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Unformatted text preview: 1 Instrumentation Electronic measurement instruments convert a physical quantity into an elec trical signal, which can then be processed and displayed. The component that initially converts the physical quantity into an electrical signal is called a trans ducer. Many transducers work by creating a potential difference across their terminals that is proportional to the physical quantity under consideration. For instance, the potential difference across a thermocouple is proportional to the heat gradient to which the thermocouple is exposed. It is particularly impor tant to note that the quantity of interest for measurement instruments is the potential difference between the terminals of the transducer, rather than the voltage (relative to ground) of either terminal individually. Thus, effective mea surement instruments require a means of precisely determining the potential difference between the terminals of the transducer. 2 Differential Amplifier A first attempt at creating a circuit to measure voltage differences is the differ ential amplifier (see Fig. 1 ). Figure 1: A differential amplifier. The output of this circuit can easily be determined using superposition. Suppressing v 1 gives a noninverting amplifier configuration, with v 2 as the 1 input. v out, 2 = 1 + R f R i R 2 R 1 + R 2 v 2 Similarly, suppressing v 2 gives an inverting amplifier configuration, with v 1 as the input. v out, 1 = R f R i v 1 Using the principle of superposition, we can sum v out, 2 and v out, 1 to obtain v out . v out = 1 + R f R i R 2 R 1 + R 2 v 2 R f R i v 1 Ideally, the output of the differential amplifier should be proportional to the difference between the inputs. We can define the commonmode and difference mode voltages of the inputs. v cm = 1 2 ( v 1 + v 2 ) v d = v 2 v 1 This system of two equations can be solved to find v 1 and v 2 in terms of v cm and v d . v 1 = v cm 1 2 v d v 2 = v cm + 1 2 v d Having decomposed the inputs into the sum of their common and difference modes, we can substitute these decompositions into the expression for v out . v out = 1 + R f R i R 2 R 1 + R 2 v 2 R f R i v 1 = 1 + R f R i R 2 R 1 + R 2 v cm + 1 2 v d R f R i v cm 1 2 v d = R 2 R 1 + R 2 1 + R f R i R f R i v cm + 1 2 R 2 R 1 + R 2 1 + R f R i + R f R i v d = A cm v cm + A d v d where A cm and A d are the commonmode and differencemode gains, respec tively. Our design goal is A cm = 0. From this goal, we can derive a condition 2 on the resistor values. A cm = 0 R 2 R 1 + R 2 1 + R f R i R f R i = 0 R 2 R 1 + R 2 = 1 R 2 R 1 + R 2 R f R i R 2 R 1 + R 2 = R 1 R 1 + R 2 R f R i R 2 R 1 = R f R i From this condition on the resistor values, we can derive an expression for the differencemode gain, assuming that our design satisfies the constraint....
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 Fall '07
 HurKoser

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