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This circuit provides 30dB gain to a 100kHz signal with a 10k input impedance. Is capacitor C2 necessary to suppress DC gain or is it redundant with the AC coupled input?

I would also like to make this circuit single-supply and I don't see how include the C2 effect. I also need to minimize noise around 100kHz.

EDIT: The bandwidth required is 100kHz ± 5kHz and it may be driven by an inductive source with C1, R1 tuned for a 100kHz resonance.

enter image description here

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  • \$\begingroup\$ You might want to mention the bandwidth you require near your signal frequency. \$\endgroup\$ – jonk Nov 21 '17 at 21:33
  • \$\begingroup\$ Without C2 any DC voltage present at the non-inverting input (opamp dc-offset) will be amplified by a factor of 1+R2/R3. \$\endgroup\$ – G36 Nov 21 '17 at 21:39
  • \$\begingroup\$ @G36 The only voltage at the non-inverting input is the opamp offset voltage or current across R1. The resulting output offset even at high gain would likely be small enough that it does not affect dynamic range. So as long as the next stage is AC coupled I presume that this DC gain would have little influence on the signal in-band. \$\endgroup\$ – Mike Nov 21 '17 at 23:25
  • \$\begingroup\$ You can use the standard op-amp formula to calculate the gain. The gain, referenced to the non-inverting input, is 1 + R2/(R3 + wC2). The term "wC2" is meant to represent omega * C2. Omega is the input frequency in radians/sec. C1 and R1 also form a high pass filter. It is not that hard to write out the transfer function of this circuit. \$\endgroup\$ – mkeith Nov 22 '17 at 4:35
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For the case where Vdd is at a positive DC voltage, and Vss is at a negative DC voltage, C2 could be removed...if you don't mind some DC offset at the output caused by amplified offset voltage.
For the case where Vdd is at a positive DC voltage, and Vss is ground (unipolar power supply), C2 is required. The usual circuit goes like this, requiring a very clean noise-free Vdd supply:

schematic

simulate this circuit – Schematic created using CircuitLab

You'll want an op-amp with gain-bandwidth product greater than 5 MHz. The default TL081 shown has insufficient 3 MHz. You have two high-pass filter corner frequencies: \$ F1 = \frac{1}{(2 \pi (C1) \frac{R1aR1b}{(R1a+R1b)})} \$ \$ F2 = \frac{1}{2 \pi (C2 R3)} \$
If your bandwidth is 100 kHz, and center frequency is 100 kHz, then these two corner frequencies should be set well below 50 kHz. C1 is unlikely to resonate easily with an inductive source with this high input impedance amplifier.

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  • \$\begingroup\$ Interesting, the non-inverting example I was looking at has the bias voltage applied in place of C2 instead of biasing the non-inverting input like you have. I suppose the most obvious implication is that you don't need a low-impedance Vdd/2 reference, but you do need two high-pass filters. \$\endgroup\$ – Mike Nov 22 '17 at 1:49
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The TL081 has only 3MHz UGBW; to expect an accurate and stable closed loop gain of 30 dB is an error. Pick a faster opamp. What error budget have you constructed? What gain stability must you have?

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  • \$\begingroup\$ I am using an OP37 with 63MHz GBP. Honestly haven't thought about gain stability, but that's a great point because I am worried about picking up amplitude noise. The gain is set with 1% resistors. \$\endgroup\$ – Mike Nov 22 '17 at 18:00
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The purpose of C2 is not to suppress DC gain. The purpose of C2 and R3 is to form a high pass filter for U1 which will pass the 100kHz signal and block the 60 Hz hum from the power supply. You will see that the given values produce a cut off frequency of 33.9 kHz. This is purposely less than half of the signal frequency but far above the unwanted 60Hz signal. If you use a split supply, then you might get away without C2 because the 60Hz signal is in both supplies and will mostly cancel out in the amplifier because of differential amplification. However, with a single supply, you will be amplifying the 60Hz hum and mixing it into your signal. The effects will depend on many factors relating to the power supply layout and decoupling. For the price of one capacitor it is best to add this part in the design. After all, we're talking about a ceramic 4.7nF to filter 60Hz hum which would otherwise require more expensive electrolytics.

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  • \$\begingroup\$ Should I care about 60 Hz hum if I only intend to demodulate a 10 kHz band around the 100 kHz carrier? Won't the 60 Hz only bother in baseband stages? \$\endgroup\$ – Mike Nov 22 '17 at 18:07

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