Analog frontends for DMMs approaching 8.5 digits - Discussions

[Echo88]

Some questions arose about the 3458A current source:
Which effects limit the internal source resistance of the 3458A current source and how could it be approximately calculated?
-> Is it the limited open loop gain of the OP or some other effect?

https://xdevs.com/doc/HP_Agilent_Keysight/3458A/doc/3458A%20CLIP.pdf page 51

Whats an appropriate way to actually measure the current source impedance of the 3458A?
Is it like Vref comparison and id need an even more stable current sink to substract most of the source current and then measure the remaining current change while changing a load resistor?
Attached is a simulation for both polarity variants where the current source only deviates due to the lightly changing leakages across Q305 and Q304 when changing the load resistance.
Also an idea with analog switches, not yet tested though and maybe there are better variants. 

[Kleinstein]

The ouput resistance (or probably better look at the conductance as it's inverse) is already pretty good. There are a few parts to contribute:
1) the leakage on the PCB and E300
2) Q304 as conductance of the gate in reverse   (should not be much)
3) the resistance in the current source time the OP-amp (U300) gain times the isolation from Q305/306.
The isolation that Q305/306 provide should be roughtly on the order of  100 to maybe 1000.
4) some gate leakage contribution of Q305 ( should be small)
5) some thermal effect from heat at Q305/Q306

One could improve on it by bootstrapping the voltage to the diode connected Q304 - though this needs a way to get the BS voltage.
Another point is a high gain op-amp for U300. For not too low a current an AZ amplifier could be an option here.
Another option could be cascode like circuit with another FET.
A higher ref. voltage (instead of 10/13 V) for the ohms part could also help as it would leave more voltage for the JFET Q305 and thus better isolation there.

The 3rd part should already be on the order of 30 Kohm * 1E6 * 100 = 3 Tohm for the 100 µA range.
For the actual use to measure resistance, there would also be the input resistance of the input amplifier / front end.

I would not expect the switching part to contribute, as they see an exxentially constant voltage.
For the new plan with bath polarities the switch U115 may have trouble providing 10 mA - not sure about that. It could at least be boarderline.
The resistors (e.g. R298-302) in the protection need to be higher value, more like 150 K. One should get away with only 4 transistors and thus loose less voltage there.

Measuring the actual output resistance in the Tohm range is challanging. The idea would be to measure the current accurately and than switch in and out some extra drop (e.g. LEDs shunted with a relay (photomos switch).  With an extra voltage drop of maybe 5 V is would be about detecting a change in the current of the 5 pA for 1 TOhm. Already with only 1 µA to start with this is hard.