Would an Opamp work wired up like this?

[Dextrine]

I am trying to measure the voltage out of this dc-dc converter and am wondering if this method would work.

The ground center point of the op amp bias is not the same reference as the return for the dc-dc output.

Thanks for any help

 

[Dextrine]

I'm really rather sure that it would, just that multiple times now something makes sense mathematically only for the real world application to be very different because of chip limitations or something.

 

[Averagesupernova]

I don't think so. If there is nothing to hold the two ground references at some particular voltage between them (0 for instance) then there is no guarantee what this amplifier will do.
-
Do you have a need to keep the grounds isolated? If so, then you will need to look at an opto-isolator scheme of some sort.

 

[AlexCaledin]

It will work if the resistors have the correct values so that it be a correct differential amplifier. Calculating those values, the resistance of the filter coil ought to be taken into account; and the input current must be kept acceptable for the signal source, at any possible voltage between the grounds.

 

[Svein]

Such a circuit always set my teeth on edge. You have a possible resonant circuit on the input, and together with some stray capacitance from the output you could very likely have an unintended oscillator. Murphy's rule for electronic circuits is: "Amplifiers oscillate and oscillators amplify".

 

[AlexCaledin]

The resonator can easily be damped - just add, across the capacitor, another one with a calculated series resistor. The input LC circuit is not obliged to do all the filtering, it's just for some general noise frequency limit, so it can be altered a bit. The accurate filter ought to be placed after the opamp.

 

[Dextrine]

Thanks for all of your replies, fortunately at least for now, it seems to be working well with my application. I do have an opto-isolator circuit after this one too.

 

[berkeman]

I am trying to measure the voltage out of this dc-dc converter and am wondering if this method would work.

The ground center point of the op amp bias is not the same reference as the return for the dc-dc output.

Thanks for any help

Why are you using that circuit instead of a simple non-inverting opamp circuit?

 

[Averagesupernova]

This circuit will not work as it is shown. Imagine putting a small dry cell (AA, AAA, etc.) across the input. The output cannot swing in such a way to bring the two inputs on the op-amp back together.

 

[AlexCaledin]

Just believe that the Op-amp's inputs are together ― then you'll see it can do it ― if you feed it well ― but, indeed, the Op-amp on the picture looks unfed...

 

[jim hardy]

I am trying to measure the voltage out of this dc-dc converter and am wondering if this method would work.

The ground center point of the op amp bias is not the same reference as the return for the dc-dc output.

How much is their difference E1 ?

The way to solve these is write KVL equations for voltage at each of the opamp's inputs ,
assuming all R's are equal :
E at + input = (E1)/2
E at - input = (E2 + E1 -E3)/2 (edit ?? this morning that looks askew somehow... (E2 + E1 + E3)/2 looks better _{old jim})

just set them equal and solve for voltage at output, E3.

Then ask two questions:
1. "Is voltage at inputs within this amplifier's common mode voltage limits ? " They're in its datasheet.
Don't expect an amplifier with 10 volt supply to handle 10 volts at its inputs unless it's specified "rail to rail", and even then don't expect it to handle 11 volts.
2. "Is voltage at output within this amplifier's drive capability for its power supply? " That's in its datasheet too.
Don't expect 15 volts output from an opamp with only 10 volt supply. Even rail-to-rail opamps lose drive capacity as output approaches power supply voltage.

See ? It's not so hard.

old jim

 

[NascentOxygen]

I am trying to measure the voltage out of this dc-dc converter and am wondering if this method would work.

The ground center point of the op amp bias is not the same reference as the return for the dc-dc output.

Your arrangement shows the + input of the op-amp being held at a potential partway between that of the two grounds. For the amplifier results to be steady and predictable the potential difference between the two grounds must be fixed, known, and stable.

 

[AlexCaledin]

Your arrangement shows the + input of the op-amp being held at a potential partway between that of the two grounds. For the amplifier results to be steady and predictable the potential difference between the two grounds must be fixed, known, and stable.

If the resistors are correct, the opamp will "calculate" the (voltage proportional to the) exact difference between the input and input's ground. It works like an instrumentation amplifier

https://en.wikipedia.org/wiki/Instrumentation_amplifier

― only without the input buffers which are not always necessary. The voltage between grounds is, typically, some unknown (but expected to be not too strong) noise (― whether we like it or not).

 

[Averagesupernova]

I will change my story. :) The extra ground threw me. The OPs schematic is a standard differential amplifier.

 

[NascentOxygen]

If the resistors are correct, the opamp will "calculate" the (voltage proportional to the) exact difference between the input and input's ground. It works like an instrumentation amplifier

https://en.wikipedia.org/wiki/Instrumentation_amplifier

Have you established what the resistance relation must be in order for this to be true?

 

[AlexCaledin]

Have you established what the resistance relation must be in order for this to be true?

Why, the only basic requirement is obvious: If the measured input is at the signal-ground potential (that is, the measured voltage is zero), then the resistors connected to the negative input of the op-amp must make there, when the output is zero, exactly the same potential as that of the positive input ― whatever the signal-ground potential is.
That simply means, the ratio of the upper resistors must be the same as below.
But to choose the resistors practically, one have to consider many other practical details about the op-amp parameters including the supply, and the input signal source parameters including the coil resistance, and the expected between-ground voltage, and the required precision and gain. (Sometimes this choice is more difficult than to build that whole thing with input buffers from Wikipedia )

 

[Svein]

Why, the only basic requirement is obvious: If the measured input is at the signal-ground potential (that is, the measured voltage is zero), then the resistors connected to the negative input of the op-amp must make there, when the output is zero, exactly the same potential as that of the positive input ― whatever the signal-ground potential is.

That is the theoretical requirement. You also need to ensure that the op-amp inputs are within the active range and that the op-amp output is within the specified range.

 

[AlexCaledin]

That is the theoretical requirement. You also need to ensure that the op-amp inputs are within the active range and that the op-amp output is within the specified range.

― oh yes I need, but that's not necessarily change the resistors (― and the question was about resistors' relation ) ― I can, in principle, increase those ranges choosing/building a high-voltage opamp.

 

[Svein]

If you need to handle a high common-mode voltage, consider a flying capacitor circuit. See http://www.linear.com/solutions/1569.