FDNR 5th order Butterworth LPF at 10,000 rad/s

FDNR

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 is realised with a GIC containing two capacitors.

Avoiding the inductor

Bruton transformation: divide every impedance of the RLC ladder by $s$ (more generally by $ks$). A voltage transfer function is a ratio of impedances, so it does not change:

In a lowpass ladder the inductors are in the series arms (often floating) and the capacitors are shunt (grounded). After the transformation the inductors become plain resistors and the grounded capacitors become grounded FDNRs, which one GIC can realise. So no inductor is left.

In the Antoniou GIC (two op-amps, five impedances $Z_1$ to $Z_5$ in a chain with $Z_5$ grounded, so that $Z_{in} = Z_1Z_3Z_5/(Z_2Z_4)$), put capacitors in positions 1 and 5 ($Z_1 = 1/sC_1$, $Z_5 = 1/sC_5$) and resistors in 2, 3, 4:

$$Z_{in} = \frac{R_3}{s^2 C_1 C_5 R_2 R_4} = \frac{1}{s^2 D}, \qquad D = \frac{C_1 C_5 R_2 R_4}{R_3}$$

With $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R$: $D = C^2 R$. At $s = j\omega$, $Z_{in} = -1/(\omega^2 D)$, a negative resistance whose value depends on frequency: an FDNR (also called a D-element or supercapacitor).

Design: 5th-order Butterworth LPF, $\omega_0$ = 10,000 rad/s

Prototype from the table: source resistor $R_s = R_1 = 1\ \Omega$, $L_1 = 0.618$ H, $C_2 = 1.618$ F, $L_3 = 2$ H, $C_4 = 1.618$ F, $L_5 = 0.618$ H, load resistor $R_L = R_2 = 1\ \Omega$.

Step 1: Bruton transformation (divide all impedances by $s$): source and load resistors (1 Ω) become 1 F capacitors, each inductor $L$ becomes a resistor of $L$ Ω, each shunt capacitor $C$ becomes a grounded FDNR with $D = C$.

Step 2: Scaling. Frequency scaling factor $\omega_0 = 10^4\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so

$$k_z = \frac{1}{\omega_0 C_t} = \frac{1}{10000 \times 1 \times 10^{-8}} = 10000\ \Omega$$ $$\begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned}$$

Step 3: Element values. Each FDNR is a GIC with $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R_D$, so $D = C^2R_D$. Taking $C = C_t$ gives $R_D = D_{new}/C_t^2 = k_z D$.

Step 4: DC path. The source and load capacitors block DC, and the op-amps in the GICs need a DC bias path. Put $R_a$ across the source capacitor and $R_b$ across the load capacitor. They must be much larger than $k_z$ (the capacitor impedance at $\omega_0$). At DC the FDNRs are open, so the DC gain is $R_b/(R_a + \Sigma R + R_b)$. To keep the prototype DC gain of 1/2, set $R_b = R_a + \Sigma R$:

$R_a$ = 1 MΩ, $\Sigma R$ = 32.36 kΩ, so $R_b$ = 1.032 MΩ.

Final circuit ($C_s \| R_a$ in series from the source, $C_L \| R_b$ as the load, each $D_k$ a GIC FDNR to ground):

Vi o--[Cs||Ra]--[R1]-+----[R3]-+----[R5]-+----o Vo
                     |         |         |
                    [D2]      [D4]    [CL||Rb]
GND ----------------------------------------------

Answer: terminating capacitors 10 nF; series resistors 6.18 kΩ, 20 kΩ, 6.18 kΩ; two FDNRs, each a GIC with C = 10 nF and $R_D$ = 16.18 kΩ; $R_a$ = 1 MΩ, $R_b$ = 1.032 MΩ.