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This is not homework. I was trying to come up with a circuit where the diode's state is not uniquely determined. This is the "on" case. Can you tell me what is the mistake in my analysis?

All resistors are 1 ohm and the source is 0.7 volts.

EDIT: Sorry about the incomplete description, I was posting from my cellphone.

This is the original cicuit. The greek text translates "constant voltage drop model 0.7V":

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This is assuming the diode is forward biased:

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This is assuming the diode is reverse biased:

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Since both assumptions work (do not lead to contradiction) they are equally valid. I was trying to make the point that the constant voltage drop model is insufficient for the analysis of some circuits, i.e. does not admit a unique solution. My professor told me that my calculations in the forward biased circuit are wrong.

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    \$\begingroup\$ It's not obvious how the diode relates. No diode shown. Can you please explain with more words how this elates and what you are trying to achieve. \$\endgroup\$ – Russell McMahon Feb 6 '12 at 10:26
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    \$\begingroup\$ The circuit shown is a traditional opamp circuit doing exactly what is expected. Gain is 2. Vin = -0.7V so Vo = 2 x -0.7 = -1.4V. \$\endgroup\$ – Russell McMahon Feb 6 '12 at 10:29
  • \$\begingroup\$ Yeap the diode is presented in the on state as a voltage source. \$\endgroup\$ – Panayiotis Karabassis Feb 6 '12 at 11:27
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    \$\begingroup\$ If you can, embed the images (maybe at lower resolution) in your question: note that you have to upload them in the editor, since the hosting site doesn't accept the url you are using \$\endgroup\$ – clabacchio Feb 6 '12 at 12:58
  • \$\begingroup\$ Thanks. Sorry but it seems like the editor does not accept embedding more than one image. Let me know if this is not the case. I did upload all three in the editor. \$\endgroup\$ – Panayiotis Karabassis Feb 6 '12 at 13:15
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The mistake depends by what you are trying to achieve.

In this case, you have the voltage source that is suppose to simulate the forward-biased diode. This means that, since the source is connected with the plus to ground, you are generating -0.7V at the non-inverting input of the OPAMP. So, a current is flowing from ground across the source, and that current depends on the output voltage and the value of the top resistance (perhaps 1 Ohm).

Then, let's look at the inverting input and the voltage divider. Since between the two OPAMP inputs there is a virtual short circuit, the central point of the divider will be -0.7V. Using two equal (1 Ohm I guess) resistors, you are causing the output to be at -1.4V. Again, the current will flow out of the ground.

Now, back again to our generator. We said that the non-inverting input is at -0.7, and the output is at -1.4V. Hence we will have a 0.7V drop over the resistor, and since (as guessed before) it's a 1 Ohm resistor, the current across the resistor and the generator/diode (since the OPAMP inputs are ideally open circuits) will be 0.7A

Conclusion

If you are trying to simulate the 0.7V drop of the forward-biased diode, it's what the supply is doing. If you are expecting to see positive voltages, it's not because of negative resistors, but because the supply has to be flipped.

Update

There are two cases, depending on the initial state:

  1. The output of the OPAMP is HIGH: then, the diode is reverse biased, no current is flowing in the upper branch, and the non-inverting input is at a higher voltage than the inverting, that is always at half the output voltage. Hence, the OPAMP goes into positive saturation;

  2. The output is LOW: then, the diode is forward biased, the voltage at the non-inverting pin is -0.7, and the situation is the aforementioned.

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  • \$\begingroup\$ So the analysis is correct, right? I was trying to come up with a contradiction to prove that this circuit is impossible. The circuit minus the source is taken from the wikipedia article "negative resistor" and is supposed to behave like a -1 ohm resistor. I'll update the question with more info. \$\endgroup\$ – Panayiotis Karabassis Feb 6 '12 at 11:34
  • \$\begingroup\$ Scratch that. I was trying to find whether the current across the diode is positive (up). Because if it isn't the diode is off. \$\endgroup\$ – Panayiotis Karabassis Feb 6 '12 at 11:43
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    \$\begingroup\$ @PanayiotisKarabassis the current goes up indeed, because the lower potential is up. \$\endgroup\$ – clabacchio Feb 6 '12 at 11:59
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    \$\begingroup\$ @PanayiotisKarabassis that should be your answer \$\endgroup\$ – clabacchio Feb 6 '12 at 12:56
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    \$\begingroup\$ @PanayiotisKarabassis The ideal opamp saturates, too. It does when the inputs are forced to be unbalanced, as in this case. And when the OPAMP output is zero, it would remains with all nodes to 0V, but it's an unstable status that will end in one of the other two (basically noise will unbalance this status). \$\endgroup\$ – clabacchio Feb 6 '12 at 13:39

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