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I've to design an input stage for signal conditioning of a voltage signal coming from an high output impedance sensor. The sensor output impedance is 100k, and the amplitude of the voltage signal is 0-5mV. The frequency of the signal is between 50-150 kHz. I'd like to minimize the op-amp used in the circuit. In terms of "circuit blocks" what architecture do you suggest?

I was thinking of using a voltage buffer for the input stage, in order to have a low impedence at the buffer output and then amplify the signal with a non inverting op-amp.

Is it a good idea?

If I had to use an opamp with a small gain bandwidth product, is there a way to externaly compensate the op-amp dominant pole and increase the GBW product?

Thanks for your help!

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  • \$\begingroup\$ A link to the sensor should clear this up. \$\endgroup\$ – Andy aka Oct 29 '16 at 13:31
  • \$\begingroup\$ I hope you can place the buffer/amp at the sensor, not remotely. A far-away connection will degrade your high-frequency response. Look for op-amps that are not unity-gain stable (not many are available) - these tend to have extended GBW, but may tend to oscillate in some circuits. \$\endgroup\$ – glen_geek Oct 29 '16 at 15:15
  • \$\begingroup\$ Well, it's not a practical question. It comes out from an exercise I've to solve. The exercise states that the op amp I have are ideal, so with infinite input impedance. This solve the problem of "matching" the source impedance with the input impedance of the amplifier. Now I'm just asking myself why they gave me the Rs=100kOhm specification, it's useless with ideal opamp. \$\endgroup\$ – FataMadrina Oct 29 '16 at 16:23
  • \$\begingroup\$ The design situation you pose is very similar to a capacitance-microphone preamplifier circuit. Professional audio-electronics engineers would be familiar with this kind of circuit. Your only difference is that the frequency response needs to be higher, and you don't need a DC bias on the sensor. You might take a look at "condenser" and "tube" microphone schematics of famous designs from Neumann, Schoeps, Telefunken, Sony, AKG, Sony, Sennheiser – all available somewhere on the web. \$\endgroup\$ – Rich S Oct 30 '16 at 15:57
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There is no reason to use an op-amp just to buffer the input- use a non-inverting configuration, for example with a JFET input op-amp.

If you want to use low GBW op-amps, you can just use more stages with less gain each.

The big problem may be that an input capacitance of only 10pF will have an impedance of about 100K at 150kHz, assuming a sine wave (worse if there are harmonics). There are techniques to get down way sub-pF using discrete parts and bootstrapping.

Minimizing the amplifier cost may not be ideal since cheap fast-ish op-amps tend to want dual supplies. Looking at the problem at a system level may yield a more optimal solution, but for cheap consider jellybean JFET amplifiers.

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  • \$\begingroup\$ The problem in my opinion is exactly the one you pointed out. Why the use of a voltage buffer doesn't help me? With a buffer I'd have a low source impedance and I could use a non inverting stage without concerning about the impedance. \$\endgroup\$ – FataMadrina Oct 29 '16 at 12:52
  • \$\begingroup\$ I don't see why you would not ask the buffer to provide gain, if it is capable of voltage gain. If you are postulating a discrete part front end then it might make sense, but you did not indicate that. \$\endgroup\$ – Spehro Pefhany Oct 29 '16 at 12:59
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Your idea is fine. One thing to consider is the first stage after a sensor is usually the most critical for noise performance. The buffer will add noise but not increase the level of the signal. So here is an alternative. Consider your sensor is a 50nA source with 100K resistance. Assuming it has no DC component it could be put directly into the summing junction of a low noise inverting jfet opamp, like an OPA827. Using a feedback resistance of 1Meg would give you a gain of 10. This means the first stage get you lower output impedance with some gain. If you don't like 1 Meg feed back resistors, who would, you can replace the feedback resistor with a T to get move desirable values. You can see how this is done by searching opamp T feedback. Without knowing what level you are trying to get to and the impact of noise performance, I can't say more.

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  • \$\begingroup\$ Thank you. As I have said in a comment, the question I've made is from an exercise, where I have to use ideal opamp. But I was interested even in a practical case, where opamps are not ideal. With an ideal of opamp I don't have the issue of the impedance, but I don't know why they gave me the Rs=100k ohm data. The T solution is very interesting btw. \$\endgroup\$ – FataMadrina Oct 29 '16 at 16:26

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