I trying to build a simple differential measurement circuit to feed a analog to digital chip to monitor current consumption of a board with an Arduino. So I decided to go with a "traditional" differential amplifier with op-amp in front to get a nice high impedance for this improvise differential probe. Here's a simple schematic of my circuit: (Op-amp have 12v supply)


To my surprise, the reading on the 5 ohm shunt resistor is way off the real reading (measure with a multimeter). Instead of having 2.02V which correspond to a 404mA consumption (which is what is expected), I got 43mV!? I used this probe (pin 3 & 5) to check voltage around others components on my board and the reading are correct. But on the shunt R1, it's completely off. I tough that since it's a really low resistance that my high-impedance will be ok be it seems not. I look around on the net and the solution seems to go with this approach but that's not working for me right now... Any suggestion?

  • 1
    \$\begingroup\$ You're powering the TLC272s off the same 12V rail? If so you are violating the common mode input range for IC1A and IC1B -- their inputs need to be about 1V below the positive rail, according to the datasheet. \$\endgroup\$ – Null Jul 11 '15 at 21:26
  • \$\begingroup\$ Even worse, you may be powering the 272 from the +5 line, in which case you're lucky the chip is still working - if it is still working. \$\endgroup\$ – WhatRoughBeast Jul 11 '15 at 21:33
  • \$\begingroup\$ @Null, which spec are you talking about? Page 4 : Differential input voltage, VID = +- VDD and Input voltage range, Vi is -0.3 to VDD. ?? Anyway, I made a test and and 2 x 2 diode in series to drop the voltage and it's the same. \$\endgroup\$ – Steve S. Jul 11 '15 at 21:47
  • \$\begingroup\$ @WhatRoughBeast My question say : Op-amp have 12v supply. ?? \$\endgroup\$ – Steve S. Jul 11 '15 at 21:48
  • \$\begingroup\$ \$V_{\text{ICR}}\$, p. 6. With a 10V supply the common mode input range is up to 9V, so you need about 1V of headroom. \$\endgroup\$ – Null Jul 11 '15 at 21:52

This is what I end up doing. And it work. Thx to Null for pointing in the right direction about Vicm and Chu for the swap on R1.

enter image description here


The output voltage (to ADC) is trying to go negative, since the left side of \$R_1\$ is more positive than the right side. Try reversing the connections to \$R_1\$

  • \$\begingroup\$ Good point! Probably changing that + lower voltage input did solve the case. So it a 50/50 answer ;-) \$\endgroup\$ – Steve S. Jul 12 '15 at 14:29

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.