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I'm working on building a seismometer using this geophone. I found a filtering and amplifying circuit as shown below: Geophone Circuit

More readable schematic here

Currently, I'm failing pretty hard at analyzing the circuit (really, really embarrassing considering I went to school for EE). There are some parts of the circuit where I am absolutely stumped.

  1. The first thing I'm curious about is the "feed forward" circuit with C9 connected to R11. Why would there be a connection like this? Actually, I'm really confused as to why the feedback network in opamp 2 is the way it is (with that line of resistors and capacitor)

  2. I'm not sure why the 2.2M resistor, R7 is there. I took this out of the circuit and found that there wasn't any effect to the circuit operation. I'm thinking that resistor is there for stability but stability for what?

  3. Normally I would assume that the other lead for the geophone would be connected to ground but it instead connected to the 2.5 V line. I'm confused to why this is and I feel like it's a really obvious thing.

  4. I want to confirm that because the op amps have a single supply, 2.5 V is applied to the noninverting input of opamp 1 (and inverting input of opamp 2) to provide a bias.

That's all the questions I have. The analysis seems really trivial and admittedly, I'm really just starting out in electronics. I've got to start reading the Horowitz-Hill book pronto!

Thanks for your help.

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  • \$\begingroup\$ What do you want to measure? I can expand in an answer. Geophone frequency response is not good for detecting distant earthquakes and are designed to detect vertical motion - also not so good. If you are going to pound on the ground with a sledge hammer or shotgun slugs and get echos from rock layers, OK. For natural seismic activity you will need gain of a million or so. You can do it with two opamps of gain 1000 each and low pass filtered (cap across the feedback resistor). If you want earthquakes, build a simple seismometer with a period of at least a second. \$\endgroup\$
    – C. Towne Springer
    Commented Jan 14, 2014 at 8:13
  • \$\begingroup\$ I don't expect to catch any distant earthquakes, although the circuit is pretty darn sensitive. The output is clipping with normal footsteps. I plan on just having the geophone pick up cars driving by, people walking/running. That sort of stuff. \$\endgroup\$
    – superkittens
    Commented Jan 14, 2014 at 23:03

1 Answer 1

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Try this, hopefully I haven't done my math wrong: -

enter image description here

  1. If you imagine R11 is inside the op-amp, the feedback components R12 and R13 act to produce a gain of two for the cap C9. This keeps the sallen key filter stable but the bonus is a 10x higher gain where the actual op-amp output is (because R11 is approximately 10x R12+R13)
  2. R7 sets the dc bias for the final stage but as the filter is extremely slow (0.0015Hz) you might not have noticed things sliding towards a power rail due to leakage currents.
  3. The junction of the 2x 220k resistors is the analogue reference point for the circuit therefore to keep noise low the sensor has a return to that point.
  4. Correct.
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