Q-meter experiment - Page 1
Forum section: electronics / rf-microwave

author: RoGeorge
author_rank: Super Contributor
timestamp: August 13, 2025, 01:59:01 pm
content: Couple of weeks ago I've scrolled too much, then felt chatty and rambled a little about negative resistance inhttps://www.eevblog.com/forum/chat/secrets-from-the-ee-scrolls/msg5993389/#msg5993389This was the example used there, to understand the idea of a circuit that is behaving like a negative resistor:That is one way of doing it. Can be many other ways, depending on how the circuit is achieving negative resistance:- by inverting the voltage- by inverting the current- by swapping the voltage and current between them (in the V/I characteristic plot of a circuit)Depending on those, the negative resistance region of the V/I plot can resemble either the shape of a letter N, or the shape of an S. This will also determine if the circuit is stable when the input of the circuit is left open, or when the input is short-circuited. A negative resistance achieved by an S-like V/I curve will be stable with its input left open, while an N-like V/I curve will be stable with its input short-circuited.Let's take an example for the V/I characteristic:In such an V/I plot, the region of negative resistance is anywhere where the I/V plot line is descending (has negative slope). Positive resistance regions is everywhere else, where the I/V plot line is ascending (positive slope). A perfectly horizontal line in an I/V plot would represent resistance zero (or short-circuit), while a perfectly vertical line would represent an infinite resistance (or open-circuit).Notice in the second panel how the plot line is first ascending (positive slope, so positive resistance), then the plot line is descending (negative slope, so negative resistance), then ascending again at the end. There is a region of negative resistance in the middle. Any source connected at 'in' will see the node 'in' as a negative resistor (between 'in' and GND, and only if the input voltage is in the negative resistance region, so between about -7V and +7V according to those plots).Also, notice how the plot in the middle panel has a shape resembling the capital letter N, so an N-like V/I characteristic, which tells this particular schematic will be stable with 'in' tied to GND, but might not be stable with the 'in' left open. So, this circuit will be stable with a short-circuit at its input (or a very low impedance), and will tend to oscillate if the input is left open (or when the input is connected to a high impedance source).This so far was about circuits that behave like a negative resistance, in general, and about one example of such a possible circuit in particular, implemented with an opamp.So far nothing yet about the Q-meter in the title.Now, let's consider a parallel LC, with ideal L and C, connected at the input of the opamp circuit above. At resonance, the impedance of the ideal parallel LC will be infinite, or else said, the LC will be like the input of the opamp was left open. The V/I plot of the opamp circuit, in the shape of an N, indicates that the LC+opamp will oscillate, simply because the opamp will see an infinite impedance at its input.What if L and C are not ideal? Usually there are resistive loses in both the L and the C of an LC tank. In practice, most of the losses are expected to happen in the coil. These resistive loses will tend to dampen the oscillations. The resistive losses will continually dissipate some of the energy stored in the LC tank. Eventually, when the resistive loses are too high, and they dissipate more energy than it is put into the LC tank, then the oscillations will stop.It can be shown that the limit between oscillating / not-oscillating is when the parallel losses resistance of the LC tank, let's call it Rp, has the same value as the negative resistance of the opamp circuit, Rn.And that is the Q-meter idea I want to try.The working principle:- get a negative impedance circuit, in which the value of the negative impedance can be controlled (adjustable)- connect it to a parallel LC tank- consider the C known, both in capacity and in lossesAdjust the variable negative resistance until the oscillations start/or stop, so to identify the region where the parallel losses resistance Rp of the LC tank is compensated by the negative resistance of the opamp circuit, Rn.In theory, this should work, but in order to get an adjustable negative resistance, a potentiometer will be required in the opamp circuit. The LC is expected to oscillate, and for small value coils, it might oscillate at high frequency, maybe hundreds of MHz. A potentiometer can not work at such high frequencies.Sure, there are other ways to get an adjustable resistor without using a mechanical potentiometer, but aside from working at high frequency, the controlled resistance should be well known, in order to calculate back the Rn and Rp and eventually the quality factor, Q.The implementation:- with a 250MHz analog multiplier AD835https://www.analog.com/media/en/technical-documentation/data-sheets/AD835.pdfDrafted a block schematic already inhttps://www.eevblog.com/forum/chat/secrets-from-the-ee-scrolls/msg5998985/#msg5998985Before going into even more details about how was that supposed to be working (maybe later), will go and solder an AD835 instead.

author: mawyatt
author_rank: Super Contributor
timestamp: August 13, 2025, 02:51:27 pm
content: In the V/I simulation, if V1 is an ideal voltage source, then Rn has no effect as it can't influence the voltage at + input to U1, so the output from U1 is just 2X V1 regardless on Rn. V1 must have a source impedance, or an added series resistance for the + feedback to have any effects.Best

author: RoGeorge
author_rank: Super Contributor
timestamp: August 13, 2025, 03:53:56 pm
content: Yes, the Rx will always see 2*Vin-Vin, so Rx (or Rn in the second embedded pic) will always see at its terminals a voltage equal with Vin. That will force a current through Rx, current equal to Ix=Vin/Rx, which current can only flow into the V1 source, because the opamp '+' input doesn't take any current.As a result, instead ofsourcingsome current into a load (here the load being the opamp circuit), V1 will have tosinkto GND the current coming from Rx (so a change of sign in the current seen from the perspective of V1). Therefore, from the perspective of V1, the entire opamp circuit will appear as a negative resistor of value -Rx.Of course, for the end application (here the Q-meter), a parallel LC will have a very high impedance in parallel with an Rp, the parallel resistive losses of the LC, and that Rp can be considered (the now missing resistor) in series with V1.

author: jwet
author_rank: Frequent Contributor
timestamp: September 06, 2025, 03:37:23 pm
content: RoGeorgeI don't have a complete answer but have a couple of tidbits that might improve your experimenting.The technique that you have arrived at is basically how regeneration works. In a more sophisticated case, the "oscillator" can be be pushed in an out of oscillation at some audio frequency usually. This is called super-regeneration. On the way into instability, you get loads of gain. There are some ancient papers by Armstrong that studied this. Tom Lee covers some of this in his great book- CMOS RF Integrated Circuit design. Super regen receivers were used for simple radios like toys and even key fob receivers in the not so distant past.There is a kind of famous standard analysis technique of a colpitts oscillator that models it as an LC tank feeding a negative resistance. Its an alternate way of looking at many oscillators and may open your eyes. I don't remember where I saw it but will post if you can't find it.Good luck.

author: RoGeorge
author_rank: Super Contributor
timestamp: September 21, 2025, 09:07:27 pm
content: Thank you, had to take a break last month, so for now this one became yet another parked project, but already have a drafted schematic and the key parts put aside in a ziplock bag.The idea I want to try is scribbled in the attached pic, not sure if it makes much sense, but added the pic anyway to not forget where I left.It uses an AD835 analog multiplier and a DC voltage Vctrl, to control the amplification A of an equivalent (imaginary) opamp formed by the analog multiplier, so in the end it should behave like a voltage controlled negative resistor.A microcontroller is supposed to browse the Vctrl DC control voltage until the Lx+Cref+AD835 start to oscillate, and thus deduce the Q, or at least that's the idea. I guess this should work, but will have to experiment with it first, in order to put all the ducks in a row.