- #1
kirikinny
- 5
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Hi, I'm looking at a circuit of a classical instrumental amplifier having 3 opamps, 2 for buffer, with Rg to adjust the gain. The output from this buffer + gain circuit is then feed into a third opamp, in a circuit that is the 'basic' differential amplifier, but whose function now that Rg is included in the buffer circuit, is only to match the resistances to have no common mode gain.
(this is the circuit I'm referring to: http://upload.wikimedia.org/wikipedia/commons/e/ed/Op-Amp_Instrumentation_Amplifier.svg)
My question is: why do we need the opamp on the right? If we take the output directly from the buffer circuit (we take voltage difference between the 2 outputs of the 2 opamps on the left), we obtain the same gain, and no common mode gain even if there is a resistors mismatch, right? because: Vout = (V2-V1) (1 + Ra/Rg + Rb/Rg), being Rg the gain controlling resistance and Ra Rb the other 2 resistors in the buffer circuit.
Thanks
(this is the circuit I'm referring to: http://upload.wikimedia.org/wikipedia/commons/e/ed/Op-Amp_Instrumentation_Amplifier.svg)
My question is: why do we need the opamp on the right? If we take the output directly from the buffer circuit (we take voltage difference between the 2 outputs of the 2 opamps on the left), we obtain the same gain, and no common mode gain even if there is a resistors mismatch, right? because: Vout = (V2-V1) (1 + Ra/Rg + Rb/Rg), being Rg the gain controlling resistance and Ra Rb the other 2 resistors in the buffer circuit.
Thanks