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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.

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  • \$\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
    Commented 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
    Commented 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
    Commented 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
    Commented Mar 18, 2021 at 22:40
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    \$\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
    Commented Mar 18, 2021 at 23:53

1 Answer 1

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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.

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