If I understand the intended behaviour of the circuit of post #13 correctly, its input is supposed to be a current input, so the lower the impedance the better. Then why put 1 megohm in series with the virtual ground?
The 10 megohm is straight in series with one of the op-amp inputs and should theoretically do nothing (except adding some noise).
All I could think of, is that the 1 megohm keeps things stable when you connect something capacitive to the input and that the 10 megohm limits the current through the op-amp input during overload or ESD discharge.
I overlooked it at first, but R1 has no impact on the current gain from the input to the meter. Maybe it limits the maximum current through the current meter, or it sets the gain if you also want to look at the op-amp's output voltage? You could connect a scope straight to the op-amp's output without any buffer.
As I said earlier, this topology allows a direct reading display on a cheap voltmeter module.
Also, the constant voltage method has a relative uncertainty that increases proportionally with the resistance, and an absolute uncertainty proportional to the square of the value
No, and anyway I don't want to restart the project from scratch. If I cannot find a fix, I will scale back my objectives to 10G
The - input is currently lifted and goes directly to the source of the transistor and the 100M resistor. The leakages are probably collected by this resistor and the reed relay, but practically everything in the vicinty is tied to the +12V, which is even "better" than the +input, since leakages from there should increase the CCS current, not reduce it
I understand the motivation for a simple solution, my brain operates that way!
I wouldn't call myself an authority in these measurement circles, but I am familiar with the process of making and taking high accuracy GigOhm and well as uOhm measurements, due to the nature of my profession, as well as calibration methodology (I have an old Keithley low level measurement book, that was mentioned, and several other very useful texts on the subject, as well as time and access to calibration standards and equipment.)
What is difficult overcome in the reverse methology of forcing amps measuring volts:
An accurate and temp co stable high resistance measurement shunt for IV conversion.
Prevention of induced noise in the circuit and influence on operational stability.
Component leakage currents....
You must consider that even the ancient and venerable BAT50 pico amp diode, and its rated leakage current, low as it is, influencing the circuit - no issue perhaps, but that is for a very very low leakage device, difficult to match or approach matching in other devices.
as an aside, modern low leakage diodes are not anything of the kind- typically they have an order of magnitude or more reverse leakage current, typically in the nA rather than pA.
At this level, a nanoAmp matters.
so perhaps, thinking aloud and trying to be helpful to this topology, even though I disagree with it....
one could use a diode reverse leakage to determine the excitation current for the circuit, by calibrating a pA leakage diode at a range of temps and measuring actual leakage. Then using this diode as a known within the circuit.
just an idea (I have a resistance bridge that uses the leakage of 741 as a diode current reference, that partially inspired it- note though, tht this is for low resistances 🤣)
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The reverse current of a semiconductor junction is heavily (but probably predictably) dependent on temperature, and other factors, some controllable, like light, some more random, like crystal defects progression, accumulated dose of background radiation, etc.
The input current of an opamp might be more predictable: the temperature dependence of the input transistors beta could compensate the leakage increase, but the device used must be selected carefully.
The forced current method has another advantage: the cold terminal is the ground, which can be advantageous to test bulky devices, or permanently earthed devices
The input current of an opamp might be more predictable: the temperature dependence of the input transistors beta could compensate the leakage increase, but the device used must be selected carefully.
The forced current method has another advantage: the cold terminal is the ground, which can be advantageous to test bulky devices, or permanently earthed devices
Certainly, light and temps.
Forced current does not have an inherent advantage, that I can see (I can't see why a voltage driven system cannot be used on earthed devices as well - since its generally this way, in real life, with installed plant)
The input bias current of an op amp could be used, provided you can find one low and consistent enough. Again maybe MOSFET, but 1pA gate charge current, I'm not confident that is good enough- especially in the presence of noise, gate capacitance, and the likely huge DUT insulation resistance.
I wonder if you are seeing issues from the time needed to charge the gate Capacitance, through the DUT resistance - and hence not approaching anything close to threshold volts?
Also my point mentioning reverse diode leakage, is to say that this will be occurring within the MOSFET gate leakage, and also the inherent structural diode, and its own reverse leakage current.
Some questions
What is the device under test? Is it a resistor or are you measuring insulation resistance of something?
How many significant digits do you require?
I have access to a calibrated DMM7510. Measuring 1G ohm has 9000 ppm uncertainty. If you look at the rise of uncertainty per decade, the uncertainty at 100G would using FIMV would be at least 10%. Measuring a 1G shielded resistor-in-a-box yields 4 Meg ohm RMS of noise. So 3 good digits at best.
The DMM7510 uses a 10M resistance in parallel with 0.69 uA current source to measure a 1G resistor.
Have you tried placing a 10M resistor in parallel with your device under test? You can do the math and correct your voltage reading. Voltage will reduce slightly when you place your DUT across the 10M resistor.
What is the device under test? Is it a resistor or are you measuring insulation resistance of something?
How many significant digits do you require?
I have access to a calibrated DMM7510. Measuring 1G ohm has 9000 ppm uncertainty. If you look at the rise of uncertainty per decade, the uncertainty at 100G would using FIMV would be at least 10%. Measuring a 1G shielded resistor-in-a-box yields 4 Meg ohm RMS of noise. So 3 good digits at best.
The DMM7510 uses a 10M resistance in parallel with 0.69 uA current source to measure a 1G resistor.
Have you tried placing a 10M resistor in parallel with your device under test? You can do the math and correct your voltage reading. Voltage will reduce slightly when you place your DUT across the 10M resistor.
You should study basic op amp amplifier implementations. The T-network is a variation on ordinary op amp amplifier topology. The - and + inputs are virtually the same voltage and no current flows into either node for an ideal op amp. The output drives its node to make the + and - inputs the same voltage.
The resistors in the picoammeter circuit form voltage dividers between the input and the output such that current amplification is much greater than when not using a T network.
A T network can be used to create a faux 1 Terra ohm resistor using two 1 meg resistors and a 1 ohm resistor. It is an instructional exercise to make a spreadsheet to calculate the effective resistance for various combination of resistors. From inspection the input of this T network is a 1,000,000 to 1 divider. 10 volts in produces 10 uV across the 1 ohm resistor. The output 1 meg resistor creates 10pA from the 10uV across the 1 ohm resistor. Hence these 3 resistors behave the same as a 1T resistor.
I'll have to draw a schematic and a couple of equations to explain what I mean. To be continued.
Indeed.
If it's an op-amp with an MOS input, the leakage is dominated by ESD protection network leakage, which is largely or completely diode reverse leakage.
If it's a JFET input op-amp, it's the gate leakage, which is also mainly junction leakage.
If it's a bipolar op-amp, the input bias current is usually way too high.
I know that, but as the meter measures the current, R1 drops out of the equation - unless you also want to connect the op-amp's output to some kind of voltage measuring device, of course. The transfer from input current to current through the meter is simply 1 + R5/R2 when the 10 Gohm resistor is called R5, with the sign conventions as in this sketch.
Hence my question about the purpose of the other resistors, R1, R3 and R4.
It could be anything, resistor, insulation resistance to GND, etc
My crude DPM has 3 digits: as I said, it is a quick and dirty tester
Yes, of course, I know the tricks, but my goal is not to measure a specific resistor, I want to build an easy to use instrument, usable without math or mental gymnastic: connect the test terminals and read the value
Ok. That is what I wanted to know.
An ideal current source has infinite output impedance. Your 100G resistance is much closer to infinity than 0 ohms. A high impedance current source driving some cabling to a high impedance load is going to pick up electrical and magnetic noise. If a simple and elegant FIMV megger could be built, you would find them on eBay from China.
An ideal voltage source has 0 ohms impedance. A low impedance source driving some cabling to a high resistance load will be much more immune to stray electric and/or magnetic fields. This is why elctrometers are FVMI and not FIMV. Another reason why FVMI is superior is that leakage at the high side is not measured. With FIMV, any leakage on the high side is an error term.
If I wanted to build one of these, I would make a simple op amp amplifier with the DUT in the feedback loop and the gain setting resistor at 1G or 10G. The op amp output driving your meter directly. This is the simplest current source.
An ideal current source has infinite output impedance. Your 100G resistance is much closer to infinity than 0 ohms. A high impedance current source driving some cabling to a high impedance load is going to pick up electrical and magnetic noise. If a simple and elegant FIMV megger could be built, you would find them on eBay from China.
An ideal voltage source has 0 ohms impedance. A low impedance source driving some cabling to a high resistance load will be much more immune to stray electric and/or magnetic fields. This is why elctrometers are FVMI and not FIMV. Another reason why FVMI is superior is that leakage at the high side is not measured. With FIMV, any leakage on the high side is an error term.
If I wanted to build one of these, I would make a simple op amp amplifier with the DUT in the feedback loop and the gain setting resistor at 1G or 10G. The op amp output driving your meter directly. This is the simplest current source.
After some reflection, I have noticed an inconsistency in the values I measured.
This is the circuit in test mode:
IL are the possible unknown current leakages.
What is strange is that I1 and I4 seem to track quite closely: with R2 at 250M, the currents are almost identical, ~250pA.
With the resistor at 900M, they both fall to zero.
That's not logical: if IL steals some current from the source node, I1 will remain the same unless IL is much too large. However, I4 should decrease by IL, but it doesn't, and I cannot find a satisfactory explanation. The circuit behaves more or less as if the 100mV reference was reduced, which it isn't of course, it comes from a low impedance and is rock-stable
One thing is certain: the problem is caused by leakages, since cutting through the FR4 improved the situation
This is the circuit in test mode:
IL are the possible unknown current leakages.
What is strange is that I1 and I4 seem to track quite closely: with R2 at 250M, the currents are almost identical, ~250pA.
With the resistor at 900M, they both fall to zero.
That's not logical: if IL steals some current from the source node, I1 will remain the same unless IL is much too large. However, I4 should decrease by IL, but it doesn't, and I cannot find a satisfactory explanation. The circuit behaves more or less as if the 100mV reference was reduced, which it isn't of course, it comes from a low impedance and is rock-stable
One thing is certain: the problem is caused by leakages, since cutting through the FR4 improved the situation
Yes, I studied the possibilty, and Marcel also proposed it.
In the end I rejected it because none of the test terminals is grounded, and it requires a bipolar supply. This version just requires a 15V unipolar supply (the voltmeter module only accepts a positive input)
How do you measure these currents? Are you sure both op-amps are stable for all combinations you tried? Does the output voltage of U1 go outside its specified range? (It shouldn't, but clearly there is something happening that shouldn't happen.)