4 ms·

It's apparently hard to measure zero vs extremely low resistance with the two probe setup you are imagining (I guess because the probes and wires aren't superco

It's apparently hard to measure zero vs extremely low resistance with the two probe setup you are imagining (I guess because the probes and wires aren't superconducting, so most of the resistance in the circuit is not in the sample). The graphs I have seen are all made with a four probe setup [1], where a constant current is run through the outer probe pair and the voltage across the inner pair is measured. If the inner probes have contact issues, the voltage (and inferred resistance) drops, potentially to zero. [1] https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds/4wireres.html https://www.ni.com/docs/en-US/bundle/ni-daqmx/page/measfunds...

applied_heat 3y ago
That’s what he said. You inject a known current and measure the voltage drop across the item you want to measure the resistance of, and then use ohms law
- SV_BubbleTime 3y ago
  And what he saying, and you weren’t listening to is it all your testing equipment is introducing resistance. You can mesaure current by electrical fields, but have the same issues as before. Your testing equipment and your very tiny sample sizes.
  - fnordpiglet 3y ago
    I sort of don’t get this either. If your testing equipment is introducing resistance can you not “tare” or calibrate your measurements by measuring the equipments current flow in a direct circuit between probes to determine its base resistance then introduce your sample and measure the difference? The resistance in a serial circuit is additive, no?
    - deleted 3y ago
      [deleted]
    - proto_lambda 3y ago
      The parasitic resistance will be orders of magnitude larger than your sample resistance, any detector capable of detecting the sample resistance would be completely swamped out by it.
      - jacquesm 3y ago
        They are measuring voltage, not resistance (the resistance is computed). It's impossible to measure resistance without current and some voltage across the resistance, and while the parasitic effects are small the resistance we're talking about (and hence the voltage across that resistance) is so small that second order effects introduced by the probe wires can have a real effect, for instance if the probe wires have a higher resistance that makes them more susceptible to electric fields, which in turn would show up as a voltage. So this is anything but trivial.
    - jacquesm 3y ago
      If you pick your reference current properly this effect should be extremely small. It will have a little bit of effect but nothing that you should normally have to worry about.
  - jacquesm 3y ago
    In a four probe setup you are measuring voltage, not resistance. But you are right that the measurement can still influence the results when the voltages measured are tiny. The way to think of this is simple: you can't measure anything without subtle joining the circuitry that you are measuring and that has an effect on the properties of the circuit as a whole for which you have to compensate. In this case: the voltage measurement is going to consume a tiny bit of power and that is due to the resistance of the measurement apparatus even if it isn't in the main current path but a secondary one. But the people that do these kinds of measurements tend to be well aware of this and will pick their measurement gear and reference current to minimize the chances of that happening.
jacquesm 3y ago
I am not imagining a 'two probe' setup (see other comment in this thread).

applied_heat 3y ago

That’s what he said. You inject a known current and measure the voltage drop across the item you want to measure the resistance of, and then use ohms law

SV_BubbleTime 3y ago
And what he saying, and you weren’t listening to is it all your testing equipment is introducing resistance. You can mesaure current by electrical fields, but have the same issues as before. Your testing equipment and your very tiny sample sizes.
- fnordpiglet 3y ago
  I sort of don’t get this either. If your testing equipment is introducing resistance can you not “tare” or calibrate your measurements by measuring the equipments current flow in a direct circuit between probes to determine its base resistance then introduce your sample and measure the difference? The resistance in a serial circuit is additive, no?
  - deleted 3y ago
    [deleted]
  - proto_lambda 3y ago
    The parasitic resistance will be orders of magnitude larger than your sample resistance, any detector capable of detecting the sample resistance would be completely swamped out by it.
    - jacquesm 3y ago
      They are measuring voltage, not resistance (the resistance is computed). It's impossible to measure resistance without current and some voltage across the resistance, and while the parasitic effects are small the resistance we're talking about (and hence the voltage across that resistance) is so small that second order effects introduced by the probe wires can have a real effect, for instance if the probe wires have a higher resistance that makes them more susceptible to electric fields, which in turn would show up as a voltage. So this is anything but trivial.
  - jacquesm 3y ago
    If you pick your reference current properly this effect should be extremely small. It will have a little bit of effect but nothing that you should normally have to worry about.
- jacquesm 3y ago
  In a four probe setup you are measuring voltage, not resistance. But you are right that the measurement can still influence the results when the voltages measured are tiny. The way to think of this is simple: you can't measure anything without subtle joining the circuitry that you are measuring and that has an effect on the properties of the circuit as a whole for which you have to compensate. In this case: the voltage measurement is going to consume a tiny bit of power and that is due to the resistance of the measurement apparatus even if it isn't in the main current path but a secondary one. But the people that do these kinds of measurements tend to be well aware of this and will pick their measurement gear and reference current to minimize the chances of that happening.

jacquesm 3y ago

I am not imagining a 'two probe' setup (see other comment in this thread).