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On a board I'm working with, there's a 120 ohm termination resistor. I need to measure the voltage across it and compare it against a reference voltage.

I'm wondering how to make the connections.

The op-amp has 2 inputs. The inverting input will be connected to the fixed reference voltage.

The non-inverting input should be hooked to the point where I can measure the voltage across the resistor. Voltage is usually measured between a point and the ground reference, but the voltage I need to acquire is between the two leads of the resistor.

How would I be able to connect it to the non-inverting input? Should I use a differential to single ended buffer or something of this nature?

Edit: ignore the switch.

enter image description here

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    \$\begingroup\$ You can't just "connect an opamp" to this and get it to work. You need a more complete circuit. The fact that you write "opamp comparator" in your title makes me think that you do not have much circuit design experience. I suggest that you study how a differential amplifier works, example: electronics-tutorials.ws/opamp/opamp_5.html \$\endgroup\$ Dec 6 '20 at 12:27
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    \$\begingroup\$ I guess that the simplest is to use a differential amplifier to amplify the signal (amplification = 1, so no amplification, is also possible) so that the signal becomes single ended. Then a comparator circuit can be used to do the comparison. \$\endgroup\$ Dec 6 '20 at 12:27
  • \$\begingroup\$ Hey there! Thanks for the quick reply, yeah I just sort of hoped for a quick fix for this. Differential to Single Ended conversion it is! \$\endgroup\$ Dec 6 '20 at 12:31
  • \$\begingroup\$ Look at this question asked a little earlier today - I would like to answer it but I can't but, the main thing here is what you need to do - use a differential amplifier. \$\endgroup\$
    – Andy aka
    Dec 6 '20 at 12:33
  • \$\begingroup\$ When you write that you need to "compare" the voltage against a reference, what exactly do you mean? What should be the output of the comparison? For example, If the voltage across the resistor is 1V and the reference voltage is 1.25V, what do you want the circuit to output? 0V (referenced to ground)? -0.25V? -5V, TTL Low? Similarly, if the voltages were reversed, what should be the output? \$\endgroup\$ Dec 6 '20 at 14:31
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enter image description here

I hope this circuit configuration helps. The first two opamp acts as a voltage buffer so that extra current doesn't flow through 120 Resistor, which may alter the output. After the first stage, both inputs are passed to the differential amplifier configuration of OpAmp. The resistors R2, R3, R4, R5 are chosen according to the gain required. After altering a little bit with OpAmp IC and other values, you may adapt to your need. I think buffer stages are required to properly measure the voltage.

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  • \$\begingroup\$ And here's the problem. You haven't shown power supplies to the op-amps and there is no information how the 0 volts of the op-amp power supply might be related to the common-mode voltage on the isr_detect pins. Anyone can throw a circuit like this at the problem without thinking or delving deeper and get it WRONG. I suggest, if you are to make an answer that has any merit, you think about what I'm saying. \$\endgroup\$
    – Andy aka
    Dec 6 '20 at 16:32
  • \$\begingroup\$ The OpAmp IC I showed is just for representation, the actual model may differ greatly due to circuit parameters, since so circuit information is given, I've given a general solution. This is a general circuit, every value needs to be calculated separately. The question doesn't mention any of those. So how'd am I going to get the values? \$\endgroup\$ Dec 6 '20 at 19:00
  • \$\begingroup\$ I’m not saying you are wrong but I am saying this question cannot be answered this simply. You are giving a false impression that there is a simple solution but you are not reading between the lines of the question and understanding the under stated problem. \$\endgroup\$
    – Andy aka
    Dec 6 '20 at 23:05
  • \$\begingroup\$ Ok. Should I then remove my answer? Or it does any help to the question. I don't want to mislead anyone, I'm also a beginner student. Maybe I'll learn by these mistakes. \$\endgroup\$ Dec 7 '20 at 4:35
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The following circuit does what I think you want.

Op amp OA1 serves as a level shifter. The output of OA1 should be equal to the voltage across \$R_{term}\$, but referenced to ground. Op amp OA2 acts as a comparator, and gives a output near \$V_{cc}\$ if the output from the previous stage is above 3.3V, and it gives an output near 0V if the voltage from the previous stage is less than 3.3V.

The resistor values are not critical, but they should be matched if accuracy is at all important. There exist monolithic resistor arrays that contain multiple matched resistors. My guess is that you may not have to worry about such things.

You didn't specify \$V_{cc}\$. In the schematic, I used 15V, but 5V works just as well.

The op amps probably do not need to be anything special. Your tripping point is 3.3V and that is in the range of virtually every op amp available. If using "dual-supply" op-amps in this application, such as 741s, you may have trouble with a 5V \$V_{cc}\$, but just about any rail-to-rail op amp should work.

The jumper around the switch can be ignored. I added it so that I could test the circuit. (You can too. There is a little option to the left below the schematic to "simulate this circuit".)

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What if the common mode voltage of the Vcable 5 volt supply is 100 volt different to measurement ground? What will happen then? I tried in comments under the question to get the OP to realize that his question is under defined and I have also said the same to the person making the other answer. You just can't throw a non-galvanically-isolated diff amp at the problem without determining these things first. Then you have to worry about AC common mode noise and how balanced the differential amplifier is. A diff amp is probably the right solution but the devil's in the detail. \$\endgroup\$
    – Andy aka
    Dec 7 '20 at 10:10
  • \$\begingroup\$ @Andyaka I agree with you that the question is under defined. I also think that my solution is probably is what he is looking for. Regarding common mode noise, I think it deserves serious consideration, perhaps more than can be easily squeezed into 600 word comments. In telegraphic style, what is the impedance for the common mode noise? If the impedance and amplitude are such that they will affect this circuit, (whose purpose is only to detect presence/absence) does that not indicate a problem that is better solved elsewhere, rather than in the detection circuitry? \$\endgroup\$ Dec 7 '20 at 14:57

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