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The following circuit is given:

AC op-amp circuit

The 2 diodes are idealised in the following way:

idealised diodes

All other parts have the following properties:

part properties

I've been trying to figure out what the voltage at the entrance of the inverted op-amp is in respect to time (u2 in the schematic below).

I know that the voltage u0(t) is essentially the same as the input voltage u1(t) without the DC part.

So u0(t) = 1,4V cos(wt - 0,5π) = 1,4V sin(wt)

So I am essentially looking for u2(t) here:

schematic

simulate this circuit – Schematic created using CircuitLab

I know that the voltage over the diodes can only be between 0V and 0,7V. But do the 2 diodes even allow a voltage drop over any of them because they are essentially blocking in both directions? If the Diode is blocking, will the voltage over R0 be the entrance voltage to the op-amp? What would be the correct way to analyse this circuit?

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  • \$\begingroup\$ It's unclear what voltage you mean when you say "I've been trying to figure out what the voltage at the entrance of the inverted op-amp is in respect to time." \$\endgroup\$ – Andy aka Jan 29 at 12:38
  • \$\begingroup\$ ..because they are essentially blocking in both directions? Ignoring the circuit around the opamp: then your conclusion would be correct. But the opamp circuit is there so you cannot ignore it. If U1 is negative then the left diode can still pull current out of the opamp circuit. Start by thinking how the circuit behaves when Vin < - 0.7 V, -0.7 V < Vin < 0 V, 0 < Vin < +0.7 V and +0.7 V < Vin \$\endgroup\$ – Bimpelrekkie Jan 29 at 12:39
  • \$\begingroup\$ To me, it is not clear if you speak about a (small-signal, linear) ac analysis or a TRAN analysis in the time domain. \$\endgroup\$ – LvW Jan 29 at 16:25

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