How to Determine Vd in an OP AMP Circuit with Diode?

[pyroknife]

I have attached the circuit of interest.
The question is to find Vd (drop across diode as shown in the attached image).

Sorry about the handwriting on there (those can be ignored).
The answer was given to be Vd=-5V.

I can't figure out how they got that answer. I don't feel there is enough information given to be bale to determine the drop.

The voltage @ the left node of diode=+5V. You don't know the voltage @ the right node of the diode though. The voltage @ right node should be 10V-1KΩ*(I4).
If the answer is -5V, then I4 must be 0, but I don't see how you could have solved for I4 with the amount of info that was given.

 

[phyzguy]

I4 must be zero. Any current that flows through that resistor has to flow from left to right, as indicated, because otherwise it is flowing backwards through the diode, which it can't do. So the diode has 5V on the left side, and 10V + I4*1K on the right side (not -I4*1K as you wrote). So, the diode is reverse biased and no current is flowing in it, so I4 = 0.

 

[pyroknife]

I4 must be zero. Any current that flows through that resistor has to flow from left to right, as indicated, because otherwise it is flowing backwards through the diode, which it can't do. So the diode has 5V on the left side, and 10V + I4*1K on the right side (not -I4*1K as you wrote). So, the diode is reverse biased and no current is flowing in it, so I4 = 0.

Oops, yeah it is +I4*1k.

Doesn't the left to right current flow as indicated mean the diode is 'forward' biased?
Right to left current flow for this diode should be reversed biased, no?

 

[phyzguy]

In order for the current to flow left to right as shown, the diode would have to be forward biased, yes, but in fact it is reverse biased so there is no current flow. Just because I draw an arrow doesn't mean current is flowing.

 

[joebuba]

phyzguy is correct. The diode is reverse-biased so that there is no current flowing: I4 = 0. Therefore, the voltage across the resistor is 0 V. The voltage across the diode from left to right(by the shown polarities on the schematic) is 5 V - 10 V = -5 V.