Back to back diodes act as protection as long as meter can survive about 0.7V across its terminals/
Back to back silicon diodes will not affect readings substantially below about 0.3V across meter terminals.
Use of a voltmeter on a 200 mV range and a series sense resistor will often allow destruction proof current metering plus superior burden voltages compared to typical good quality ammeters. (eg 1 mV/ma with 0.1 mA resolution or 0.1 mV/mA with 1 mA resolution).
Suggestion provided for superior test method.
Reverse parallel diodes do potentially work, but it depends somewhat on your meter characteristics. Having a "poor" meter spec may make diodes work better than having a good spec. As a guide (see below) a meter with protection diodes should not be used above about 300 mV dop across the metetr (if that) as as diode conduction is approached current rises exponentially.
In the article cited by Vicatu they mention figures of 1.8 mV/mA for a Fluke 87v meter. And they quote 70 uV/mA for their solution on the 300 mA range.
The term "burden voltage' is just jargon for saying that the meter has resistance, so that when current flows through it there is a voltage drop. Just Ohms law V = I x R.
1.8 mV/mA translates to 1.8 ohms meter resistance.
70 uV/mA translates to 0.07 ohm.
An acceptably good alternative can be had with multimeters which have a 200 mV range. In a 3.5 digit meter this gives 100 uV resolution in the LSD.
If you pass current through a 1 ohm resistor it drops 1 mV per mA. If you measure that voltage with a 200 mV meter range yu get a resolution of 0.1 mA and a maximum current of 200 mA. BUT if you then apply a high current you may fry the resistor BUT the meter, being on a volts range will survive.
If you use a 0.1 ohm resistor (= 0.1 mV/mA = 18 x better than the fluke meter) then at 1 mA you get 100 uV and the 200 mV range will resolve 1 mA. Note as aever that resolution and mV are not the same. This arrangement will now read up to 2 amps in 1 mA steps. Again, meter is safe against overload.
Using the 1 ohm resistor, you get 1 mV/mA so at say 2 mA you get 2 mV droo which is highly acceptable for testing almost anything. At that current you could often use 10 ohms for 10 mV / mA, 10 uA resolution (notionally), a drop of 100 mV at 10 mA and a full scale deflection of 20 mA on a 200 mV range. And you will not damage your meter when you attempt to draw an amp.
At 1 ohm resistance = 1 mV/mA to get 700 mV drop to activate a diode you need 700 mA. With 10 ohms series resistance you need 70 mA. With 0.1 ohm series resistance you need 7 amps. And in each case the meter will not be damaged with or without diodes.
Using the Fluke 87V at 1.8 mV/mA you need 700/1.8 =~ 390 mA to produce 700 mV for diodes to bite on. One would hope that a Fluke set 20 200 mA range with say 0.1 mA resolution (and maybe 0.01 mA) would survive a 2:1 overload.
The above article suggsts that other meters are far worse than the 87V for burden voltage. Which just means that they would be protected at even lower mAs of overload.
A problem is that a diode does not conduct sharply. The exponential knee will cause substantial current fklow at 500 mV and discernible flow at even 300 mV. So the effective mA range of a meter with diodes may be say half of 700 mV drop maximum.
I have a meter which is excellent for this.
It has a 10A range with auti ranging - so you get massively low shuntresistance combined with autoswitching from 10A to 200 mA ranges.The 200 mA range gives me a 100 uA resolution. Whether that is good enough is up to you.
You may b tempted to us Schottky diodes to reduce the forward voltage drop. Unfortunately these have very poor (_= high) reverse current specs and this gets worse or much worse as temperature goes up. So Schottky f=diodes will not usually be useful as protection diodes/
I have not seen the circuit of the micro-current adaptor but assume that it is essentially a suitable low offset voltage and low bias current amplifier - possibly just a single hi spec opamp. eg an 0.01 ohm resistor returns 10 uV per mA - amplifying the voltage across this with 100x gain gives 200 mA measurement on a 200 mV range with 0.1 mA resolution. Stepping upto a still acceptable 0.1 ohm sense resistor and still 100x gain gives 0.01 mA resolution on a 200 mV range.
Aha - they gave the circuit ...
Yes, just a single amplifier. The LMV321 is just used to set the 1/2 supply virtual ground point. The MAX4239ASA opamp datasheet here.
$2.47/1 at Digikey. A bargain. They say -
The MAX4238/MAX4239 are low-noise, low-drift, ultrahigh precision amplifiers that offer near-zero DC offset
and drift through the use of patented autocorrelating
zeroing techniques. This method constantly measures
and compensates the input offset, eliminating drift over
time and temperature and the effect of 1/f noise.
Both devices feature rail-to-rail outputs, operate from a single
2.7V to 5.5V supply, and consume only 600µA. An activelow shutdown mode decreases supply current to 0.1µA.
The MAX4238 is unity-gain stable with a gain-bandwidth product of 1MHz, whi le the decompensated
MAX4239 is stable with AV ≥ 10V/V and a GBWP of
6.5MHz. The MAX4238/MAX4239 are available in 8-pin
narrow SO, 6-pin TDFN and SOT23 package
Current through a forward biased PN junction at low Vf :
Current versus voltage across a PN junction can be approximated as exponentially increasing current with voltage (or logarithmically increasing voltage with current.) The underlying theory is "complex" (to put it mildly).
This website provides a superb explanation (as part of a full text book on "Princuples of semiconductor devices") in as much detail as you are liable to find anytwhere (and more than you may want) but in as accessible a form as such material can be.
A reasonably good summary is given by the chart below.
"n" is a diode ideality factor (often assumed to be 1) which varies between 1 & 2 in the different regions.
We are usually interested in the "high injectuon region" from about Vf = 0.5V to 0.8V. As a rough guide this indicates that at Vf = 0.3V current will be about I_0.6V/10,000 and at Vf = 0.1V current ~= I_0.6V/1,000,000 . ie you can expect minimal effects on readings at Vf = 0.3V and utterly minimal effects at Vf=0.1V at the sort of accuracies liable to be of interest.
I use the following method of providing a testing power supply with good success. I've shown the concept using an LM317 but I actually use a P Channel MOSFET, opamp and voltage reference in place of the LM317. The effect is much the same EXCEPT that the LM317 will add it's operating current to the measured current whereas my actual arrangement adds essentially nothing. I can draw up my actual arrangement and add it if there is any interest.
Provide an adjustable power supply
On the "upstream" side of the voltage regulated supply provide a series "current sense" resistor whose ** voltage drop** will be an indication of load current.
Measure voltage drop across the series resistor to determine current drain.
- Provide a filter capacitor Ci at the input to the voltage regulator. Large rvalues will smooth the current meter response but improve the output stability under step load changes.
The current sense resistor can be eg 0.1 ohm or 1 ohm or 10 ohms or other.
The value of this resistor is NOT reflected in the stability of the output under load variations BUT it does affect the required input voltage.
Vin min = vout + Vregulator_dropout + Imax x Rsense.
Rsense = 1 ohm.
Vsense = 1mv per mA or V per amp (not surprisingly.)
Power in R sense = I^2R = 1 Watt at 1 amp, so easily done.
200 mV meter will easily resolve 0.1 mA.
Rsense = 10 ohm.
Vsense = 10 mV/mA.
Power in R sense = I^2R = 10 Watt at 1 amp, so design needed.
Power = 0.1 Watt at 100 mA.
1 mA = 10 V.
10 uA = 100 uV so this will resolve 10 uA on a 2 mv, 3.5 digit meter.
I max measured on 200 mV range = 20 mA BUT meter will not be damaged by 100 mA or 1A overloading.
Use of an autoranging meter allows currents from say 10's of uV to say 1A to be measured.
.
200 mV meter will easily resolve 0.1 mA.
It is not difficult to integrate current. If you're willing to roll your own you'll have complete control over the specs.
As you probably know, a capacitor has the relationships Q = CV and \$Q = \int i \cdot dt\$ .
One way I see off the top of my head to do this is to create a current mirror to charge a cap. Reading off the voltage of the cap is all that is needed. You can get caps as accurate as you need and there are many accurate current mirror configurations.
With such a method you can really get any amount of complexity that you need. You could have multiple resolutions(multiple mirrors and caps of different sizes). You can use op amps to improve the resolution and create a simple reset.
Of course it's not as simple as using a chip but as you have already stated, you can't find any chips that suit your needs.
It may be possible use current sensing(even proximity) but I'm not sure the accuracy you will get. For example, if your load is fairly low you could stick a 1ohm resistor in series. The voltage across the resistor is then equal to the current. Integrate this(say, using an op amp) and you have the charge. The efficiency here would be much larger, almost near unity while the current mirror method will be slightly less than 50%.
Best Answer
What is the origin of these currents?: Every cable is an antenna. I think you measure noise (FM radio, cell phone, WIFI, Bluetooth, DECT, ...)
How do I measure currents with nA accuracy? Use shielded cables; shield your DUT; Use a battery as supply for your DUT; Eventually use a shielding enclosure for the whole setup (DUT + DMM); increase the number of samples the DMM uses to build the RMS value;
Maybe this free book helps: "Precision DC Current, Voltage,and Resistance Measurements" https://download.tek.com/document/LowLevelHandbook_7Ed.pdf