0
\$\begingroup\$

I want to measure current as small as 0.1mA

I tried looking for current sense amplifiers but I didn't find any which can work at this range.

Please suggest some DIY solution.

Note:

  1. Please don't ask "Why do I need this?", "What is the current source?"
  2. Please don't suggest ready made expensive market ammeters.

Current is DC, 10-15 uA error is tolerable, steps can be as large as 10 uA, max current will be less than 5mA. Voltage drop of 5% is acceptable (applied biased on my load will vary from 10 V DC to 30V DC) Budget is not an issue, I want to make it myself. No preference regarding recording.

\$\endgroup\$
10
  • 9
    \$\begingroup\$ What resolution and accuracy? What is the maximum current you wish to measure? What type of current; DC or AC. If AC (or has significant spectral content) what range of frequencies. Is this a true RMS measurement or just an average measurement? What is the thing you wish to use to display/record the value? \$\endgroup\$ – Andy aka Mar 25 at 8:22
  • 5
    \$\begingroup\$ "Please don't suggest ready made expensive market ammeters." - what is your budget? \$\endgroup\$ – Bruce Abbott Mar 25 at 8:33
  • 2
    \$\begingroup\$ yeah, if all devices that you can buy for a specific purpose (and you've quite a few to choose from here, don't you) are expensive, then probably because building one isn't cheap. We can't even start to work something out without you answering all of Andy's questions above, so until you do: vote to close as too broad. (I'd have more questions to throw in, but honestly, let's start with these. It's probably also a very good idea if you describe the overall purpose of this,a lot of the things can be inferred from that. Edit your question to include the required detail,don't just comment) \$\endgroup\$ – Marcus Müller Mar 25 at 8:37
  • 1
    \$\begingroup\$ For perspective, I'm a site moderator. You can usually get vastly better answers by providing as much relevant information as you can. What is "expensive". What voltage drop is acceptable across the current sensor - this can make a vast difference. What is your source impedance. Load. ... . What accuracy do you want? || If say 0.1 volt drop is acceptable then a sense resistor of R = V/I = 0.1 / 0.0002 = 500 Ohms will allow a very simple and low cost opamp circuit to provide a voltage level of your choice. Whether that voltage drop is acceptable is unknown to us because you haven't told us. \$\endgroup\$ – Russell McMahon Mar 25 at 9:20
  • 1
    \$\begingroup\$ It's almost always cheaper to buy than build something like this, especially since you'd need to calibrate it, and how are you going to do that without a reference meter? \$\endgroup\$ – pjc50 Mar 25 at 9:41
13
\$\begingroup\$

Since you don't care much about voltage drop you can use a resistor and an ordinary 3-1/2 digit multimeter with a 200mV FS range and 10M\$\Omega\$ input resistance.

A 1K resistor will drop 100mV at 100uA and give 100nA resolution (the display will read directly in uA with the decimal point in the correct position).

\$\endgroup\$
12
\$\begingroup\$

You can check out the uCurrent project from the EEVBlog, Schematics and BOM are available and they also sell the unit.

enter image description here

You could use the schematic, get the DIL version of the components and build that on a breadboard.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ This is a good start. Indeed 100 µA is not particularly low a current, that can be easily detected using a transimpedance amplifier as shown here. However, this is not really a current measurement but just a current to voltage converter. The more difficult part of the project is not really adressed here: how to convert the analog reading to a display or communication interface. But this cannot be answered unless more information is posted in the question. \$\endgroup\$ – tobalt Mar 25 at 11:02
  • \$\begingroup\$ @tobalt indeed it doesn't address the conversion part, but given the question supposedly this is open to OP. Hooking up a voltmeter to the output is quite the simplest way, having an ADC/MCU is also a solution and isn't necessarily difficult, but would be the scope of a separate question, that is why I did not include it there. \$\endgroup\$ – Damien Mar 26 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.