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The gain of a charge-sensitive amplifier (CSA) is Vo/Qi~1/Cf, where Vo is the output voltage, Qi is input charge, and Cf is the feedback capacitor.

Why can't I get rid of the whole op-amp part of the circuit and directly use a voltmeter to measure the voltage drop, V, on Cf?

That V is calculated as Qi/Cf, exactly the same as Vo.

The internal resistance Rf of the voltmeter can be large so that most charges will be collected by Cf. The slow discharge of Cf over Rf will bring the voltage back to zero gradually so that the circuit is ready for the next measurement.

My specific application of this circuit is for radiation detection. The radiation detector can be seen as a combination of a capacitor and a current source if it is hit by radiation. We normally use a feedback resistor to bring back the baseline gradually.

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4 Answers 4

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Your voltmeter would contain an amplifier anyway. By adding a transimpedance amp in the front end, you relax requirements for downstream routing and signal handling.

The TIA also keeps the input voltage at a stable level.

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  • \$\begingroup\$ Thanks for the answer. A needle voltmeter is simply made of a coil and a resistor, which is much simpler than an op-amp. There must be some big advantage of using op-amp instead of a simple voltmeter here. You mentioned downstream routing and signal handling and a stable input voltage. Are those determination factors? Could you please elaborate more? \$\endgroup\$
    – Jing
    Jun 22, 2022 at 10:10
  • \$\begingroup\$ Charge amps typically need extremely high input impedance >>GOhm. You can't achieve this with a coil input, and routing across a board will also add too much parasitic leakage and capacitance. \$\endgroup\$
    – tobalt
    Jun 22, 2022 at 10:43
  • \$\begingroup\$ Please see my comment on kruemi's answer. We use a feedback resistor to effectively reduce the input impedance of a CSA. The internal impedance of a voltmeter works as a feedback resistor. \$\endgroup\$
    – Jing
    Jun 22, 2022 at 10:48
  • \$\begingroup\$ @Jing please draw a schematic in your question. It is not clear to me what you mean. \$\endgroup\$
    – tobalt
    Jun 22, 2022 at 11:07
  • \$\begingroup\$ the feedback resister is connected in parallel to Cf and op-amp if I have to add it to the plot in my original post. \$\endgroup\$
    – Jing
    Jun 23, 2022 at 3:52
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Short answer: It's because you want to measure a charge, not a voltage.

Longer answer: There is a saying in german: "Wer misst, misst Mist" ("Who measures, measures crap"). Generally speaking you'll never measure the true signal because in many cases your measurement setup influences the signal. It's important to know your devices and what influence they have, if this influence can be neglected or you have to take it into account.

The input impedance of a voltmeter is not infinite. DVM have an input impedance in the range of 10 Megaohms. An many cases, the resistance depends on the selected range of the voltmeter. The more sensitive, the lower the input impedance. So a charge is not a voltage. But it can be converted to a voltage by using it to charge a capacitor. But as soon as you connect the Voltmeter to the capacitor you'll start to discharge it, which is not what oyu want. That's way you have a charge amplifier which has an input impedance several oders of magnitude higher than the on on a common voltmeter.

I hope this answers your gestion. Otherwise ask away in the comments (or open a new question if it's more than just a clarifiaction that's needed)

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  • \$\begingroup\$ Thanks for the answer. I am aware of the influence of a voltmeter parallel to the capacitor. Yes, it will discharge the capacitor, but we do it anyhow through a feedback resistor in a real CSA to bring the baseline back to zero. The internal impedance of the parallel voltmeter can be regarded as the feedback resistor, can't it? \$\endgroup\$
    – Jing
    Jun 22, 2022 at 10:04
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A typical use case may clarify this. If we have a force sensor containing a crystal with, say, 50 pF, this provides a charge once a force is applied. A voltmeter would discharge this sensor signal as all RC elements do. The reading tends to zero after some time, but the force is still present.

Removing the force would give a negative reading until discharged again. Calculating the force from this measurement is possible, but not very precise. And you need to know what has been before to interpret near zero volt readings, could be full force or nothing.

A voltmeter with very high input resistance can hold the voltage, but the sensor charge signal will be reduced by the not well defined input capacitance of the circuit. Again not very precise and difficult to calibrate.

The integrator circuit with the OpAmp shows the same stable output voltage as long as the force is present and returns to zero if removed. The input impedance is very low, the input voltage is zero. The leakage currents on the board are very low because there is no voltage difference.

Those amplifiers don't have a resistor parallel to Cf. A relay contact is used to zero out the amplifier during initialization and in between, when the system knows, that the force is gone or will be used as new reference.

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  • \$\begingroup\$ Jens, thanks for the nice example. I now know that CSA can also be used for force sensors. Yes, I see the advantage of using op-amp if there is no feedback resistor in the circuit. However, in my application, that is, reading out charges from a radiation detector, we do need a feedback resistor to bring back the baseline. I agree that the capacitance is not well defined if a voltmeter is connected. However, I don't see why this is hard to calibrate. In case of radiation detector, we know the energy deposit in the detector and can use that to calibrate the voltage output. \$\endgroup\$
    – Jing
    Jun 23, 2022 at 3:43
  • \$\begingroup\$ @Jing The calibration issue arises when you just change the connection wire between the charge source and the CSA. If they are on the same board, I agree, it is not a big problem. \$\endgroup\$
    – Jens
    Jun 23, 2022 at 13:11
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Because one of the requirements of the charge sensitive amplifier is that it keeps the input pin at 0V. So the usual implementation is a transimpedance amplifier, whose input is a virtual ground.

If it did not do that, then the charge to be measured would have to charge the whole capacitance connected to it in order to create a measurable voltage, including the capacitance of cables, and the capacitance of the probe... Using a virtual ground input avoids having to measure and calibrate out all that stuff.

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