown
Reply to
Martin Brown

Drawings generally need more contrast too, and maybe some cropping, sometimes sharpening. I use Irfanview.

Reply to
John Larkin

You can add them.

Reply to
Steve Wilson

I'm not sure those numbers work. For 10nH, for example, the L/R constant is 200ps.

But you could digitally drive a constant current source, that works.

20mA 100kHz sine into 1uH makes 25mV p-p, or 25uV for 1nH.

Cheers, James

Reply to
dagmargoodboat

You measure it by watching the follower oscillate with no external feedback. Put a small pot in series with the base and watch where the oscillation stops.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com
Reply to
Phil Hobbs

1 volt across 50 ohms is 20 mA.

One would of course want a sine wave drive at a few (or many) different frequencies. Hz or KHz for power magnetics, many MHz for nanohenry inductors.

Reply to
John Larkin

I know. I was thinking the signal level would be too small for an ADC and the timing differences too small to resolve -- ADCs aren't good at high-res 1MHz 25uV measurements.

But some appropriate gain stages fix that, and the 50 ohms is enough larger than most interesting inductors' e.s.r. that 1V drive becomes effectively a 40mA p-p current source. That's not terrible.

The original scheme resolved 0.1nH with 100kHz x 20mA excitation. That's pretty elegant. Not bad for two quad op-amps and one MC1496.

A 'digital' version with a DAC, ADC, and a uC still needs the gain stages, but saves the multiplier and a few discrete hairballs.

Cheers, James Arthur

Reply to
dagmargoodboat

the circuit as it is above is, as you say, awfully complicated...

Reply to
bitrex

Signal average. Noise is cheap and plentiful.

If the frequency can be run up, you don't need so much gain.

Reply to
jlarkin

Thanks. This has been very illuminating. First, I find the term negative resistance has nothing to do with the classical definition, where an increasing voltage causes decreasing current. An example is tunnel diode oscillators.

In this application, there is absolutely no negative resistance in the classical sense. The term is a complete misnomer in this useage.

Second, I find a base resistance of 14.96 Ohms in the circuit I gave you is sufficient to basically stop the oscillations.

Third, adding Darlington increases the required resistance to 46.58 Ohms.

These findings are of tremendous importance in everyday electronics.

It explains why a small bead or resistor in the base of a transistor is so effective at stopping parasitic oscillations. It also explains why parasitic oscillations are so hard to kill in Darlingstons.

The next problem is to find out exactly how the small base resistance works. This opens a completely new field of investigation where I am certain the new knowledge gained will be worth the effort.

Thanks again.

Reply to
Steve Wilson

Given an LC tank or an equivalent 1-port passive resonator, only seeing a negative resistance will make it oscillate. There is a class of such oscillators that are analyzed based on negative resistance.

The Mini-Circuits type MMICS are unconditionally stable. They are Darlingtons.

Reply to
jlarkin

Gain's cheap, speed isn't.

I understand the urge to clean up all the discretes, but it seems a bit campy to throw a million transistors + software at it.

I fell victim to that cleaning urge with my 'upgrade.' I drove the inductor with a triangle-wave current excitation, since that was stable, easily calibrated, and easily generated from my variable-frequency digital source. No DAC required.

Triangular current-drive changes the inductor voltage to a squarewave proportional to inductance, with e.s.r. ramps instead of flat tops and bottoms.

The e.s.r. ramp starts at -i excitation and ends with +i excitation, so if you in-phase demodulate, the e.s.r. component cancels and you're left with the pure inductive component.

I replaced the original Jim Thompson(?) MC1496 analog multiplier with CMOS switches. That saved a bunch of biasing and tweaking. De-modulating in-phase eliminated the earlier design's quadrature phase-shifters and associated adjustments.

(I'm basically in software hell at the moment, trying to get a daisychain of undocumented, layered, script-kiddie IDE abstraction-heaps going, so three analog ICs and a probe-able discrete hairball seems pretty attractive at the moment, along with a carburetor, points, and a distributor.)

Cheers, James Arthur

Reply to
dagmargoodboat

Well, it's negative under some conditions in the small signal analysis.

The input impedance of an emitter follower is something like Zb + Ze + ZbZe, with Zb=a'/S // Z(Cbe), the impedance from base to emitter and Ze=Re // Z(Ce), the emitter load impedance. Under certain conditions, this ZbZe term can go negative enough to make the overall input impedance negative.

I include below a simple LTspice AC equivalent circuit model of an emitter follower. Plot V(in)/I(Rin) in Cartesian coordinates and you'll see the real part go negative around

200MHz.

Jeroen Belleman

============================== Version 4 SHEET 1 880 820 WIRE 240 128 240 96 WIRE -128 144 -192 144 WIRE -16 144 -48 144 WIRE 32 144 -16 144 WIRE 192 144 32 144 WIRE 32 192 32 144 WIRE 64 192 32 192 WIRE 176 192 144 192 WIRE 192 192 176 192 WIRE -192 208 -192 144 WIRE 32 288 32 192 WIRE 64 288 32 288 WIRE 176 288 176 192 WIRE 176 288 128 288 WIRE 208 288 176 288 WIRE 240 288 240 208 WIRE 240 288 208 288 WIRE 368 288 240 288 WIRE -192 304 -192 288 WIRE 240 352 240 288 WIRE 368 368 368 288 WIRE 240 496 240 432 WIRE 368 496 368 432 FLAG -192 304 0 FLAG -16 144 in FLAG 240 496 0 FLAG 208 288 e FLAG 240 96 0 FLAG 368 496 0 SYMBOL voltage -192 192 R0 WINDOW 123 24 124 Left 2 WINDOW 39 0 0 Left 2 SYMATTR Value2 AC 1 SYMATTR InstName V1 SYMATTR Value SINE(0 1 1k) SYMBOL res -144 128 M90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName Rin SYMATTR Value 50 SYMBOL g2 240 224 M180 SYMATTR InstName G1 SYMATTR Value .4 SYMBOL res 224 336 R0 SYMATTR InstName R1 SYMATTR Value 100 SYMBOL res 160 176 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R3 SYMATTR Value 250 SYMBOL cap 384 432 R180 WINDOW 0 24 56 Left 2 WINDOW 3 24 8 Left 2 SYMATTR InstName C1 SYMATTR Value 10p SYMBOL cap 128 272 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C2 SYMATTR Value 10p TEXT -264 48 Left 2 !.ac dec 100 10 1G

Reply to
Jeroen Belleman

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