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In the above inverting amplifier, we assume -ve node to be at 0V approx (virtual ground), because the open loop gain Aol is very high and hence the differential voltage (Vd = V+ - V- = Vsat/Aol) has to be 0V approx. But why this concept is not applied in the below schmitt trigger ? Should not the differential voltage Vd be = Vsat/Aol here too, despite the positive feedback ?

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2 Answers 2

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The virtual ground principle applies to the classical inverting opamp circuit (with NEGATIVE DC feedback) only. Such a negative DC feedback ensures a fixed DC operating point around which the amplified signal can swing.

For dual DC voltage supply this operational point is identical to a very small DC voltage difference directly across the opamp terminals (µV range). Assuming an IDEAL opamp (infinite DC gain)this voltage difference approaches zero (virtual ground principle).

Such an operation (amplification of an applied input signal around a fixed DC operational point ) requires linear operation of the whole amplifier.

In contrast, an opamp with positive feedback also has a fixed DC output voltage (saturation at app. one of the power rails), which, however, is not called "operational point". In this case, the opamp is not biased for linear operation and cannot be regarded as a linear amplifier with a very large voltafe gain. For this reason, the voltage at the inverting terminal is not very small resp. zero. Instead, this voltage is simply Vout,dc*R1/(R1+R2).

MITURAJ - does this answer the question?

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  • \$\begingroup\$ Yes ..got it :-) \$\endgroup\$
    – Mitu Raj
    Commented Dec 1, 2017 at 16:54
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In the inverting amplifier the feedback from input to output is negative, which stabilizes the output (to make the - input the same level as the + input).

In the schmitt-trigger there is positive feedback, hence the output is driven to the extreme (in the absense of supply voltage limitations, to + or - infinte).

The 'virtual ground' concept applies onl;y when the opamp drives itself to stability (both inputs at the same level). The S.T. does exactly the opposite: it drives itself to on eof the extremes.

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  • \$\begingroup\$ For my opinion, the term "stability" does not apply in this context. The shown circuit with positive resistive feedback is stable. It fulfills all stability criteria. Instead, the the property of LINEAR operation (around a fixed DC operating point) is the key to the answer. \$\endgroup\$
    – LvW
    Commented Nov 29, 2017 at 21:11
  • \$\begingroup\$ @LvW can you expand your opinion as an answer ? \$\endgroup\$
    – Mitu Raj
    Commented Dec 1, 2017 at 1:07

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