# Negative Impedance Converters

“Negative resistance” may seem like a purely academic concept, but can be easily realized in practice with a handful of common components. By adding a single resistor to a standard non-inverting op amp circuit, we can create a negative impedance converter, which has applications in load cancellation, oscillator circuits, and more.

#### Derivation of NIC Input Impedance

Here we will derive the input resistance of the following Negative Impedance Converter (NIC) and show that it is equal to:

$$\mathbf{R}_{IN} = \mathbf{-} \frac{\mathbf{R}_{1}\mathbf{R}_{3}}{\mathbf{R}_{2}}\\$$

Negative Impedance Converter (NIC)

The applied input voltage, Vin, causes some resulting current, I1. If we can predict I1 for a given Vin, then we can use Ohm’s law to calculate the effective resistance, Rin, seen by the voltage source at Vin.

No current flows into an (ideal) op amp’s inputs, so all of current I1 must pass through resistor R1.

Applying Ohm’s law, the I1 current is equal to the voltage drop across R1 divided by its resistance:

$$\mathbf{I}_{1} = \frac{\mathbf{V}_{IN} – \mathbf{V}_{OUT}}{\mathbf{R}_{1}} \tag{1}\\$$

Knowing that the op amp is in a non-inverting configuration, and assuming an ideal voltage source for Vin, we know that the output voltage Vout is:

$$\mathbf{V}_{OUT} = \mathbf{V}_{IN} (1 + \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}}) \tag{2}\\$$

Substituting equation #2 into equation #1, we can factor Vin out of the numerator and simplify:

$$\mathbf{I}_{1} = \frac{\mathbf{V}_{IN} – \mathbf{V}_{IN} (1 + \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}})}{\mathbf{R}_{1}} \tag{3}\\$$

$$\mathbf{I}_{1} = \frac{\mathbf{V}_{IN} [1 – 1 (1 + \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}})]}{\mathbf{R}_{1}} \tag{4}\\$$

$$\mathbf{I}_{1} = \frac{\mathbf{V}_{IN} [1 – (1 + \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}})]}{\mathbf{R}_{1}} \tag{5}\\$$

$$\mathbf{I}_{1} = \frac{\mathbf{V}_{IN} [1 – 1 – \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}}]}{\mathbf{R}_{1}} \tag{6}\\$$

$$\mathbf{I}_{1} = \frac{-\mathbf{V}_{IN} \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}}}{\mathbf{R}_{1}} \tag{7}\\$$

We can now divide both sides by Vin and simplify the complex fraction on the right-hand side of the equation:

$$\frac{\mathbf{I}_{1}}{\mathbf{V}_{IN}} = \frac{- \frac{\mathbf{R}_{2}}{\mathbf{R}_{3}}}{\mathbf{R}_{1}} \tag{8}\\$$

$$\frac{\mathbf{I}_{1}}{\mathbf{V}_{IN}} = \mathbf{-} \frac{\mathbf{R}_{2}}{\mathbf{R}_{1}\mathbf{R}_{3}} \tag{9}\\$$

Since Ohm’s law defines resistance as voltage divided by current, we just have to flip both sides of the equation to finally arrive at:

$$\mathbf{R}_{IN} = \frac{\mathbf{V}_{IN}}{\mathbf{I}_{1}} = \mathbf{-} \frac{\mathbf{R}_{1}\mathbf{R}_{3}}{\mathbf{R}_{2}} \tag{10}\\$$

This is the input resistance seen looking into the input of the NIC circuit.
If R2 and R3 are made equal to each other, the input resistance is simply equal to -R1.
If R1 and R2 are made equal to each other, the input resistance is simply equal to -R3.