Here is an exercise that I am not quite understand.

What is the value of the voltage Vo if V = 3V?
For me, I always think that with the 3V, at first, it has to drop 2V in the device.
From this I can compute other variables but it is wrong in this case.
(I think that because I think it is simillar as BJT, between B and E has to drop a voltage Vbe = constant)
Could you tell me why it is wrong and the proper way to solve it?

 

Keep in mind that the voltage drop is 2V if current through the device is larger then 2A.
But your Vin cannot provide more then 3V/2Ω = 1.5A of a current.
So you need to use a graphical method to solve this problem.
http://en.wikipedia.org/wiki/Diode_modelling#Graphical_solution

 

Thanks Jony
I didn't recognize it because I only focused on voltage not current!

 

You can either solve it graphically, which is known as a load-line analysis, or you can see if you can solve it analytically by determining which region of the device's characteristic you are in. If the magnitude of the voltage is less that 2V or the magnitude of the current is less than 2A, then you are in the slanted part of the characteristic and can ignore what happens at |V| >= 2V since you aren't there. So if you put on a set of blinders and just look at the part of the characteristic in which |V| < 2V, you see that it is indistinguishable from that of a 1Ω resistor. Thus, you can replace the device with a 1Ω resistor and analyze the equivalent circuit. Once you are done, you just need to verify that the voltage and current are both consistent with being on that part of the graph.

This is actually just like determining if a diode or a transistor in a circuit is forward biased, cutoff, saturated, or whatever.

In general, you assume a region of operation and analyze the circuit under that assumption. When done, you check to see if the results are consistent with the assumption. If they are, you are done (unless there is a chance that the circuit is multistable). If they aren't, then your assumption was wrong, so pick another one.

In this case, you have three regions of operation. At V=-2V the thing looks like a -2V battery with the constraint that the current must turn out to be <-2A. At V=+2V the thing looks like a +2V battery with the constrain that the current must turn out to be >+2A. In between, the thing looks like a 1Ω resistor with the constraint that the current must be between -2A and +2A.

So let's take your first assumption, that Vo=2V.

That means that you have 1V across the other two 1Ω resistors, meaning that Io=0.5A, which violates the constraints. So you know this is wrong.