--- title: "GIC current relation and grounded FDNR" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:gic-grounded-fdnr" source: https://www.bctengineer.com/q/semester_5/filter_design/gic-grounded-fdnr site: BCT Engineering --- # GIC current relation and grounded FDNR Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2072 Chaitra, 5 marks - Video: [Making Frequency Dependent Resistor using GIC Topology.](https://www.youtube.com/watch?v=GGspLuLo8WU) — The Sage of Science - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/gic-grounded-fdnr - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Draw the circuit diagram of a generalized impedance converter. Derive the relationship between input and output current. How can it be used to simulate a grounded FDNR? Explain. ## Short answer A GIC uses two op-amps and five impedances; it transmits voltage unchanged and scales current as I_in = (Z2Z4/Z1Z3) I_out. To simulate a grounded FDNR, place capacitors in positions Z1 and Z5 and resistors in Z2, Z3, Z4; the input impedance is Zin = 1/(s²D) with D = C1C5R2R4/R3. ## Answer #### GIC circuit The **generalized impedance converter (GIC)** (Antoniou circuit) uses two op-amps and five impedances in a chain from the input to ground: ```text 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 ``` #### Relation between input and output current Treat node 1 as port 1 ($V_1$, $I_1$) and node 5 as port 2, where the load $Z_5$ draws current $I_2$ ($V_2 = I_2 Z_5$). Ideal op-amps force node 1, node 3 and node 5 to the same voltage and draw no input current, so $V_1 = V_2$. Let $V_{N2}$ and $V_{N4}$ be the voltages of nodes 2 and 4 (op-amp outputs). $$ \begin{aligned} \text{Node 5: } & V_{N4} = V_2 + Z_4 I_2 \\ \text{Node 3: } & \frac{V_{N2} - V_1}{Z_2} = \frac{V_1 - V_{N4}}{Z_3} = -\frac{Z_4}{Z_3}I_2 \\ \text{Node 1: } & I_1 = \frac{V_1 - V_{N2}}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3}\, I_2 \end{aligned} $$ So the GIC passes voltage unchanged and scales current: $$ V_1 = V_2, \qquad I_1 = \frac{Z_2Z_4}{Z_1Z_3}\, I_2, \qquad Z_{in} = \frac{V_1}{I_1} = \frac{Z_1Z_3}{Z_2Z_4}\,Z_5 = k(s)\,Z_5 $$ #### Grounded FDNR 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 design: choose $C$ (e.g. 10 nF), then $R = D/C^2$. Example: $D = 10^{-12}\ \text{F}^2\Omega$ with $C$ = 10 nF gives $R = 10^{-12}/10^{-16}$ = 10 kΩ. Because port 1 is referred to ground, this FDNR is grounded, which is exactly what the Bruton-transformed lowpass ladder needs (shunt capacitors become grounded FDNRs). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the ideal GIC, how are port voltages and currents related? A. V1 = V2 and I1 = [Z2Z4/(Z1Z3)] I2 B. V1 = V2 and I1 = I2 C. V1 = -V2 and I1 = I2 D. V1 = k V2 and I1 = I2 Answer: A. The GIC passes voltage unchanged and scales current by Z2Z4/(Z1Z3) = 1/k. (Key point: Voltage and current relation) 2. Why are nodes 1, 3 and 5 at the same voltage? A. The load Z5 is zero B. The outputs are grounded C. Ideal op-amp inputs sit at equal voltage and draw no current D. All the impedances are equal Answer: C. Virtual short between op-amp inputs with no input current forces V1 = V3 = V5. (Key point: Why voltages are equal) 3. Which element placement gives a grounded FDNR? A. Capacitor in position 4 only B. Capacitors in positions 2 and 4 C. All resistors D. Capacitors in positions 1 and 5, resistors in 2, 3, 4 Answer: D. Z1 and Z5 capacitive give Zin = R3/(s^2 C1 C5 R2 R4) = 1/(s^2 D). (Key point: FDNR arrangement) 4. At s = jω an FDNR with Zin = 1/(s^2 D) behaves as: A. An inductor B. A capacitor C. A positive resistance independent of frequency D. A negative resistance equal to -1/(ω^2 D) Answer: D. s^2 = -ω^2, so Zin = -1/(ω^2 D), a frequency-dependent negative resistance. 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