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GIC simulation of a grounded inductor

Short answer

Antoniou's GIC is a two-op-amp circuit with five impedances giving Z_in = Z1 Z3 Z5 / (Z2 Z4). To simulate a grounded inductor, set Z4 = 1/(sC4) and the others as resistors, yielding Z_in = sL with L = C4 R1 R3 R5 / R2 (or Z2 capacitive with L = C2 R1 R3 R5 / R4).

What is GIC (Antoniou's GIC)? How can a GIC be used to simulate a grounded inductor (in the passive filter)? Explain with necessary figures and derivations.

GIC (Antoniou's GIC)

The GIC (Antoniou's generalized impedance converter) is a two-op-amp circuit with five impedances Z1…Z5Z_1 \dots Z_5 in a chain. Its input impedance is

Zin=Z1Z3Z5Z2Z4Z_{in} = \frac{Z_1Z_3Z_5}{Z_2Z_4}

By choosing which ZZ's are resistors and which are capacitors, it can simulate a grounded inductor or an FDNR.

 Iin -->
 V1 o-----+---------- A1 input
          |
          Z1
          |
          +---------- A1 output (Va)
          Z2
          |
     (3)  +---------- A1 and A2 inputs
          Z3
          |
          +---------- A2 output (Vb)
          Z4
          |
     (5)  +---------- A2 input
          Z5
          |
         GND

Op-amp A1 has its inputs at nodes 1 and 3; A2 has its inputs at nodes 3 and 5.

Derivation (ideal op-amps: virtual short gives V1=V3=V5=VV_1 = V_3 = V_5 = V; no input current):

IZ5=VZ5(flows through Z4 too)Vb=V+Z4VZ5IZ3=V−VbZ3=−Z4Z3Z5V(flows through Z2 too)Va=V+Z2IZ3=V−Z2Z4Z3Z5VIin=V−VaZ1=Z2Z4Z1Z3Z5VZin=VIin=Z1Z3Z5Z2Z4\begin{aligned} I_{Z5} &= \frac{V}{Z_5} \quad\text{(flows through } Z_4 \text{ too)} \\ V_b &= V + Z_4\frac{V}{Z_5} \\ I_{Z3} &= \frac{V - V_b}{Z_3} = -\frac{Z_4}{Z_3Z_5}V \quad\text{(flows through } Z_2 \text{ too)} \\ V_a &= V + Z_2 I_{Z3} = V - \frac{Z_2Z_4}{Z_3Z_5}V \\ I_{in} &= \frac{V - V_a}{Z_1} = \frac{Z_2Z_4}{Z_1Z_3Z_5}V \\ Z_{in} &= \frac{V}{I_{in}} = \frac{Z_1Z_3Z_5}{Z_2Z_4} \end{aligned}

Simulating a grounded inductor

Grounded inductor simulation: make Z4=1sC4Z_4 = \dfrac{1}{sC_4} and Z1,Z2,Z3,Z5Z_1, Z_2, Z_3, Z_5 resistors:

Zin=R1R3R5R2 sC4=sL,L=C4R1R3R5R2Z_{in} = \frac{R_1R_3R_5}{R_2}\,sC_4 = sL, \qquad L = \frac{C_4R_1R_3R_5}{R_2}

(Equally, Z2Z_2 can be the capacitor, giving L=C2R1R3R5/R4L = C_2R_1R_3R_5/R_4.) With all resistors equal to RR, L=R2CL = R^2C. One end of the simulated inductor is ground, so the GIC replaces the shunt (grounded) inductors of an LC ladder, e.g. in a high-pass ladder.

Example: L=0.5L = 0.5 H with C4=0.1 μC_4 = 0.1\ \muF and R1=R2=R3=R5=RR_1 = R_2 = R_3 = R_5 = R: R=L/C=0.5/10−7=2.236 kΩR = \sqrt{L/C} = \sqrt{0.5/10^{-7}} = 2.236\ \text{k}\Omega.

Advantages: the realised inductor has a good Q over a useful frequency range (limited by the op-amp gain-bandwidth); it uses only R, C and op-amps, so the low-sensitivity property of the passive LC ladder is kept.

✓ Answer checked by Claude Sonnet 5.5

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