Can anyone please explain, with sufficient mathematics, why the virtual short concept is not applicable to an operational amplifier in positive feedback.
Electronic – Positive feedback and virtual short in Operational Amplifiers
mathmodelingoperational-amplifier
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See my answer here for an example of why you might want to use negative and positive feedback at the same time. The 598.3K resistor in the positive feedback path maintains a constant current through the variable resistor and the negative feedback path determines the gain (output volts per ohm of resistance of the variable resistor).
To see how this yields a constant current, consider Scott's generalized NIC answer. This is also a NIC- creating a negative resistance to cancel out the incremental effect of the 100K resistor R2, so the current remains constant.
Looking into the non-inverting input of the op-amp (without R2 and the variable resistor connected, but with R4 in place), the resistance looks like:
Referring to the NIC schematic, we have:-
R1 = 100K || 25.68K = 20.43K R2 = 122.2K R3 = 598.3K
\$ R_{IN} = \$ - 598.3K \$ \cdot \$ \$ 20.43K \over 122.2K\$ = -100K, which exactly cancels out the effect of the 100K resistor R2 (R2 on the original schematic).
I know that Opamp trys to hold the input voltages the same
Not quite. The correct statement is
"When negative feedback is present, the voltage across the (ideal) op-amp input terminals is zero"
why we're not getting a virtual ground on the + node if we connect - node to ground?
In the typical configuration, the op-amp output is connected in some way to the inverting input and, for the ideal op-amp, the output voltage will be whatever it needs to be such that the inverting input voltage is the same as the non-inverting input voltage.
If one grounds the inverting input instead and connects the output in some way to the non-inverting input, it is true that mathematically, one can show that there is an output voltage that will make the non-inverting input voltage zero.
However, it is easy to show that this is an unstable situation - positive feedback - and that if the non-inverting input voltage is disturbed, the output voltage 'runs away', amplifying the disturbance rather than attenuating it as is the case with negative feedback.
Best Answer
At first, I assume that you speak about operational amplifiers and the virtual short across the opamps input terminals, right? In this case, your statement - in this general form - is not correct. Let me explain:
The term "virtual short" applies to amplifier units with a very large open-loop gain which may be set to infinity (during mathematical manipulations/calculations). However, this assumption is true if the opamp is dynamically stable and operated in its linear region only. Normally, this is the case for negative feedback. However, there are some other applications which use negative and positive feedback at the same time. As long as the negative feedback is dominating (negative feedback factor larger than the pos. feedback factor) the circuit remains stable - and the "virtual short" principle continues to apply.
More than that, there are active filter circuits - Sallen-Key topologies, for example - which need positive feedback for Q enhancement. These circuits have negative feedback for DC (stable operating point) and in addition positive feedback for some specific frequencies (pole frequency region). Of course, in case of large Q values the circuit operates closer to the stability limit than simple amplifiers - but as long as negative feedback dominates the active filter circuit is stable and working as desired (and the virtual short principle applies).
In summary, the "virtual short" scenario does not apply for opamps with a feedback arrangement that causes dynamic instability. This is the case - for example - if we have a pure resistive feedback to the non-inv. terminal with a loop gain larger than unity (however, slight positive feedback leading to a loop gain below unity is stable!).
However, there may be an exception to this rule during switching: The classical Schmitt trigger is such a circuit with resistive positive feedback. When this circuit is used to build a squarewave oscillator, the opamp output is switched between both supply voltage limits - and during the switching phase the circuit crosses the region where we have a linear relationship between input and output. During this very short time period, the voltage between both opamp input nodes is negligible snmall (virtual short circuit).