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In the pictured schematic, is there a band pass filter in place? I’m having trouble understanding the function of R4, R6, C3 & C4. Any help or resources would be appreciated. Schematic

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If your not fond of simulators, just calculate what the impedance of the capacitors would be at a few different frequencies, as the frequency increases the impedance of both capacitors drop, so the gain of the op amps changes at that frequency.

C3 means it has a high pass filter, as that is reducing the gain of low frequency signals, C4 means it has a low pass filter, as that reduces the gain of higher frequency signals So I would call it a band pass filter.

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  • \$\begingroup\$ So from a signal perspective, is it flowing from the output through to the inverting input or to ground? \$\endgroup\$ – user245187 Apr 16 at 13:07
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    \$\begingroup\$ From a signal perspective, you have 2 impedance groups that set the gain, this gain follows the equation 1+ (Rf / Rin) call R6 + P1 in parallel with C4 Rf, and call R4 in series with C3 Rin, the output is that signal multiplied by the gain from that relationship. for DC, the circuit behaves as a voltage follower due to C3, so gain of 1, at AC, current flows through C3 to ground \$\endgroup\$ – Reroute Apr 16 at 13:13
  • \$\begingroup\$ that makes more sense. Thankyou. \$\endgroup\$ – user245187 Apr 16 at 13:16
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Bandpass filters often has some way to set the peaking, which adjusts the -3dB bandwidth.

This circuit is simpler, since the opamp does its best to make its Vin- always track its Vin+, which cleanly separates the RC groups.

In this circuit, a more meaningful description is: has a LowPass F3dB set by the network from opamp's output to Vin-, and has a HighPass F3dB set by the network from Vin- to Ground.

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  • \$\begingroup\$ Lowpass? Did you overlook the role of C3>>C4 ? \$\endgroup\$ – LvW Apr 16 at 16:09

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