Bruton transformation; FDNR using GIC

Importance of the Bruton transformation

The Bruton transformation divides every impedance of an RLC ladder by $s$: $R \to$ capacitor $1/R$, $L \to$ resistor $L$, $C \to$ FDNR $D = C$. Because a voltage transfer function is a ratio of impedances, it is unchanged. Its importance:

1. Removes all inductors, including floating (series) ones in lowpass ladders, which are hard to simulate directly.
2. The only active elements needed are grounded FDNRs, each made with one GIC; floating elements are avoided.
3. The design starts from a doubly terminated LC ladder, so the very low sensitivity of the ladder to element changes is kept.
4. The final circuit has only resistors, capacitors and op-amps, suitable for integration and for low frequencies where inductors would be huge.
5. Extra care: a DC path must be added (large resistors across the terminating capacitors).

FDNR using a GIC

   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

Each op-amp has negative feedback (A1 through $Z_2$, A2 through $Z_4$). With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$.

$$\begin{aligned} \text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\ \text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\ \text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in} \end{aligned}$$ $$Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4}$$

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).

Example

Third-order Butterworth LPF (prototype $R_s = 1$, $L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H, $R_L = 1$) at $\omega_0$ = 10,000 rad/s.

1. Bruton: $C_s = 1$ F, $R_1 = 1\ \Omega$, $D_2 = 2$, $R_3 = 1\ \Omega$, $C_L = 1$ F.
2. Choose $C_t$ = 10 nF: $k_z = 1/(10^4 \times 10^{-8}) = 10^4\ \Omega$.
3. Resistors: $R_1 = R_3$ = 10 kΩ. Terminating capacitors 10 nF.
4. $D_2 = 2/(10^4 \times 10^8) = 2 \times 10^{-12}$; with GIC capacitors 10 nF, $R_D = D/C^2 = 2 \times 10^{-12}/10^{-16}$ = 20 kΩ.
5. DC path: $R_a$ = 1 MΩ, $R_b = R_a + 20$ kΩ = 1.02 MΩ.

 Vi o--[10nF||Ra]--[10k]--+--[10k]--+---o Vo
                          |         |
                       [GIC D2] [10nF||Rb]
 GND ---------------------+---------+