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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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DESIGN PROBLEM : 4-bit increment by 2 circuit

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Problem: Derive the logical expression for a 4-bit increment by 2 circuit and draw the architecture of it.

Solution: The task here is to design a circuit that increments its count by two. Since, it is a 4-bit circuit, the total number of possible states is 16. Each state transitions to the state which has a binary value two greater than it. Now, there are two possible scenarios based upon the initial state that the counter gets into:

1. It can count 0 -> 2 -> 4 -> 6 -> 8 -> 10 -> 12 -> 14 -> 0 (their binary equivalents)

2. It can count 1 -> 3 -> 5 -> 7 -> 9 -> 11 -> 13 -> 15 -> 1 (their binary equivalents)

The state transition table can be represented as shown below:



We can find the expression for outputs using K-maps as below.

Expression for D3(next): Let us first derive the expression for D3(next). The K-map can be represented as below:

The expression for D3(next) as derived from K-map is:
D3(next) = D3.D2' + D3.D1' + D3'.D2.D1
D3(next) = D3.(D2'+D1') + D3'.D2.D1.
D3(next) = D3.(D2.D1)'+D3'.(D2.D1)
D3(next) = D3 (exor) (D2.D1) 

Expression for D2(next): Given below is the K-map derived from state transition table for D2(next).


The expression for D2(next) as derived from K-map is:
D2(next) = D2'.D1 + D2.D1' = D2 (exor) D1

Expression for D1(next):  Given below is the K-map derived from state transition table for D1(next).

The expression for D1(next) is derived from K-map as:
D1(next) = D1'

Expression for D0(next): Given below is the K-map for D0(next).

The expression for D0(next) is:
D0(next) = D0

Combining all the expressions, the circuit is as given below:



Can you come up with a better solution for this problem? Let us know your views in comments.

This question was asked by one Himadri Roy Pramanik on our post your query page. You can also post your queries there. We will try to answer using our limited knowledge.

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