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I have a sensor that gives me an output between 1 and 4V, and I wanted to filter out noise present on the output to give me a more stable and accurate reading.

I set up an active 2nd-order Butterworth filter using a Sallen-key topology to try to filter out any signals above ~42Hz with 0dB gain. I'm pretty sure I've constructed the circuit correctly since it's only 5 components, but the output is remaining at 2.7V regardless of input which I've been varying. What might be an issue, is that the rail to rail voltage of the op-amp is 0-5V, the same supply used for the sensors. The Op amp I'm using is an LT1013, which I've checked is a single supply op-amp.

It's been a while since I've done any analogue design, so I'm a little rusty... Any idea where my problem might lie here?

enter image description here

Component values
R1 = 0
R2 = 10k
C1 = 220nF
C2 = 470nF

Sensor datasheet. From what I can gather the output impedance it 10kOhms and 20nF.

The output from the filter will feed into an ADC (MCP3202).

EDIT: I tried to take the output impedance into account, using it as my R1 value and adjusting the other component values accordingly. However the problem persists mostly, but I am now getting a very slight movement in the output, though no where near the magnitude of the input.

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    \$\begingroup\$ Seriously!? How do you imagine we are supposed to know what R1, R2, etc are in your circuit? Show the schematic of exactly how you hooked this up. Also, what is the output impedance of the sensor? What is the input impedance requirement of whatever is downstream from this filter? \$\endgroup\$
    – Olin Lathrop
    Apr 8, 2013 at 11:51
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    \$\begingroup\$ Well a Sallen-key topology is fairly well known for those who have designed active filters before, but I guess I could have been clearer with my information. Updated the post to include more info now. \$\endgroup\$
    – Sensors
    Apr 8, 2013 at 12:01
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    \$\begingroup\$ The LT1013 has a non-standard pinout and a few obsolete packages. Which package are you using, and have you checked you are using the correct pin diagram? \$\endgroup\$
    – apalopohapa
    Apr 8, 2013 at 17:14
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    \$\begingroup\$ I don't think the output impedance is 10k: the datasheet says "Analog Resistive Output Load" and lists 10k as minimum. In other words, it mustn't drive a load stronger (lower) than 10k. So you shouldn't be omitting R1. \$\endgroup\$
    – apalopohapa
    Apr 8, 2013 at 17:34
  • \$\begingroup\$ I'm using the datasheet provided from the same page that I ordered it from, but I'll check to make sure. Also, I wasn't sure of that myself, it was just the closest piece of info I found that seemed like output impedance. I've never come across a characteristic stated like that before. Thanks, I'll redesign with some adjustments to the filter resistance. \$\endgroup\$
    – Sensors
    Apr 9, 2013 at 8:16

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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: -

enter image description here

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    \$\begingroup\$ Thanks for the information, very informative indeed. I tried what you suggested, but my output is still remains at around 50-60% of the supply voltage regardless of variations in the input voltage between 1-4v. I've tested the op-amp as a voltage follower too and that seems to work as expected, so it's not the op-amp \$\endgroup\$
    – Sensors
    Apr 9, 2013 at 10:25
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    \$\begingroup\$ The sensor has a 10Hz low pass bandwidth and it probably needs setting up via the SPI bus. Are you aware of this? Maybe also you should post your exact circuit diagram including pin numbers of the op-amp and what supplies you are using for it. \$\endgroup\$
    – Andy aka
    Apr 9, 2013 at 10:33
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    \$\begingroup\$ What are you measuring your output with - do you have an oscilloscope to view the waveforms? Do you have a signal generator to hand? \$\endgroup\$
    – Andy aka
    Apr 9, 2013 at 10:35
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    \$\begingroup\$ You say the circuit works as a voltage follower. The circuit you have is a voltage follower for DC voltage. The sensor spec says it produces a quiescent output of half Vsupply which is 2.5V for a 5V supply - can you determine that this is true? If it is true, is the output from the op-amp also at 2.5V quiescent - it should be. Your circuit diagram: are you definitely sure you have no gain built into your op-amp - this is typically achieved by a resistor in place of the short from the output back to -V input. Can you confirm this or post a "true" circuit? \$\endgroup\$
    – Andy aka
    Apr 9, 2013 at 10:45
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    \$\begingroup\$ @Kureigu I'm glad to help and if you think this exercise was useful to you then please consider selecting this answer as "accepted". I think you press a star button somewhere close to the up-down vote counter next to the top of my answer. \$\endgroup\$
    – Andy aka
    Apr 9, 2013 at 11:26

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