simulate this circuit – Schematic created using CircuitLab

I don't understand the reasoning behind the analysis of this circuit. The professor started by saying that if diodes D2 and D5 are ON and D4 and D3 are OFF than there is no global feedback in the circuit and V+ and V- of the first ideal op amp are not the same because the I1 is independent and Vo=R2*IO. Same goes for the other two diodes with the output being Vo=-R2*I0. When all 4 diodes are working then there is feedback and Vo=V2 because V+=V-.

He than proceeded to draw a diagram that is confirmed by the simulation. He also added that if there were a resistor in parallel with the I1 there would be feedback in all cases but the feedback would be weak and not much would change.

How do I know that there is no feedback in the case where 2 diodes are working and the other 2 are off and how does that change when the resistor in parallel with I1 is added. Also how do all 4 working diodes introduce the feedback here?

Isn't feedback a connection between the input and the output, and adding diodes doesn't seem to do anything with the input signal as far as i can see.

  • \$\begingroup\$ For a start: remove R1, V2 and opamps OA1 and OA2, from your circuit, and connect the wire where the output of opamp OA1 originally was, to ground. (So, cathode D4 = anode D2 = GND). That should give insight what happens when 1) enabling D2/D5 and disabling D3/D4, 2) enabling D3/D4 and disabling D2/D5 and 3) enabling all diodes. \$\endgroup\$ – Huisman Mar 26 at 22:17
  • \$\begingroup\$ Next, insert a voltage source between cathode D4 and ground and check what happens for different voltages for step 1), 2) and 3) ... Step 4) Place a resistor parallel with I1 and determine what happens in that situation. \$\endgroup\$ – Huisman Mar 26 at 22:52

Consider just the current source I1 and the four diodes around it. With no other current imposed on it, all four diodes are conducting, with half of I1's current flowing around each of the side loops.

Any current applied in the horizontal direction must flow through D2+D5 or D3+D4, depending on its sign. This current will add/subtract from the current supplied by I1 as appropriate, reducing the current in some diodes while increasing the current in the others.

Now, what happens if that external current tries to rise above 1 mA? Two of the diodes will cut off altogether, and the external current can ONLY flow through I1, steered by the remaining diodes. This means that the current CANNOT rise above 1 mA, and that the effective impedance of this circuit becomes infinite, effectively breaking the feedback loop.

  • \$\begingroup\$ This current will add/subtract from the current supplied by I1 You cannot 2 current sources in series, though? \$\endgroup\$ – Huisman Mar 26 at 22:48
  • \$\begingroup\$ No, you can't put two different current sources directly in series. However, this circuit allows the "difference" current to circulate around the loop of diodes, which removes this limitation. \$\endgroup\$ – Dave Tweed Mar 26 at 22:56
  • \$\begingroup\$ Ah, your answer only applies when all diodes are "ON", I was also still considering OP's options where only 2 of 4 diodes were "ON" a priori. \$\endgroup\$ – Huisman Mar 26 at 23:01

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