I am receiving a DC signal through a device called a Faraday cup. It is effectively a conductive piece of copper and when a gas is ionized nearby and the ions collide with the cup it is accepted as a current.

I want to do the following:

  1. Convert this current into a voltage, the current is low so the resistance of the cup itself isn't quite enough to convert it all into voltage so I've included a resistor in the circuit as shown in the figure. The other end of the cup is tied to ground through a small bleed resistor as shown.

  2. Take this low voltage input and amplify it, preferably 100 times.

  3. Minimize noise and impedance issues whenever possible.

So to this end I was thinking of the following circuit.


simulate this circuit – Schematic created using CircuitLab

I will have a low pass filter set up as the input to the first opamp, the buffer, so I'm able to allow the lower (DC) current inputs into the op amp since that is what I desire. I want the resistor to be high enough to convert the current to voltage as I mentioned before, but capacitance can be changed to make the cutoff frequency decently low (So I can get as close to DC as possible).

On the other hand I'll have a high pass filter going to ground because I want all those high frequency noise and other stuff to be removed from my circuit.

Then the output of the buffer is fed into a non inverting amplifier with a gain of 10, then another with a gain of 10. The reason for this is because of bandwidth limitations on the LM358. So when all is said and done I want to have a decently high output voltage in the 1-3V range with as minimal noise as I can get.

Does this seem like a decent set up or are there ways for me to improve it? I had considered using an instrumentation amplifier but currently am not because:

  1. I've haven't been able to get the gain to work, ie I'd simulate 100 gain but hooking that up I'd only be getting 20-40.

  2. Single chip InstrAmp are a bit pricey

  3. I just need to compare the faraday to ground so the two buffered inputs wouldn't benefit me as I'd just need to ground the inverting input.

  • 1
    \$\begingroup\$ Take a look at transimpedance amplifiers. \$\endgroup\$
    – Golaž
    Oct 7, 2015 at 11:20
  • \$\begingroup\$ What do you mean "the resistance of the cup itself isn't quite enough to convert it all into voltage"? That doesn't make a great deal of sense. You also need to clarify what frequency you're talking about. You say DC, but if it really was DC you wouldn't have a signal at all; how fast can the actual signal change? \$\endgroup\$ Oct 7, 2015 at 11:33
  • \$\begingroup\$ To Golaz: I've initially used a transipedence but I don't think it was enough on its own. Where would you suggest I place it in my circuit? To Nick: Sorry if I wasn't clear, what I mean is that if the current being input is very low (1mA or less) and the resistance of the cup is only 200 ohms then I am not able to convert the available current to voltage, if that makes sense. \$\endgroup\$
    – FrankerZ
    Oct 7, 2015 at 11:36
  • 1
    \$\begingroup\$ That circuit cannot work, the + input of the first opamp has no DC connection to anything. Basically the input is floating. This might work in a simulator but in practice this is not going to work. Also the the 2 resistors and 2 capacitors at the input will do nothing you think they provide filtering but they do not if you connect them like that. Please study a book about how to use opamps first, then design something. \$\endgroup\$ Oct 7, 2015 at 11:37
  • \$\begingroup\$ I see what you mean about the filters not working but what do you mean about the input being floating? \$\endgroup\$
    – FrankerZ
    Oct 7, 2015 at 11:53

1 Answer 1


The preferred technique for measuring the current of an ion beam is using a transimpedance amplifier (TIA) configuration. I'm think here of gas mass spectrometers but generally it is a good idea because it keeps the target potential at near zero volts. Imagine what happens to your ion beam as the "cup" starts charging up - will it deflect the beam and make measurement errors? Possibly so. Anyway this is the sort of circuit I believe makes most sense: -

enter image description here

The output voltage is R x input current and you might find that R needs to be in the order of giga ohms. There are specialist op-amps that suit this application too because leakage currents from the op-amp inputs can produce a significant error given the size that R needs to be. Check out TI and ADI's portfolios of specialist op-amps for what is available that has low bias currents.

The input impedance seen by the ion beam is nominally zero and the op-amp achieves that with negative feedback thus maintaining the inverting input at the same potential as the non-inverting input i.e. 0V.

  • \$\begingroup\$ This might be a silly question but imaging the Faraday as some sort of material like a copper plate. Would I need to ground it? To clarify in that circuit it shows the cup directly set as the input to the circuit but would I need to connect the other end of the cup to ground? \$\endgroup\$
    – FrankerZ
    Oct 7, 2015 at 12:59
  • \$\begingroup\$ If you ground it, how can the charge go into this circuit ? It needs to enter the circuit as the current marked "I". If you ground it, - and + input of the opmap will be... yes, shorted. You would measure nothing. \$\endgroup\$ Oct 7, 2015 at 13:55
  • \$\begingroup\$ @FrankerZ - it is grounded because the op-amp output keeps is at 0V by the action of negative feedback - it's called a virtual earth for that very reason. \$\endgroup\$
    – Andy aka
    Oct 7, 2015 at 14:42
  • \$\begingroup\$ Alright that makes sense. Negative feedback makes the impedence difference of the inputs close to 0 and the output of the op amp close to zero as well (Since current * 0 = 0). Is there any way to improve the design with the use of more op amps? The low pass filter exists to remove any sort of non-DC input noise from static but would I be better served in any way of reducing overall system noise with additional op amps? \$\endgroup\$
    – FrankerZ
    Oct 7, 2015 at 15:20
  • \$\begingroup\$ There can be an improvement but usually only in a few cases. It goes like this - the noise at the output is as a result of thermal noise in the feedback resistor. It's high in value so makes a lot of thermal noise but, that noise is proportional to the square root of resistance. Let's say you chose a resistor that was one-ninth the value needed to get the signal output voltage you need. You would then need a 2nd stage gain of nine but, reducing the resistor by 9 only reduces the noise by 3 then, a gain of nine increases noise by 27! No argument but in extreme cases it is needed. \$\endgroup\$
    – Andy aka
    Oct 7, 2015 at 16:48

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