User:Circuit dreamer/Negative resistance Talk 20-02-09

Reconstructing the introductory part
''I have the intention of replacing the introductory part (Concepts) of the article with the text below (see also the full version of my suggestions for improving the article to see my reasons). I'm waiting for your opinion about the content and the implementation of the introductory part.'' Circuit-fantasist (talk) 17:09, 20 January 2009 (UTC)

Negative versus "positive" impedance
The nature of electrical negative impedance is clarified below by comparing an ordinary "positive" resistor having a resistance R with a true negative resistor having a resistance -R. For this purpose, two pairs of equivalent electrical circuits are used, in which the resistors are connected in series with the loads so that the same current passes through them (Fig. 1) or in parallel to the loads so that the same voltage is applied across them (Fig. 3).

What a "positive" impedance is
In electrical circuits, passive elements (resistors, capacitors and inductors) impede current by their inherent resistance, capacitance and inductance (impedance is the combination of them). As a result, voltage drops appear across the passive elements that represent the energy losses in them. Passive elements absorb this energy from the exciting electrical source: resistors dissipate energy from them to outside environment while capacitors and inductors accumulate energy into themselves. From the negative impedance viewpoint it is not important if passive elements dissipate or accumulate energy; it is only important that they absorb energy.









Electrical elements may be connected in series, parallel or mixed. If a resistor R is connected in series with a load (Fig. 1a), a voltage drop VR = R.I that is proportional to the current appears across the resistor. When capacitors or inductors are placed at this place, voltage drops changing through time appear across these elements.

Conversely, if the resistor R is connected in parallel with the load (Fig. 1b), a current IR = VL/R  that is proportional to the voltage flows through the resistor. When capacitors or inductors are placed there, currents changing through time flow through them.

A grapho-analytical interpretation (by using superimposed IV curves) of the "positive" resistance phenomenon is shown on Fig. 2. It is supposed the resistor is supplied by a real current source (Fig. 2a) or by a real voltage source (Fig. 2b). When the input current/voltage varies, the crossing operating point slides over an IV curve representing the "positive" resistance. It is a real IV curve having a positive slope and passing through the origin of the coordinate system.



What a negative impedance is
Elements with negative impedance do the opposite - they inject energy into electrical circuits. Whereas passive elements with "positive" impedance absorb energy from the input source (they are loads), the elements with negative impedance add energy to the input source (they are sources ). While voltage drops appear across "positive" elements, negative elements produce voltage; while "positive" elements sink current, negative elements produce current. However, they are not ordinary, steady electrical sources; they are varying sources (voltage sources whose voltage across them depends on the current through them or current sources whose current depends on the voltage across them).





On Fig. 3a a negative resistor NR with a resistance -R is connected in series with the load so that a voltage VNR = R.I that is proportional to the current appears across the negative resistor. However, while above the "positive" resistor detracts the voltage VR from the input voltage (VR is a voltage drop), here the negative resistor adds a voltage VNR to the input voltage (here VNR is a voltage). The element named "resistor" is really a resistor while the "negative resistor" here is actually a voltage source, whose voltage is proportional to the current passing through it. If negative capacitors or inductors are placed at this place, they produce voltages changing through time.

''A negative resistor can be implemented as a varying (dynamic) voltage source, whose voltage is proportional to the current passing through it. This two-terminal current-controlled voltage source acts as a current-driven negative resistor.''





Conversely, when the negative resistor is connected in parallel with the load (Fig. 3b), a current INR = VL/R that is proportional to the voltage drop across the load flows through the negative resistor. However, while above the "positive" resistor sinks the current from the input current (diverts it from the load current), here the negative resistor adds a current to the input current (injects an additional current into the load). If negative capacitors or inductors are placed there, they inject currents changing through time.

''A negative resistor can be implemented also as a varying (dynamic) current source, whose current is proportional to the voltage across it. This two-terminal voltage-controlled current source acts as a voltage-driven negative resistor.''

The graphoanalytical interpretation of the negative resistance phenomenon is shown on Fig. 4. It is supposed the negative resistor is supplied by a real current source (Fig. 4a) or by a real voltage source (Fig. 4b). When the input current/voltage varies, the voltage/current source representing the negative "resistor" changes its voltage/current. As a result, its IV curve moves and the crossing operating point slides over a new dynamic IV curve representing the negative resistance. It is not a real IV curve; it is an artificial, imaginary IV curve having a negative slope and passing through the origin of the coordinate system.

Circuit-fantasist (talk) 08:10, 24 January 2009 (UTC) 

An extract from Archive_4
I have copied the insertion below (in italic) from Archive_4. Circuit-fantasist (talk) 22:08, 22 February 2009 (UTC)

''I told you: I read the edits and the reason I removed the circuit (Basic Initial Circuit below as C-F has so helpfully added it to the talk page again) is because it is a bad example and there is a better example right after it. In fact that circuit is a Transimpedance amplifier TIA. C-F's hand-waving about the gain "(typically, A = 2)" is just pulled out of the air. After more contortions, he summarizes with "The circuit behavior depends on the value of the internal resistance Ri of the input voltage source that constitues a voltage divider with the resistance R. If R/Ri > A, the circuit is stable and operates in linear mode (this is the case when an ordinary amplifier with small but steady gain is used). If R/Ri < A, the circuit is unstable and operates in bi-stable mode"  Where is Ri in the schematic? It has to be included as the input impedance, since he is using it elsewhere in the analysis(?). Instead of a negative impedance -R at the input, there is an impedance of Ri-R, possibly not negative at all. This circuit is very speculative as a negative resistor example. Hand-waving and bending op-amp behavior to make the analysis(?) fit the claim doesn't fly in my book''

''Other parts I removed that never made any sense can be found under "Current-driven negative impedance elements". It begins:''

"Real voltage-supplied electrical circuits contain at least two elements with positive impedance (see Fig. 1a again): one useful (the load) and another undesired (for example, line resistance between the source and the load) that has to be eliminated by including an additional current-driven negative impedance element (Fig. 5)."

and ends:

"However, for this purpose, the op-amp needs to "sense" the voltage drop across the positive impedance element; sometimes this is impossible or inconvenient, for example if it is a long telephone line."

Again, with a lot of self-induced contortions betwixed the two and with no realistic conclusion.

''I think this article is basically done. In a few days it has gone from a mish-mash of confused thinking about negative resistance; something that probably wouldn't even last on Keelynet very long, to a decent article by Wikipedia standards. Instead of encouraging the backsliding of this article to its earlier state of excessive OR, POV and all the other bad things that give Wikipedians angst we should move on to some of the other articles on our hit list. I see the happy stick men are still inhabiting Negative_differential_resistance. Why is that article still there? I thought all forks must go. So let's move forward and start by deleting that awful mess of POV, OR, etc. That done I think Current-to-voltage_converter needs a re-write like we have done here (collective pat on the back). But the title has to change to Transimpedance Amplifier with TIA redirecting to it. I have worked in the IR field (building FTIR spectrometers, liquid He cooled focal plane arrays, and designing multi-spectral IR cameras) for over 30 years and it would be nice to be able to refer a Wikipedia TIA article that is readable to someone who is just getting started in this field. So let's move forward. Zen-in (talk) 20:42, 22 February 2009 (UTC)''




 * Zen-in, please think before writing! What do you talk? A transimpedance amplifier is a circuit with negative feedback where the resistor R is connected between the output and the inverting input! Regarding to Ri, I have written "The circuit behavior depends on the value of the internal resistance Ri of the input voltage source that constitutes a voltage divider with the resistance R." What is the problem? Ri is the total equivalent resistance of the external circuit; it does not belong to this circuit (sorry, add 1 to R/Ri, it may be important). Look at one of my "graffiti" that I have so helpfully added to the talk page to see what I have meant (here, Ri = r1||r2||RL). I hope this colorful and informative picture will help you to see the great idea behind this bare circuit and the happy stick men will cheer up you. If you do not like it, read this EDN article and my comment after it to understand what they want to say.


 * I am sorry that you can't understand the great idea of voltage compensation implemented by a transimpedance amplifier and by a current-driven negative resistor and that you can't see the advantages of the second solution in telephony line repeaters. I would clarify it here again but, as I can see, this is of no use in your case.


 * Regarding to "...mish-mash of confused thinking about negative resistance...", I would like to note that you are already the only Wikipedian contributing this page that has no idea what true negative resistance, differential negative resistance and negative impedance are. Circuit-fantasist (talk) 22:55, 22 February 2009 (UTC)


 * C-f, it is pointless archiving if you are just going to empty the archive back on to the page. What you need to do is find quality references for the things you want to add to the article.  What you do not need to do is explain in great detail what you want to add.  No one will take any notice unless their are sources to back it up, then we can read the sources and you do not need to explain.  The circuit you are pointing to in the link above is a NIC which is already in the article.  Sp in ni  ng  Spark  22:59, 22 February 2009 (UTC)


 * Right you are about that op-amp circuit not being a TIA C-F. It really is just a non-linear bistable circuit.  Maybe it has a negative resistance regime but no-one has proven this.  What input voltages and currents will make the input be a negative resistor?  Can you point me to a reference that has an incremental analysis of this circuit?  I'm playing with a digital filter chip right now, the MF10 and recently made a prototype board for the part.  I'll wire up an op-amp circuit like we have been talking about and see what it does.  Spice is a good tool to answer questions like this.  but you may find it hard to do a simulation on Spice because you are trying to generate a non-zero voltage across the inputs.  If there is a stick-man inside an op-amp, his job is to prevent that.  Zen-in (talk) 01:10, 23 February 2009 (UTC)

How positive feedback and true negative resistance are related
(a reply to the Zen-in's insertion above)

Zen-in, you are welcome to discuss clever circuit ideas with me; I will do it with the greatest of pleasure. Below I expose my insights about the connection between two great phenomena - positive feedback and true negative resistance. Circuit-fantasist (talk) 17:13, 24 February 2009 (UTC)



Linear or bistable circuit?
Here I have commented your speculation, "It really is just a non-linear bistable circuit."

The discussed circuit is a non-inverting amplifier with positive feedback where two devices with opposite behavior, one amplifying and other attenuating, are consecutively connected in a loop. We may look at it from the classical negative feedback theory viewpoint. As you know, depending on the proportion between the values of the amplification A and the attenuation β = Ri/(Ri + R), i.e. on the value of the loop gain βA, a circuit with positive feedback may be stable or may be not. As I have already said above, we assume the circuit is driven by a real voltage source with internal resistance Ri. If it was a perfect voltage source with Ri = 0, there will not a feedback and the circuit will behave as a comparator. If βA < 1, the signal traveling along the feedback loop attenuates and the circuit is stable and linear (it acts as an amplifier); if βA > 1, the signal traveling along the feedback loop increases continuously up to saturation and the circuit is unstable (bistable, non-linear) and acts as a comparator with hysteresis (Schmidt trigger). These considerations are general; they comprise all kinds of circuits with positive feedback.

Parallel or series positive feedback?
But is only the presence of positive feedback sufficient to obtain a negative resistance? No! As you probably know, there are two kinds of feedback depending on the way of connecting the input source and the amplifier output - parallel and series. In this circuit a parallel kind of positive feedback is applied that means there is a connection between the input source and the amplifier output by the resistor R acting as a "bridge". Because of this connection, the output can affect (in this case, "help") the real input voltage source by injecting or sinking a current through the "bridge" resistor R (for example, if the input voltage is positive, the op-amp injects a current into the input source). As a result, the input resistance of the circuit changes (in this case, increases). That is why, circuits with parallel feedback (no matter positive or negative) have different than initial resistance (decreased, zeroed, increased, infinite or negative) and, as a result, they affect the input source; this is a common property of circuits with parallel feedback. So, we may make a conclusion:

''Every circuit with negative resistance is a circuit with positive feedback but not every circuit with positive feedback is a circuit with negative resistance. Only circuits with parallel positive feedback exhibit negative resistance.''

In the dual circuits with series positive feedback there is no connection (a "bridge") between the op-amp output and the input source; they are detached by the high input impedance of the op-amp. The output does not affect the input source even it is a real, imperfect voltage source with considerable internal resistance Ri. As a result, the input resistance of the circuit is the high op-amp input impedance.

''Circuits with series positive feedback do not exhibit negative resistance. We cannot make negative resistance circuits based on series positive feedback.''

How the circuit behaves depending on the gain
Here I have commented your speculation, "Maybe it has a negative resistance regime but no-one has proven this."

It is interesting to see how the circuit behavior depends on the gain A of the amplifier. For this purpose, let's vary the gain from zero to large value and observe how the circuit input resistance changes from positive to negative (this is a story about dynamic resistance). According to the old version], the input resistance is $$ R_\text{in} = {v \over i} = \frac{R}{1-A} $$. Let's consider the most typical cases:

A = 0. In this case, the resistor R is grounded from the right; figuratively speaking, the right side of resistor R is "immovable". As a result, the circuit acts as a positive resistor with resistance R.

0 < A < 1. The op-amp "moves" slightly the right end of the resistor R in the same direction as the input source "moves" the left side of the resistor; the circuit acts as a positive resistor with inreased resistance > R.

A = 1. The op-amp "moves" the right end of the resistor R exactly as the input source "moves" the left side of the resistor. As a result, the famous bootstrapping phenomenon occurs: no current flows and the circuit behaves as a positive resistor with infinite resistance.

1 < A < 2. The op-amp "moves" (in the same direction) the right end of the resistor R more vigorously than the input source "moves" the left side of the resistor. As a result, the current changes its direction: if the input voltage is positive, the circuit (the op-amp) injects a current into the input source and v.v., if the input voltage is negative, the circuit (the op-amp) absorbs a current from the input source. As a result, the circuit acts as a negative "resistor" with bigger |negative resistance| > |R|. But is the input source really a source in this case or it is a load? Yes, it is a load; the circuit with negative resistance treats the input source as a load.

A = 2. The circuit is a negative "resistor" with equivalent |negative resistance| = |R|.

A > 2. The circuit is a negative "resistor" with small |negative resistance| < |R|.

What is the voltage across the amp input?
Here I have commented your thought, "...you are trying to generate a non-zero voltage across the inputs."

This is a non-inverting amplifier with positive feedback; so the voltage across the amplifier input is the full input voltage. Only negative feedback can zero the voltage across an op-amp input (the so-called Golden rule 1).

The end
Zen-in, I am not interested in incremental analysis of this circuit. It would be interesting for me to conduct experiments with a real circuit. I used Howland current source to drive the transistor under test in a curve tracer years ago. Also, I investigated it in the laboratory with my students last year. Maybe, this year my students and I will reinvent and investigate this odd circuit in the laboratory. Then we will tell all this on Circuit idea. Congratulations for playing with digital filters; they are to abstract for me. Circuit-fantasist (talk) 17:41, 24 February 2009 (UTC)


 * Circuit-fantasist, If you have an op-amp circuit where there is a voltage across the inputs, because of the high D.C. gain an op-amp has, the output will be railed. It will be entirely non-linear.  Your theoretical negative resistance region will be infinitesimally small.  Take a look at this article - Schmitt_trigger and go to the schematic under Comparator implementation.  In this article the authors show that by using current conservation (Kirchoff's current law: the sum of all the currents flowing out of a node is 0), and treating the + input as a virtual ground, they derive the transfer characteristics of the circuit.  They show it is bistable and has hysteresis.  When the + input VIN goes from a negative value to +R1/R2*VS (where VS is the positive railed output voltage), The output goes from -Vs to +Vs. As it does that the voltage across the inputs (V+ - V-) goes from 0 V to some value greater than 0 V. Where you think the negative resistance should be, there is a hysteresis region.  Now if R1 is made very small, the hysteresis region just gets smaller and the transition happens closer to 0 V.  The circuit is still bistable.   No mention is made of the gain of this circuit in the analysis.  But it is in the analysis because they say the transition occurs when op-amp input V+ = 0 V.   That is how DC or low frequency op-amp analysis is done.  At higher frequencies, closer to the bandwidth of a particular op-amp, the gain is much reduced, because of the frequency compensation most op-amps have.  That is a different matter.  You are advertising a negative resistor, not a negative impedance. And the last time I checked, resistance was measured with D.C. currents and voltages.  Zen-in (talk) 04:27, 25 February 2009 (UTC)


 * Zen-in, I have indulged in copying your comments above from my talk page as it would be helpful for other wikipedians and visitors to see and to join the discussion. IMO, revealing the truth about circuits is the most important activity, in order to create valuable pages (before to expose it encyclopedically). For this purpose, it is worth to "waste" efforts, time and space of talk pages. Later, we might archive, move these discussions to the related pages or just delete if they are useless. Circuit-fantasist (talk) 09:21, 25 February 2009 (UTC)

More about the elementary circuit with positive feedback
Zen-in, your thoughts are extremely interesting since they give us another idea about presenting the negative impedance converter. By solving step-by-step the problems that you have posed we may show the circuit evolution and reinvent the circuit. You are right and they are right as well (although I have some objections to the contents of this page); I would like only to elaborate and generalize this idea.

This is the same circuit


As you can see, this is exactly our circuit above; only it is driven here by a perfect voltage source with Ri = 0. Although they have not defined its name this circuit is a non-inverting comparator with hysteresis (the article is not well structured, the dual inverting comparator with hysteresis is shown below as a part of an RC relaxation oscillator). It is interesting their negative feedback alternatives have opposite names - accordingly inverting amplifier and non-inverting amplifier. The first of them use a parallel feedback that, as I have considered above, has changed their input resistance; the second use a series feedback and they have a constant high input resistance (equivalent to the original op-amp input resistance). So the non-inverting comparator with hysteresis (more precisely speaking, its right part including the resistor R2 and the op-amp) possesses a negative resistance while the dual inverting comparator with hysteresis does not exhibit negative resistance.



What do resistors R1 and R2 actually do?
Now, please let me show first what is the idea of the two resistors R1 and R2 in circuits with parallel feedback (no matter positive or negative).

In resistive circuits with parallel feedback the two resistors R1 and R2 constitute the extremely useful circuit of a parallel voltage summer. From this viewpoint, our non-inverting comparator with hysteresis and the inverting amplifier consist not of two resistors and an op-amp but of a parallel voltage summer and an op-amp. Looking from this "systematic" viewpoint at these circuits we can present them as feedback systems where the role of this two-resistor circuit is to add (positive feedback) or to subtract (negative feedback) the output and the input voltages. In the remote 1986, I was impressed by this humble circuit as a component of the op-amp inverting amplifier and I have been thinking about it thereafter. In 1989 I created an interactive on-line animated voltage diagram based on Apple computer that I have been using up to the present in the laboratory. In 2002, I created two Flash movies about the passive and active parallel voltage summers. In 2007, I showed how to build it in a Circuit idea story. Last year, my students and I reinvented it by reproducing Ohm's experiment in the laboratory (by the way, tomorrow I will have the first meeting witn my students in the laboratory and I will discuss this experiment). Imagine I have created all these stories about... only two resistors connected in series!

Choosing an amplifier for our comparator with hysteresis
What you are written about this circuit is right. Yes, it is a circuit with very, very... infinitesimal (as an absolute value) input negative resistance RIN = R2/(1 - A) because of the huge op-amp gain. But this enormous gain is of no use here. In order to act as a bistable comparator with hysteresis, only a gain A that is a little bigger than 1 + R2/R1 is sufficient (in order βA > 1). For example, if we use typical R1 = 1 kΩ and R2 = 10 kΩ, we need only a gain of 12 to make the circuit act in bistable mode as a comparator with hysteresis. In this case, the right part of the circuit including R2 and the op-amp will act as a negative resistor with negative input resistance RIN = 10/(1 - 12) = - 900 Ω but this is no matter here since we want to create a comparator, not a negative resistor. Nowadays, we have not such amplifiers with so small gain; we have op-amps with enormous (e.g., 100000) gain. OK, there is no problem and we put some op-amp to solve our problem... and we solve it. Now the right part of the circuit including R2 and the op-amp will act as a negative resistor with an infinitesimal input negative resistance RIN = 10/(1 - 100000) = - 0.0001 Ω. It is no matter again; we needn't it but it does not disturb us. Only... is it really right? Maybe it's time to consider this situation.

Reducing the op-amp gain to obtain a NIC


But if we want to make this circuit act in a linear mode, that is quite a different matter... If we make just an amplifier with positive feedback, we have to satisfy only A < 1 + R2/R1 = 11, in order to keep it stable. If we want the right part of the circuit including R2 and the op-amp to act as a negative resistor, we have to calculate the needed negative resistance RIN = R2/(1 - A) obeying A < 1 + R2/R1 to keep the circuit stable. So we need to reduce the op-amp gain...

How can we decrease a gain? Of course, by applying a negative feedback from the output to the inverting input... by connecting a voltage divider... by making an op-amp non-inverting amplifier.

Thus we have finally obtained the well-known circuit of a negative impedance converter... we have reinvented it... we have known the role of every circuit element... of each of the four resistors and the very op-amp. Circuit-fantasist (talk) 21:13, 25 February 2009 (UTC)



How do voltage-driven true negative resistors affect input sources?
Our circuit (a non-inverting comparator with hysteresis) affects the input source through the seriesly connected positive feedback elements. If we consider it, we will find out the circuit has very, very strange behavior...

Thevenin's source


This is the arrangement above - a perfect voltage source connected in series to the resistor R1 (Ri). When the input source has a positive voltage, it "wants" to "push" a current into the circuit treating it as a load. Only, the circuit does not allow it to do that; it impedes the input source in its "desire" to produce an exit current. It even reverts the current direction thus making the input source act as a load! For this purpose, the amplifier increases its output voltage more than the input one and the current begins flowing from the positive amp's power supply to the input voltage source.

And v.v., when the input source has a negative voltage, it "wants" to "suck" a current from the circuit treating it as a load. Again, the circuit does not allow it to do that; it impedes the input source in its "desire" to produce an enter current. It reverts the current direction as above thus making the input source act as a load again! For this purpose, the amplifier drops its output voltage under the input one and the current begins flowing from the input voltage source to the negaitive amp's power supply (the amp's power supply "sucks" a current from the input negative voltage source instead the opposite).

More figuratively speaking, when the input voltage source "attacks" our circuit trying to pass a current through it, the circuit "counter-attacks" the input source by passing a current through it in an opposite direction.

All this is true but I do not see what the use of this arrangement is...



Norton's source


Here, a current source drives a resistor (a load); the circuit is connected in parallel to the load and "helps" the input source in its "desire" to pass a current through the load. We have considered this arrangement above for the purpose of compensating a heavy load. The famous circuit of Howland current source is based also on this arrangement.

(to be continued) Circuit-fantasist (talk) 19:27, 26 February 2009 (UTC)

