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I want to find the voltage V, the exercise suggests using Thevenin, I didn't learn it officially in class so I'm trying my best. The diode is ideal. This is the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Since I'm trying to find V with respect to ground, this is my open circuit:

schematic

simulate this circuit

There is no current so V is 15, which is obviously wrong, I know I made a mistake, just want to know how to proceed if the voltage I want is inside parallel wires...

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    \$\begingroup\$ When you say ideal, do you mean ideal as in 0.7V drop or ideal as in 0V drop? \$\endgroup\$ – BeB00 Sep 25 '16 at 23:07
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    \$\begingroup\$ Assuming you mean 0V drop, the diode is irrelevant (AFAIK). \$\endgroup\$ – BeB00 Sep 25 '16 at 23:10
  • \$\begingroup\$ When you start by taking the diode out of the picture, R1 should still be connected to (B) in your second picture. Calculated the Thevinin equivalent of the V1+R1+R2 for point (A)-(B). Now put the diode back in, you should get a simple loop with: (B) -- Vth -- Rth -- (A) -- D1 -- R2 -- (B) \$\endgroup\$ – rioraxe Sep 26 '16 at 9:22
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First, ask yourself if the diode is forward biased. You can tell this by shorting the diode and seeing which way the current flows. In this case, the diode is in fact forward biased.

Now, with the diode forward biased, what it its voltage drop? Since this is an ideal diode, that should be easy. So, can you treat the two 20k resistors as if they are in parallel? If so, can you derive the voltage at the junction of R1 and R3? How about D1 and R2? Does that solve your problem?

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  • \$\begingroup\$ This problem is not complicated but I want to train Thevenin and I don't know how I could apply it here. Thanks. \$\endgroup\$ – João Pedro Sep 25 '16 at 23:17
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    \$\begingroup\$ @JoãoPedro:I don't know either - extra info is required. A Thevenin equivalent divides a circuit into source and load. Which of those four components is source (associated with V1)? Where would the output terminals be envisioned? You must establish answers to these first before you can calculate Thevenin voltage, and Thevenin resistance. It is unclear from your schematic where the "output terminals" are located. \$\endgroup\$ – glen_geek Sep 26 '16 at 22:57
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This is how to solve the problem using Thevenin equivalent.

We should find the Thevenin equivalent at A,B . Since your diode is forward biased let's replace it with a normal wire like this

schematic

simulate this circuit – Schematic created using CircuitLab

To find Vth let's take R2 off the circuit:

schematic

simulate this circuit

To calculate Vth (which is the voltage at A), now you have a voltage divider consisted of R1 and R3. You can find that Vth equals 10v.

To find Rth, short the voltage power source.

schematic

simulate this circuit

you can find that R1 and R3 are in parallel and Rth = 6.66 kΩ. Now putting back R2, the final thevenin equivalent is:

schematic

simulate this circuit

Using the equivalent circuit you can calculate the volatage needed at A (or at V as you have denoted). It is a simple volatage divider. By doing the calculations you can find that the voltage at A equals 7.5V.

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The diode makes that the thevenin equivalent has two resistive equivalents depending on the polarity of the supply. In the conductive state the diode can be seen as a short circuit ( ideal diode). Making it possible to derive the Rth The same is possible for the Vth. Diode is still to be seen as a short circuit.

If the polarity of the supply reverses you have to consider the diode as open so the branch with R2 has in that case no connection and again the Rth and Vth can be derived

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  • KVL equations use loop voltage sum =0 (not open loop)
  • KCL loop equations use sum node current =0

    • but since this example is simple.
  • Vab = 1/2*V1 for +ve signal since 20k // 20k = 10k ( parallel)
  • Vab = 2/3*V1 for -ve signal since 20k/(10k+20k)= 2/3
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The last reasoning seems to be correct except that it seems to forget to put back the diode at its place which will have potential drop of 0.7 V.

If you do so, then you will have to calculate the current through the entire circuit, which is equal to 0.3488 mA. From there calculate the voltage at point A with respect to ground which is equal to 20k*0.488 mA = 6.976 V.

I may be wrong. I just did a quick lecture of the solution.

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  • \$\begingroup\$ When you say the last reasoning, which reasoning are you referring to? The order of the answers will change depending upon the votes, which may vary over time. So, it is better to link to either the answer you are referring to (if it is one of the answers that you are referring to)... or are you referring to the last reasoning of the OP? \$\endgroup\$ – Greenonline Oct 13 '18 at 21:43

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