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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)

schematic

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?

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    \$\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
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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

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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\$

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  • \$\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

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