How to explain the peak frequency and resistance relationship in this op amp circuit?

Well, I am trying to analyze the following circuit (assuming an ideal opamp):

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$ begin{cases} text{I}_3=text{I}_1+text{I}_2\ \ text{I}_3=text{I}_4\ \ text{I}_2=text{I}_5\ \ text{I}_4=text{I}_1+text{I}_5 end{cases}tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ begin{cases} text{I}_1=frac{text{V}_text{i}-text{V}_1}{text{R}_1}\ \ text{I}_2=frac{text{V}_3-text{V}_1}{text{R}_2}\ \ text{I}_3=frac{text{V}_1-text{V}_2}{text{R}_3}\ \ text{I}_4=frac{text{V}_2}{text{R}_4}\ \ text{I}_5=frac{text{V}_text{o}-text{V}_3}{text{R}_5} end{cases}tag2 $$

Substitute $(2)$ into $(1)$, in order to get:

$$ begin{cases} frac{text{V}_1-text{V}_2}{text{R}_3}=frac{text{V}_text{i}-text{V}_1}{text{R}_1}+frac{text{V}_3-text{V}_1}{text{R}_2}\ \ frac{text{V}_1-text{V}_2}{text{R}_3}=frac{text{V}_2}{text{R}_4}\ \ frac{text{V}_3-text{V}_1}{text{R}_2}=frac{text{V}_text{o}-text{V}_3}{text{R}_5}\ \ frac{text{V}_2}{text{R}_4}=frac{text{V}_text{i}-text{V}_1}{text{R}_1}+frac{text{V}_text{o}-text{V}_3}{text{R}_5} end{cases}tag3 $$

Now, Using an ideal opamp, We know that $text{V}_x:=text{V}_+=text{V}_-=text{V}_2=text{V}_3$. So we can rewrite equation $(3)$ as follows:

$$ begin{cases} frac{text{V}_1-text{V}_x}{text{R}_3}=frac{text{V}_text{i}-text{V}_1}{text{R}_1}+frac{text{V}_x-text{V}_1}{text{R}_2}\ \ frac{text{V}_1-text{V}_x}{text{R}_3}=frac{text{V}_x}{text{R}_4}\ \ frac{text{V}_x-text{V}_1}{text{R}_2}=frac{text{V}_text{o}-text{V}_x}{text{R}_5}\ \ frac{text{V}_x}{text{R}_4}=frac{text{V}_text{i}-text{V}_1}{text{R}_1}+frac{text{V}_text{o}-text{V}_x}{text{R}_5} end{cases}tag4 $$

Now, we can solve for the transfer function:

$$mathcal{H}:=frac{text{V}_text{o}}{text{V}_text{i}}=frac{text{R}_2text{R}_4-text{R}_3text{R}_5}{text{R}_1left(text{R}_2+text{R}_3right)+text{R}_2left(text{R}_3+text{R}_4right)}tag5$$

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Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$text{R}_2=frac{1}{text{sC}_1}tag6$$
• $$text{R}_4=frac{1}{text{sC}_2}tag7$$

So, the transfer function becomes:

$$mathcal{h}left(text{s}right)=frac{text{v}_text{o}left(text{s}right)}{text{v}_text{i}left(text{s}right)}=frac{frac{1}{text{sC}_1}frac{1}{text{sC}_2}-text{R}_3text{R}_5}{text{R}_1left(frac{1}{text{sC}_1}+text{R}_3right)+frac{1}{text{sC}_1}left(text{R}_3+frac{1}{text{sC}_2}right)}=$$ $$frac{1-text{C}_1text{C}_2text{R}_3text{R}_5text{s}^2}{text{C}_1text{C}_2text{R}_1text{R}_3text{s}^2+text{C}_2left(text{R}_1+text{R}_3right)text{s}+1}tag8$$

Because we are working with sinusodial signals, we can write $text{s}=text{j}omega$ where $text{j}^2=-1$ and $omega=2pitext{f}$ with $text{f}$ is the frequency in Hertz. So, we can write:

$$underline{mathcal{h}}left(text{j}omegaright)=frac{underline{text{v}}_text{o}left(text{j}omegaright)}{underline{text{v}}_text{i}left(text{j}omegaright)}=frac{1-text{C}_1text{C}_2text{R}_3text{R}_5left(text{j}omegaright)^2}{text{C}_1text{C}_2text{R}_1text{R}_3left(text{j}omegaright)^2+text{C}_2left(text{R}_1+text{R}_3right)left(text{j}omegaright)+1}=$$ $$frac{1+text{C}_1text{C}_2text{R}_3text{R}_5omega^2}{1-text{C}_1text{C}_2text{R}_1text{R}_3omega^2+text{C}_2left(text{R}_1+text{R}_3right)omegatext{j}}tag9$$

Now, we can find the amplitude by finding the absolute value of $(9)$:

$$left|underline{mathcal{h}}left(text{j}omegaright)right|=frac{1+text{C}_1text{C}_2text{R}_3text{R}_5omega^2}{sqrt{left(1-text{C}_1text{C}_2text{R}_1text{R}_3omega^2right)^2+left(text{C}_2left(text{R}_1+text{R}_3right)omegaright)^2}}tag{10}$$

When we want to find where the peak frequency is, we need to find:

$$frac{partialleft|underline{mathcal{h}}left(text{j}hat{omega}right)right|}{partialhat{omega}}=0spaceLongrightarrowspacehat{omega}=dotstag{11}$$

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When $text{R}:=text{R}_1=text{R}_3=text{R}_5$ we get the following transfer function (see $(10)$):

$$left|underline{mathcal{h}}left(text{j}omegaright)right|=frac{1+text{C}_1text{C}_2text{R}^2omega^2}{sqrt{left(1-text{C}_1text{C}_2text{R}^2omega^2right)^2+left(2text{C}_2text{R}omegaright)^2}}tag{12}$$

When we want to find where the peak frequency is, we get:

$$frac{partialleft|underline{mathcal{h}}left(text{j}hat{omega}right)right|}{partialhat{omega}}=0spaceLongrightarrowspacehat{omega}=frac{1}{text{R}}cdotfrac{1}{sqrt{text{C}_1text{C}_2}}tag{13}$$

So, for the frequency we get:

$$hat{omega}=2pihat{text{f}}=frac{1}{text{R}}cdotfrac{1}{sqrt{text{C}_1text{C}_2}}spaceLongleftrightarrowspacehat{text{f}}=frac{1}{2pitext{R}}cdotfrac{1}{sqrt{text{C}_1text{C}_2}}tag{14}$$

So, we get:

1. When $text{C}_1=224cdot10^{-9}spacetext{F}$,$text{C}_1=112cdot10^{-9}spacetext{F}$, and $text{R}=1000spaceOmega$: $$hat{text{f}}=frac{1}{2picdot1000}cdotfrac{1}{sqrt{224cdot10^{-9}cdot112cdot10^{-9}}}=frac{15625 sqrt{2}}{7 pi }approx1004.817spacetext{Hz}tag{15}$$
2. When $text{C}_1=224cdot10^{-9}spacetext{F}$,$text{C}_1=112cdot10^{-9}spacetext{F}$, and $text{R}=10000spaceOmega$: $$hat{text{f}}=frac{1}{2picdot10000}cdotfrac{1}{sqrt{224cdot10^{-9}cdot112cdot10^{-9}}}=frac{3125}{7 sqrt{2} pi }approx100.482spacetext{Hz}tag{16}$$
3. When $text{C}_1=224cdot10^{-9}spacetext{F}$,$text{C}_1=112cdot10^{-9}spacetext{F}$, and $text{R}=100000spaceOmega$: $$hat{text{f}}=frac{1}{2picdot100000}cdotfrac{1}{sqrt{224cdot10^{-9}cdot112cdot10^{-9}}}=frac{625}{14 sqrt{2} pi }approx10.048spacetext{Hz}tag{17}$$

EDIT:

Also notice that when $text{C}_1=224cdot10^{-9}spacetext{F}$ and $text{C}_1=112cdot10^{-9}spacetext{F}$, we have:

$$hat{text{f}}=frac{1}{2pitext{R}}cdotfrac{1}{sqrt{224cdot10^{-9}cdot112cdot10^{-9}}}=$$ $$frac{1}{text{R}}cdotfrac{15625000 sqrt{2}}{7 pi }approx1.004817317139628cdot10^6cdotfrac{1}{text{R}}tag{18}$$

So, when we multiply both sides by $text{R}$, we get:

$$hat{text{f}}cdottext{R}=frac{15625000 sqrt{2}}{7 pi }approx1.004817317139628cdot10^6approx10^6tag{19}$$