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I am trying to calculate the current of the circuit but I am unable to find the correct answer. I was able to solve the same circuit without the diode witch gave me 2mA. I am unsure of what to do with the diode added to the circuit. Here is my circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The diode polarization is direct with a voltage of 0,7V. How do I solve the current of this circuit? Any help would be greatly appreciated.

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    \$\begingroup\$ You solve the circuit for both cases where you assume the diode is and isn't conducting. Then you pick the one without the inconsistency (by checking the voltage drop across the diode and seeing if it actually makes sense). For example, If the diode is not supposed to be conducting then R3 has no current in it, which means R3 has no voltage drop which means the voltage on the left terminal of R3 is equal to the voltage on the right terminal of R2. Use that to check if the voltage drop across the diode is actually reversing biasing the diode. \$\endgroup\$ – Toor Feb 27 at 17:45
  • \$\begingroup\$ Do you know how to use nodal analysis? It's not necessary (Thevenin is good enough.) I'm just curious if you've gotten that far, yet. \$\endgroup\$ – jonk Feb 28 at 4:14
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For problems like this, the first step is to assume the state of the diode. Is it conducting or not?

Once you assume a state, then you redraw your circuit to match the assumption. For example, if you assume the diode is not conducting, what can you replace it in the schematic with? How about if you assume it is conducting (hint: remember you said it has a voltage of 0.7V)?

After making these assumptions, re-drawing and solving, one of the solved circuits will result in a contradiction with the original assumption. Perhaps you calculate the voltage drop across the diode as something smaller than 0.7V by using KVL around the loop formed by R2 and R3. This contradicts the diode being at 0.7V when conducting, which means that assumption was wrong.

The circuit that doesn't result in any contradictions is the correct assumption, and the values from that circuit are the "right" answer.

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  • \$\begingroup\$ If i understand correctly I have to pretend that the diode is a resistor and I need to find it resistance value. Vd=RI2 And then I need to find I2 with KVL I =I2+I3 I am not able to find I since I of the circuit since I do not know the resistance of Vd what should I do \$\endgroup\$ – Junior Cortenbach Feb 27 at 18:45
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    \$\begingroup\$ @JuniorCortenbach No, you don't need to know it's resistance value. You can use KVL or KCL directly without having to calculate \$V_D\$ or \$I_D\$ based on your assumptions. If you assume the diode is conducting, \$V_D = 0.7\$ and then you can use any circuit analysis methods (superposition, pure KVL/KCL) to solve the rest of the circuit and check your assumption. \$\endgroup\$ – Shamtam Feb 27 at 19:50
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    \$\begingroup\$ Analogously, if you assume the diode is not conducting, then you know \$I_D = 0\$, which allows you to solve the rest of the circuit (again with KVL/KCL). \$\endgroup\$ – Shamtam Feb 27 at 19:51
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    \$\begingroup\$ If the diode isn’t conducting, then it is just a voltage divider using just R1 and R2. Because they are equal, the voltage at the node between R1 and R2 would be 15 volts. This makes it easy to see that the diode must be conducting. \$\endgroup\$ – Joe Mac Feb 28 at 5:29
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    \$\begingroup\$ In this specific case, yes, it's nearly trivial to solve. I was detailing a general approach to handle nonlinear elements. \$\endgroup\$ – Shamtam Feb 28 at 14:15
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You might consider converting your voltage source and the \$R_1\$ and \$R_2\$ resistor divider pair into an equivalent Thevenin voltage and Thevenin source resistance. Once replaced that way, the remaining circuit is trivial to understand and quite easy to solve. (It also instantly avoids all this, "assume one way, then assume the other way" testing process.)

schematic

simulate this circuit – Schematic created using CircuitLab

All you need to do is work out \$V_\text{TH}\$ and \$R_\text{TH}\$ and decide if \$V_\text{TH}\$ is large enough. I'm sure you can work out the details from there.

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    \$\begingroup\$ This is definitely the easiest approach. \$\endgroup\$ – Joe Mac Feb 28 at 5:36
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Not sure how you got 2mA - Without the diode, using simple circuit analysis and Ohm's Law, the total current would be 1.2mA.

Since the diode is forward-biased, R3 will still be in the DC circuit and the total current will still be 1.2mA with the voltage drop across R3 will just be reduced by 0.7 Volts.

If you turn the diode around, that is a different story where no current would flow through R3 and you would have a simple series circuit with a total resistance of 20kohms and therefore a total current of 1.5mA.

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    \$\begingroup\$ With a short where the diode is, you get 2mA out of the source (\$\frac{30 V}{10k\Omega + 5k\Omega} = 2 mA\$)... I don't see how you figure 1.2mA in any permutation of this circuit. \$\endgroup\$ – Shamtam Feb 27 at 18:27
  • \$\begingroup\$ Of the three parts in your "answer", two are wrong. \$\endgroup\$ – AnalogKid Feb 28 at 13:30

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