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enter image description here

All the active filters I've found 0n books and google are inverting type (i.e. input signal is applied at the negative input pin of the op-amp).

I thought if I apply the input signal at the positive terminal, it may work as a non-inverting high pass filter. But the frequency response (Proteus 8.6 simulation) looks like a low pass filter.

As my power supply is either 0-5 or 0-3.3 volts, I need to design a filter that does not create a negative output voltage. Plus I'll connect the output to a microcontroller ( or Arduino), which also can't have negative voltage.

Can anyone help me by providing a circuit diagram of Active Non-Inverting High Pass Filter?

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    \$\begingroup\$ Your title says band pass. \$\endgroup\$ – Andy aka Sep 7 at 19:25
  • \$\begingroup\$ Oh sorry. Typing mistake. \$\endgroup\$ – Sadat Rafi Sep 7 at 19:28
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    \$\begingroup\$ Sure. Classic unity gain Sallen & Key for one. \$\endgroup\$ – Brian Drummond Sep 7 at 19:33
  • \$\begingroup\$ Thanks @BrianDrummond. \$\endgroup\$ – Sadat Rafi Sep 7 at 19:46
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    \$\begingroup\$ Your upper circuit should have a gain of 1 at DC, breaking upward at about 8kHz, then breaking downward at 16kHz to settle out at a gain of 2. So something is wrong -- perhaps because you have Vee = ground instead of some negative voltage? It's still not what you want, of course. \$\endgroup\$ – TimWescott Sep 7 at 20:19
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Just use a standard passive high pass filter and stick a non-inverting op-amp buffer on the output. Use two resistors for the filter with one to Vcc and the other to ground to add the required DC offset. Job done.

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One of the reasons involve the changing of the transfer function. Considering \$Z_f\$ as the feedback equivalent impedance and \$Z_i\$ as the input equivalent impedance, the transfer function for an inverting opamp is \$-Z_f/Z_i\$, while for a non-inverting opamp, \$Z_i\$ is the grounded equivalent impedance, and the gain is \$1+Z_f/Z_i\$.

Now let's consider a simple lowpass, \$\omega/(s+\omega)\$. If you were to use a capacitor (\$C_1\$) in parallel with a resistor (\$R_1\$) as \$Z_f\$, and a resistor (\$R_2\$) as \$Z_i\$, you'd have this resultant transfer function:

$$ H(s)=-\frac{C_1||R_1}{R_2}=-\frac{\frac{1}{sC_1+\frac{1}{R_1}}}{R_2}=-\frac{R_1}{R_1R_2Cs+R_2}=-\frac{R_1}{R_2}\frac{\frac{1}{R_1C}}{s+\frac{1}{R_1C}} $$

Which is the lowpass transfer function with a gain. If you apply the same elements to the non-inverting configuration, this is what you get:

$$ H(s)=\left(1+\frac{R_1}{R_2}\right)\frac{\frac{1}{R_1C}}{s+\frac{1}{R_1C}}=\left(1+\frac{R_1}{R_2}\right)\frac{\frac{R_2}{R_1+R_2}s+\frac{1}{R_1C}}{s+\frac{1}{R_1C}} $$

Which is a shelf-lowpass, of the form \$(s+\omega_z)/(s+\omega_p)\$. That \$1\$ changes everything: it adds the denominator to the numerator.

What you have there is a high-pass, as an inverting configuration, which transforms into a shelf-highpass for the non-inverting configuration. The reason why you don't see this response in your picture is because you're using unipolar supply for the inverting opamp, while using bipolar supply for the non-inverting one. If you'll make both opamps suplied from the same sources, the results should change.

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