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## Theory of Common Mode Rejection Ratio II

Mismatch in the Gain Determining Resistors:Consider the standard 3 op-amp instrumentation amplifier as shown in figure below:

Figure: Schematic diagram of a Standard Instrumentation Amplifier

If resistors are mismatched due to the manufacturing tolerances, ΔR, in such a way that:

R

_{2A }= R_{2}(1 + Δ_{R}), R_{2B}= R_{2}(1 - Δ_{R}),R

_{3A }= R_{3 }(1 - Δ_{R}), R_{3B}= R_{3}(1 + Δ_{R}),R

_{4A}= R_{4}(1 + Δ_{R}), R_{4B}= R_{4}(1 - Δ_{R})Whenever this is permitted for the analysis of gain of the two phases we have for the first phase:

V

_{o1}= [1 + (R_{2A}/R_{1})] V_{1}– [(R_{2A}/R_{1})] V_{2}] = [1 + {R_{2}(1 + Δ_{R})}/R_{1}] V_{1}- [{R_{2}(1 + Δ_{R})}/R_{1}] V_{2}V

_{o2}= [1 + (R_{2B}/R_{1})] V_{2}– [(R_{2B}/R_{1})] V_{1}] = [1 + {R_{2}(1 - Δ_{R})}/R_{1}] V_{2}- [{R_{2}(1 - Δ_{R})}/R_{1}] V_{1}Once again when we take the input voltages as shown:

V

_{1}= V_{ic }+ (V_{id}/2) and V_{2}= V_{ic}– (V_{id}/2)We then get:

Note that the error in resistor values, ΔR, doesn’t change the gain given to the common-mode component that remains at unity. Whenever expanded further:

V

_{o1}= V_{ic}+ [1 + 2(R_{2}/R_{1})] (V_{id}/2) + [(2 Δ_{R}) (R_{2}/R_{1}) (V_{id}/2)]V

_{o2}= V_{ic}- [1 + 2(R_{2}/R_{1})] (V_{id}/2) + [(2 Δ_{R}) (R_{2}/R_{1}) (V_{id}/2)]Note that beneath such circumstances:

V

_{o1}- V_{o2}= [1 + 2(R_{2}/R_{1})] V_{id}This exhibits that the errors in the resistor values tend to cancel one other out whenever they are of similar magnitudes keeping the differential gain near to its design value.

The gain determining resistors in an output phase can as well be mismatched.

The prior analysis of the gain of this phase established the output as:

V

_{o}= [(R_{3B}+ R_{4B})/R_{3B}] [R_{4B}/(R_{3A}+ R_{4A})] V_{o1}– [(R_{4B}/ R_{3B})/ V_{o2}]With the mismatches involved as outlined above this becomes:

By using the Binomial theorem on denominator factors and ignoring higher order terms provides:

The norm with this stage is to make the nominal values R

_{3}= R_{4}for a balanced phase and hence the coefficients of V_{O1}including Δ_{R}become unity. Then neglecting the second order terms gives:V

_{o}= [(R_{4}/R_{3}) V_{o1}] – [(R_{4}/R_{3}) (1 - 2Δ_{R}) Vo_{2}]Substituting the voltages V

_{O1}and V_{O2}from the above gives:Expanding provides:

This exhibits that CMRR is directly proportional to the gain of cross-coupled input phase and the mismatch in the resistors in the output phase.

Finite CMRR of an Operational Amplifier:The gain provided a signal which is applied to non-inverting input of an op-amp which is always slightly dissimilar than the gain provided to a signal applied to inverting input, that is, there is a mismatch in the input-to-output gain among the two channels of op-amp. This signifies that when a common-mode signal is applied to both input terminal of the op-amp, the output is not zero however some small finite value. This implies that the CMRR of op-amp itself is finite. The small output voltage that exists can in effect be termed to the input as a differential signal by taking it as the input common-mode voltage divided by CMRR of the op-amp. When this input voltage is applied as a differential signal it gives similar error voltage at the op-amp output as is produced by the common-mode input voltage beneath the influence of finite CMRR of the op-amp itself. This is convenient for the analysis of closed loop negative-feedback configurations as the consequence of limited CMRR of the op-amps can be accounted for the analysis, whereas treating the op-amps as ideal and also the other properties of the configuration as suitable.

Figure: The Equivalent effect of finite CMRR of the Operational Amplifier

The effect of limited CMRR of the amplifiers in standard 3 op-amp instrumentation amplifier can be measured by using the above principle as in the circuit configuration of figure shown below. In this case, the common-mode voltage, V

_{ic}, is still applied at each and every input as it produces a resultant voltage at the output of each and every op-amp even when the op-amp CMRRs were infinite and this should be accounted for in establishing the common-mode input voltage to the second phase of the amplifier.The input voltages to each and every side of the amplifier can then be explained as:

V

_{1}= V_{ic }+ (V_{id}/2) ± (V_{ic}/CMRR_{1}) and V_{2}= V_{ic }- (V_{id}/2) ± (V_{ic}/CMRR_{2})The amplifier circuit can then be analyzed in the standard way.

For the initial phase, the output voltages are obtained as follows:

The utilization of the ± symbol in connection with op-amp CMRRs is just to point out that the channel imbalance can act in either direction and there are several combinations that will outcome in the worst case. Thus, the CMRR values can in effect be treated as the positive magnitudes and their degrading effects on the total CMRR of instrumentation amplifier considered as cumulative. Then the expression can be taken as of the form of:

V

_{o}= A_{d}V_{id }+ A_{C}V_{ic}Where:

A

_{d }= (R_{4}/R_{3}) [1 + 2(R_{2}/R_{1})] and A_{C}= (R_{4}/R_{3}) [{1 + 2(R_{2}/R_{1})} {(1/CMRR_{1}) + (1/CMRR_{2})} + (1/CMRR_{3})]From this the total CMRR of the amplifier due to finite CMRRs in op-amps can be received as:

The CMRR can as well be expressed in dBs as:

CMRR

_{OP}|_{dB}= 20 log_{10}CMRR_{OP}The expression above is frequently inverted and written more suitably as:

1/CMRR

_{OP}= (1/ CMRR_{1}) + (1/CMRR_{2}) + 1/[1+ 2(R_{2}/R_{1})] CMRR_{3}The analysis above exhibits that CMRR of the instrumentation amplifier is limited by finite CMRRs of the op-amps, with such of the input phase being more significant, whereas the effect of CMRR of the second phase is decreased by a factor equivalent to the gain of the initial phase, G

_{1}. If the CMRRs of the op amps containing similar magnitude then:1/CMRR

_{OP}= (1/CMRR_{1}) + (1/CMRR_{1}) + (1/G CMRR_{1}) = [2 + (1/G)] (1/CMRR_{1})Then, when G >> 1 then the CMRR is approximately:

CMRR

_{OP}= (1/2) CMRR_{1}The total CMRR of instrumentation amplifier resultant from the combined effects of the three influencing factors observed can be obtained by treating the individual factors as influencing the total CMRR in a parallel way. Then:

1/CMRR = (1/CMRR

_{ΔZ}) + (1/CMRR_{ΔR}) + (1/CMRR_{OP})And,

CMRR|

_{dB}= 20 log_{10 }CMRRWhen one CMRR factor is low, then this will outcome in an overall low of CMRR for the amplifier. The total CMRR can never be higher than that of lowest individual factor.

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