subtract two AC voltages with instrumentation amps

Question

I'm implementing the so-called "3 omega technique", where a thin metal line is excited with a sine current. The metal has a high TCR which causes a 3rd harmonic oscillation, and I need to measure the amplitude of this voltage using a lock-in amp. I use a Keithley 6221 current function generator, which has high harmonic noise above 5 kHz, which interferes with the measurement of the 3rd harmonic.

In the litterature, several people have used a circuit such as the one illustrated in fig 2.6, pg 47 of this document : http://repository.lib.ncsu.edu/ir/bitstream/1840.16/5418/1/etd.pdf

The metal line is placed in series with a variable resistor with a low TCR, which is adjusted to the same nominal resistance as the metal line. Therefore by subtracting the 2 voltages, this will remove the fundamental signal, and harmonics caused by noise from the function generator, and the only thing left should be the 3 omega oscillation from the metal line.

However, I haven't found any detailed description of the actual circuit, other than the basic diagram in the link above. The only extra detail given by the authors is that they use AD624 amps.

So my question is, how do I actually implement this circuit ?

The 3rd harmonic is measured versus excitation frequency, which can vary between 1 Hz and 30 kHz. However harmonic noise from the generator isn't a problem at low frequency. The resistance of the metal line is typically 50 - 100 Ohm.

Thanks !

EDIT : my reputation is now high enough to post the diagram here directly

Answer 1

An instrumentation amp is essentially a difference amplifier. The output is directly proportional to the difference between the two input (essentially, which is what you need) and so any common mode signals, ie, those which exist in both inputs, should in principle be eliminated. It looks like a fairly straightforward application of an inamp to me.

Simply tap the voltages at both points and dump them into the two inputs of an inamp. You have to make sure that the inamp has sufficient bandwidth to respond to your harmonics, and inherent noise and stability at the frequency of interest that comes from the inamp itself should be lower than your signal.

EDIT:

On looking at your circuit, I will posit one of three possibilities :

1. The circuit is incorrect, and the label "Lock-In" should be on the rightmost corner, as if the lockin is not shown.

2. The circuit is incorrect, and the output of the first opamp(the triangle, left top) goes to the second (left bottom) instead of the third, and to its angled side. (as ref input)

3. The circuit is correct but the text is not, and the signal from the resistor is used as the lock in reference instead of something nicer coming out of the current source. The subtraction then occurs inside the lockin implicitly. From what I know of lockins, I'd bet against this possibility.

Using an instrumentation amp is easy. A difference amp is slightly more complicated. In both cases, look at the datasheet for a typical application circuit. For an inamp, the output is generally given by G x (In+ - In-) + Ref, where G is generally set by a resistor or two. Ref is typically ground unless otherwise specified, and lets you add an offset to the output.

From what I gather, both the left two triangles can be instrumentation or difference amplifiers. The output of the top one is what you want to subtract from that of the bottom one. In order to do that, one way is to send both into a third instrumentation or difference amplifier, as in possibility 1 with the lockin mislabelled. The inamp will subtract the two signals and let you give it to the locking with impunity.

Another way go perform this subtraction is to send the output of the top inamp into the ref input of the bottom inamp. I suspect you'd want one of the two inamps to be in reverse polarity for that to work, but I'll have to think about it. This would correspond to possibility 2.

Answer 2

Here is a great piece of reference material for understanding instrumentation amplifiers. I have a paper copy on my desk so I know it's good. It's produced by Analog devices. Don't read it yet....

The circuit you have shown may have problems and these are to do with common mode AC rejection. The top amp in your diagram will have a different common-mode voltage on its inputs than the bottom amp. Common mode means the average of the two inputs - the average signal on the top amp relative to ground is different to the bottom amp. Unless you have really, really good instro amps you will get errors so I recommend a conventional wheatstone bridge approach with 1 instrumentation amp - I'd use the AD8221 from Analog devices. Here is a snippet from the book linked above that shows it being used in an AC excitation scheme: -

Ignore the AD630 that follows; the important part is the AD8221 and the full-wheatstone bridge that is on its left. This is for a strain gauge amplifier but the principle and component values are virtually the same - 50 or 100 ohms as the bridge components doesn't alter the principle or the circuitry.

Notice that the instro amp will have exactly Vac/2 on both its inputs and because it is a combined amp there won't be the common-mode issues you'd get with your original circuit.

Alternative idea I would also consider, as an alternative to using this type of circuitry, a high-order low pass filter on the output of the generator to reduce the 3rd harmonic by at least a factor of ten. Just doing this and not using a wheatstone bridge will probably give you the accuracy needed for your lock-in amplifier. I'll not go into details here but suffice to say you should be able to get a decent one implemented from passive components such as inductors, capacitors and resistors. You could even make a notch filter out of these components to achieve a decent measure of 3rd harmonic reduction.