0

Let's say that I have a periodic charge build-up across the capacitor C1.

I could either choose to measure the current using a transimpedance amplifier as shown below or could measure the voltage instead.

I've been told it's probably better to measure the current because of the parasitic capacitance coming from the BNC cable (represented as a capacitor in the image below).

I do see that the voltage you measure is going to be less for high frequency voltage signal because there will be voltage drop across the parasitic capacitance.

However, don't you also lose current to the parasitic capacitance of the BNC cable? Is the loss miniscule because the parasitic capacitance is on the order of pF? (1st question) So, the parasitic capacitor will get charged up quick and most of the current will still flow to the transimpedance amplifier?

The second question I have is, when using a transimpedance amplifier, as shown below, I was told to put a 1pF capacitor in parallel with the gain resistor to stabilize the feedback.

What is the typical frequency range of the undesired oscillation that we want to get rid of by having this capacitor in parallel with the resistor (to effectively remove the parasitic capacitance of the BNC cable)? Don't you sort of have to know what the unwanted signal's frequency is going to be to choose the right capacitance value to remove that signal? Could someone explain to me what's the best way to choose the right value for the capacitor?

schematic

simulate this circuit – Schematic created using CircuitLab

ADDED

schematic

simulate this circuit

Blackwidow
  • 521
  • 5
  • 15

2 Answers2

0

Answer

Your Transimpedance Amp, TIA, feedback RC=T values must match the source RC values for best results. The spectral voltage gain, the impedance ratios are matched ratios for the broad-spectrum flat response. This is like tuning a scope 10:1 probe for a square wave response.

Other info

No long skinny ground leads are allowed as this adds a potential LC resonance > 10MHz ( usually around 30MHz) which is usually outside the OA bandwidth but might not. This is a common fault using scope 10:1 probe with excessive ground clip inductance. Ground L=1uH/m approx gives series resonance with cable 100pF/m approx. Your values may vary.

Consider a source is a piezoelectric (PZ) charge crystal for example of known C and G=1/R leakage and cable is a known impedance with typ. 100pF/m depending on impedance and type.

The BNC connector C is negligible ~1pF compared to cable capacitance, but any motional change in the cable will vary C some xxx ppm ( parts per million) which will appear as a piezo effective noise so the cable must be semi-rigid not standard RG58 for lowest noise. Perhaps double braided stiff RG58 taped down is OK with special low pF coax as we used for crystal accelerometers.

Tony Stewart EE75
  • 1
  • 3
  • 54
  • 182
0

The transimpedance amplifier produces a virtual earth at the inverting input and this means it is, in effect, an amplifier with zero ohms input impedance. The 1st knock-on from this is that charge that might have been built up in C1 is immediately converted to current. This is because no charge is allowed to build in C1 because it is effectively shorted by the input circuit's virtual earth. Q = CV and V = 0.

The 2nd knock-on is, that because the input is a short-circuit, it also shorts the parasitic capacitance in the cable and therefore, to a great extent, any effect the parasitic capacitance might have is eradicated.

For your 2nd question the feedback capacitor is not placed there to stabilize the feedback; in fact on many op-amps it can destabilize the circuit. It is there to reduce noise gain at high frequencies; the noise effectively being that produced by the op-amp itself.

See this answer I gave for a better explanation of the problem of noise gain. And here's another relevant answer I gave.

Andy aka
  • 434,556
  • 28
  • 351
  • 777
  • Thank you for the response. I think everything you said about the benefits of measuring current makes sense. In case we do choose to measure voltage instead though, is the problem the fact that a BNC cable works like a voltage divider (as drawn in the image I just added) and the voltage across the parasitic capacitor ends up becoming the voltage you end up measuring? Thank you! – Blackwidow Sep 27 '18 at 16:53
  • When measuring voltage you get a false impression of charge due to the parasitic capacitance taking current proportional to the rate of change of voltage that appears across it. I would recommend using a simulator to get your head around this. – Andy aka Sep 27 '18 at 17:03
  • Thank you again. However, if you don't have the resistor that I drew and if the whole thing can be effectively modeled just with a capacitor to the ground, wouldn't the voltage value you measure still be the same as the input voltage? I think without having the voltage divider picture, I don't see how the parasitic capacitance messes up your voltage reading. – Blackwidow Sep 27 '18 at 17:09
  • If your voltage measurement is trying to infer charge then adding an unknown capacitor in parallel with C1 means you have an unknown scaling factor. If you know the cable capacitance then measuring voltage is fine. If you measure current you don’t have that unknown scaling factor to deal with nor do you lose sensitivity when the parasitic capacitance is significant. – Andy aka Sep 27 '18 at 17:15