1
\$\begingroup\$

This is my current sense circuit that I connect to Arduino in order to measure how much current a load draws.

enter image description here

As you can see at the bottom of the image there is a -5 V voltage source connected to the instrumentation amplifier.

The -5 V voltage source I design is this using TLE2426CD.

enter image description here

I get stuck trying to connect that -5 V to the instrumentation amplifier. I try different connections on the simulator but I not get same results as the first picture. I think my problem is in the concept of virtual ground.

\$\endgroup\$
7
  • \$\begingroup\$ Ehhhmmm, you've got -5 V on XMM1 because you have connected the meter's leads backwards on a +5 V supply. -(-5 V) = +5 V. \$\endgroup\$
    – Transistor
    Mar 18, 2021 at 22:33
  • \$\begingroup\$ Transistor, as you said I put the multimeter on the photo in order to show that my "virtual ground" is pin 1 of U3. \$\endgroup\$
    – marcosbc
    Mar 18, 2021 at 22:35
  • \$\begingroup\$ Is the intent to actually make the -5V reference isopotential with the negative input on the instrumentation amplifier or did you put those ground symbols there to make the simulation converge? \$\endgroup\$
    – crasic
    Mar 18, 2021 at 22:35
  • \$\begingroup\$ Crasic, in the first photo I put the ground simbol between AD623 and MCP6042 in order simulation to work. When I try to connect the -5v power supply I design the ground simbol was removed in order to use my "virtual ground" that is pin 1 of U3. \$\endgroup\$
    – marcosbc
    Mar 18, 2021 at 22:40
  • 2
    \$\begingroup\$ Your TLE2426CD does not provide -5V. Its takes the +10V you give it and splits it in half - so it's giving you +5V. The point of using something like a TLE2426CD is that you use its output as your 0V ground point in the rest of your circuit. Once you do that, your original 0V ground becomes -5V because it's 5V less than the output of the TLE2426CD. \$\endgroup\$
    – brhans
    Mar 18, 2021 at 23:53

1 Answer 1

Reset to default
1
\$\begingroup\$

The "0V reference" is created by the TLE2426D:

enter image description here

So, initially, to match only your simulation, the power rails would become:

enter image description here

The 0V connection to the input (if needed) can't be determined without knowing more about that you are measuring.

But, according to your text I assume you are powering your arduino from the 12V (which is correct) and not from the TLE2426, and you want your output varying from 0 to 5V according to the power supply. From a previous post, I also assume you need to measure V4 with both polarities. Hence, the INA output, according to the 12 V power supply and your gain and ref:

$$V_4=-50mV \rightarrow V_{INA} \approx 2.25 V$$

$$V_4=50mV \rightarrow V_{INA} \approx 7.25 V$$

Is this correct? If it is, you need to "shift down" the output by 2.25V, and not multiply it by two. You could provide this shift by applying a different reference to the INA, using the op. amp. as a voltage follower connected to a voltage divider.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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