I want to analyze the cutoff frequencies, order and Qs of filters inside a complicated electronic circuit. Instead of solving equations, as a rookie I find using SPICE a lot easier. However, the results so far don't speak for themselves, for example take the band-pass filter:


After connecting the IN and OUT to an AC battery and then running a SPICE simulation in order to graph the frequency and phase response, the results aren't even indicative of a band-pass filter. What needs to be done in order to make the approach work?

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    \$\begingroup\$ Your opamp does not seem to be supplied wit power \$\endgroup\$ – PlasmaHH Jul 13 '16 at 10:28
  • \$\begingroup\$ Are you really connecting OUT to an ac source (=battery?)? \$\endgroup\$ – Chu Jul 13 '16 at 10:43
  • \$\begingroup\$ Agree with previous comments. There is no power to the op-amp. Also, you must not connect the output to anything (or at least make sure it doesn't overload the op amp, like a resistor over several hundred ohms) \$\endgroup\$ – Digiproc Jul 13 '16 at 11:37
  • \$\begingroup\$ @Chu I don't know how else would you run the simulation... \$\endgroup\$ – Tony Jul 13 '16 at 11:42
  • \$\begingroup\$ First, get some power to the op-amp. Then, leave the output disconnected. It has very low impedance, anyway. \$\endgroup\$ – Digiproc Jul 13 '16 at 12:17

Here is how you would set this up in LTspice. Click for full-size.

Filter Response

Note, 0.0047 for C1 and C2 is ambiguous - is that micro-farads? So 4.7nF? I'm assuming so.

You have to look at the LT1007 Datasheet to determine it's properties (is it rail-to-rail, can it operate with the negative supply grounded or do you need to supply it with + and - voltages?, etc.) This part is NOT a rail-to-rail opamp, so must be supplied with +/- voltages, because it's output cannot swing all the way to the power rail voltages.

Design the circuit, press "Run" button, choose AC Sweep, set ranges.

  • \$\begingroup\$ Thank you! The value of the capacitor is a bit of a mystery. According to the schematic: oi66.tinypic.com/nb20jn.jpg, it is indeed 0.0047. The result won't make any sense, but inputting the value of 0.47u gives the right result for center frequency. The bandwidth then still doesn't make sense. Any ideas how to fix? \$\endgroup\$ – Tony Jul 13 '16 at 14:54
  • \$\begingroup\$ The bandwidth is going to be highly dependent on IC26B. Any idea what it is? Also note that there is a 47k input resistor in the schematic. \$\endgroup\$ – rdtsc Jul 13 '16 at 16:13

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