How does this T-network op amp function as a boost filter?

Question

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).

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?

Answer 1

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.