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High CMRR Instrumentation Amplifier (Schematic and Layout) design for biomedical applications

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Instrumentation amplifiers are intended to be used whenever acquisition of a useful signal is difficult. IA’s must have extremely high input impedances because source impedances may be high and/or unbalanced. bias and offset currents are low and relatively stable so that the source impedance need not be constant. Balanced differential inputs are provided so that the signal source may be referenced to any reasonable level independent of the IA output load reference. Common mode rejection, a measure of input balance, is very high so that noise pickup and ground drops, characteristic of remote sensor applications, are minimized.Care is taken to provide high, well characterized stability of critical parameters under varying conditions, such as changing temperatures and supply voltages. Finally, all components that are critical to the performance of the IA are internal to the device. The precision of an IA is provided at the expense of flexibility. By committing to the one specific task of
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Series - Parallel Resistances the Combination

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Whenever we work with circuits in the real world, they are seldom as straightforward as a simple series or parallel circuit. Normally, they are a combination of the two, called a SERIES-PARALLEL circuit. While they look forbidding at first, you must keep in mind that ALL circuits can be broken down into smaller parts. They can be made simpler to work with. Such is the case with the Series-Parallel circuit.

If you look at the example on the right, it has 3 resistors and 1 battery. R 1 and R 2 are both 10 Ω in parallel.
          1
    ------------
      1       1
    ---- + ---- 
     R 1      R 2
If we do the math (reciprocating the reciprocals), we come up with a total of 5Ω for these two.
We can say that:
R 1&2 = 5Ω.


We also have R 3 in the circuit, which is 20Ω. Once we have combined the 2 parallel resistors, we have a simpler circuit.... 2 series resistors. R 1&2 and R 3 . If we add the value of these two resistors, we come up with
R Total =R 1&2 +R 3 .


So R Total =5Ω+20Ω=25Ω.


Then if we know the voltage, we can find the current through the entire circuit, and through each individual resistor. Go ahead and try plugging in a voltage (like 25V) and finding the currents. You'll be surprised at how simple it is.

Let's try another example. In the circuit on the right, we have 3 resistors again. But this time, they are configured differently. Do you see how you would combine them for the total resistance? First, you must add the 10 Ω resistors by adding them. This is simple because they are in Series.
    10 Ω + 10 Ω = 20 Ω.
Using our parallel circuit formula then:
          1
    ------------
      1       1
    ---- + ---- 
     R 1      R 2
We find that our total resistance for the circuit = 10Ω. Note that if we have the same battery (25V), our current turns out much different through the entire circuit. So the same components, configured differently, will create a vast difference in the way the circuit works.