The instrumentation amplifier amplifies the difference between the two inputs. In Industrial application there are many unwanted noise that are picked up by the circuit, the instrumentation amplifier has the ability to remove these unwanted noise. The CMRR of instrumentation amplifier is very high.
The instrumentation amplifiers are distinctive and the resistor value are Selected Such that one resistor can change the overall gain of the circut This can be done by choosing proper value of each resistor.
The first state is a balanced "P, balanced op amplifier formed by $A_1$ & $A_2$ which amplifies the differential signal but passes the common mode signal without amplification. The second stage formed by $A_3$ is differential amplifier which largely removes the common mode signal.
The voltage $V_{01}$ consists of two components. the voltage due $V_1$ and voltage due to $V_2$.
If $V_2$ = 0 then virtual ground and amplifier $A_2$ will act as an inverting amplifier with a gain of,
$$V_{01} = V_1 (\frac{R_1 + R_2}{R_2})$$
And, If $V_1$ = 0 then virtual ground and amplifier $A_1$ will act as an inverting amplifier with a gain of,
$$V_{01} = - (\frac{R_1}{R_2}) V_2$$
The output from amplifier $A_1$ with respect to ground will be -
$$V_{01} = (\frac{R_1 + R_2}{R_2}) V_1 - (\frac{R_1}{R_2}) V_2$$
$$V_{01} = \frac{(R_1 + R_2) V_1 - R_1(V_2)}{R_2}$$
$$V_{01} = (\frac{R_1}{R_2} + 1) V_1 - (\frac{R_1}{R_2})V_2$$
$$V_{01} = \frac{R_1}{R_2}[V_1 - V_2] + V_1\ --- (i)$$
Similarly the output from the amplifier is given as,
$$V_{02} = \frac{R_1}{R_2} (V_2 - V_1) + V_2\ --- (ii)$$
The gain of the amplifier is given as,
$$V_0 = \frac{R_4}{R_3} (V_{02} - V_{01})\ --- (iii)$$
Putting equation (i) and (ii) in the equation (iii) we get,
$$V_0 = \frac{R_4}{R_3} ([\frac{R_1}{R_2}(V_2 - V_1)] + V_2) - \frac{R_1}{R_2} (V_1 - V_2) + V_1$$
This can be simplified as,
$$V_0 = \frac{R_4}{R_3} [\frac{R_1 V_2}{R_2} - \frac{R_1 V_1}{R_2} + V_2 - \frac{R_1 V_1}{R_2} - \frac{R_1 V_2}{R_2} + V_1 ] $$
$$V_0 = \frac{R_4}{R_3} [\frac{2R_1 V_2}{R_2} + V_2 - \frac{2R_1 V_1}{R_2}+ V_1 ] $$
$$V_0 = \frac{R_4}{R_3} (V_2 [1 + \frac{2R_1}{R_2}] - V_1 [1 + \frac{2R_1}{R_2}])$$
Therefore, $$V_0 = (V_2 - V_1) \frac{R_4}{R_3} (1 + \frac{2R_1}{R_2})$$
So, the differential gain G is,
$$G = \frac{R_4}{R_3} (1 + \frac{2R_1}{R_2})$$
For common R gain is given as,
$$Gain\ =\ A\ =\ 1 + \frac{2R_3}{R_4}\ --- (iv)$$
$$Because\ V_0 = A(V_2 - V_1)$$
If the gain is given as |0| then from equation,
$$A = 1 + \frac{2R_3}{R_4}$$
$$|0| = 1 + \frac{2R_3}{R_4}$$
$$|0| -1 = \frac{2R_3}{R_4}$$
$$Therefore,\ 2R_3 = R_4 100$$
$$R_3 = R_4 \frac{100}{2}$$
$$R_3 = R_4 50$$
Therefore, Resistor R can be computed as,
$$R_3 = 50 R_4$$
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Krishna_Agrawal
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