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I am trying to build an amplifier for a strain gauge sensor using the 2 op amps instrumentation amplifier design provided in the Texas Instruments note: https://www.ti.com/lit/an/sboa247/sboa247.pdf?ts=1605266073942&ref_url=https%253A%252F%252Fwww.google.com%252F. enter image description here

I was able to derivate the expression for voltage gain of the amplifier, but I have troubles understanding the expression for the common voltage gain and also why the total resistance of the bridge is 4 X 120. enter image description here

Can someone provide an explanation for my dilemmas regarding the common mode voltage of this amplifier design? I appreciate your help!

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  • \$\begingroup\$ I don't understand your dilemma and why you are confused. It's certainly confusing the 4 x 120 ohm bit but it's not clear whether you think it should be 4 x 120 or someone or something is telling you it's that and, a bridge doesn't really have a total resistance parameter because there are four terminals so, that's a little confusing too. \$\endgroup\$
    – Andy aka
    Nov 13 '20 at 15:32
  • \$\begingroup\$ I don't think that the resistance should be 4 X 120 ohm, this is stated in the link that I provided. The expression for the common voltage is similar to that of a voltage divider, but I cannot work out the resistors that form the divider because of the strange term R_bridge. \$\endgroup\$ Nov 13 '20 at 15:43
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Well, just set R9 to be zero, and all bridge resistors to be 120 and the potential divider produces a ratio of 360/480 and clearly that's wrong because it should be 0.5. At this point, if I were you I'd ignore the document and find another I could rely on.

Moving on... just look at the glaring error putting R5 in the numerator when they meant R9 so, if I used R9 = 0 in that formula instead of R5 then it comes up with the right answer but, I'd be tossing this document in the bin for such glaring mistakes: -

enter image description here

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  • \$\begingroup\$ Thank you for your reply sir! I also thought that the expression was quite strange, but I guess I was so focused on trying to understand the meaning behind it that I missed the obvious error :). Personally, I don't find any particular reason for including the resistor R9 in addition to the current limiting resistor R8. \$\endgroup\$ Nov 13 '20 at 16:12
  • \$\begingroup\$ @IlieRazvan - for a quarter bridge circuit you'd be daft to use either R8 or R9 when a pure voltage source (or better still a current source) would perform better - you'd get a stabler output voltage and a more guaranteed relationship between Vout and the change in value of R7 if you do and the maths show that a current source is slightly more linear than a voltage source hence, the circuit using R8, R9 is pretty naive. Do you know what to do next if you are satisfied with the answer. If not place another comment for clarification. \$\endgroup\$
    – Andy aka
    Nov 13 '20 at 16:37

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