Dr Kanaan,
I think that it's clear that, depending on the magnitudes of R2, R3, and R1, this might be a negative or positive feedback amplifier.
thank you sir, James Garry , i have reading materials that said it depends on source if it is current or voltage , voltage source mean negative feedback .whereas positive feedback mean positive. but i didn't reached to conclusion why ?
sincerely
Dr Kanaan,
As drawn, there is a constant voltage applied to the positive terminal of the op-amp. The current that flows through the resistor network, R3,R2,R1, leads to a voltage at the negative input. Imagine now if R1 is an open circuit, or a short - then two very different behaviours arise.
I can't quite see how changing the nature of the source (constant current or constant voltage) changes matters - maybe worth modelling in SPICE?
Dear James Garry , you are alright , excellent conclusion and discussion , yes i think it is mater to do also some simulation for source type,
if you have any further information please let me know.
thank you for your efforts and time
Hello Tariq,
I sense that simulation will exactly create the outcome described - that voltage will not proceed in 'lockstep' with current - but will oppose it.
But the derivation is, I gather, a good test for an engineering class.
The voltage source has a very low output impedance Rin . The output is unpredictable . You have to measure the voltage sources in order to make a model ! It's an ordinary voltage negative feedback, wich can go to a Schmit triger .
Let say that AmpOP is not saturated and V° is the voltage at its inputs. Let say your source voltage is perfect, so Rin->0 . V°=Vin and
Vout =Vin(R1+R2)/R1 . It is a voltage negative feedback with gain >1 .
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