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I have the following circuit: enter image description here

I have a hard time determining the mathematics of this circuit. If we ignore the Wheatstone Bridge to the left, focusing on the differential amplifier, i know that the amplification is given by:

(V2-V1)*(R6/R5)

Given that R6=R8 and R5=R7. But what happens when I add my Wheatstone bridge, which doesn't have the same resistor values on the two sides of it? (10k ohm at the left side and 1k ohm on the right side).

In the picture i have managed to balance the R5 and R7 so that the output is ~0V when there's no difference in the bridge - but this is just by lucky guessing. I would like to understand the mathematical formulas for the circuit :-)

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  • \$\begingroup\$ R5 is not same as R7. Which is correct? \$\endgroup\$ Feb 28 '15 at 16:04
  • \$\begingroup\$ Yes because R5 is connected to R1 and R2 while R7 is connected to R4 and R3. Those two nets must equal the same resistance somehow? My problem is that I don't know how to calculate R5 if I change R1 and R2 for example. \$\endgroup\$
    – Jolle
    Feb 28 '15 at 16:08
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I'm going to state this up front - a differential amplifier made from an op-amp and four resistors is perfectly fine when (and only when) the input sources are much, much lower impedance than the resistors used. You will not get this when a bridge is attached because, as the bridge is off-balanced by whatever it is you are measuring, its output resistance changes, generating significant errors in a diff amp configured for a gain of 50.

The junction of R4 and R3 sees a high loading impedance - basically it's R7 in series with R8 and of course the non-inverting input impedance of the op-amp can be ignored. On the other hand, the loading impedance seen by the junction of R1 and R2 is just R5 because the inverting input is forced always to be at the same potential as the non-inverting input. This is natural for op-amps in linear applications like this.

So you have an imbalanced circuit in terms of loading and this totally causes the wheatstone bridge to be unbalanced. You've tried to correct this by modifying R5 and R7 and you are now confused.

I'm not going to bother giving you the mathematical formula for your circuit because it's pointless and the reason it's pointless is that the circuit can be reduced in the number of components and nobody would build this circuit for a real life project because it would be terrible to control and have really awful common-mode rejection. Read on....

Firstly, consider using an instrumentation amplifier - it makes life a whole lot easier and you get massively superior performance. If you don't use an INAMP then use two op-amps wired like an INAMP: -

enter image description here

Taken from here.

Or use 3 op-amps wired like an INAMP: -

enter image description here

Either solution proposed is going to outperform your circuit by a mile.

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  • \$\begingroup\$ OK, it makes sense - I am learning something here (or rather refreshing). Thank you Andy! In the solution with 2 OP AMPS I see that there are a two trimmers due to resistor tolerances. Could these trimmers be implemented with a DAC maybe? \$\endgroup\$
    – Jolle
    Feb 28 '15 at 16:28
  • \$\begingroup\$ They could be - you could use a digital pot but please do consider the industry standard way of doing this sort of thing - get an instrumentation amplifier like an AD620 (e.g.) \$\endgroup\$
    – Andy aka
    Feb 28 '15 at 16:30
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    \$\begingroup\$ OK, I will look into that. Problem with AD620 is that it is too expensive for my application. I think that I'll go for the INAMP implemented by two OP AMPS like you suggested, and then try to find a way around the trimmer resistors. :-) . Thanks again! \$\endgroup\$
    – Jolle
    Feb 28 '15 at 16:36
  • \$\begingroup\$ I've got to disagree, @Andyaka , about your analysis in your first paragraph. Although he did it wrong, he attempted to compensate for bridge impedances. He forgot (or did not know) about Thevenin equivalents. If he changes R5 to 4k, and R7 to 9.5k, his circuit will work just fine. \$\endgroup\$ Feb 28 '15 at 17:40
  • \$\begingroup\$ @whatroughbeast the bridge impedance will change and this will cause common mode errors that cannot be separated from real signal. It would be really difficult to re-linearize this especially at a gain of 50. For instance, if this were a pressure bridge you might expect all the impedance to rise with temperature and this will cause errors of a significant level. \$\endgroup\$
    – Andy aka
    Feb 28 '15 at 19:19

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