I wanted to measure the phase difference between the current (magnetic field due to the solenoid) and the system response. I am doing this by measuring the voltage drop across the resistor in series with the solenoid and using this as the reference frequency of the lockin amplifier. Experimental Setup

On increasing the frequency of the source (function generator), the impedance of the circuit is increasing and decreasing the current in the circuit. This reduces the amplitude of the reference frequency to the point where the lockin amplifier is not able to lock to the reference.

I wanted to amplify the drop across AB using an op-amp. I am not sure if a differential amplifier will do the job for this purpose. One this that caught me off-guard while designing the differential amplifier was what do i consider as ground here, if i make V1 as A and V2 as B.

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I am not sure if i can even do this here. There is another doubt I got while looking at some datasheet like the one for OP27 (snippet below). It says the maximum differential input voltage between the terminals is just 0.7V. This can't possibly mean that the op-amp can only amplify voltage difference between the terminals that are below 0.7V.enter image description here.

I do get the feeling that I am doing a lot of stuff wrong. Any help would be greatly appreciated. Cheers! :)


I got a hint from Bruce's comment and Metacollins answer. So now I have a common ground for the entire experimental setup. Is this the way i should go about it? edit:

  • \$\begingroup\$ can you ground one end of the resistor? \$\endgroup\$ Sep 7, 2019 at 7:47
  • 1
    \$\begingroup\$ What is supplying the AC, and at what point is it grounded (to your lockin amp)? \$\endgroup\$ Sep 7, 2019 at 8:20
  • \$\begingroup\$ @BruceAbbott I have updated the diagram, is the grounding in the setup ok? \$\endgroup\$ Sep 9, 2019 at 6:21

1 Answer 1


This can't possibly mean that the op-amp can only amplify voltage difference between the terminals that are below 0.7V

I'm afraid that's exactly what it means. Well, actually, it doesn't mean it can't amplify a voltage greater than 0.7 volts across the inputs per se... it just means that 0.7V exceeds the voltage drop of some internal diodes, so above 0.7V, these diodes will turn on. The footnote lets you know that due to the high performance nature of the op amp, there are no current limiting resistors so it will definitely explode or catch fire (sometimes at the same time) if whatever is across the inputs can dump some current if back to back diodes have turned on, effectively shorting the inputs to each other. Through 700mV of heat dissipating voltage drop inside the op amp of course.

You have the input resistors in series though, but its just something to be aware of. Don't connect anything directly on accident. As long as the current is limited though, then it just, as you were hoping it wouldn't, limits the differential input voltage to 0.7V.

On the other hand, the voltage across the inputs does not necessarily equal the voltage your measuring. Remember, you have negative feedback. The op amp will be actively forcing the difference between its inputs to 0V - that's what differential amplifiers do. And unless something is horribly wrong with your resistor values, even if the voltage going into the two input resistors is more than 0.7V,

That said, you have a feed back resistor, so the op amp will be forcing its inputs to 0V. The diodes will turn on briefly if the rise time of the differential voltage exceeds the op amp's slew rate, but its fine as long as you have high value resistors to limit any current. You can try it using an op27 model in ltspice if you want. 5V into 10k resistors for all 4 resistors doesn't even result in any distortion in the output, even though the differential voltage briefly exceeds 1.8V while the op amp catches up. What it DOES to is really destroy your input bias current while those diodes are turned on.

You can also use a voltage divider on the inputs and up the gain, at the cost of noise.

As for what ground is for the differential inputs, its whatever you want it to be. Remember, the differential signal has no ground, they're a voltage with respect to each other. So they don't care what you ground them to, the grounding is entirely for the op amp. The bias current has to return somewhere. And the bigger picture is that since it is for the op amp bias current, that means you connect the input ground to the potential you want the output referred to. So in your case, you'd probably want your actual ground there. Assuming you had, for example, 12V and -12V on the OP27's + and - power rails, and you grounded the input to 0V, and your gain was -2, then 1V voltage differential would result in -2V on the output, referenced to that ground you chose. If you grounded it to some third rail that was -3V, then nothing would change, except the output would now be referred to that as ground, so 1V in would produce -5V, and 0V in would let the op amp idle at -3V.

So don't worry about it and just ground it to what you want ground for the output to be, its that easy!

  • \$\begingroup\$ Thank you for your answer, it cleared a lot of my doubts. Can you have a look at the updated signal chain. Cheers! \$\endgroup\$ Sep 9, 2019 at 6:22

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