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schematic

simulate this circuit – Schematic created using CircuitLab

$$U_{DF}=0.7V$$ Calculate the currents of diodes. I've used KVL and KCL in 3 nodes, and I got $$I_{D1}=0.2mA$$

$$I_{D2}=-0.5mA$$ but i don't think that this second current should be negative.

loop 3 $$3V - R_1I-0.7-0,7=0 $$ which gives $$I =1.6mA$$ loop 1 $$3V-R_1I-R_2I_0=0$$ $$I_0=1.4mA$$ from KCL $$I_{D1}=I-I_0=0.2mA$$ loop 2 $$3V-R_1I-0.7-R_3I_1=0$$ $$I_1=0.7mA$$ from KCL $$I_{D2}=I_{D1}-I_1=-0.5mA$$

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  • \$\begingroup\$ Draw in the current loops and see if that helps. (You can use the arrow symbol. Make it some colour other than black.) \$\endgroup\$
    – Transistor
    May 16 '18 at 18:57
  • \$\begingroup\$ Label all components so the 1k resistors are R1, R2 and R3 then either label your nodes or currents, It wouldn't hurt to do both. \$\endgroup\$ May 16 '18 at 19:17
  • \$\begingroup\$ That was a tricky question. You can make sure ur assumption that : diode D2 is conducting , is wrong right when you got -ve value there. \$\endgroup\$ May 17 '18 at 12:41
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Using KVL and KCl here seems to confuse you from getting the correct answers. If you consider 0 V as the negative side of your voltage source, what must the voltage be on R2 in the middle? Answer:1.4V. Why ?

Now work out the rest easily .

Change R1,R2,V1 to 1.5V+0.5k then V(R3) becomesChange R1,R2,V1 to 1.5V+0.5k is the equivalent source and KVL indicates if you remove D2, (to test for potential V(D2))=V((R2))

I(D1)=(1.5V-0.7V)/(0.5k+1k)=0.8V/1.5k= 0.53...mA Thus if V(R2) cannot be 1.4V since V(D2)=0.53V is not conducting significantly (<<0.7V)

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  • \$\begingroup\$ It would be great, if you could look at my calculations and say what is wrong in this approach \$\endgroup\$
    – Kornel_S
    May 16 '18 at 19:42
  • \$\begingroup\$ You must apply superposition to get I so it is the correct sum \$\endgroup\$ May 16 '18 at 19:46
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    \$\begingroup\$ @Kornel_S - You must keep in mind that "UDF=0.7V" is only true for diodes which are conducting. Is that true in this case? For Loop 2, "3V−R1I−0.7−R3I1=0, I1=0.7mA" assumes that D2 is conducting. But notice that the current through R2 can be a maximum of 0.2 mA, so the voltage across R2 can be no more than 0.2 volts. Therefor D2 is not conducting, and the equation becomes "3V−R1I-R3I1=0. I1=0.2mA". \$\endgroup\$ May 16 '18 at 21:16
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Diodes are non linear devices and they may be approximated with a boxcar behavior, where if conduction 0.7V drop whatever the current, if no conduction zero current whatever the voltage across them. From an assumption of conduction for both of them you can start calc. R3 has 0.7V across, thus 0.7 mA. R2 sees 2 diodes, 1.4V, thus 1.4mA. R1 sees 3V - 1.4V and the current flowing is too small to allow diodes in conduction plus resistors. If D2 off, you see that with about 0.53V of voltage drop you get 1.23V before D1 and the current sourced by the generator is about 1.75 mA. Of course, in reality diodes start conduction at about 0.5V and both will be somehow conducting; in eg PSpice you get a slightly larger current from the generator.

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  • \$\begingroup\$ you mean that voltage drop on R3 (I've added numeration) is equal to the voltage drop on the diode d2? I don't really understand why. \$\endgroup\$
    – Kornel_S
    May 16 '18 at 19:44
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    \$\begingroup\$ @Kornel_S it is in parallel. \$\endgroup\$
    – Arsenal
    May 16 '18 at 19:44
  • \$\begingroup\$ @Kornel_S - How can it be anything else? \$\endgroup\$ May 16 '18 at 21:05

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