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I am designing a circuit which takes an 3uV at 10-10kHz input that filters, applies gain and integrates the analogue signal. I'm thinking of using the circuit below for the topology using two stages consisting of a Sallen-key filter and active RC integrator.

schematic

simulate this circuit – Schematic created using CircuitLab

Vin has a source impedance of 3.4Kohm.

In this low power level application, would a buffer be required to feed the sallen-key or would the circuit effectively operate in this system?

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    \$\begingroup\$ If Vin has a source impedance of 3.4 KOhm, the value of R1 should be much bigger. \$\endgroup\$ – Uwe Jul 7 '18 at 20:50
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    \$\begingroup\$ High R values have high noise. So you must amplify the signal with very low noise design. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 7 '18 at 20:57
  • \$\begingroup\$ I was planning on using R3 and R4 to amplify the signal before passing the integrator. So are you saying this is not feasible and the signal needs an additional buffer/gain input stage? \$\endgroup\$ – user160063 Jul 7 '18 at 21:09
  • \$\begingroup\$ @sidA30 Given the source impedance and the extremely low signal levels, I think you may require a discrete JFET pre-amplifier stage. (Maybe BJT. If the source impedance were lower, say below \$500\:\Omega\$, then I'd say a discrete BJT stage would be better. But that's not the case. BJT's shot noise can be reduced with lower quiescent currents but this is where JFETs can actually give somewhat better noise performance.) Just a thought to consider. \$\endgroup\$ – jonk Jul 7 '18 at 21:44
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    \$\begingroup\$ @jonk This old appnote also suggests low noise as you say: onsemi.com/pub/Collateral/AN-6602.pdf.pdf . I must admit, that I have usually been dealing with low source Z's <<1k \$\endgroup\$ – Henry Crun Jul 8 '18 at 3:38
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Your issue is probably noise. You should start by calculating the thermal noise voltage in 10kHz bandwidth @3.4kohm, and deciding what signal/noise ratio you need to get. That will probably tell you that you need a low noise preamp/buffer before the filter.

An active filter is very noisy. You have R1,2,4 all adding thermal noise. You have the input signal attenuated by R1C1 R2C2. Then you have opamps which are mostly fairly noisy. To get the thermal noise voltage down, you need to make the R's much lower than your source R - which means you must buffer. But it also means that you need an opamp which has a low equivalent noise resistance. The best opamps (AD797) have about 500ohms ENR - so you can't make the filter Rs much lower than this or again, noise figure gets worse.

This active filter arrangement is noisier than one which has only a single RC per filter stage. If you put OA2 before OA1 (with gain) it would be the preamp, and the whole would be quieter.

If you have out of band signal that needs filtering before the preamp, an LC low pass filter would be best. You will need a preamp before the active filter, with significant gain (40dB / 100x) to get good SNR. LC filters are well worth considering. This whole arrangement performs worse than one L and 2 C's.

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Does the filter "require" buffers? "Require" and "operate effectively" are strong words. Theoretically, if the input source has an impedance different from zero, and the next stage has an impedance less than infinity, the shape of transfer function of this filter will change. So the answer depends on how much change you can tolerate.

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In this low power level application, would a buffer be required to feed the sallen-key or would the circuit effectively operate in this system?

No, it wont (most of the time). When you compare the values of R1 and R2 to the input resistance of the amplifier it will normally be much smaller and will not affect the input current of the amplifier. If you get into the Mega-Ohm range with your filter resistance, then it might need to be considered, this will not be the case with most filters.

Since the source impedance is 3.4k and the total resistance of the sallen key circuit is in all likelihood in the Gigaohm range (depending on the selection of the amplifier) it shouldn't make a huge difference. Almost no current, nA to fA of input bias current of the op amp is needed, so a few mA for source impedance should be sufficient.

Noise is also added by adding in a voltage follower stage (and other costs of power, footprint of the op amp on the PCB, and additional money), so there is a trade off between low noise and impedance buffering.

In almost all cases I would avoid the additional voltage follower buffer.

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  • \$\begingroup\$ While I would generally agree with the sentiment about additional follower/buffer, I would strongly disagree with the rest. Output impedance of the source will be a part of resistor R1, so all filter characteristics will be screwed if R1 is not much higher than 3.4k. Or R1 should be correspondingly reduced to maintain filter's design characteristics. \$\endgroup\$ – Ale..chenski Jul 7 '18 at 21:58
  • \$\begingroup\$ I'll maybe simulate it later, I wrote this from mainly a DC perspective. I think the source impedance is sufficiently high, if you calculated the current draw from the amplifier it would be below the nA range in most cases that is several orders of magnitude higher than what is required and probably negligible \$\endgroup\$ – Voltage Spike♦ Jul 7 '18 at 22:42
  • \$\begingroup\$ I would also disagree. The input impedance is not the DC resistance it is a sum of the various R and Xc values at any particular frequency. In the stopband (hiF) XC1,XC2 are very slow, so input R~=R1. For correct frequency response, source R must be <<R1. ie. R1 should be 33k or more. (but this blows your noise out of the water) \$\endgroup\$ – Henry Crun Jul 8 '18 at 1:25
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1st you need a minimum SNR spec for 10kHz BW at 3uV input. This leads to a requirement of an low noise amplifier with a current noise density in nA/√Hz and voltage noise density in nV/√Hz.

Presumably you want > 60:1 SNR so noise bandwidth from resistors and amplifier for 10^4 Hz BW requires noise to be > 3uV/50/100 = 0.67 nV/√Hz

This will be a challenge as low Noise Op Amps tend to be 10x this amount.

https://training.ti.com/ti-precision-labs-op-amps-noise-lab?cu=14685

Rohm has a new LNA for audio. enter image description here

You are asking the wrong question. It should be how can I measure the following signal that is below thermal noise?

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Vin has a source impedance of 3.4Kohm.

As with most filters, the source impedance needs to be known. Most filters assume that the source is near-enough zero ohms but, if you source impedance is 3.4 kohm then the value for R1 (as calculated when assuming a zero source impedance) should be lowered by 3.4 kohm when fed from a source impedance of 3.4 kohm.

Of course if R1 is naturally quite high in value (100s of kohm or more) then lowering R1 won't be absolutely nesessary. With R1 = R2 you maximize the Q of the circuit and so altering one with respect to the other actually lowers the Q. This method is used to modify Q to what is needed in the filter to obtain the desired performance.

I am designing a circuit which takes an 3uV at 10-10kHz input

If your op-amp is specified as having an equivalent noise density of 1 \$nV/\sqrt{Hz}\$, over a 10 kHz bandwidth (your signal range) the equivalent noise will be 1 nV x \$\sqrt{10000}\$ = 1 uV RMS. So be aware that you need very good op-amps for this plane to fly.

I have ignored low frequency noise values and the tendency for equivalent noise to be higher in the DC to 1 kHz range. You need to carefully select your op-amp and possibly down-grade your expectations if working to a tight budget. I have also ignored equivalent input current noise density.

You need to consider all these things.

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