Resistor Matcher

Would this work?

https://www.circuitlab.com/circuit/7v4sy2/lm741-test/

My lame meter only has 20k/V of resistance, so at .1V range, it's only got ~2K of resistance, so I cannot match resistors of large resistance with just a wheatstone bridge.

The only connections not shown in the diagram is the Null offset, which is not available in CL. I still need to figure out the exact values for R 14, 15, 16, but basically they would set my input impedance and thus the sensitivity of the circuit (for different test resistor ranges).

by revolver_365
February 26, 2013

No.

It won't work.

You can't use an opamp without feedback and expect to get anything sensible out of it. Having an input offset trim makes matters worse: how would you know if you are measuring resistor error or input offset drift?

You have misunderstood how a wheatstone bridge works.

At balance, you are looking for a null. Therefore no current flows through the meter across the bridge.

http://en.wikipedia.org/wiki/Wheatstone_bridge

You do not need a high impedance voltmeter. In fact you would be better off replacing the voltmeter with a sensitive ammeter or galvanometer:

http://en.wikipedia.org/wiki/Galvanometer

by signality
February 26, 2013

On my meter, the .1V range and 25uA range is the same setting. And yes, basically if I don't measure anything, then they're matched. For an ideal meter (infinity resistance), you would measure about 2.3mV per 0.1% mismatch. This diminishes to only a few uV when testing resistors with large resistance because your load resistance (the meter, approx 2k ohm) is tiny when compared to source (the resistors 1M+ ohms).

In my setup, the op amp inputs would function as the bottom two resistors in the wheatstone setup. To calibrate it, I would add a 100ohm resistor to the top side of V3 (to limit current) then short out R11 and R12 with copper wires. I would use multi-turn trims for my null offset, I'm just not sure if that would be good enough. Adding gain control without loosing differential would require more op amps and I'm trying to keep things simple. (And would STILL require null offset) I understand it would require recalibration every now and then, but that's fine.

The 741 has about 60,000x gain so 2.3mV (.1% mismatch) would be 138V, but if I lower the input impedance by putting resistors across it (the unattached resistors in the middle, say 100k for example), for a .1% mismatch of 10M resistors would produce approx. 3.7uV across the input and 757mV on my MM. Try running some simulations on it :) I'm just not sure if my calibration/ null offset methods would be sufficient.

by revolver_365
February 26, 2013

When you look inside the 741 opamp, the inputs are two transistor bases. the emitter currents, controlled by a current source, force the base currents to be (emitter current)/HFE. when the opamp is in its linear range, assuming you can ride the offset input adjustment adequately, the inputs are low value current sources rather than base input resistances. You notice the data sheet mentions the input bias current as if this were the case. The differential input resistance doesn't help model the two missing bridge resistors, but rather shunts the measuring nodes. The resistance you want to be part of your bridge is the common mode input resistance, which is not well characterized.

If you have some kind of sensitive detector, exchanging the two resistors to be matched, allows matching in the presence of offset: since the offset should not change, the difference between the two resistor exchanges will be an indication of match.

Depending on the precision you expect, the existence of thermal voltages and the temperature changes due to even touching the resistors will affect the outcome. What kind of resistors deserve such treatment? Ordinary carbon film resistors have tempco in the order of 100ppm and up.

Sometimes a judiciously designed adjustment in an otherwise carefully matched network will help. See some application notes from National Semiconductor (now TI) in the design of instrumentation amplifiers.

The wheatstone bridge with the traditional four resistors is well established in metrology for matching resistors.

Bridges usually operate nulling the two intermediate branch voltages, however, there are some bridges that use a nonzero or off balance indication, but near balance.

Old textbooks on electrical measurements guide you through the various schemes and error contributions due to imperfections in the measurement. Particular authors I recommend are M Stout, W. Michels, F. Laws, and F Harris.

I find the opamp has a seductively high open loop voltage gain, which is an attractive nuisance to people not familiar with the accepted application of negative feedback.

The 741 is an example of the situation where gain is very cheap, but its proper utilization is in the stability it permits in feedback networks.

Historically, Harold Black reasoned that negative feedback while burning up possible high gain, could be traded off for the even more useful high stability. No one accepted at that time that feedback would result in anything but oscillation or instability.

by jfurman
February 26, 2013

@jfurman,

You have pointed out one of the main reasons that @revolver_365's circuit won't work as required.

As I explained before, you cannot use an opamp open loop and expect to get any sensible answers out of it. The whole point about an opamp is that it is far from perfect. It is only the huge open loop gain (Avol) together with the application of negative feedback that make it look like an near ideal linear device.

Sure it may have low input bias current and offset voltages which don't improve with feedback but the linearity and the repeatability of opamp designs and the measurements made using those design only comes from the Avol and negative feedback.

The rest of your post is interesting too.

You point about the effects of temperature differences and resistor temperature coefficients is particularly important for making higher precision measurements. It's one thing to make a bridge, match up all the bits that have to be, do all the calibration but it's quite another to be able to make the assembly repeatable from one hour to the next.

@revolver_365,

You might like to have a look at:

then:

by signality
February 26, 2013

@jfurman

Thanks for your explanation! Yes, I'm a bit new. And yes, I understand negative feedback, but as I mentioned, I would need another pair of opAmps (perhaps a 082) and would prefer keeping things simple, even if I had to trade it for some accuracy (0.1% match is my goal is my goal). Do you think it would work? Would the common mode input resistance/ current bias be matched (or close enough) for the + and - input?

I had a similar setup, with feedback (not possible with this design?), to match fets https://www.circuitlab.com/circuit/5uq585/fetmatching_2/

I had a different idea, but this one doesn't work for sure, as current is draining through the opAmp input. https://www.circuitlab.com/circuit/4dk936/new-rmatch/

And yes, the tmp thing was apparent to me. But I also want to mention that I had two matched 100 ohm resistor (that I matched with a bridge) to match more resistors, and even with 10v source (they were running quite warm), their ratios remained relatively stable!

by revolver_365
February 26, 2013

@signality

Cool! You said something while I was typing up my reply. I'll check out your circuit.

Thanks!

:)

by revolver_365
February 26, 2013

@signality

That's similar to the setup I simulated in another design (but without OA2), but it still suffers from the input bias problem (but I'm starting to see that this is unavoidable). I live in a small town in Thailand, and I'm not sure if I would be able to find the LMP7721 or AD8551 (I couldn't even get a 2N3819!) But, I will check them out and grab a few if they have any to offer.

Although, (sometimes being a fool) it just occurred to me that it's okay if perfectly balanced doesn't = 0V/A, as long as I amplify the reading a bit (so that it's big enough for my MM to read) and I get the same reading for all the resistors (I would reference them to the the same ref resistor of course, which was how I matched my 100 ohm pair) and as @jfurman was saying about bridges that use non-zero indication.

So, I guess my problem is solved?

by revolver_365
February 26, 2013

Open an account with Farnell?

http://th.element14.com/

RS (not RadioShack!)

http://thailand.rs-online.com/web/?&cm_mmc=World-Selector-Page--Online-Referral--MainWorldMap-201211-_-MainWorldMap

Or DigiKey?

http://www.digikey.co.th/

Sometimes you can get parts as samples direct from the manufacturer or occasionally from a distributor.

Do you have a work or educational establishment email address & phone number?

That helps with sample requests.

by signality
February 27, 2013

I will certainly check them out! :D Looks interesting!

I learnt a lot! Thanks again! :D

by revolver_365
February 27, 2013

"So, I guess my problem is solved?"

If yes, then would you like to mark this thread as Solved?

(No-one else can!)

:)

Thanks

by signality
February 27, 2013

I don't see this button anywhere!

by revolver_365
February 27, 2013

Mark as solved at the bottom left of the posting window?

by signality
February 28, 2013

Silly me.

There isn't one. It's only in the Support forum.

by signality
February 28, 2013

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