# Circuit Idea/Linear Mode of Current Inversion NIC

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Investigating the Linear Mode of Negative Impedance Converters with Current Inversion

Circuit idea: Making the negative feedback dominate over the positive one.

INIC operating in linear mode is a voltage-driven N-shaped true negative resistor.

## Introduction

### What a INIC operating in linear mode means

In order to make a current-driven negative resistor, we have produced a 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. A negative impedance converter with current inversion (INIC) is such an op-amp implementation of a true negative resistor with N-shaped IV curve. This composed circuit consists of two components (Fig. 1b): a "positive" resistor R and a non-inverting amplifier with gain of 2 that amplifies the input voltage. The amplifier is implemented by the op-amp OA and the two equal resistors R1 = R2 = R constituting a voltage divider with ratio 0.5 (see also the “b” figures below). The whole circuit acts as a "helping" current source that injects a current IOUT through the resistor R into the input circuit and tries to raise its voltage. Thus, the output voltage affects the input voltage (this great phenomenon is referred to as positive feedback). The whole op-amp circuit behaves as a current source producing a current IOUT that is equivalent to the current flowing through the corresponding “positive” resistor R (IOUT = VIN/R). The op-amp as though converts the "positive" resistance R into a negative -R; that is why, this circuit is named negative impedance converter. As the current flowing through the circuit has an opposite direction to the current flowing through the initial "positive" resistor R, it is named negative impedance converter with current inversion.

### How to investigate the linear mode of INIC

INIC, like VNIC, is a two-terminal circuit; so, to give an idea of how it operates, we have to measure its IV curve. For this purpose, we have to apply a varying input voltage across the circuit (the constant voltage source is the ideal input source for INIC acting in linear mode) and to measure the output current flowing between its terminals. There are four typical for the circuit operation points over the IV curve representing the corresponding three circuit states. For each of them, we may draw the corresponding picture of partial IV curve (from beginning to the current point) and the corresponding circuit diagram representing the current state. Let's label the first sequence of pictures with extension "a" and place them on the left; accordingly, to label the second sequence of pictures with extension "b" and place them on the right. Because the purpose of this paper is rather to reveal the basic idea behind this exotic circuit solution than to calculate its parameters a qualitative heuristic approach is used.

Fig. 1a: An N-shaped IV curve of current inversion NIC.
Fig. 1b: Enriching the classic circuit diagram of current inversion NIC by superimposed voltage bars and current loops.

## Investigating the circuit at ideal driving conditions

The ideal input source for an INIC acting in linear mode is the constant voltage source (CVS). If we apply a negative voltage to INIC (Fig. 2a), the input voltage source "pulls down" the bottom end of the resistor R (connected to the non-inverting input) toward the ground. The op-amp decreases its output voltage trying to restore the equilibrium (it "pulls down" the top end of the resistor R toward the negative supply rail).

### Left positive resistance region

#### Negative input voltage, negative output current

From the initial point 0 to point 1 (Fig. 2a), we have applied high enough negative voltage |VIN| > |VSAT-| to INIC so that the voltage of the non-inverting input is more negative than the voltage of the inverting input. As a result, the op-amp stays in negative saturation. The circuit has positive resistance; it behaves as a real voltage source with a constant voltage VSAT- and an internal resistance R (see the equivalent circuit on the left side of Fig. 2b). It is represented by a linear IV curve on Fig. 2a (the part from point 0 to point 1).

Fig. 2a: Scanning the left bottom part of the curve (0-1).
Fig. 2b: Applying a high negative voltage |VIN| > |VSAT-| to INIC.

#### Negative input voltage, zero output current

At point 1 (Fig. 3a), we apply a relatively moderate negative voltage |VIN| > |VSAT-| to INIC so that the voltage of the non-inverting input is more negative than the voltage of the inverting input. As a result, the op-amp continues staying in negative saturation. The circuit has positive resistance; it continues behaving as a real voltage source.

Fig. 3a: Investigating the point 1 of the curve.
Fig. 3b: Applying a relatively moderate negative voltage |VIN| = |VSAT-| to INIC.

#### Negative input voltage, positive output current

From point 1 to point 2 (Fig. 4a), we apply a moderate negative voltage |VSAT-|/2 < |VIN| < |VSAT-| to INIC so that the voltage of the non-inverting input remains more negative than the voltage of the inverting input. As a result, the op-amp continues staying in negative saturation. The circuit has positive resistance; it continues behaving as a real voltage source.

Fig. 4a: Scanning the left top part (1 - 2) of the curve.
Fig. 4b: Applying a moderate negative voltage |VSAT-|/2 < |VIN| < |VSAT-|.

### Middle negative resistance region

#### Negative input voltage, positive output current

The “magic” of negative resistance begins at point 2. From point 2 to point 3 (Fig. 3b), we apply a low voltage |VIN| < |VSAT-|/2 from INIC (Fig. 3a). The op-amp goes out of saturation and begins operating in active region. The voltage of the non-inverting input is almost equal to the voltage of the inverting input. The input voltage decreases continuously and the output current decreases as well. As a result, the op-amp IV curve moves to the right and the crossing operating point slides over a new dynamic IV curve representing the negative resistance. Note 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. The whole circuit behaves as a "helping" dynamic current source that is connected to the common load in parallel and in the same direction with the input current source.

Fig. 5a: Scanning the left middle part (2 - 3) of the curve.
Fig. 5b: Applying a low negative voltage |VIN| < |VSAT-|/2 to INIC.

#### Zero input voltage, zero output current

At point 3 there is no input voltage and the op-amp produces zero output voltage. The whole circuit behaves just as a resistor R. On the graphical presentation, the op-amp IV curve crosses the origin.

Fig. 6a: Investigating the point 3 of the curve.
Fig. 6b: Applying zero voltage VIN = 0 to INIC.

#### Positive input voltage, negative output current

From point 3 to point 4 (Fig. 4b), we begin applying low positive voltage VIN < VSAT+/2 to INIC. The op-amp continues operating in active region and the voltage of the inverting input is almost equal to the voltage of the non-inverting input. The input voltage increases continuously and the output current increases as well. The op-amp IV curve continues moving to the right and the crossing operating point continues sliding over the negative resistance IV curve. The whole circuit continues behaving as a "helping" dynamic current source that is connected to the common load in parallel and in the same direction with the input current source.

Fig. 7a: Scanning the right middle part (3 - 4) of the curve.
Fig. 7b: Applying a low positive voltage 0 < VIN < VSAT+/2 to INIC.

### Right positive resistance region

#### Positive input voltage, negative output current

From point 4 to point 5 (Fig. 8a), we apply a moderate positive voltage VSAT+|/2 < VIN| < VSAT+ to INIC so that the voltage of the non-inverting input remains more positive than the voltage of the inverting input. As a result, the op-amp continues staying in positive saturation. The circuit has positive resistance; it continues behaving as a real voltage source.

Fig. 8a: Scanning the right bottom part (4 - 5) of the curve.
Fig. 8b: Applying a moderate positive voltage VSAT+/2 < VIN < VSAT+) to INIC.

#### Positive input voltage, zero output current

At point 5 (Fig. 9a), we apply a relatively moderate positive voltage VIN = VSAT- to INIC so that the voltage of the non-inverting input remains more positive than the voltage of the inverting input (Fig. 9b). As a result, the op-amp continues staying in positive saturation. The circuit has positive resistance; it continues behaving as a real voltage source.

Fig. 9a: Investigating the point 5 of the curve.
Fig. 9b: Applying a relatively moderate positive voltage VIN = VSAT+ to INIC.

#### Positive input voltage, positive output current

From point 5 to point 6 (Fig. 10a), we apply high positive voltage VIN > VSAT+ to INIC. As the voltage of the non-inverting input is more positive than the voltage of the inverting input, the op-amp stays in positive saturation. The circuit has positive resistance; it behaves as a real voltage source with a constant voltage VSAT+ and an internal resistance R (see the equivalent circuit on the left side of Fig. 10b). It is represented by a linear IV curve on Fig. 10b (the part from point 5 to point 6).

Fig. 10a: Scanning the right top part (5 - 6) of the curve.
Fig. 10b: Applying a high positive voltage VIN > VSAT+ to INIC.