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Ideal gyrator and floating inductor simulation

Short answer

An ideal gyrator is a lossless two-port with Zin = r^2/ZL. A floating inductor is realized by cascading two identical gyrators with a grounded capacitor between them, yielding L = r^2 C.

What is ideal gyrator? How can you simulate floating inductor using gyrator? Explain with necessary figure and derivations.

Ideal gyrator

An ideal gyrator is a lossless two-port that inverts impedance: a load ZLZ_L on port 2 appears at port 1 as Zin=r2/ZLZ_{in} = r^2/Z_L, where rr is the gyration resistance. Its defining equations are

V1=−rI2,V2=rI1⟹Zin=V1I1=r2ZLV_1 = -r I_2, \qquad V_2 = r I_1 \quad\Longrightarrow\quad Z_{in} = \frac{V_1}{I_1} = \frac{r^2}{Z_L}

So a capacitor CC on port 2 gives Zin=r2sC=sLZ_{in} = r^2 sC = sL with L=r2CL = r^2C. A gyrator is built with op-amps (e.g. two voltage-controlled current sources of opposite sign, I1=gV2I_1 = gV_2, I2=−gV1I_2 = -gV_1, r=1/gr = 1/g).

Floating inductor using gyrators

One gyrator with a grounded capacitor gives only a grounded inductor. A floating inductor is made by cascading two identical gyrators with a grounded capacitor between them:

        +-----------+          +-----------+
 a o----| Gyrator 1 |----+-----| Gyrator 2 |----o b
        |    (r)    |    |     |    (r)    |
        +-----------+   === C  +-----------+
                         |
 GND ---------------------------------------------- GND

Each gyrator and the capacitor share the common ground. The transmission (ABCD) matrix of an ideal gyrator (V1=rI2V_1 = rI_2, I1=V2/rI_1 = V_2/r, with I2I_2 leaving port 2) and of a shunt capacitor are

G=[0r1/r0],YC=[10sC1]G = \begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix}, \qquad Y_C = \begin{bmatrix} 1 & 0 \\ sC & 1 \end{bmatrix}

For the cascade:

G YC G=[0r1/r0][10sC1][0r1/r0]=[rsCr1/r0][0r1/r0]=[1sCr201]\begin{aligned} G\,Y_C\,G &= \begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ sC & 1 \end{bmatrix}\begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix} \\ &= \begin{bmatrix} rsC & r \\ 1/r & 0 \end{bmatrix}\begin{bmatrix} 0 & r \\ 1/r & 0 \end{bmatrix} = \begin{bmatrix} 1 & sCr^2 \\ 0 & 1 \end{bmatrix} \end{aligned}

The matrix [1Z01]\begin{bmatrix} 1 & Z \\ 0 & 1 \end{bmatrix} is that of a single series impedance ZZ between the two terminals. Here Z=s r2CZ = s\,r^2C, so the circuit between a and b is a floating inductor

L=r2CL = r^2 C

Example: rr = 10 kΩ and CC = 10 nF give L=108×10−8L = 10^8 \times 10^{-8} = 1 H. The two gyrators must have equal gyration resistance; any mismatch leaves an unwanted grounded element. The capacitor is grounded, which suits IC realisation.

✓ Answer checked by Claude Sonnet 5.5

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