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I want to measure resistances in the range of 0.001 Ω, to 0.5 Ω, at a rate of 80 kHz, with a power usage of less than 30 mA @ 24 V. To do this, I've tried implementing an op-amp in a few 'classical' configurations (inverting amplifier, non inverting, adding, and so on), and the power needed for such a measurement is incredible (on the order of 10s of watts).

Eventually I came to the conclusion that running a 1 mA current through the resistor could provide an adequate response to measure the resistance, however, I'm still encountering issues in incorporating feedback in the op-amp circuit, so I'm not sure this approach is viable.

What sorts of methods exist for measuring such a low resistance on a power budget?

Here is my current attempt in LTSpice:

enter image description here

In this case, U1 is a texas instruments TL084. There is 4 decades of amplification here, which means that It's going to amplify any noise on pin 1 by 1000x, meaning that RF and thermal signals will be quite significant.

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  • \$\begingroup\$ Do you want to measure the resistances without some kind of auto range? Are they connected to something else? \$\endgroup\$
    – Jens
    Commented Nov 26, 2022 at 3:40
  • \$\begingroup\$ @Jens I'd prefer if there was a single range, but 2 ranges, or a logarithmic output is great too. \$\endgroup\$
    – tuskiomi
    Commented Nov 26, 2022 at 4:16
  • \$\begingroup\$ What format does this measurement have to be in? \$\endgroup\$ Commented Nov 26, 2022 at 4:45
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    \$\begingroup\$ @tuskiomi I'm glad to read that a log output would be acceptable. Given the input dynamic range is 1:500 it's probably a good idea. But I've no idea what kind of tolerance your 'resistance' has for pulsed currents, for example. Can it withstand a high pulse from a capacitive discharge, for example, without destroying it? what about your 80 kHz ADC? How long must the signal be stable to capture it in the sample and hold? How long to take a reading? Anyway, all kinds of ideas and way too little detail about the limitations and boundaries that surround your measurement target. \$\endgroup\$
    – jonk
    Commented Nov 26, 2022 at 6:02
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    \$\begingroup\$ @tuskiomi So it is fine if it reads 10 mOhm when the resistance is actually 1 mOhm, so long as when presented with 1 mOhm it again produces 10 mOhm. So is the <5% with respect to 1 mOhm 'true' value (which without NIST traceable and accurate equipment and methods you have no way of really knowing), or the reading itself, or the 0.5 Ohm maximum range value? It you mean precision to 5% of the minimum detectable value, 1 mOhm, then that is 50 millionths of an Ohm instrumental precision. I think there's a lot that's not being disclosed that may make all this more reasonable. But I'm done for now. \$\endgroup\$
    – jonk
    Commented Nov 26, 2022 at 7:49

1 Answer 1

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You should be able to make a buck converter from your 24 VDC to provide 100 mA at about 1 volt, so that you can have 100 uV on the 0.001 ohm load. An instrumentation amplifier can read the test signal and amplify it to 100 mV. If that's not enough, you could even get 1 amp at 500 mV, or total of 500 mW which is within your limit of 24V 30 mA or 720 mW. A 1 ohm resistor in series with the load could provide 500 mV to be used with a ratiometic ADC circuit.

I don't know why you need to measure at 80 kHz, but many microcontrollers can do that with 8 or 10 bit resolution.

Please add a schematic or block diagram showing your best try at a design based on these guidelines, and we can help getting it to work to your specifications and accuracy requirements.

(edit) Another method to reduce power usage for a Digital Low Resistance Ohmmeter (DLRO) is to charge a capacitor (or battery) to about 1.2 VDC and then discharge it through a 1 ohm (or 100 mOhm) resistor and the load to be measured, so it will have a test current of 1 or 10 amps, while drawing an average of only a few hundred mW. But this is probably unsuited to your 80 kHz requirement.

(edit2) Here is a simulation of a buck converter with an output of 100 mA into 10 ohms (1V), and two op-amps providing a total gain of 20,000, so a 100 uOhm test resistor provides 200 mV, and 1 mOhm provides 2.0 V, for an ADC. I had to provide separate +/- 5V supplies for the AD822 dual op-amp. Even though the AD822 is supposed to be rail-to-rail input and output. These supplies could be implemented with a 78L05 regulator and a TC7660 charge pump.

LTC3631 24V to 1V 100 mA Buck Converter

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  • \$\begingroup\$ I'll take these into account and post my attempts \$\endgroup\$
    – tuskiomi
    Commented Nov 26, 2022 at 6:28
  • \$\begingroup\$ I updated my question- I'm not sure buck converters can reliably go to 100uV alone, can they? I'll need some sort of linear filtering? \$\endgroup\$
    – tuskiomi
    Commented Nov 26, 2022 at 22:14
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    \$\begingroup\$ It should be possible to make a current regulated buck converter to get 1 ampere, from a 24 VDC source, although it might require a very low duty cycle using just a series inductor. It might be more practical to design a step-down transformer. But you will need to take into account the forward voltage drop of rectifiers, which will mean about 500 mV at 1 amp, or 500 mW, so an active rectifier is probably needed to get power below 700 mW. \$\endgroup\$
    – PStechPaul
    Commented Nov 26, 2022 at 23:19

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