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I am using an op-amp as a differential amplifier to find the current draw of a load by measuring the voltage across a 0.1 Ω resistor that the current goes through. I set it up using this schematic:

diff-amp circuit

R1 and R2 are 1.1 kΩ and R3 and R4 are 10 kΩ for a 10× gain. There is a 0.1 Ω resistor between V1 and V2, the load is connected to V1 and operates at 12 V, pulling a range of 370 mA to 3 A. I am using an OP177, powered off +24 V.

For the most part the circuit works across the range I need it to: from a Vdiff of 90-290 mV, Vout is 0.88-2.94 V. However, any Vdiff under ~90 mV results in the same Vout of 0.83 V. It seems like the op-amp has a certain voltage differential threshold between V1 and V2 required to operate correctly, but I cannot find anything in the datasheet that talks about this.

Can someone verify that this is an actual spec rather than something I did wrong in my circuit, and if so, how do I find this value in the datasheet?

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  • \$\begingroup\$ Nope, not an "actual spec". What's the tolerance on your resistors? Can you show a picture of your layout? What's the amplifier driving? Have you looked at the output with a scope? \$\endgroup\$
    – John D
    Feb 10, 2023 at 19:34
  • \$\begingroup\$ How are you powering your op amp, and what op amp is it?? It may not be capable of going under 0.83V \$\endgroup\$ Feb 10, 2023 at 19:37
  • \$\begingroup\$ @JohnD The 0.1ohm is 1%, the other four are basic breadboard guys so probably 5% or 10%. I attribute the tolerance on the gain value to that, but don't see how the tolerances would cause this. I can't show you the physical testing layout. The amplifier is not driving anything, I'm using it as a current sensing device. The scope shows the output as a slightly noisy 0.83V. \$\endgroup\$
    – InBedded16
    Feb 10, 2023 at 19:40
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    \$\begingroup\$ @ScottSeidman It is an OP177 as shown in the question. Unusual input stage topology and unusually low offset voltage specs and pretty low offset/bias current, too. \$\endgroup\$ Feb 10, 2023 at 19:41
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    \$\begingroup\$ @InBedded16 If the resistor values are not tightly matched the common-mode rejection will be poor, and the output will pass some of the DC common mode voltage at the input. If I recall correctly 1% resistors will result in a CMRR of around 48dB, but 5 or 10 % resistors will be way worse. There's also the issue of the part not being rail-to-rail output, so if your supplies are not bipolar you can run into the output swing limitation. \$\endgroup\$
    – John D
    Feb 10, 2023 at 19:48

2 Answers 2

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The OP177 isn't a rail-to-rail op amp. The datasheet specifies behaviour at ±15 V power, and all it guarantees is that it will get within 1.5 V of the rails (saturating at ±13.5 V) if the output load is light enough.

enter image description here

So getting 0.83 V output when the negative rail is 0 V is actually a decent amount better than the specifications would imply.

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    \$\begingroup\$ Why on Earth would anyone use a precision device like this without a bipolar supply? But I see the 24 V and I don't see the questioner mentioning that this is +/- 12 V. Hopefully, they will say they are using +/- 12 V? \$\endgroup\$ Feb 10, 2023 at 19:44
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    \$\begingroup\$ @periblepsis I've seen enough people do that to suspect, at least. \$\endgroup\$
    – Hearth
    Feb 10, 2023 at 19:47
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    \$\begingroup\$ @Hearth so basically, it outlines power supply at +- 15V, and output voltage range at +-14V (depending on RL). So you can surmise that the output voltage can only get to within 1V of the rail, and possibly better depending on certain conditions which I must have luckily created in my circuit? Do I have that right? \$\endgroup\$
    – InBedded16
    Feb 10, 2023 at 20:01
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    \$\begingroup\$ @JohnD oh duh. Because if I use a bipolar power supply, a 0.50mV differential for example would be well above the negative rail, and thus valid. But its only "bad" because I want to sense voltages near 0V - if I cared about 2-5V for example, it would be just fine. \$\endgroup\$
    – InBedded16
    Feb 10, 2023 at 20:04
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    \$\begingroup\$ @InBedded16 I would say the reason it's better isn't just due to circuit conditions (though it is partially; it'd probably be very different at extreme temperatures, for instance), but also just that the individual chip you have could be better than the rating. The ratings only have a minimum and a typical--they guarantee that the device will be no worse than that, but not that it won't be better. They want it to be better, in fact, to have a margin of error so you don't come to them with lawyers and claim they didn't meet the guaranteed performance! \$\endgroup\$
    – Hearth
    Feb 10, 2023 at 20:24
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It sounds like you are running the operational amplifier from a single supply of +24V (VDD=24V, VSS=0V)

Op-Amps have a range of allowable input voltages that it can work at for normal operation in a linear manor. This is often called the input common mode voltage.

In the differential Op-Amp circuit above, I am assuming that V2 is also grounded and so V- is sitting at or close to 0V. In order to handle inputs so close to either supply rail (in this case VSS), you need a rail to rail input Op-Amp, which the OP177 is not.

The OP177 has an input voltage range (Table 1, OP177 datasheet) of +/-14V (Typ) for a +/-15V supply. In other words, the Op-Amp inputs need to be +/-1V above the rail for the Op-Amp to work reliably. This is a typical case, minimum is +/-13V.

Coming back to your circuit, I suspect that there is some additional parasitic resistance between V2 and Ground. At higher currents flowing through the 0.1R measurement resistor, V2 rises just enough for the op-amp to work as expected. At lower measurement currents, V2 is too close to the VSS negative rail, and so V+ is too close to the VSS negative rail, so the Op-Amp doesn't work as expected.

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