Bruton transformation and FDNR

Bruton transformation

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.

Simulating the inductor problem with FDNR

A lowpass LC ladder has inductors in the series arms. Inductors are bulky, lossy and cannot be integrated. After the Bruton transformation (all impedances divided by $s$):

• each inductor $L$ becomes a resistor $R = L$ (no inductor left),
• each grounded capacitor $C$ becomes a grounded FDNR $D = C$,
• each terminating resistor $R$ becomes a capacitor $1/R$.

So the inductor is "simulated" indirectly: the whole network is converted to R, C and grounded FDNRs with the same voltage transfer function.

Grounded inductors specifically. A grounded inductor $L$ (for example in a highpass ladder) becomes a grounded resistor $R = L$. The series capacitors of such a ladder would become floating FDNRs, which are hard to build, so the Bruton method is used mainly for lowpass ladders, whose shunt capacitors become grounded FDNRs.

 Original:  o-[Rs]-(L1)-+-(L3)-+---o
                        C2     C4 [RL]
 Bruton:    o-[Cs]-[R1]-+-[R3]-+---o
                        D2     D4 [CL]

Realising the FDNR with 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).

Practical points

• Because the terminations become capacitors, there is no DC path; resistors $R_a$ (across the source capacitor) and $R_b$ (across the load capacitor) are added, much larger than the capacitor impedances in the passband, with $R_b = R_a + \Sigma R$ to keep the DC gain at 1/2.
• Only grounded FDNRs are needed, so one GIC per shunt capacitor is enough, and the low sensitivity of the ladder is kept.