Active passband filter with higher gain than open loop

Question

I'm desgining an active band pass filter using a LM324 op amp. The signal I need to amplify is centered on 20kHz and has a width of around 1kHz.

Looking at the open loop gain of the LM324 I see that at 20kHz the open loop gain is about 40dB. Thus, I would expect that a closed loop gain (when narrowly centered on 20kHz) shall not exceed 40dB.

However, when running some simulations on LTspice, I was surprised to achieve gain values that are higher than 40dB. Does is mean that the simulation must be wrong for some reason or is it actualy possible?

In other words - is it possible to get, at a certain frequency, a higher gain than the open loop gain at that same frequency? (using passive components only)

EDIT: Calculating the theoretical transfer function (as Jan Eerland posted) and plotting it with the open gain loop:

Answer 1

Well, the transfer function of your circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\displaystyle\text{V}_\text{o}\left(\text{s}\right)}{\displaystyle\text{V}_\text{i}\left(\text{s}\right)}=1+\frac{\displaystyle\text{R}_1}{\displaystyle\text{R}_2+\frac{\displaystyle1}{\displaystyle\text{sC}}}=1+\frac{\displaystyle\text{sCR}_1}{\displaystyle1+\text{sCR}_2}\tag1$$

So, for the amplitude we get:

$$ \begin{alignat*}{1} \left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\left|1+\frac{\displaystyle\text{j}\omega\text{CR}_1}{\displaystyle1+\text{j}\omega\text{CR}_2}\right|\\ \\ &=\left|1+\frac{\displaystyle\text{j}\omega\text{CR}_1}{\displaystyle1+\text{j}\omega\text{CR}_2}\cdot\frac{\displaystyle1-\text{j}\omega\text{CR}_2}{\displaystyle1-\text{j}\omega\text{CR}_2}\right|\\ \\ &=\left|1+\frac{\displaystyle\text{j}\omega\text{CR}_1\left(1-\text{j}\omega\text{CR}_2\right)}{\displaystyle1^2+\left(\omega\text{CR}_2\right)^2}\right|\\ \\ &=\left|1+\frac{\displaystyle\text{j}\omega\text{CR}_1-\text{j}\omega\text{CR}_1\text{j}\omega\text{CR}_2}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}\right|\\ \\ &=\left|1+\frac{\displaystyle\text{R}_1\text{R}_2\left(\text{C}\omega\right)^2+\text{j}\omega\text{CR}_1}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}\right|\\ \\ &=\sqrt{\left(1+\frac{\displaystyle\text{R}_1\text{R}_2\left(\text{C}\omega\right)^2}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}\right)^2+\left(\frac{\displaystyle\omega\text{CR}_1}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}\right)^2} \end{alignat*} \tag2 $$

And the argument is given by:

$$ \begin{alignat*}{1} \arg\left(\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(1+\frac{\displaystyle\text{R}_1\text{R}_2\left(\text{C}\omega\right)^2}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}+\frac{\displaystyle\omega\text{CR}_1}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}\cdot\text{j}\right)\\ \\ &=\arctan\left(\frac{\displaystyle\frac{\displaystyle\omega\text{CR}_1}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}}{\displaystyle1+\frac{\displaystyle\text{R}_1\text{R}_2\left(\text{C}\omega\right)^2}{\displaystyle1+\left(\omega\text{CR}_2\right)^2}}\right)\\ \\ &=\arctan\left(\frac{\displaystyle\omega\text{CR}_1}{\displaystyle1+\left(\omega\text{CR}_2\right)^2+\text{R}_1\text{R}_2\left(\text{C}\omega\right)^2}\right) \end{alignat*} \tag3 $$

Answer 2

Yes it can. This demonstrates the difficulty implementing a true differentiator with op-amps. As the sweep indicates R1 in series with the capacitor is necessary to prevent oscillation. The sweep can be used to find a minimum value for R1.

Where the closed loop curve nears the open loop curve, the high open loop gain simplification can not be used. The open loop TF must be included when deriving the closed loop TF.