Precision, low DA and low value capacitor (1-10pF)?

[Kleinstein]

The reset at the ouput of the integrator usually does not work that way. It would only work kind of it there are protection diodes to ground at the input. It would still leave some residual charge.

[Atomillo]

Current would flow to the capacitor when the switch is closed to ground, but due to the low currents and the low time it would be turned on, such charge would be small.

And because I would be measuring the slope, a small leftover residual charge would be unimportant.

At least that's how I planned it.

[Kleinstein]

One would discharge the capacitor only if there is also a path for the current to flow also on the left side (integrator input). This could work, e.g. if the signal source is a photodiode, but not really well with a more resistive source.

[Atomillo]

Oh I think I understand.
If the switch closes, then because the op amp saturates there is no virtual ground. Thus the voltage across the capacitor reverses sign (before it was positive in the output of the op amp, and now it is the other way around) because charge conserves.

Thus unless there is a diode in the input it would not discharge.

It would work only with some op amps that include protection diodes (in the X Chapters in page 291 Chapter 4x3.7 they mention this trick)

[jonpaul]

Atomillo, a few notes:


Low current: Study JFET and valve electrometers, eg from Kiethley to see the usual techniques. You can buy old electrometers off epay for perhaps $100.
The manual has all info perhaps you can build.

 for example the excelled  KEITHLEY  600B, 602, 604  SOLID STATE ELECTROMETER 
 

Hi Value  R: We  purchased  10,000 pcs  1 Gig R from Victoreen, USA in 1980s..1990s,   OEM part t ~ $1.50 in 1 K qty.

The precision  1..1000 gig  R were used in electromenters, Radiation instruments,    special glass enclosed by Victoreen, etc.


Prices: Nowadays all components are made offshore eg China, even by USA firms, and these parts are special order and high MOQ.

The cost is following the ancient Eco 101 rule of supply and demand.

We see common parts cost increasing 20..200% per year since last 2 years, factory files and shutdown, supply chain logistic, Covid,  US admin dollar debasement policies.

Be patient and keep searching Epay for what you need eventually you may find it.

Philo:  Nothing is fast, easy or cheap! Do the work and be patient, you will be rewarded and learn a lot!


Bon Chance,

Jon

[magic]

I think at those current levels your feedback capacitor could simply be the parasitic capacitance of opamp pins and a reed switch going from OUT to IN-. Wave a magnet nearby and the capacitor discharges. Electromagnet could also work if its coil is screened electrostatically to prevent capacitive charge injection into the input node.

Any integrator resetting technique which involves shorting the output and discharging through protection diodes obviously sends a 0.7V spike to the circuit under test. I hope you are OK with that. If you are OK with even greater spikes, opamps without differential input protection diodes still have diodes clamping each input to the rails. If you short the output side to each rail in turn, the capacitor is guaranteed to hold less than ±0.7V afterwards. With an RRO opamp, such a maneuver could be performed by simply driving IN+ to each rail in turn.(It can't). It could be a good idea to include some resistance in series with the capacitor to limit peak current through the diodes. A few kΩ ought to have negligible effect on normal operation.

As for characterizing the gain of a capacitive TIA, why can't you simply connect a know value ~1000x larger capacitor to the input and drive its other end with a very low noise, low amplitude AC voltage source? Then your TIA is a normal inverting opamp stage and closed loop gain is the ratio of capacitances so the feedback capacitance including all parasitics can be calculated. A very low noise, low amplitude AC source is made by dividing down a low noise, high amplitude AC source with low value resistors.

[Atomillo]

Atomillo, a few notes:


Low current: Study JFET and valve electrometers, eg from Kiethley to see the usual techniques. You can buy old electrometers off epay for perhaps $100.
The manual has all info perhaps you can build.

 for example the excelled  KEITHLEY  600B, 602, 604  SOLID STATE ELECTROMETER 
 

Hi Value  R: We  purchased  10,000 pcs  1 Gig R from Victoreen, USA in 1980s..1990s,   OEM part t ~ $1.50 in 1 K qty.

The precision  1..1000 gig  R were used in electromenters, Radiation instruments,    special glass enclosed by Victoreen, etc.


Prices: Nowadays all components are made offshore eg China, even by USA firms, and these parts are special order and high MOQ.

The cost is following the ancient Eco 101 rule of supply and demand.

We see common parts cost increasing 20..200% per year since last 2 years, factory files and shutdown, supply chain logistic, Covid,  US admin dollar debasement policies.

Be patient and keep searching Epay for what you need eventually you may find it.

Philo:  Nothing is fast, easy or cheap! Do the work and be patient, you will be rewarded and learn a lot!


Bon Chance,

Jon

The "Nothing is fast, easy or cheap" phrase is very true. Now that I have a bit more experience with eBay I see that often the best course of action is to just wait. A lot of the times it seems, an item seems to have an extremely high price because all the ones with reasonable prices get eventually bought and only the unresonable ones remain. In that situation, the most sensible thing is to wait for another reasonable seller to appear. I've also learned you can see the prices of selled items, this gives you a much better insight.

I've printed and I'm reading the manual of K617. All the electrometers seem to use a pair of low leakage JFETs as front end. Nowadays with specialized op amps like the amaizing ADA4622 or such this seems to be mostly obsolete. The problem with buying old used electrometers is that then you also have to buy the necessary triaxial accesories. This hobby can get expensive quite fast!

[Atomillo]

I think at those current levels your feedback capacitor could simply be the parasitic capacitance of opamp pins and a reed switch going from OUT to IN-. Wave a magnet nearby and the capacitor discharges. Electromagnet could also work if its coil is screened electrostatically to prevent capacitive charge injection into the input node.

Any integrator resetting technique which involves shorting the output and discharging through protection diodes obviously sends a 0.7V spike to the circuit under test. I hope you are OK with that. If you are OK with even greater spikes, opamps without differential input protection diodes still have diodes clamping each input to the rails. If you short the output side to each rail in turn, the capacitor is guaranteed to hold less than ±0.7V afterwards. With an RRO opamp, such a maneuver could be performed by simply driving IN+ to each rail in turn.(It can't). It could be a good idea to include some resistance in series with the capacitor to limit peak current through the diodes. A few kΩ ought to have negligible effect on normal operation.

As for characterizing the gain of a capacitive TIA, why can't you simply connect a know value ~1000x larger capacitor to the input and drive its other end with a very low noise, low amplitude AC voltage source? Then your TIA is a normal inverting opamp stage and closed loop gain is the ratio of capacitances so the feedback capacitance including all parasitics can be calculated. A very low noise, low amplitude AC source is made by dividing down a low noise, high amplitude AC source with low value resistors.

I actually didn't even think about the reed switch option. Because it would be a vacuum capacitor there would be no concerns about DA or leakage currents through the inside. And the guard terminal could simply be a piece of copper tape tapep (is that the correct verb?) to the crystal body.

Also, I didn't think about that option. Accurately measuring nF is much easier than measuring pF. Thanks for this ideas they are very interesting.

I actually found an example of reed switch being used as a capacitor: https://xdevs.com/doc/_Metrology/femto_ampere_current_source.pdf by forum member Doktor Pyta. It also shows the importance of using a guard terminal.

[magic]

Leakage is still possible through contamination in the glass or on its surface

But there are switches rated for ludicrously high isolation resistance (teraohms) if you look around. Even a cheaper switch could be tested - run the integrator with open circuit input and you are measuring the sum of all leakages in the circuit. External surface can be washed with IPA to remove fingerprints and crap like that if it becomes a problem.

I'm reading that PDF. I would use a soundcard to generate and measure sinewaves instead of making a triangle generator. Note that you don't need to know anything about their absolute amplitude (within reason), only the accurate ratio of TIA output to the input into the voltage divider driving the nF capacitor.

[Kleinstein]

Reed switches can still have leakage: the glass is not a perfect isolator in a humid environment it can become relatively large leakage current. One could still get away with this with 2 switches in series and a ground link in between. So the relay / reed contact for the input discharge would see essentially zero voltage when open.

For protection there usually is a large resistor (e.g. > 100 K) right in front of the OPs input anyway. So the current during a discharge from the output and protection diodes is limited in current already.

[magic]

That's a good point and the input conducts even less current than the feedback network so it's better that way.
It has also occurred to me that with pF capacitors there isn't much energy to destroy those diodes anyway.

[Atomillo]

Leakage across the reed switch would only occur through the glass surface itselft. The guard terminal would eliminate leakage through the surface and I don't see how leakage could occur inside the evacuated enclosure

Kleinstein, could you sketch that circuit please? I'm having difficulty imagining it.

Many thanks!

[Kleinstein]

Attachted is a short drawing. The right of the 2 relays could as well be a CMOS switch.
The OP could of cause be a lower leakage one


Leakage can be on the outer surface, but also on the inner surface of the glass and even the bulk glass itself is not a perfect isolator.

[Atomillo]

I understand that the feedback capacitor would be another reed switch left always open?

[Kleinstein]

For the integration cap one has multiple options, like an air cap / the shield , a C0G or  polystyree cap.
A reed relay as capacitor would no sense in this case as than could just use 1 reed relay with nothing extra.

[Berni]

If you think that is expensive wait til you find out how much a low leakage triaxial connector costs.

But yeah you can make your own low current source by simply getting a really large value resistor and applying a fixed voltage across it