Input Impedance of a Low Pass Filter... - Page 1
Forum section: electronics / beginners

author: Shredhead
author_rank: Regular Contributor
timestamp: June 09, 2016, 05:41:33 pm
content: I'm having no success in finding any information about how to figure out what the input impedance is of an active 2nd order low pass filter like the attached schematic.Would the input impedance change if the values of R3&R4 were changed, -for example if I were to use a dual pot for these to make the filter's frequency adjustable?Thanks.

author: Yansi
author_rank: Super Contributor
timestamp: June 09, 2016, 05:54:52 pm
content: You can make a quick analysis yourself, if you know how to compute with complex impedances.Such filter as drawn on the image has very imbalanced component values. Those resistors should be increased in resistance one order of magnitude and the caps decreased the same amount.There is a whole writing on this topic on wikipedia too:https://en.wikipedia.org/wiki/Sallen%E2%80%93Key_topology#Input_Impedance

author: TimFox
author_rank: Super Contributor
timestamp: June 09, 2016, 07:08:06 pm
content: Assuming that you have a Spice model corresponding to that drawing, "measure" the current through the first resistor (or the voltage across it) with a nominal 1 V voltage AC source driving the circuit. The real and imaginary components of that current give the real and imaginary components of the input admittance, which is the reciprocal of the impedance. The easiest way to convert that to impedance is to use magnitude and phase of the admittance: the impedance's magnitude is the reciprocal of the magnitude and the impedance's phase is the negative of the phase.

author: The Electrician
author_rank: Frequent Contributor
timestamp: June 09, 2016, 08:27:33 pm
content: You could do it like this:

author: BobsURuncle
author_rank: Regular Contributor
timestamp: June 09, 2016, 08:31:51 pm
content: Zin = R4 + R3||1/sC3 + 1/sC4= R4 + (R3/sC3) / (R3+1/sC3) + 1/sC4= R4 + R3/(1+R3C3s) + 1/sC4

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 09, 2016, 10:19:34 pm
content: Note that such filters are only valid with avoltage input. They are derived in terms of voltages. If the voltage you apply to the input node, doesn't happen to be exactly the voltage you think it should be... that's your problem.The usual solution is to buffer it with a follower.Tim

author: Audioguru
author_rank: Super Contributor
timestamp: June 10, 2016, 01:29:52 am
content: I agree that the resistor values are much too low at only 100 ohms that needs a power amplifier to drive instead of an opamp. If the resistors are 10k ohms then the capacitors can be 10nF instead of 1uF.Because the capacitor values are the same then the response is a droopy Bessel instead of a sharp Butterworth that has the feedback capacitor value double the value of the capacitor to ground.If you use a dual pot then the response will be messed up a bit by their differences because the pots will not be the same. For an opamp to drive this lowpass filter then the resistors should not be less than 1k each.

author: LvW
author_rank: Frequent Contributor
timestamp: June 10, 2016, 08:18:23 am
content: The quality factor of a unity gain equal-component lowpass (Sallen-Key) is Qp=1/(3-Acl) with the closed-loop gain Acl=1.Hence, we have only Q=1/2.This is a very poor lowpass - even worse than a Bessel response which has Q=0.577.

author: danadak
author_rank: Super Contributor
timestamp: June 10, 2016, 03:08:19 pm
content: V4 in simulation appears to be polarity reversed, - should be connected to- pin on part.Regards, Dana.

author: Shredhead
author_rank: Regular Contributor
timestamp: June 11, 2016, 01:24:11 am
content: @The Electrician -Thanks very much for that plot.@Audioguru -I had no idea the same cap values affected Q like that. Is there a source of info that you'd recommend reading on that?@LvW -Would increasing the gain help the Q?@danadak -That is the only way that I could get ltspice to run a simulation. Flipping polarity gives back an error.I'm not sure what to do from a design perspective now because I wanted to make the frequency adjustable with a dual gang pot adjusting it up in value to decrease the frequency of the low pass. It looks like from The Electrician's plot, the worst case impedance would be the value of R3 & R4. If the op amp gets loaded above the cutoff where there is barely any signal passing, how would that affect the chip and the signal out of it?

author: bson
author_rank: Unknown
timestamp: June 11, 2016, 02:07:53 am
content: Keep in mind there is passive coupling between the input and output, and the opamp operates under feedback, so the input impedance will be affected by the load impedance. This is usually characterized as S12, the reverse isolation.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: June 11, 2016, 10:30:11 am
content: Quote from: danadak on June 10, 2016, 03:08:19 pmV4 is -15V, so V4+ is the more negative side and is connected correctly in the sim.