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GIC current relation and grounded FDNR

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.

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.

GIC circuit

The generalized impedance converter (GIC) (Antoniou circuit) uses two op-amps and five impedances in a chain from the input to ground:

   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 (V1V_1, I1I_1) and node 5 as port 2, where the load Z5Z_5 draws current I2I_2 (V2=I2Z5V_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 V1=V2V_1 = V_2. Let VN2V_{N2} and VN4V_{N4} be the voltages of nodes 2 and 4 (op-amp outputs).

Node 5: VN4=V2+Z4I2Node 3: VN2−V1Z2=V1−VN4Z3=−Z4Z3I2Node 1: I1=V1−VN2Z1=Z2Z4Z1Z3 I2\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:

V1=V2,I1=Z2Z4Z1Z3 I2,Zin=V1I1=Z1Z3Z2Z4 Z5=k(s) Z5V_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 (Z1=1/sC1Z_1 = 1/sC_1, Z5=1/sC5Z_5 = 1/sC_5) and resistors in 2, 3, 4:

Zin=R3s2C1C5R2R4=1s2D,D=C1C5R2R4R3Z_{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 C1=C5=CC_1 = C_5 = C and R2=R3=R4=RR_2 = R_3 = R_4 = R: D=C2RD = C^2 R. At s=jωs = j\omega, Zin=−1/(ω2D)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 CC (e.g. 10 nF), then R=D/C2R = D/C^2. Example: D=10−12 F2ΩD = 10^{-12}\ \text{F}^2\Omega with CC = 10 nF gives R=10−12/10−16R = 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).

✓ Answer checked by Claude Sonnet 5.5

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