The instrumentation amplifier

Posted on Feb 6, 2014

As suggested before, it is beneficial to be able to adjust the gain of the amplifier circuit without having to change more than one resistor value, as is necessary with the previous design of differential amplifier. The so-called instrumentation builds on the last version of differential amplifier to give us that capability: This intimidating circ

The instrumentation amplifier
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uit is constructed from a buffered differential amplifier stage with three new resistors linking the two buffer circuits together. Consider all resistors to be of equal value except for Rgain. The negative feedback of the upper-left op-amp causes the voltage at point 1 (top of Rgain) to be equal to V1. Likewise, the voltage at point 2 (bottom of Rgain) is held to a value equal to V2. This establishes a voltage drop across Rgain equal to the voltage difference between V1 and V2. That voltage drop causes a current through Rgain, and since the feedback loops of the two input op-amps draw no current, that same amount of current through Rgain must be going through the two "R" resistors above and below it. This produces a voltage drop between points 3 and 4 equal to: The regular differential amplifier on the right-hand side of the circuit then takes this voltage drop between points 3 and 4, and amplifies it by a gain of 1 (assuming again that all "R" resistors are of equal value). Though this looks like a cumbersome way to build a differential amplifier, it has the distinct advantages of possessing extremely high input impedances on the V1 and V2 inputs (because they connect straight into the noninverting inputs of their respective op-amps), and adjustable gain that can be set by a single resistor. Manipulating the above formula a bit, we have a general expression for overall voltage gain in the instrumentation amplifier: Though...

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