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I'm working on a project that requires me to amplify 70 pA to read out at 10 V (the digitizer I'm using to collect data has a range of 0 to 10 V readout). On top of this, I need to sample my voltage at either 2.5 MHz or 5 MHz. I'm greatly limited by the GBW of op amps since I can't get a reasonable gain at the frequency that I require without multiple stages. I am considering 5+ stage amplification but a friend of mine who studies EE said he thinks the noise generated from the additional connections would greatly impact the precision of my circuit. I've attached an example circuit by TI that I'm basing my design off of, in this example, they claim only a GBW of >9.5 MHz is required to achieve 53.6K gain (94.58 db) at 1 MHz, which doesn't make sense by my understanding of how GBW works (GBW = gain * frequency). enter image description here

Why is only 9.5 MHz of GBW is required in this case? I would think that 94.58 MHz of GBW would be needed to achieve the given gain at the given frequency.

Will using 5+ stages of op amps be an issue? I don't have much experience working with circuits so I'm not sure what the standard method is.

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Winston Wilhere
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  • In this case you're driving the amplifier with a current-source-like input. So the "noise gain" of the op-amp is near unity and your current amplification bandwidth is closer to the unity gain bandwidth of the op-amp. (Parasitic capacitance will limit the BW to less than the ideal GBW.) – John D May 23 '23 at 21:40
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    Since this is charges crossing a junction the inherent noise will be \$\sqrt{2 q B I}\$ or about 11 pA at your 5 MHz bandwidth. – periblepsis May 23 '23 at 21:49
  • And I would think that 53600MHz GBW would be needed if the gain is 53600 as you say – user253751 May 23 '23 at 21:50
  • You can model the diode as a current source in parallel with a junction capacitance. Then write the transfer function including the feedback capacitance. You can see that BW will not be the GBW divided by the transimpedance. You can also see that the op-amp noise will peak due to the junction capacitance, which is somewhat mitigated by the feedback capacitor. – John D May 23 '23 at 21:58
  • Winston, I need to see more theoretical work here. The basics are that Poisson statistics are in play with photodiodes. Especially at high acceptance bandwidth. Your signal is only 6.6 to 9.3 times higher than the noise floor. And that's assuming a perfect system before and after the photodiode. This is more in a 'detection' regime than in a measurement one. This means ***all*** of your effort needs to be placed at the front end transducer, optics, and first stage electronics. Nothing else matters much. Once it's gotten into the system all you can do is trade bandwidth for S/N. – periblepsis May 23 '23 at 22:18
  • periblepsis, I really wish I could show you more theoretical stuff, I study aerospace engineering, this is for a project in plasmas, so I'm am completely out of my element with this kind of circuit design. I'm not too worried about the noise, this is for a ratio pyrometer so I will be calibrating the system with the noise included, and it should mostly cancel itself out. I just need to figure out a reasonable way to achieve the gain I need at a 5 MHz sampling frequency and everything else should work itself out I hope. – Winston Wilhere May 23 '23 at 23:05
  • user253751, I was under the impression that GBW was calculated with gain in decibels. Although this still wouldn't align with the circuit analysis I've been using. – Winston Wilhere May 23 '23 at 23:07
  • @WinstonWilhere It's a significant part of what I did -- ratio pyrometry using thin film filters in the near IR bands to achieve NIST traceable 5 mK accuracy over a wide range -- at a place called Accufiber. I know I don't appreciate what you need to do for aerospace engineering with plasmas. I dealt only between 400 C and 3500 C and a max of about 1800 C for direct contact sputtered platinum sapphire rods (before they melt.) So I'm projecting. But you will care about noise at these levels. You just may not yet know it. – periblepsis May 27 '23 at 20:59

1 Answers1

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I'm working on a project that requires me to amplify 70 pA to read out at 10 V (the digitizer I'm using to collect data has a range of 0 to 10 V readout). On top of this, I need to sample my voltage at either 2.5 MHz or 5 MHz.

That's on the order of a few hundred million photons per second. While it's probably not impossible to do that with a photodiode given enough experience in the arcane art of transimpedance amplifier design, usually you would use a detector with photoelectron gain such as an APD or a SiPM. These devices generate between 100 to 1,000,000 electrons per photon, meaning for the same photon flux you'd have 7nA to 70 uA. That is much easier to amplify and will make electronic noise basically irrelevant compared to shot noise.

As for your main question:

Why is only 9.5 MHz of GBW is required in this case? I would think that 94.58 MHz of GBW would be needed to achieve the given gain at the given frequency.

Look at the units in your calculations. Your "gain" has units of ohms, so really you have 94.58 ohms/second, which isn't a physically meaningful quantity. Gain should be unitless, but your input is current and your output is volts, so that simple calculation is not going to work.

But I would step back and ask if you can buy a SiPM or similar detector that will give you a huge current from low light. Then you can amplify that with a simple circuit or even a resistor and a voltage amp.

user1850479
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    Ohm/s might not be very meaningful, but capacitance is F == s/ohm; we might invert this in the same way that a transconductance amplifier is measured in reciprocal ohms. It would only manifest as a capacitance (or whatever) if the amplifier's loop is closed such that an impedance can be measured at a single node, of course (i.e. a gyrator, or generally, impedance converter circuits). – Tim Williams May 24 '23 at 10:46