Inductors have a bad reputation as filter components—not only do they transmit electromagnetic interference, but they also act as antennas for receiving EMI signals generated elsewhere.

To avoid these problems, you can simulate an inductor's impedance by combining
two wideband transconductance amplifiers (WTAs) and a capacitor (*Fig.
1a*). The combined circuit then acts as a synthetic inductor (L_{SYN})
with one end connected to ground.

By forcing current at L_{SYN} and measuring the resulting voltage,
you can determine the equivalent impedance Z_{EQ} of the circuit:

Z_{EQ} = ΩC_{SYN}/gm1 × gm2

where gm = transconductance.

The equivalent inductance, therefore, is:

L_{EQ} = C_{SYN} /gm1 × gm2

This single-port network clearly offers the frequency-proportional impedance of an inductor, and the inductance value can be large if gm1 × gm2 is much less than 1. The only limitation is that the network must always connect to ground.

High-pass, all-pole ladder filters make good applications because all of their inductors connect to ground. Two WTAs and a capacitor must be substituted for each one, so you should choose a configuration with the minimum number of inductors.

To be cost-effective, your design should feature a series capacitor at each
end of the filter, with the simulated inductor acting as a shunt between them
(*Fig. 1b*). The input capacitor blocks
any dc applied to the filter, and the output capacitor blocks any dc offset
introduced by the synthetic inductor. Though constructed with active components,
the filter does retain some of the advantages of a passive filter.

In an actual circuit (*Fig. 2*), C_{1}
and C_{3} become bypass capacitors and C_{2} is part of the
simulated inductor. The transconductance for each WTA is set by an external
resistor (R_{1} or R_{3}) according to the relationship gm =
8/R (where R = R_{1} or R_{3}). Because the simulated inductance
depends on the product of these transconductances, it may appear that you have
a range of choices for each. But the optimum circuit for a given application
restricts gm values by allowing the full range of output swing for each WTA.

To determine these optimal gm values, start with equal transconductance and simulate the filter in Spice using "g" elements (voltag-controlled current sources) for the amplifiers. While sweeping the frequency at least one decade above and below the filter's corner frequency, observe each WTA output for its peak voltage magnitude (the two peaks may occur at different frequencies).

At the synthetic inductor's port (pin 13 of IC2), the peak value is demanded by the filter and can't be changed. A real inductor would produce the same peak. Therefore, you adjust the other peak to match. Let K equal the ratio of gm2 to gm1, and then, since gain is proportional to transconductance, you divide gm1 by K and multiply gm2 by K. Finally, rerun the Spice simulation with these new gm values to verify that the peaks are equal and that the filter shape has not changed.

The filter exhibits a maximum attenuation of 58.6 dB/decade. The slope decreases at lower frequency because the synthetic inductor's Q is affected by its series resistance (comparable 1.25-mH inductors also have an appreciable resistance of 53 \[OHM\] or so). At 10 Hz, for instance, the attenuation for an ideal filter is -90 dB. For this circuit, the attenuation is -80 dB.