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I am trying to solve this problem using the ideal diode model.

I first assumed that D1 will be off and D2 will be on, and got this:

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Usually the negative voltage means that the diode is indeed off and our assumption is correct. However, I'm confused in this example as given D1's orientation, it looks like a negative voltage at the diode's negative node should turn this diode on.

Am I misunderstanding something or are there more things going on in this circuit?

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  • \$\begingroup\$ You are indeed correct. That node between the diodes can't be negative. Thus the ideal diode D1 will "clamp" that negative voltage to GND. And thus that node will be at 0V with respect to GND. Knowing that you can re-calculate all the currents and solve for the voltages. \$\endgroup\$
    – Blup1980
    Commented Feb 9, 2022 at 7:55
  • \$\begingroup\$ ??? Diodes D1 and D2 are On in both cases. \$\endgroup\$
    – Antonio51
    Commented Feb 9, 2022 at 11:52

3 Answers 3

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Voltages always are relative. And the choice what node to label as "0 V" is arbitrary, as far as the physics is concerned.

In most digital circuits, the lowest voltage is chosen as ground (so that all voltages are positive); in many analog circuits, the middle between the two power rails is chosen as ground. This makes understanding them easier.

In some exercises/homework questions, the ground is chosen so as to confuse the student.

The diode itself does not 'know' how the voltages are labeled (i.e., relative to which node voltages are measured). It 'knows' only the voltage drop between its anode and its cathode. In the ideal diode model, when the anode tries to have a higher voltage than the cathode, a current flows from the anode to the cathode. And in this circuit, 0 V is higher than −1.5 V.

You could add 9 V to all voltages; this would change the absolute voltages, but not any voltage drops over any components, and the behaviour of this circuit would not change. And then you would have 9 V at the anode and 7.5 V at the cathode, with the obvious consequences.

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  • \$\begingroup\$ This is what I was going to write, upvoted. @Amanda, look at the voltages as 15 V, 9 V and 0 V instead of 6 V, 0 V and -9 V. It'll use the same maths procedure and get the same results but might give you a clearer view of what's going on. \$\endgroup\$
    – TonyM
    Commented Feb 9, 2022 at 8:54
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I often recommend redrawing the schematic. For example, as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

The most positive supply is at the top, the most negative at the bottom, and the orientation of the diodes is so they are anode towards top and cathode towards bottom. Finally, because it's fine to do so, I swapped the positions of \$D_1\$ and \$R_1\$. (There's a good reason for that.)

From this diagram, you can pretty much recognize that \$D_1\$ is on. If \$D_2\$ is off then \$D_1\$ is on because of the \$6\:\text{V}\$ supply through \$R_2\$ and \$R_1\$. If \$D_2\$ is on then \$D_1\$ is on because of ground through \$D_2\$ and \$R_1\$. Either way, it's on. You don't need to guess. It's a fact.

So \$V_1\$ is either \$-9\:\text{V}\$ (case (a)) or else \$-8.4\:\text{V}\$ (case (b).)

We can now simplify the schematic another step or two:

schematic

simulate this circuit

On the left, I've removed \$D_1\$ since we've decided it is always on. And we already know that \$V_1\$ is one of two possible values. On the right, I've converted the two voltage sources and the two resistors (which form a divider) into their Thevenin equivalent, where: \$V_{_\text{TH}}=\frac{V_1\,R_2+6\:\text{V}\,R_1}{R_1+R_2}\$ and \$R_{_\text{TH}}=\frac{R_1\,R_2}{R_1+R_2}\$. (Note that the value of \$V_1\$ depends upon whether case (a) or case (b) is being considered.)

Two possibilities:

  • case (a): If you find that \$V_{_\text{TH}} \le 0\:\text{V}\$ then \$D_2\$ is on. In this case, \$V_{_\text{TH}}=-1.5\:\text{V}\$. So \$D_2\$ is on.
  • case (b): If you find that \$V_{_\text{TH}} \le -600\:\text{mV}\$ then \$D_2\$ is on. In this case, \$V_{_\text{TH}}=-1.2\:\text{V}\$. So \$D_2\$ is again on.

So both cases find the condition to be true and therefore \$D_2\$ is also always on.

This means that \$V_2\$ is either \$0\:\text{V}\$ in case (a) or else \$-600\:\text{mV}\$ in case (b).

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Little problem on Q-point !
Diodes replaced by 1 Ohm -> or 1 pOhm ...

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