Capacity Embedded in PCB for High-Sensitive Electrometer Legitimate?

Question

For a highly-sensitive electrometer one needs not only high resistance but also a really small capacitance to achieve certain AC behavior e.g. stability of the opamp.

The figure below shows a typical example of a electrometer circuit with an opamp and can be extracted from this Analog Devices application note.

This is equal to the following circuit.

^{simulate this circuit – Schematic created using CircuitLab}

The values given in the schematics plus the noise budget calculations I did, result in a output range (worst-case, \$T_A = 25°C\$) from \$320\, µ V\$ (noise floor) up to \$ 3.3 \, V\$ when a dual power supply (\$V_+=3.3 \, V\$ and \$V_-=-3.3 \, V\$) is used. According to simulation a \$3 dB\$ bandwidth of \$ 530.52 \, Hz\$ works well for a pulsing of the input signal with \$ 200 \, Hz\$. The idea is suppressing noise below this frequency as you mix your signal up from \$0 \, Hz\$ to a higher \$f_{IF}\$. The bandwidth of the system is mainly determined (or from a ideal POV only determined) by the feedback capacitor and can be calculated with

$$f_{3dB} = \frac{1}{R_f C_f 2 \pi}$$

So the capacitance needed is in the femtofarad range (\$30 \,fF\$). Now my question is if you have a feedback path (cf. picture), where you want to avoid coupling impedances (to somewhere outside of this low-current path) of any kind (you use all kind of fancy guarding techniques e.g. via fencing) would you say it is legitimate to use an embedded, self-made pcb cap? You know like basically two plates on TOP and BOTTOM.

I mean the calculations say, that for \$30 \, fF\$ you need two plates with an area of approx \$1.153 \, mm^2\$ when you use FR4 (\$\epsilon_r = 4.7\$) and a PCB with \$1.6 \, mm\$ thickness.

So you have

$$ C = \frac{\epsilon A}{d} $$

which can be rewritten as

$$ A = \frac{C d}{\epsilon} $$

If I got it correctly you can just draw two squares on the PCB on both sides, each of them has the following dimensions $$\sqrt{A} \cdot \sqrt{A}.$$

Conclusion: Can I use an embedded capacitance when working in the femtoampere (YES \$10^{-15}\$ A) range or am I missing out on something here? It is quite hard to get such small capacitances, which is the reason why a embedded capacitor shall be used.

Answer 1

If you define your sensor as a Charge Meter with a sensor of 100 fF =\${10^{-13}}~~\$ [F] and a constant current of \$10^{-15} [A]~~\$ then

\$\frac{dV}{dt} = \frac{Ic}{C} = \frac{10^{-15}[A]}{10^{-13}[F]}={}{10^{-2}}{}[\frac{V}{s}] = t_R~~=\frac{0.35}{f_{-3dB}}\$ from 10% to 90% [V]

thus \$ f_{-3dB} = \frac{0.35}{10^{-2}}~=35 ~ \$ [Hz]

This \$ f_{-3dB}~~\$may be too high for your expectations.

Potential Solutions:

• reduce stray C by using active guarding around tracks with same CM voltage buffered adjacent to signal tracks such that ΔV=0 thus ΔIc=0 between guard tracks and even less to stray conductors.
• use chopper stabilized methods in datasheet.
• use lower input bias current Op Amps.
• reduce stray leakage R by using air tracks to reduce surface dust creepage R as explained in datasheet
• Compute track PCB capacitance which is 4x air to reduce conductor capacitance to sensors using wider separation and air tracks perhaps with AWG 30~40 magnet wire.
• use gold leaf sensors and measure displacement like Coulomb did which had a long decay time [s]
• use circuits employed for ESD meters to measure discharge or charge with S&H with suitable protection. They often use a charge amplifier (TIA) commonly used for large piezo accelerometers to measure vibration charges to convert to voltage but maybe also BW limited.

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