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enter image description here

Supposedly this circuit is supposed to be a bass frequency booster.

I've been using this article as a reference, although the circuit they analyze is a different layout than mine.

I apologize if my work is hard to read or follow, but I'm getting something in a (1 - BPF) form for a transfer function (I struggled to get this into anything slightly resembling standard form, so there may be some calculation errors).

enter image description here

Struggled getting this in standard form, so there may be errors

This looks like a notch filter to me, rather than a boost.

The article the original image is from says

A bridged-T feedback network is basically a low-pass filter mixed with a high-pass filter. It's usually a passive filter that creates a notch (cuts the frequencies) but when it's in a feedback loop it becomes a boost.

But I'm not seeing why or how that's the case. They also link to a site with the same T-network form that has the transfer function in a different form than I have mine in, but also concludes that it's a notch filter.

Is someone able to help me understand why the feedback network turns this notch into a boost?

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You're missing the action of the op-amp.

Let's start with your transfer function so far. We'll find out that the notation is a bit unfortunate, but bear with me. We need to do the good ol' op-amp thing where we set \$v_- = v_+\$. The T filter is driven by the op-amp output, and feeds the op-amp input. So your "\$v_o\$" is \$v_-\$. Similarly, your "\$v_i\$" is the op-amp output (hence the comment about unfortunate notation). So let $$H(s) = \frac{V_o}{V_i}$$

Translating to my terms, and using \$V_{out}\$ for the op-amp output, in hopes of avoiding completely hopeless difficulties in notation: \$V_- = V_{out}H(s)\$

Let \$V_+ = V_-\$. Then \$V_+ = V_{out}H(s)\$. Now solve for \$V_{out}\$: $$V_{out}(s) = \frac{V_+(s)}{H(s)}$$

So because the feedback is a notch, the overall circuit boosts.

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  • \$\begingroup\$ Oh... so simple, yet I completely missed that. Thank you! \$\endgroup\$
    – Nick Nagy
    Jan 29 at 2:02
  • \$\begingroup\$ Actually I'm still confused on this. Why is "my vi" the op-amp output? That implies that, in my original T-bridge drawing, if I swapped the labels for vi and vo then I'd find the correct transfer function? But wouldn't that contradict the layout of the T-network in the attached diagram? \$\endgroup\$
    – Nick Nagy
    Jan 30 at 19:09
  • \$\begingroup\$ I clarified in the answer. Your \$V_i\$ is the op-amp output, because the op-amp drives the filter there. In this case swapping the input and output in your math would give the same answer, but only by chance -- you're looking at a feedback loop, with an op-amp in it; you have to take that into account in your computations. \$\endgroup\$
    – TimWescott
    Jan 30 at 19:57
  • \$\begingroup\$ So because Vout is driving the feedback circuit, it functions as the voltage input/source? I'm thrown by the filter itself being symmetrical, and the choice of direction for current flow being arbitrary \$\endgroup\$
    – Nick Nagy
    Jan 30 at 20:05
  • \$\begingroup\$ The direction of signal flow isn't arbitrary -- it's dictated by the fact that the op-amp has well-defined inputs and outputs. Networks of passive components will inherently allow signals to go in either direction, so if one is just sitting there in space you can just designate which is the input or output. That leaves the op-amp to determine which is the filter's input and which is it's output. \$\endgroup\$
    – TimWescott
    Jan 30 at 20:14

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