“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:

\(

\begin{equation}

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

\end{equation}

\)

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:

\(

\begin{equation}

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

\end{equation}

\)

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:

\(

\begin{equation}

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

\end{equation}

\)

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

\(

\begin{equation}

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

\end{equation}

\)

\(

\begin{equation}

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

\end{equation}

\)

\(

\begin{equation}

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

\end{equation}

\)

\(

\begin{equation}

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

\end{equation}

\)

\(

\begin{equation}

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

\end{equation}

\)

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

\(

\begin{equation}

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

\end{equation}

\)

\(

\begin{equation}

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

\end{equation}

\)

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:

\(

\begin{equation}

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

\end{equation}

\)

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.

**References and Additional Reading**

Description | Reference |
---|---|

Negative Impedance Converters | Negative Impedance Converter, Wikipedia |

Use of NIC as an active load | Negative Resistor Cancels Op Amp Load, Maxim Application Note 1868 |

Chua chaotic oscillator | Improved Implementation of Chua’s Chaotic Oscillator Using Current Feedback Op Amp, A.S. Elwakil & M.P. Kennedy |

The use of impedance converters in active filters | The Filter Wizard issue 18: Gee, I see! The Ins and Outs of Generalized Impedance Converters, Kendall Castor-Perry |