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After introducing the two-stages op-amp and frequency compensation for this circuit, during lecture the Professor asked us to find as an exercise a possible polarization of the following circuit:

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in which by hypothesis Vdd=1.8V and the input dc common mode is 0.9V. While solving together this exercise, the Professor said that, in order to use this circuit in the unity gain buffer configuration, the output dc voltage must be set equal to 0.9V. In this way the input dc common mode coincides with the output dc voltage, as required by the unity gain buffer configuration.

Question: why do I have to "artificially" impose that the output dc voltage is equal to the input dc common mode in order to have unity gain buffer configuration? Is not the negative feedback which imposes Vout=Vin?

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I agree with you that the Professor's statement is unclear and confusing.

When used as a voltage buffer, the DC voltage at inverting and non-inverting inputs and the output are determined by the DC voltage at the non-inverting input. As you mention, the feedback loop takes care of that.

Designing the output to be biased at Vdd/2 by itself (without feedback) isn't even possible with this circuit as the output's DC voltage depends on the output current of two current sources (M5 and M6) into a high impedance point. Even if the drain currents of M5 and M6 would be exactly identical, the voltage at the output would be undefined. In the real world these currents are never identical so the output will clip to Vdd or ground.

Only a feedback loop can control the output voltage at the output such that it will have a predictable value.

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  • \$\begingroup\$ Thank you for your answer. Then did Professor find the bias voltages as if unity negative feedback was applied? Moreover, you said that it's impossible to set the output voltage exactly at Vdd/2 (and I agree: there are systematic and for sure random offset). Should I then suppose, every time I design an op-amp, that the output dc voltage value will be exactly set only by negative feedback (and thus my design is just an "help" to facilitate negative feedback to set Vdd/2 at the output)? \$\endgroup\$
    – Stefanino
    Commented Sep 3, 2019 at 7:50
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    \$\begingroup\$ Without the feedback in place, this circuit would behave like a comparator so the output is either Vdd or ground. So yes, the feedback is needed to design /bias the circuit properly so that it works as intended. The feedback is an essential part of how this circuit works when used as a voltage buffer so the feedback needs to be included when designing the circuit. If you want to use the circuit as a comparator then you would not include the feedback. \$\endgroup\$
    – Bimpelrekkie
    Commented Sep 3, 2019 at 8:39

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