I am designing low pass active filter (Sallen-Key structure) and I have come to a point where I have to decide what should be the value of C (C1 = C2 = C) and then calculate values of resistances. I know that capacitance can not be very low because then I would have to take into consideration parasitic capacitances of other elements on the PCB. I also know that it is not recommended to take high value of C but I do not know why.
Electronic – Why are high values of capacitance in low pass active filters not recommended
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First, \$\alpha = 2 \zeta\$, \$\alpha\$ (damping constant) is used in filters, and typically \$\zeta\$ is used in controls.
Q is defined as \$ Q = \dfrac {f_o}{\Delta f}\ (= \dfrac{1}{\alpha}) \$. It is a band pass parameter and does not apply to first order filters.
If you want to change the step response, you're going to have to change the damping. In the case of the Sallen-Key, that involves changing the gain of the amplifier. This is where you have to start being careful. Damping is set based off the approximation you chose. As an example, for a Butterworth approximation \$\alpha = 1.414\$. If you want a faster rise time, you'll want a lower damping constant. Look at the Chebyshev =approximations. You do need to be aware that there will be overshoot with those approximations.
Cascading filters does not have the effect you think it does. For example, simply cascading two second order Butterworth low pass filters for example, does not equal a fourth order Butterworth low pass. There are tables of common approximations and the appropriate correction factors that need to be considered when cascading filters.
If you are trying to filter-out signals above 42Hz the sallen-key is fine but you've made serious errors in your calculations. Here are some pointers: -
- R1 is never 0 - it's usually the same value as R2
- C2 is hardly ever greater than C1 - it's usually the same or less
Assuming R1 = R2 = 10k and C1 and C2 (as stated) then the cut-off frequency calculates at 49.5Hz but, to obtain a decent flat pass-band (DC to 49.5Hz) and a reasonable amount of attenuation above this point, C2 is far too large.
Try C2 at 100nF and try lifting R1 and R2 both to 22k. This will be about 48.8Hz.
There are subtleties when designing 2nd order filters and the main one is called Q. Q or quality factor alters the shape of the filter from a rather sloppy pass-band (gradually and progressively attenuating frequencies) to a much sharper well-defined flatter pass-band and ultimately, it can produce big resonant peaks at the cut-off when Q is very large.
The picture below is of a 2nd order mechanical spring-balance system but the picture is good because fundamentally the same formulas apply to electronic circuits and it shows what happens when damping is high and low. Damping is proportional to the inverse of Q just in case you are wondering: -
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Best Answer
High values of C would imply low values of R, therefore high signal currents.
In simple circuits (e.g. emitter followers) when the signal current is a large proportion of the quiescent current, you have to consider the device's nonlinearity (i.e. distortion). In op-amp based circuits, as long as the opamp has sufficient bandwidth and drive capability, this is a lesser concern.
In either case, high signal current means high supply current, i.e. heat and wasted power.
However, go too far in the other direction, and low signal currents mean high resistor values, and higher noise (both shot noise from the lower currents, and Johnson noise from the higher resistances).
The right balance depends on the application.