# Circuit Idea/Negative Impedance Converter

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Revealing the Mystery of Negative Impedance Converters

Circuit idea: Adding the same voltage (current) that appears across (flows through) a corresponding "positive" impedance element.

Electrical elements with negative impedance are extremely useful in circuitry as they can compensate the losses in passive elements having the equivalent positive impedance. Unfortunately, true negative impedance elements do not exist in nature; there are only ordinary passive elements with positive impedance (resistors, capacitors and inductors). It is a powerful idea to use them as "shaping" elements for creating the according "mirror" elements with negative impedance. Negative impedance converters exploit this idea by converting the positive impedance into a negative one with the same value.

### Current-driven negative impedance elements

##### The problem
Fig. 13. Compensating the line resistance by a transimpedance amplifier.

The simple current-driven element with negative impedance is just an op-amp that "copies" the voltage drop across the already existing element with positive impedance (see again Fig. 7). For this purpose, it uses an additional third wire (a "sense") to "feel" by its inverting input the difference between the voltage drop across the positive impedance element and its output voltage. However, sometimes this is inconvenient or, unfortunately, it is just impossible. In order to clarify the situation, a well-known problem from circuitry is solved on Fig. 13 - creating a perfect voltage source with zero internal resistance[1].

There are two problems to be solved in this arrangement: first, real voltage sources have some internal resistance; second, there is some line resistance between the source and the load. As a result, the voltage across the load droops when the source is loaded as disturbing voltage drop appears across the internal resistance and across the line resistance. That is why, in electronics, there is a need of voltage sources having zero internal resistance and lines having zero line resistance. The classical solution is to buffer the imperfect voltage source by a powerful voltage follower. However, the booster has to stay at the end of the line, near the load; unfortunately, in many cases, this is just impossible. More exotic and sometimes, more useful solution is to compensate the internal resistance by an equivalent negative resistance.

If the line resistance has to be compensated, then it would be possible to connect the op-amp inverting input (the "sense") to the left side of the line "resistor" (Fig. 13) realizing the simple current-driven negative resistor from Fig. 8 (a transimpedance amplifier). In this arrangement, the op-amp compares its output voltage with the voltage drop across the line resistance Ri and change it so that to keep (almost) zero difference between them. As a result, the op-amp produces output voltage that compensates the voltage drop across the line resistor R. Note in this arrangement the op-amp has "flying" power supply, in order the load to be grounded.

##### The basic idea of current-driven negative impedance
Fig. 14. Compensating a source internal resistance by a current-driven negative resistor.

However, if an internal resistance has to be compensated, it is impossible to connect the op-amp "sense" to the left side of the internal "resistor" as the internal resistance is distributed inside the source. Or, if a distant positive resistance has to be compensated, it is inconvinient to stretch an additional third wire.

Two-terminal current-driven negative resistors use a clever trick to solve these problems (Fig. 14): as the existing positive resistor Ri is not accessible or it is not at a short distance, they use an additional positive resistor R with an equivalent resistance (a duplicate of the unaccessible resistor Ri). Since the current passing through the two series connected resistors is the same and they have equal resistances, the voltage drops across them are also equal. In this way, current-driven negative resistors use the common current to measure the voltage drop across the unaccessible or remote positive resistor (the current loop interface exploits the same idea).

However, the additional voltage drop VR across the duplicate resistor R is disturbing and has to be compensated as well. Therefore, the compensating voltage has to be two times higher; so, the doubling voltage source BH produces a compensating voltage VH= 2Ri.I. A half of this voltage compensates the voltage drop across the internal resistance Ri; the rest half compensates the voltage drop across the duplicate resistor R. As a result, only the load resistance RL remains in the circuit and the load voltage VL stays equal to the input voltage.

You may think of the duplicate resistor R as a current-to-voltage converter that converts the flowing current I into a proportional "copy" voltage drop VR = Ri.I, which drives the compensating voltage source BH (an amplifier with K = 2). From this viewpoint, this kind of negative impedance element is a current-driven voltage source consisting of a current-to-voltage converter and a voltage amplifier.

Note the overall voltage across the composed negative resistor -R has an opposite polarity to the voltage drop across the internal "positive" resistor R as though the circuit has inverted the initial voltage.

Simple negative impedance elements only inject a portion of energy into circuits equal to the losses in the positive elements while true current-driven negative impedance elements, in order to inject one portion of energy into circuits, absorb one portion and inject two portions of energy. For this purpose, they use an additional positive impedance element (simple negative impedance elements use the existing, accessible positive impedance element). One might say that true current-driven negative impedance elements convert a positive impedance element into a negative impedance one by inverting the initial voltage; so, they behave as negative impedance converters with voltage inversion (VNIC).

##### The op-amp implementation
Fig. 15. How to create an op-amp negative impedance converter with voltage inversion.

In the op-amp implementation (Fig. 15), the duplicate resistor R (below the op-amp) acts as a current-to-voltage converter that converts the flowing current I into a proportional "copy" voltage drop VR = Ri.I. A voltage divider consisting of two equal resistors R (above the op-amp) is connected between the op-amp's output and its non-inverting input. So, the op-amp compares half of its output voltage with the "copy" voltage drop across the duplicate resistor R instead with the "original" voltage drop across the internal resistance Ri and changes it so that to keep (almost) zero difference between them. As a result, the op-amp produces two times higher output voltage than the "copy" voltage drop across the duplicate resistor R = Ri. Half the voltage compensates the voltage drop across the internal resistance Ri; the rest half compensates the voltage drop across the duplicate resistor R. In this exemplary arrangement, the op-amp circuit is "floating"; it has a local, internal ground that is not connected with the common, external ground.

Actually, this op-amp circuit converts the positive resistance R of the duplicate resistor into a negative resistance -R, i.e. it acts as a negative impedance converter (NIC). As the voltage across this kind of negative resistance circuit has an opposite polarity to the voltage drop across the initial "positive" resistor, this converter is named negative impedance converter with voltage inversion (VNIC).

##### Real current-driven true negative impedance element
An IV curve of a real current-driven (S-shaped) true negative resistor.

#### What do current-driven negative resistors actually do?

A current-driven true negative resistor with resistance -R connected in series with a positive resistor with a total resistance RTOT destroys, eats, neutralizes R-part of the total positive resistance thus converting it to zero resistance. Only, in order to have a stability (see below|), some part of the positive resistance has to remain.

#### Stability (operating mode)

Current-driven true negative resistors are circuits with positive feedback where a part of the output quantity adds to the input quantity. The gain of the feedback loop is proportional to the ratio between the negative resistance RN and positive resistance RP. In order to have a stability (to operate in active mode), we need the positive resistance to dominate over the negative one (RN/RP < 1). For the op-amp INIC from Fig. 15 this means: Ri/(Ri + R) > R/(R + R) = 1/2. Otherwise, the circuit will operate in bi-stable (memory) mode.

### Voltage-driven negative impedance elements

#### Voltage-driven negative resistor ("neutralizing" a load resistance)

##### The problem
Fig. 16a. If there is no load connected, the voltage divider is perfect.

In nature, real sources (motors, beings, etc.) have a limited power. If they are not loaded, they behave perfectly. But if they are loaded (for example, if we try to raise a big weight), they droop.

A similar problem exists in electronics (electricity) when imperfect voltage sources are loaded. A more concrete example is the simplest varying voltage source on Fig. 16 that consists of a steady voltage source V and a potentiometer P (a voltage divider r1-r2). If there is no load connected (Fig. 16a), this real voltage source works well - VOUT = r2/(r1 + r2).

Fig. 16b. If a voltage divider is loaded by a heavy "positive" load, it droops.

However, when a load RL is connected (Fig. 16b), it "sucks" a current IL and the output voltage VL drops.

We may generalize the problem if we complicate a bit the dual current-supplied electrical circuit shown on Fig. 2a by adding another element with positive impedance PE2 (see Fig. 18 below). Now it contains two elements with positive impedance connected in parallel: the first element PE1 (the load) is useful; the second element PE2 (e.g., a leakage resistance, a stray capacitance, a voltmeter internal resistance, etc.) is undesired. Well, what do we do to remove the disturbance?

##### The basic idea of voltage-driven negative impedance

The classic remedy is to connect a voltage follower (a unity-gain amplifier acting as a buffer amplifier) before the load, in order to decrease the current IL (to increase the load resistance RL). Unfortunately, this solution introduces some errors inherent for this circuit [2]. Then let's look for a remedy in our routine.

Fig. 17: A powerful idea from mechanics: compensating a weght by an "anti-weight"
Fig. 18. Compensating a disturbing current by a voltage-driven negative impedance element.

What can we do in real life when some object (being, machine etc.) supplied by a real power source droops? We can just help it. For this purpose, we usually use an additional power source, which "helps" the main source by compensating the losses caused by the load. For example, if someone has to raise a heavy loaded cage, we can help it by an equivalent "anti-weght". This is the well-known powerful idea of mechanics named anti-weight or anti-load that is widely used in the lift systems, cranes etc. (Fig. 17)

According to this powerful "neutralization" idea (also mentioned in the introduction), the positive impedance of the undesired element can be eliminated (can be made infinite) by connecting in parallel an additional voltage-driven element NE with the same negative impedance. How does it happen?

A current flows through the undesired positive impedance element PE2 (Fig. 18) that depends on the voltage across it in a some definite way (linear, non-linear or time-dependent). In order to eliminate this disturbing current, the same current (depending in the same way on the voltage) has to be produced by the compensating negative impedance element NE. As a result, the disturbing element PE2 will not consume any current from the input source; the compensating negative impedance element will provide all the needed current to PE2.

##### The basic electrical circuit

As this idea is so wonderful then let's realize it[3]. How do we create the needed voltage-driven negative resistor? We can use various building "scenarios" to do it, let's begin...

Fig. 19. A voltage-driven negative resistor "neutralizes" the load influence.

Scenario 1. In order to make a current-driven negative resistor, we have produced voltage that is proportional to the current flowing through it. Now, in order to make a voltage-driven negative resistor, we have to do the opposite - to produce current that is proportional to the voltage across it. For this purpose, we connect in series a "helping" voltage source (the output of an amplifier) and a "positive" resistor R acting as a voltage-to-current converter (Fig. 19). The voltage source has to keep the same voltage VR across the resistor as the voltage VL across the load; that means it to produce two times higher "helping" voltage VH= 2VL.

Scenario 2. If a voltage VL is applied across the "positive" load resistor RL, it will consume a load current IL = VL/RL. Conversely, if we apply the same voltage VL across an identical negative resistor -R with resistance RL, it has to produce the same current IH = VL/RL = IL. So, we have to "lift" the right end of the resistor (originally connected to ground) with voltage VL toward its left end that is connected to the load. For this purpose, we connect a compensating voltage source BH (a non-inverting amplifier with K = 2) in series with the "copy" positive resistor R having the same resistance as the "original" positive resistor RL (Fig. 19).

The voltage source makes a current IH = (VH - VL)/R = (2VL - VL)/RL = VL/RL = IL that is equal to the load current IL flow through the load. In this way, the whole load current IL is provided only by the "helping" current source IH (the negative resistor -RL) instead by the real input voltage source. The load does not consume any energy from the input source since it is supplied completely by the "helping" source. Figuratively speaking, the load "pulls" the point A down toward the ground while the resistor R "pulls" the point A up toward the voltage VH. As a result of this "stretching", the point A experiences "weightlessness" (as it pulls itself up) and it follows easily the point B. There is no current flowing through the "bridge" connecting the point B and point A since the whole right part of the circuit (RL, R and VH) behaves as a load with infinite internal resistance. This is the well-known phenomenon of bootstrapping and it is put in practice for the first time by Baron Munchhausen (the legend says that he was using his own boot straps to pull himself out of the sea:)

Note the current flowing through the composed voltage-driven negative resistor -R has an opposite direction to the current flowing through the initial "positive" resistor R as though the circuit has inverted the initial current.

##### An implementation by...
Fig. 20. Creating a voltage-driven negative resistor by an amplifier.
###### ...a fixed gain amplifier...

We need a doubling voltage source; a non-inverting amplifier A having a gain of (only) +2 can act as such a "helping" voltage source (Fig. 20). We have just to connect the amp's input to the point A and the amp's output in place of the "helping" voltage source BH. The amplifier is single-supplied since here the input voltage is only positive.

The amplifier doses the voltage +V of the power supply, in order to produce the voltage needed (VA = 2VR). Actually, the steady voltage source +V and the amplifier A constitute the varying voltage source needed. The combination of this composed voltage source and the resistor R acts as a "helping" current source. It injects a current IH through the resistor R into the point A and raises its voltage; as a result, the point A "pulls itself up". The output voltage affects the input voltage as a part of the output voltage adds to the input voltage. This great phenomenon is referred to as positive feedback.

Fig. 21. Creating a negative impedance converter with current inversion (INIC) by an op-amp.
###### ...an op-amp amplifier with negative feedback

In electronics, we realize such amplifiers with fixed gain (in this case we need G = 2) by operational amplifiers. There are perfect op-amps having extremely large but unstable voltage gain (typically 200000). By applying a negative feedback we can make an op-amp amplify exactly two times needed. How do we do this magic?

Negative feedback systems have a nice feature to reverse the causality in electronic circuits. For example, if we put a passive circuit (an integrator, differentiator, attenuator, etc.) into the feedback loop, we will obtain the opposite active circuit (a differentiator, integrator, amplifier, etc.) According to this idea, let's build a voltage divider having a ratio 0.5 by connecting in series two equal resistors R1 and R2; then, let's connect it between the op-amp's output and the inverting input. As a result, we obtain an op-amp non-inverting amplifier having the stable gain of 2 needed.

Actually, this op-amp circuit converts the positive resistance R of the duplicate resistor into a negative resistance -R, i.e. it acts as a negative impedance converter (NIC). As the current flowing through this kind of negative resistance circuit has an opposite direction to the current flowing through the initial "positive" resistor, this circuit is named negative impedance converter with current inversion (INIC).

##### Real voltage-driven true negative impedance element
An N-shaped IV curve of a real voltage-driven true negative resistor.

When the op-amp output voltage approaches supply rails, it stops changing as the op-amp saturates and begins acting as an ordinary constant voltage source. The "magic" of negative resistance ceases.

##### What do voltage-driven negative resistors actually do?

A voltage-driven true negative resistor with resistance -R connected in parallel to a positive resistor with a total resistance RTOT destroys, eats, neutralizes R-part of the total positive resistance thus converting it to infinite resistance. Only, in order to have a stability (see below), some portion of negative resistance has to remain.

##### Stability (operating mode)

Voltage-driven true negative resistors are also circuits with positive feedback where a part of the output quantity adds to the input quantity. Here, the gain of the feedback loop is proportional to the ratio between the positive resistance RP and negative resistance RN. So, in order to have a stability (to operate in active mode), now we need the negative resistance to dominate over the positive one ( RP/RN < 1). For the op-amp INIC from Fig. 21 this means: RL/(R + RL) < R2/(R1 + R2). Otherwise, the circuit will operate in bi-stable (memory) mode.

#### Voltage-driven negative capacitor ("neutralizing" a stray capacitance)

So far we have been using linear ohmic resistors as initial, passive elements with "positive" resistance to make dual active elements with negative resistance (current- and voltage-controlled ones). But with the same success we can transmute every non-linear "positive" resistor into a negative one (e.g., a diode into a negative diode). Finally, by using the same technique, we can create various time-dependent elements with negative impedance, e.g. a negative capacitor.

The concept of negative capacitance is abstract enough; so, let's consider a typical application - "neutralizing" a stray capacitance by a negative capacitance. Although this brilliant idea is proposed as far back as in early 60s[4] maybe we might find its origin in Armstrong's radio times. It seems paradoxical but there are not still clear, simple and intuitive explanations of the capacitive "neutralization" idea. So, it is worth unveiling the mystery of this clever trick.

Negative capacitors are AC circuits driven by sine wave input voltage. In order to really understand how they exactly operate, we will show what the voltages are and where the currents flow in the circuits at one given (arbitrary chosen) moment of the sine wave. So, think of the pictures of voltage bars and current loops superimposed on the figures below as kinds of snapshots.

##### The problem caused by the stray capacitance

Imagine a sine wave generator with output resistance RIN drives a load with infinite input resistance (Fig. 22a). As no current flows through the resistance there is no voltage drop across it and, as a result, the output voltage is equal to the input one (VOUT = VIN).

Fig. 22a: A real voltage source working at ideal load conditions (there is no stray capacitance).
Fig. 22b: A real voltage source affected by a stray capacitance.

If the load has a significant stray capacitance CSTR, it constitutes (in conjunction with the resistance RIN) an integrating circuit (Fig. 22b). As a result, the output voltage begins lagging and thus differing from the input one.

The problem is that the capacitor draws a current from the input source; it is a passive element that absorbs energy from the exciting electrical source and accumulates the "stolen" energy into itself (see above).

##### The basic electrical circuit
Fig. 23: "Neutralizing" the stray capacitance by a negative capacitor.

Exactly in the same way as above, we may "neutralize" the positive impedance of the stray capacitance CSTR by connecting in parallel an additional voltage-driven negative capacitor with the same but negative impedance. While the ordinary "positive" capacitor consumes energy from the input source (it is a load); the negative capacitor does the opposite - it injects energy into the circuit (it is a source). Speaking more concrete, while a series connected "positive" capacitor detracts a voltage drop from the input voltage, a current-driven negative capacitor adds voltage to the input voltage (it is a voltage source); while a parallel connected "positive" capacitor "sucks" current, a voltage-driven negative capacitor produces current (it is a current source). As above, a current flows through the undesired stray capacitance (Fig. 23) that is a differential of the voltage across it. In order to eliminate this disturbing current, the negative capacitor has to produce the same current (depending in the same way on the voltage through time). As a result, the stray capacitance will not consume any current from the input source; the negative capacitor will provide all the current needed to charge the stray capacitance. It is wonderful but yet... how do we make a negative capacitor?

We may use the same trick as above - to convert a "positive" capacitor into a negative one. For this purpose, we connect a "helping" voltage source VH = 2.VSTR (a non-inverting amplifier with K = 2) in series with a "positive" capacitor C having the same as the stray capacitance CSTR. The voltage source makes a current IH flow that is equal to the current IC flowing through the stray capacitance. In this way, the whole current IC is provided only by the "helping" voltage source VH instead by the real input voltage source. The load does not consume any energy from the input source since it is supplied completely by the "helping" source. The input voltage source works at ideal load conditions; it has the "feeling" that there is not a capacitive load connected and the output voltage VSTR is equal to the input one VIN. The situation is exactly as it is shown on Fig. 22a.

Where do we take the output voltage from?

We may use, as usual, VSTR as an output voltage (OUT1 serves as an output). Only, if the load has some resistance RL, it will constitute a voltage divider with the internal resistance RIN and the output voltage will droop - VOUT = VIN.RL/(RL + RIN (note that the negative capacitor compensates only the stray capacitance; it does not compensate the load resistance). But we have the unique possibility to use the compensating voltage VH = 2VSTR as an output voltage (OUT2 serves as an output)! As a result, the stray capacitance will consume energy from the helping voltage source instead from the input voltage source and the output voltage will not droop. In addition, it will be amplified two times (whether we wish it or not).

##### Op-amp implementation
Fig. 25: "Neutralizing" the stray capacitance by an INIC (an extract from page 8 of an original paper).
Fig. 24: "Neutralizing" the stray capacitance by an INIC.

The op-amp implementation of a negative capacitor (Fig. 24) is similar to the op-amp circuit of a negative resistor (Fig. 21) with only one difference - a capacitor C is connected instead the resistor R. As above, the op-amp and the voltage divider (the resistors R1 and R2) constitute a non-inverting amplifier with gain of 2 that serves as a compensating voltage source.

Finally, let's look at the scanned image on Fig. 25 giving thanks to pioneers. It is an extract from page 8 of the remarkable genuine paper written by Dan Sheingold in the enthusiastic issue The Lightning Empiricist of Philbrick Researches in the distant 60's. As you can see, the basic idea behind this exotic circuit solution is thoroughly hidden there...and it was staying hidden as many as 45 years...and we have finally managed to reveal it relying only on our human intuition and common sense!

## References

1. How to compensate resistive losses by series connected negative resistors
2. Negative-resistance load canceller helps drive heavy loads shows a typical voltage-driven negative resistance application.
3. How to compensate resistive losses by parallel connected negative resistors
4. Impedance and admittance transformations using operational amplifiers is a genuine source from Philbrick Reserches (written by D. H. Sheingold).