Why is component spread a figure of merit for analog filters design?

Question

Sometimes I read that in analog filter design, component spread is a figure of merit. The lower the better.

What is the advantage (or disadvantage) of having a low (high) component spread?

The only limit I see for components value, is that an actual component with that value must exist. But I don't see how component spread could affect this. What's there to consider?

Note: I consider component spread as \$ C_{max}/C_{min} \$ where \$C_{max}\$ is the maximum value of the capacitors in the filter, while \$C_{min}\$ is the minimum value. The same goes for resistors and inductors. I'm not sure if any mixed component spread may be meaningful.

EDIT1

From the comments/answers I would like to precise that:

• I've got passive filters in mind, like an LC ladder structure or any other standard implementation;
• You can neglect any loading effects on the load or any effects of the source connected to the filter. We can just consider any transfer function (i.e. something independent from the source and the load) and then, among the many implementations of the same transfer function, the filter with the lower component spread (defined as above) is the best one. (of course, there are many other figures of merits in a filter, but we want to limit the analysis to this one).

This is how I interpreted the fact that component spread is a figure of merit in filters, but I don't understand the disadvantages of having it high.

Answer 1

Think in terms of tolerance and error propagation.

Your components have finite tolerances that can be 20% or worse when it comes to capacitors. Even worse than that if the capacitors are ceramic and have a DC bias on them.

If you have a large component value spread the largest components will dominate the tolerance of the filter poles and zeros, to the point of making the smaller components irrelevant.

Furthermore, a quick error/sensitivity analysis shows that the lowest sensitivity of the filter to component errors happens when the components are identical in value.

Answer 2

I suppose, you are speaking about active analog filters using integrated amplifiers like opamps ? During the design process (calculating passive parts values), these active elements are treated as ideal. Primarily, this applies to input impedances (assumed to be infinite), output impedances (asumed to be zero) and the gain (assumed to be infinite and frequency independent).

These simplifications will cause, of course, errors (deviation from the desired freqency response). And these deviations are rather small and acceptable (if compared with other error sources like passive parts tolerances) when the "neglection rules" are fullfilled as good as possible - which means: * Resistor values connected to the opamps input nodes must me much smaller than the finite (but neglected) input resistance of the opamp, * Resistor values connected to the opamps output must me much larger than the finite (but neglected) output resistance of the opamp.

These rules can best be met when the resistor values are in the lower or mid kilo-Ohm range. That is the reason we try to use values which are not too small and also not too large. This requirement leads to the term "component spread", which should be as small as possible. In this case, both mentioned (conflicting) requirenments can be sufficiently met at the same time.

As far as capacitors are concerned, there is a similar reasoning. Capacitor values should not be as small as unwanted parasitic capacitances which are in the lower pF range. On the other hand, large capacitors are room consuming and have lower Q values if compared with smaller capacitors. Hence, capacitors in the nano-Fard region are preferred.