GIC to avoid shunt inductors in LC ladder

A generalized impedance converter (GIC) is an active two-port, usually Antoniou's two-op-amp, five-impedance circuit. Its input impedance is the load impedance $Z_5$ multiplied by a conversion function $k(s) = Z_1Z_3/(Z_2Z_4)$. By choosing which positions hold capacitors, a GIC can make an inductor or an FDNR from only R, C and op-amps.

GIC circuit and input impedance

   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 $V_1 = V_3 = V_5 = V_{in}$.

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

Avoiding shunt inductors

Put a capacitor in position 4 ($Z_4 = 1/sC$) and resistors elsewhere:

$$Z_{in} = \frac{R_1 R_3 R_5}{R_2}\, sC = sL, \qquad L = \frac{C R_1 R_3 R_5}{R_2}$$

With all resistors equal to $R$: $L = CR^2$. One end of this inductor is ground, so it replaces a grounded (shunt) inductor.

In an LC ladder the shunt inductors (for example in a highpass ladder, where every inductor goes from a node to ground) are exactly such grounded inductors. The procedure is:

1. Obtain the passive ladder (prototype values, then frequency and impedance scaling).
2. Keep the resistors and capacitors as they are.
3. Replace each shunt inductor $L$ by a GIC: choose $C$ (e.g. 10 nF) and $R = \sqrt{L/C}$.

 o--[Rs]--||--+--||--+--||--o Vo
              |      |      |
            [GIC]  [GIC]   [RL]
              |      |      |
 GND ---------+------+------+

Example: a shunt inductor of 0.1 H with $C$ = 10 nF needs $R = \sqrt{0.1/10^{-8}}$ = 3.162 kΩ in positions 1, 2, 3 and 5.

Advantages: no bulky, lossy inductors; the ladder's low sensitivity is kept; elements are easy to integrate. Floating (series) inductors cannot be replaced by one grounded GIC; for those, use two GICs or the Bruton/FDNR method.