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I have this circuit:

enter image description here

I know that D1 is cut-off and ID2=1.33mA. How can I find the V?

In the answer of this excercise it says that V=VΒ=-10+10*1.33=3.3V. Is it correct? If yes, how did it come of and why is the V we are searching the same as the V in B?

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  • \$\begingroup\$ I assume we are looking at ideal diodes here. \$\endgroup\$ – RoyC Feb 2 '17 at 9:47
  • \$\begingroup\$ why is D1 in cutoff region? \$\endgroup\$ – User323693 Feb 2 '17 at 9:52
  • \$\begingroup\$ @RoyC Yes. I forgot to mention that. \$\endgroup\$ – Thanasis1101 Feb 2 '17 at 9:55
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10*1.33 is the 10k resistor voltage drop(1.33 mA). From the -10V to the B point, the voltage difference is on that resistor. V = VB=-10+voltage drop. The diode being ideal V is the same as in B.

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If you assume no current flow out of the terminal then it becomes trivial:

Given ID2 = 1.33mA then using ohms law voltage on the 5k resistor must be 5000*0.00133 = 6.65 V
If one end of the resistor is at +10V then the other end must be at 10-6.65 = 3.35 V

For non-ideal parts and a load current run a DC sweep simulation of this with a varying value for R3.

schematic

simulate this circuit – Schematic created using CircuitLab

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