User:Circuit dreamer/Negative impedance

The definition is confusing and even wrong...

"Negative resistance is an electrical property of some circuits where an increasing current in the circuit results in a decreasing voltage" is maybe the most unsuccessful definition of the negative resistance phenomenon. It is correct only for differential negative resistors at that operating in linear mode. Let's scrutinize the definition before to correct and then move it to the negative differential resistance article, where its place is. I have exposed in detail my considerations below since this is a dominating definition and it is extremely difficult to make people use their own brains instead to beleive blindly in dominating dogmata. Circuit-fantasist (talk) 06:18, 16 February 2009 (UTC)

...in the case of true negative resistance...

A general arrangement

In the case of true negative resistance the definition is confusing and completely misleading. Because of this formal formulation, when I was a student, I didn't understand how a true negative resistor (e.g., NIC) actually works. First, let's remember (again and again and again...) that a true negative resistor is a dynamic source, not a resistor while the differential negative resistor is a (over)dynamic resistor. A true negative resistor produces energy (a voltage) while a differential negative resistor consumes energy (an external source creates a voltage drop across it). I realize the latter seems absurd, paradoxical and even wrong but it is true: a differential negative resistor dissipates energy because it is a resistor (dynamic, over-dynamic but still a resistor). According to this definition (...an increasing current in the circuit results in a decreasing voltage...), a true current-driven negative resistor produces voltage that is proportional to the current passing through it while a voltage drop appears across the differential negative resistor. The voltage across a true negative resistor is its own voltage while the voltage across a differential negative resistor is else's voltage.

Fig. 1. If the current increases, the true current-driven negative resistor increases its voltage as well.

The problem is that when an ordinary human being reads this definition and especially if there is not an accompanying figure of the according IV curve he/she thinks of the current and voltage values as absolute values. For example, imagine a 12 volt battery with grounded positive terminal. If you say to some electrician, hobbyist, student, teacher and even ordinary engineer (not a designer using formal mathematical methods) that the battery has decreased its voltage, e.g. by 2 volts, they will think of this battery as a 10 volt source not as a 14 volt one.

The true current-driven negative resistor produces a voltage V_H that is a "copy" of the voltage drop V_E2 across the "positive" resistor (Fig. 1). So, if the current flowing through the circuit increases, the "original" voltage drop across the "positive" resistor increases and the negative resistor increases its "copy" voltage as well instead to decrease it as the definition affirms. Namely because of this amazing behavior the true negative resistor acting as an additional "helping" voltage source "neutralizes" the equivalent "positive" resistance - it increases its voltage as much as it is needed to compensate the voltage drop across the "positive" resistor. In this way, the overall exciting voltage in this circuit increases and the current does not change. Note the negative resistor has to increase, not to decrease its voltage (it is a voltage, not a voltage drop) to do this magic... and this is the great idea behind the legendary transimpedance amplifier, behind an op-amp inverting current source...and behind all op-amp inverting circuits with parallel negative feedback.

Of course, you might say that the potential of the upper source terminal becomes more negative when the current increases and therefore to reach a conclusion that the voltage decreases but this is formal, sterile and useless for understanding the circuit operation. And what is more, this wording is confusing and misleading for understanding the true negative resistance phenomenon.

A concrete op-amp implementation

Let's now scrutinize a concrete true negative resistor implementation - the mystic negative impedance converter (NIC or, more precisely in this case, VNIC). As there is not some satisfactorily explanation of this odd circuit on the web I will expose here my original viewpoint. If you do not believe me, browse through all these 115,000 pages dedicated to negative impedance converter and place here the better explanation that you have found. If not, sorry, you have to be satisfied with my explanations... I will copy them later to the discussion page of Negative impedance converter where I started a discussion about the famous circuit two years ago.

Fig. 2. If the current increases, the op-amp of a negative impedance converter increases its output voltage as well.

In this exemplary arrangement (Fig. 2), where a VNIC compensates the internal resistance Ri of the input voltage source V_IN, we have to see what the op-amp does when the input voltage increases (if it obeys the ubiquitous definition). Of course, we have first to have an idea how this odd circuit works. In order to grasp the great idea behind it, it is extremely useful to track where the current flows (to draw the full current loop). Then, let's begin acting as the current:) and travelling along the current path (see also Negative impedance).

The positive input voltage V_IN causes a current I_IN to pass first through the internal "positive" resistance Ri, then through the NIC's duplicate resistor R = Ri (below the op-amp), after to enter the op-amp's output, to exit its negative supply terminal, to flow through the negative supply source -V, to pass through the load R_L and finally to return to the input voltage source. The op-amp has compares half of its output voltage (because of the voltage divider consisting of two equal resistors R above the op-amp) with the "copy" voltage drop across the duplicate resistor R and changes it so that a zero difference is between them. As a result, the op-amp produces two times higher output voltage than the "copy" voltage drop across the duplicate resistor. Half the voltage compensates the voltage drop across the duplicate resistor R. The rest half is the voltage produced by this true negative resistor; it compensates the voltage drop across the internal resistance Ri. As a result, the op-amp adds a voltage V_Ri to the input voltage.

How, if the input current increases (e.g., because the input voltage V_IN increases), the voltage drop across the duplicate resistor R increases as well and the op-amp increases its output voltage (regarding to its local ground) as well instead to decrease it as the definition affirms. In this case it is still more formal, sterile and misleading to say that the voltage of the negative resistor becomes more negative since the circuit is flying. Circuit-fantasist (talk) 21:45, 15 February 2009 (UTC)

...and in the case of differential negative resistance.

I start here maybe the most interesting discussion about differential negative resistors concerning two unique operation modes - linear and bistable. Linear applications include various exotic 1-port amplifiers (e.g., legendary tunnel diode amplifier). A lot of interesting negative resistor applications are based on bistable mode: exotic switching circuits with dual-treshold action (hysteresis), memory circuits (flip-flops), etc. Later, we might copy or move this discussion to negative differential resistance page where its place is but now we have to ascertain if the differential negative resistors obey the definition. Please, join the discussion; the topic is extremely interesting and important. Circuit-fantasist (talk) 22:57, 17 February 2009 (UTC)

Differential negative resistor operating in linear mode

Obviously, this definition is formulated especially for differential negative resistance but, unfortunately, it has problems even in this case. The definition is correct only if the differential negative resistor operates in linear mode (i.e., if we drive a current-driven differential negative resistor by a current and if we drive a voltage-driven differential negative resistor by a voltage). Let's ascertain it.

Fig. 3. An increasing voltage across a voltage-driven differential negative resistance results in a decreasing current and v.v.

Fig. 4. An increasing current through a current-driven differential negative resistance results in a decreasing voltage drop and v.v.

As a rule, electronics resources show the IV curve of a differential negative resistance in its final N or S shape (e.g., see a bare curve) and note there is a section with a negative slope. But nobody shows how this magic is done and what it actually means. Only, in order to grasp the idea of differential negative resistance, it is extremely important to show exactly how and why this part of the curve is inclined to the left. Here is the secret of the phenomenon.

The trick is extremely simple, clear and intuitive: this is the great dynamizing idea that is brought here to the utmost degree. A differential negative resistor is notning more than a kind of "self-varying", dynamic resistor that changes extremely its instant (ohmic, "positive") resistance depending on the current passing through the resistor or on the voltage applied across it; a differential negative resistor is actually an over-dynamic resistor. I have shown the evolution of this great idea in a circuit story about S-shaped differential negative resistors (I have placed a reference to this story for your convenience, not to promote myself).

Applying a voltage across a voltage-driven differential negative resistance

Look at Fig. 3 to see how a voltage-driven differential resistor having an N-shaped IV curve behaves, in order to do this magic in the area of the negative resistance. In this graphical representation, I have superimposed the IV curves of the input voltage source (in a red color) and the differential negative resistor (in blue) on a common coordinate system. In addition, to make things absolutely clear, I have superimposed the IV curve of the instant ohmic resistance R (in orange) as well. Now, turn on your most human mental faculty - your imagination, and visualize in your mind's eye what happens when the input voltage begins varying.

For example, when the input voltage begins increasing, its IV curve moves horizontally from left to right remaining parallel to itself. If the resistor was "static", the crossing (operating) point would slide along the IV curve of its ohmic resistance having a positive slope. But it is a more "clever":) dynamic resistor (like my opponents in this talk page:). So, it begins changing (in this case, increasing) its instant ohmic resistance R and rotating vigorously its IV curve clockwise. As a result, a wonder happens and the operating point slides along a hew IV curve having a negative slope. Note this is not a real curve as the ohmic IV curve is. It is an artificial, a synthetic, an imaginative IV curve; it is an illusion.

Does this case obey the definition, "...an increasing current in the circuit results in a decreasing voltage..."? No, as the voltage jumps in the negative resistance region (see below).

Passing a current through a current-driven differential negative resistance

A current-driven differential resistor with an S-shaped IV curve (Fig. 4) has the opposite crafty behavior:) - when the input current increases, it decreases its instant ohmic resistance R and rotates vigorously its IV curve contraclockwise. Now, the operating point slides upwards along a hew IV curve having a negative slope as above.

Does this case obey the definition now? Yes, if we only replace "voltage" with "voltage drop":

Differential negative resistor operating in bistable mode

If the differential negative resistor operates in bistable mode (i.e., if we drive a current-driven differential negative resistor by a voltage and if we drive a voltage-driven differential negative resistor by a current), the definition is not true.

I give an opportunity to you to see why the definition has failed in these cases. If you do not manage, I will elucidate why. Circuit-fantasist (talk) 22:05, 15 February 2009 (UTC)

Applying a voltage across a current-driven differential negative resistance

Fig. 5a. When an increasing voltage reaches V_H the current increases (jumps) but the voltage stays constant during the jump.

Fig. 5b. When a decreasing voltage reaches V_L the current decreases (jumps) but the voltage stays constant during the jump.

Passing a current through a voltage-driven differential negative resistance

Fig. 6a. When an increasing current reaches I_H the voltage increases (jumps) but the current stays constant during the jump.

Fig. 6b. When a decreasing current reaches I_L the voltage decreases (jumps) but the current stays constant during the jump.