**1 Answer**

written 8.3 years ago by | • modified 8.3 years ago |

In order to measure and control some of physical quantity such as temperature, humidity, water flow etc. transducer is used to drive display unit which indicate change in the above parameters. The signal level of transducer is low which doesn’t drive display unit. Hence in order to raise signal level of transducer Instrumentation amplifier is used.

Important features of an instrumentation amplifier are as follows:

- High CMRR (Common Mode Rejection Ratio)
- Large input impedance.
- Low output impedance.
- High gain accuracy.
- Both the input voltage should not be same i.e V1 ≠ V2 .
- High gain stability with low temperature coefficient.

Derivation of output voltage equation :

Consider Figure 1 ,

Effect of $V_1$ and $V_2$ on $V_{01}$ and $V_{02}$ can be obtained by superposition theorem ,

Consider $V_1$ and $V_2$ is grounded. Hence A1 amplifier becomes as shown in Figure 2,

$$V'_{01} = (1 + \frac{R_1}{R_2})V_1$$

Similarly, due to $V_2 A_1$ becomes,

$$V^"_{01} = -\frac{R_1}{R_2}V_2$$

$$V_{01} = V'_{01} + V^"_{02}$$

$$V_{01} = (\frac{R_2 + R_1}{R_2})V_1 - \frac{R_1}{R_2}V_2$$

Similarly $V_{02}$ due to $V_1$ and $V_2$ ( Considering $A_2$), $$V_{02} = (\frac{R_2 + R_1}{R_2})V_2 - \frac{R_1}{R_2}V_1$$

Hence $V_O$ can be obtained by adding $V_{01}$ and $V_{02}$, $$V_O = V_{02} - V_{01}$$ $$V_O = (\frac{R_2 + R_1}{R_2})V_2 - \frac{R_1}{R_2}V_1 - (\frac{R_2 + R_1}{R_2})V_1 + \frac{R_1}{R_2}V_1 $$

$$V_O = V_2 + (\frac{R_1}{R_2})V_2 - \frac{R_1}{R_2}V_1 - V_1 - (\frac{R_1}{R_2})V_1 + (\frac{R_1}{R_2})V_2 $$

$$V_O = V_2 + 2(\frac{R_1}{R_2})V_2 - V_1 - 2(\frac{R_1}{R_2})V_1$$

$$V_O = V_2(1 + 2(\frac{R_1}{R_2})) - V_1(1 + 2(\frac{R_1}{R_2}))$$

$$V_O = (1 + 2(\frac{R_1}{R_2}))(V_2 - V_1)$$

$$\boxed{V_O = (1 + 2(\frac{R_1}{R_2}))(V_2 - V_1)}$$

Thus by varing $R_2$, $V_O$ can be obtained.

## Application :

- Differential instrumentation amplifier using Transducer Bridge. Figure 4 shows circuit diagram, the circuit uses a resistive transducer whose resistance changes as a function of the physical quantity to be measured. The bridge is initially balanced by a dc supply voltage $V_{dc}$ so that $V_1 = V_2$ . As the physical quantity changes, the resistance RT of the transducer also changes, causing an unbalance in the bridge ($V_1$ ≠ $V_2$ ). This differential voltage now gets amplified by the three op-amp differential instrumentation amplifier.