GIC simulation of floating inductor

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.

Input impedance of the 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 $V_1 = V_3 = V_5 = V_{in}$.

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

Grounded inductor

With $Z_4 = 1/sC$ and all other positions resistors $R$: $Z_{in} = sCR^2$, i.e. $L = CR^2$. This works only when one end of the inductor is grounded.

Floating inductor using two GICs

A series (floating) inductor, as in a lowpass ladder, has neither end grounded. It is simulated by two identical GICs connected back to back with a resistor $R_x$ between their port-2 terminals:

  a o--[GIC 1]--o x --[ Rx ]-- y o--[GIC 2]--o b
      port1  port2             port2  port1
      (GIC 2 is the same circuit, reversed)

Treat the GIC with $Z_5$ removed as a two-port (port 2 at node 5). From the analysis above, $V_1 = V_2$ and $I_1 = I_2/k(s)$, with $k(s) = Z_1Z_3/(Z_2Z_4)$ (this follows from $Z_{in} = k(s)Z_5$ and $I_2 = V_2/Z_5$). With $Z_4 = 1/sC$ and equal resistors, $k(s) = sCR$. In transmission (ABCD) form:

$$\text{GIC 1: } \begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}, \quad \text{series } R_x: \begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}, \quad \text{GIC 2 reversed: } \begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}\begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 1 & kR_x \\ 0 & 1 \end{bmatrix}$$

This is the matrix of a single series (floating) impedance $Z = k(s)R_x = s\,(CRR_x)$, so between a and b we get a floating inductor

$$L = C R R_x$$

Example: $C$ = 10 nF, $R = R_x$ = 10 kΩ gives $L = 10^{-8} \times 10^8$ = 1 H. The two GICs must be well matched; a mismatch leaves a small unwanted grounded element. (The Bruton/FDNR method is the usual alternative for lowpass ladders.)