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Re: physically how virtual short of the opamp can be explain
let's say ideal opamp which have 2 inputs (negative & posetive) and one output
and the output voltage = (v2 - v1)*A
where:
v2; posetive input voltage
v1; negative input voltage
A ; is the gain of the opamp
so (v2 - v1) = Vo / A
where Vo is the output voltage
usualy A is very big and assumed to approuch infinety so Vo / A will approch zero
and (v2 -v1) = zero & (v2 -v1) is the voltage differance between the two trminals
and when it is approach zero thats means there is no defferance between the two input terminals and it is mean short circuit in electrical concepts
Re: physically how virtual short of the opamp can be explain
see opamp has a vry high gain e.g. 741 has 2*10^5
we use amplifiers in linear region
i.e. assuming power supply +-15V let us say
o/p is below 15V
so i/p will be 14/(2*10^5) =70 micro V
( i have considered 14 V as the saturation voltage 1V less than power supply )
70 micro Volt is close to 0 V (gnd) but not exactly ground potential
hence it is called virtual ground.
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