# Capacity Embedded in PCB for High-Sensitive Electrometer Legitimate?

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.

• I'm confused. You seem to mean "capacitance", not "capacity". But, you describe this "capacity" in Amps (current), not Farads (capacitance). Please edit your question to make it more clear! Thanks. – bitsmack Jun 28 '17 at 17:14
• Sorry. I corrected my mistakes. – m3x1m0m Jun 29 '17 at 8:25

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.

• Thank you for your answer. I had another approach calculating the $f_{3dB}$. $$f_{3dB} = \frac{1}{R_f C_f 2 \pi}$$ You can get this equation by using $P = U \cdot I^*$. So this would result in approx. 160 Hz. It is nothing more than a parallel resonant circuit consisting of R and C. I can not just put C to 0 as I have a pole. So you can see in the LTSpice ac analysis, that it is escalating at some point. That is why I need a capacitance. A really small one. And my question is: Can I do this by putting an embedded capacitance in the feedback loop or will this have negative side eff.? – m3x1m0m Jun 29 '17 at 8:40
• I hope I collected all the information needed to design such a system properly now. It is not a communication system so I do not really have a SNR. But I guess, that is OK? Basically everything above the noise floor ($320 \, µ V$ worst-case at the output) is visible. – m3x1m0m Jun 30 '17 at 8:41