I have a doubt about the two green voltages in the opamp shown.
Let's assume an ideal opamp and all the resistors are equal
I noticed in the drawing that the green point V4 I indicated is a bit ambiguous, it is NOT an applied voltage .. it's simply the terminal to which I would connect the '' of the voltmeter measuring Vo .. while the '+' is Vo. R4 is floating.
In lecture V+ was calculated as Vin/2=2.5V and the (simplified) equiavalent circuit is as follows:
I tried to simulate the circuit on Everycircuit (app which I don't know how reliable it can be) and found this result:
1) Initially, I thought of performing a voltage divider since the 2 resistors are in series (and therefore the same current flows, although ideally zero in this case) .. in fact V+=Vi*1k/(1k+1k)=2.5V
2) Then, by watching the simulation, I had this doubt: there is no current circulating on the R3 and R4 as it is an open circuit, so there can be no voltage across each resistor. Why V+is not equal to 5V ?? (as the simulation shows).
Why don't I see 2.5V in simulation ??
Let's assume an ideal opamp and all the resistors are equal
I noticed in the drawing that the green point V4 I indicated is a bit ambiguous, it is NOT an applied voltage .. it's simply the terminal to which I would connect the '' of the voltmeter measuring Vo .. while the '+' is Vo. R4 is floating.
In lecture V+ was calculated as Vin/2=2.5V and the (simplified) equiavalent circuit is as follows:
I tried to simulate the circuit on Everycircuit (app which I don't know how reliable it can be) and found this result:
1) Initially, I thought of performing a voltage divider since the 2 resistors are in series (and therefore the same current flows, although ideally zero in this case) .. in fact V+=Vi*1k/(1k+1k)=2.5V
2) Then, by watching the simulation, I had this doubt: there is no current circulating on the R3 and R4 as it is an open circuit, so there can be no voltage across each resistor. Why V+is not equal to 5V ?? (as the simulation shows).
Why don't I see 2.5V in simulation ??
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