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:
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;
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.
You are right to be confused.
What is the meaning of "flyback diode"?
For a "flyback" diode in parallel with an inductance, and meant to dissipate inductor energy when current through the inductor is interrupted, larger Vf for fixed current dissipates this energy more quickly. (By the way, you should not depend only on maximum current rating to judge if the diode is a good fit. You MAY ALSO have to calculate power dissipated in the junction, the junction temperature rise over ambient, know the max ambient temperature, and ensure the max junction temperature is less than the data sheet. If the frequency of operation gets high enough, the diode will burn out even if the current is less than the maximum peak current on the data sheet.)
Now, for a "flyback diode" used as a diode in a "flyback" circuit which boosts voltage, the efficiency of the circuit will be higher for a diode with low Vf at a fixed current. Reverse recovery figures prominently into the efficiency calculation so reverse recovery is important too.
So you can see there is some confusion because there are two different answers.
In general, it is much better to discuss circuit questions with a schematic. Otherwise, improper generalizations are inevitable.
Best Answer
Different semiconductor junctions have different forward voltages (and reverse leakage currents, and reverse breakdown voltages, etc.) The forward drop of a typical small-signal silicon diode is around 0.7 volts. Same thing only germanium, around 0.3V. The forward drop of a PIN (p-type, intrinsic, n-type) power diode like a 1N4004 is more like a volt or more. The forward drop of a typical 1A power Schottky is something like 0.3V at low currents, higher for their design working currents.
Band gap has a lot to do with it -- germanium has a lower band gap than silicon, which has a lower band gap than GaAs or other LED materials. Silicon carbide has a higher band gap yet, and silicon carbide Schottky diodes have forward drops of something like 2V (check my number on that).
Aside from band gap, the doping profile of the junction has a lot to do with it, too -- a Schottky diode is an extreme example, but a PIN diode will generally have a higher forward drop (and reverse breakdown voltage) than a PN junction. LED forward drops range from about 1.5V for red LEDs to 3 for blue -- this makes sense because the LED mechanism is basically to generate one photon per electron, so the forward drop in volts has to be equal to or more than the energy of the emitted photons in electron-volts.