0
\$\begingroup\$

I am building a simple current sensor. The amp says the maximum differential voltage is 5V. I'm not getting more than 1V in differential voltage at the moment.

Is it safe to test differential voltage on a floating 24V line? What if the grounds are referenced?

It seems like if the line is floating it would work ok but if it was ground referenced it might kill the amp, seeing as the internals might see a 24V potential difference.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Look for common mode range, if I'm understanding things (which I may not be.) \$\endgroup\$
    – jonk
    Apr 25 '17 at 23:20
  • \$\begingroup\$ @jonk is right, the common mode range is the key. Why not post a link to the datasheet and add a schematic of what you are trying to do? Then we can say for sure if it will work or not. \$\endgroup\$
    – John D
    Apr 25 '17 at 23:29
  • \$\begingroup\$ @JohnD mouser.com/ds/2/609/ADA4528-1_4528-2-878254.pdf \$\endgroup\$
    – RobC
    Apr 25 '17 at 23:48
  • \$\begingroup\$ looks pretty good to me... so long as the 30V on the on the shunt falls between the supply terminalks on the OP-amp. (perhaps use a resistive divider to ensure this) - and you probably want some negative feedback unless you are using the op-amp as a comparitor. \$\endgroup\$
    – Jasen
    Apr 26 '17 at 5:30
1
\$\begingroup\$

As shown your circuit will not work. You state that the common mode input range of the amplifier is 5V, yet you have the inputs tied to a shunt on a 24V output.

You could try to build a precision divider on the input to the amplifier, but a simpler solution is to get an amplifier more suited to the task. There are lots of current shunt monitor products that can operate with a 24V common mode input. Here's one example from TI, but you can find others from TI or Maxim or LT etc.

INA303

\$\endgroup\$
0
\$\begingroup\$

As shown, your circuit may work. That's because you have labeled the current sense resistor terminals as 'floating' input, which means that there is no common ground (and the marked input voltages are therefore not ground-referenced). That means that the input impedance of the amp will set the DC level referenced to the negative rail, unless there are other connections not shown.

There MUST be other connections not shown, because the schematic lacks any negative feedback. Since you know what input pin is at lesser potential, it might suffice to ground that one and use a noninverting voltage gain amplifier feedback.

\$\endgroup\$

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.