---
title: "FDNR to avoid bulky inductors"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:fdnr-avoid-bulky-inductors"
source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-avoid-bulky-inductors
site: BCT Engineering
---

# FDNR to avoid bulky inductors

Filter Design · Chapter 8: Design of High-Order Active Filters

- Asked in: 2071 Chaitra, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-avoid-bulky-inductors
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is Frequency Dependent Negative Resistor? How can it be used to avoid bulky inductors in the design of your circuits? Explain with suitable examples.

## Short answer

A frequency-dependent negative resistor (FDNR) has impedance 1/(s²D), a negative resistance falling as 1/ω². To avoid inductors, apply the Bruton transformation (divide all impedances by s): inductors become resistors, grounded capacitors become grounded FDNRs realised by GICs with two capacitors. Example: a 10 krad/s third-order Butterworth filter replaces 1 H inductors with 10 kΩ resistors and GIC FDNRs.

## Answer

#### Frequency dependent negative resistor

A **frequency dependent negative resistor (FDNR)** is a one-port with impedance $Z(s) = \dfrac{1}{s^2D}$. At $s = j\omega$, $Z = -\dfrac{1}{\omega^2 D}$: a real, negative resistance whose size falls as $1/\omega^2$. $D$ (unit F²Ω) is its constant. It cannot be made from passive elements; a GIC with two capacitors realises it.

#### Avoiding bulky inductors

Inductors for audio and low frequencies are large, heavy, lossy (low Q), pick up magnetic noise and cannot be put on an IC. The FDNR method removes them:

1. Start from the doubly terminated LC ladder (it has low sensitivity to component changes).
2. Apply the **Bruton transformation**: divide every impedance by $s$. Then $R \to C = 1/R$, $L \to R = L$, $C \to$ FDNR $D = C$. The voltage transfer function does not change.
3. In a lowpass ladder all inductors are in series arms, so they all become ordinary resistors. The shunt capacitors become **grounded** FDNRs.
4. Realise each FDNR with one GIC: capacitors in positions 1 and 5, resistors in 2, 3, 4, giving $D = C^2R$.
5. Add large resistors across the terminating capacitors for a DC path.

```text
   Iin  node 1
 o-->-----+--------------(+) A1
 Vin      |
         [Z1]
          +--- node 2 ---- output of A1
         [Z2]
          +--- node 3 ---- (-) A1 and (-) A2
         [Z3]
          +--- node 4 ---- output of A2
         [Z4]
          +--- node 5 ---- (+) A2
         [Z5]
          |
         GND
```

#### Example

Third-order Butterworth LPF at $\omega_0$ = 10,000 rad/s. Prototype: $R_s = 1$, $L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H, $R_L = 1$. Without FDNR the scaled inductors (with $k_z = 10^4$) would be $L = k_zL/\omega_0$ = 1 H each, which is bulky.

With FDNR ($C_t$ = 10 nF, $k_z = 10^4\ \Omega$):

| Element | Bruton | Final |
|---|---|---|
| $R_s$ | 1 F | 10 nF (with $R_a$ = 1 MΩ across it) |
| $L_1$ = 1 H | 1 Ω | 10 kΩ |
| $C_2$ = 2 F | $D$ = 2 | GIC: C = 10 nF, $R_D$ = 20 kΩ |
| $L_3$ = 1 H | 1 Ω | 10 kΩ |
| $R_L$ | 1 F | 10 nF (with $R_b$ = 1.02 MΩ across it) |

The 1 H inductors are replaced by 10 kΩ resistors and one GIC.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why are inductors undesirable in low-frequency filters?
   A. They are large, heavy, low-Q and cannot be integrated on an IC
   B. They have no losses
   C. They only work at DC
   D. They are too cheap to use
   Answer: A. Low-frequency inductors are bulky, lossy, pick up magnetic noise and cannot be put on a chip. (Key point: Problems with inductors)

2. Under the Bruton transformation, what does a capacitor C become?
   A. A resistor 1/C
   B. An inductor C
   C. An FDNR with D = C
   D. A capacitor 1/C
   Answer: C. 1/(sC) divided by s is 1/(s^2 C), an FDNR with D = C. (Key point: Capacitor under Bruton)

3. Why start from a doubly terminated LC ladder?
   A. It needs no terminations
   B. It has the lowest sensitivity to element values
   C. It has the fewest elements
   D. It is the only ladder with inductors
   Answer: B. Doubly terminated LC ladders have the lowest sensitivity, and that is kept after the transformation. (Key point: Starting ladder)

4. In the 3rd-order example (kz = 10^4 Ω), what replaces each 1 H inductor?
   A. A 10 mH inductor
   B. A 10 nF capacitor
   C. A 20 kΩ GIC
   D. A 10 kΩ resistor
   Answer: D. Each inductor L becomes a resistor L x kz = 1 x 10 kΩ, and the shunt 2 F becomes one GIC FDNR. (Key point: Example result)

## Related in Filter Design

- [FDNR to avoid inductors; design from Figure 1 at 10⁴ rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-figure1-ladder-1e4) — Filter Design, chapter 8
- [FDNR to avoid inductors; 3rd order LPF at 4000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-third-order-4000) — Filter Design, chapter 8, asked 2×
- [GIC to avoid shunt inductors in LC ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Agic-avoid-shunt-inductors) — Filter Design, chapter 8
- [FDNR LPF at 10⁴ rad/s from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-lpf-1e4-from-ladder) — Filter Design, chapter 8