In simple analysis, these two circuits give the same result ... i.e. producing 1V output for a 10pA input. Is there any advantage of one over the other in practice?

I am measuring currents in the high femtoAmp / low picoAmp range from a source that is hard to fully isolate. There is no appreciable difference in linearity treating the signal as a current source vs measuring voltage across the big resistor. I have tried the transimpedance configuration and it works, but noise and random leakages cause a lot of problems. I have not tried the non-inverting amplifier configuration. Frequency response is not important as I will low-pass filter it down to about 2Hz cutoff anyway. The application is for measuring concentration of ions in the air, so the sensing elements can be shielded roughly, but ultimately are big and are exposed.

The Lmp7721 has -/+3fA typical bias and it's imput impedance is very high, so I think it acts pretty close to an ideal op-amp for these purposes.Transimpedance vs Amplifying voltage across a resistor

  • \$\begingroup\$ One involves developing voltage across the source as it generates current, the other does not. That might affect the operation or accuracy of the source. \$\endgroup\$
    – user16324
    Commented Jun 3, 2016 at 9:05

2 Answers 2


I have tried the transimpedance configuration and it works, but noise and random leakages cause a lot of problems

Here's why you get a noise problem with a TIA (it is all to do with parasitic capacitance): -

enter image description here

Every op-amp has an internal input noise source "inconveniently" located in series with either of the inputs and, for the sake of what I'm trying to demonstrate, I've shown it in series with Vin+.

So you have "ion collection plates" and these form a capacitance to ground of maybe 100 pF (guesswork). It doesn't matter what the exact value is as any capacitance will cause what is known as noise gain (the big stumbling block of some TIA circuits).

So, forget about your real ion current input and just concentrate on that noise being amplified by the following factor:

\$1+\dfrac{R_f}{X_C}\$ (that's what a non-inverting gain stage will do and that is what you have!)

Now, for the sake of simplicity let's say between DC and 10 Hz you have an average voltage noise density of ~100 nV/sqrt(Hz). Over a 10 Hz bandwidth that becomes 100 nV * sqrt(10) = ~300 nV RMS.

Xc at 1 Hz is 1.59 G ohms so your noise gain at 1 Hz is 1 + 100/1.59 = ~ 64 making your output noise approximately 19.2 uV RMS.

This is just a simplified explanation of where the basic "in-band" noise arises in a TIA due to "noise gain". Clearly at (say) 10 kHz the capacitive reactance is much, much lower and you have a lot of noise "out-of-band". So at 10 kHz you have a reactance of 159 k ohms and a noise gain of 628,000 (see "redemption" below).

This input noise (between 1 kHz and 100 kHz) is ~ 6nV * sqrt(99k) = 1.9 uV and due to the noise gain will appear at the output at a level of 1.19 V RMS and could quite easily swamp your signal massively. However, you have to remember it is out of band noise and can be easily filtered away.

The thermal noise of the 100 G ohm resistor should also be considered.

The above calculations are just ball-park rule of thumb but won't be too far off. And now a slice of reality that helps...


I see you have Cf set to zero pF. Assuming that parasitics will contribute something in the realm of 0.5 pF, your noise gain will start to be curtailed at about 3 Hz so this will significantly reduce the "out-of-band" output noise because the noise gain now reduces at higher frequencies meaning a noise gain of 628 k is impossible - I'm taking you through this bit by bit.

The upshot is that noise gain at higher frequencies becomes the ratio of the two capacitive reactances (feedback divided by input capacitance) and in my example noise gain (due to the feedback parasitic capacitance of 0.5 pF and the input capacitance of 100 pF) will be ~200. The in-band noise gain will still be 64 at 1 Hz because at this low frequency the 100 G ohm resistor is more dominant than a 0.5 pF capacitor.

Bottom circuit

Plate capacitance of 100 pF and input resistance of 100 G ohm make it a low pass filter with a cut-off of 0.016 Hz i.e. quite unsuitable for an in-band top frequency of 2 Hz. The TIA is still the best option despite the noise issues.

  • \$\begingroup\$ My head just exploded... I will have to re-read this tomorrow when I am fresh for an answer. FYI, the calculated capacitance of the input plates is about 40pf but I figure reality is -/+ 20pf. The actual capacitor I used in the Transimpedance amplifier is about 5pF. I set it to zero for purposes of the diagram and to calculate the gain at 0Hz. In practice, I am looking at a a stable physical phenomenon, so frequency response actually a negative. In my TIA version 1, I use a 2'nd order Butterworth filter at 2.4 Hz 3dB cutoff and it still jitters around too much. \$\endgroup\$ Commented Jun 3, 2016 at 15:15
  • \$\begingroup\$ Another thought... my range of jittering with the TIA circuit is about-/+ 10mv so output noize in the uV range is not important. \$\endgroup\$ Commented Jun 3, 2016 at 15:28
  • \$\begingroup\$ The actual sensors are two things.... a tube of aluminum about 50 mm in diameter and 60 cm long, and an inner conductor about 5mm diameter ans same length. Th ewhole thing shielded to ground.... You can see why it is sensitive to external noise. \$\endgroup\$ Commented Jun 3, 2016 at 15:30
  • \$\begingroup\$ The jittering could be op-amp noise but you'll have to put some figures on things to get a better answer. Could also be power supply noise. Also, have you tried isolating the plate electrostatically from the rest of the universe just to get a "zero" reading? Small changing strains in the PCB could even create enough current to give big problems - it's called triboelectric effect and it's basically what piezo devices exhibit. I built an ion amplifier that went on a mass spectrometer and tribo effects were noticeable. \$\endgroup\$
    – Andy aka
    Commented Jun 3, 2016 at 15:31
  • \$\begingroup\$ My wonder is whether under these conditions the TIA or the NIA have some inherent advantage. Otherwise, I will just have to shield where I can.. \$\endgroup\$ Commented Jun 3, 2016 at 15:31

The top circuit is the better one by far.

The advantage of the top circuit is very fundamental, it's input impedance is low which is the ideal way of measuring currents from a sensor that behaves as a current source. The opamp will keep the voltage at the input zero so parasitic capacitances are kept at the same voltage. In effect, the influence of these parasitic capacitances is removed.

The bottom circuit has an input impedance of 100 Gohm. A current from the sensor will develop a voltage across R1 but any parasitic capacitances are also charged/discharged. So part of the current from the sensor is used for this. The resulting bandwidth will be a lot lower compared to the top circuit. At 100 G ohm, only a few pF already make a large difference.

  • \$\begingroup\$ If I don't care about bandwidth for this one (the physical condition measured is constant over several minutes). In fact, capacitance that simply adds to the stability of the output. In the real circuit I used 5pF capacitance in the transimpedance amplifier and it cuts the output sensitivity roughly in half at 1Hz. Is there any advantage in noise or other issues.... My problem at first was layout... I had ground loops galore over just a few centimeters of PCB spacing and leaking from mysterious places when humidity was high. \$\endgroup\$ Commented Jun 3, 2016 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.