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I have to solve the task shown on the image below. (The diodes should be ideal with a forward voltage of 0.7V) enter image description here

I tried solving it but always got the wrong current for D1 and V. First I drew it new like the schematics below.

schematic

simulate this circuit – Schematic created using CircuitLab

How do I need to approach such a task? I don't fully understand the -5V source.

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  • \$\begingroup\$ Your schematic is wrong. The -5V source should be +5V so that the bottom terminal is -5V with respect to the top. \$\endgroup\$ – Greg d'Eon Apr 2 '15 at 14:37
  • \$\begingroup\$ ok, I changed the bottom source. \$\endgroup\$ – Pascal Apr 2 '15 at 14:39
  • \$\begingroup\$ That looks good. (Man, don't CircuitLab's ammeters look ugly?) \$\endgroup\$ – Greg d'Eon Apr 2 '15 at 14:40
  • \$\begingroup\$ It's upside down :P \$\endgroup\$ – Pascal Apr 2 '15 at 14:42
  • \$\begingroup\$ This may be of some use: www-inst.eecs.berkeley.edu/~ee40/fa09/lectures/Lec_18.pdf See slides 12 and 13 \$\endgroup\$ – Tut Apr 2 '15 at 15:41
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I'd start with the assumption that both diodes are forward biased and see what happens.

D1 connected to ground means the anode is 0.7V. D2 will have a forward drop of 0.7 meaning that it's actually creating a virtual ground at V(out).

Then it's just voltages over resistors. (5-0.7)/10k = current through 10k. (0-(-5))/5k = current through 5k.

Then use KCL to determine what I is. Current through 10k flows into node, current through 5K flows out of node, and current I flows out of node. So you should have an equation like this:
$$I_{10k}-I_{5k}-I=0$$

When you do this, you'll see that I is actually negative which can't be because the diode would then be reversed biased and would block all current. That means D1 becomes an open in this circuit.

Now you just have one series line of voltage across resistors and a diode. Ohms law states this:
$$\frac{5-(-5)-0.7}{10k+5k}=I_{series}$$

With Iseries and 5k, you can find Vout. You know that I is 0 because D1 is reverse biased.

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The diodes will obviously change the voltage drops across the resistor, but you also need to take into account that they are unidirectional and that the anode cannot be more than 0.7V above the cathode. Once you take all those into account you should be able to solve it.

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  • \$\begingroup\$ So I would calculate it like this: Ir1 = (V1-D1)/R1 = 0.43mA Ir2 = (V2 - D1 + D2) / R2 = 0.86mA So that would mean, AM1 should show -0.43mA \$\endgroup\$ – Pascal Apr 2 '15 at 14:45
  • \$\begingroup\$ Except that I can't be less than 0A. And the voltage at D1's anode can't be higher than 0.7V. Which means that my answer is obviously wrong. Let me fix it... \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 2 '15 at 14:50
  • \$\begingroup\$ Should VM1 be equal to 0V? \$\endgroup\$ – Pascal Apr 2 '15 at 14:59
  • \$\begingroup\$ It can't be, because R1 will drop more voltage than R2 regardless of the diode. \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 2 '15 at 15:00

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