# Thevenin/Norton? (Diode)

1. Mar 7, 2013

### ElijahRockers

1. The problem statement, all variables and given/known data

Graph V_out as a function of I_in

Attempt:

I was able to complete the previous three questions by doing norton-thevenin source transformations and getting a single voltage source, which let me figure out the appropriate value of I_in for which the diode operates.

But with the diode in parallel I'm not sure I can transform. Can I do this?:

If so, then that will allow me to draw a graph, but the diode in parallel threw me off.

EDIT: The voltage in the last diagram shouldn't have a value of 1V associated with it, this exercise involves no specific values for any components

2. Mar 7, 2013

### Staff: Mentor

It's a bit tricky. You can think of the diode as a resistor in parallel with R1, but it's a resistor that changes value depending upon the direction of the current that wants to pass through it. Your transformed source model will have two versions: one where the diode is conducting and one where the diode is not conducting. Bit of a catch-22 there, since you need to know which model to apply when, in order to find out which model to apply...

It might be simpler to consider the direction of current flow imposed by I_in over its domain and model the diode accordingly in the different conditions.

Alternatively, you might think of the diode as a "voltage clamp" across R1 which allows the voltage to increase in the usual Ohm's law fashion in one polarity, but clamps it to the diode forward voltage (0V for ideal diode) in the other. Sketch the voltage across R1 without the diode present, then add the effect of the clamp. Vout is just the VR1 voltage curve shifted up by Vb.

3. Mar 7, 2013

### ElijahRockers

So when I_in > 0, V_out is just the voltage drop across Vb and R1, and if I_in < 0 it would be Vd + Vb?

Also, I'm assuming my attempt at source transforming is wrong, because of the diode, right?

Last edited: Mar 7, 2013
4. Mar 7, 2013

### Staff: Mentor

That's the big picture, yes. You might have to pay a bit of attention to the details when I_in is negative but small so that the potential drop across R1 is less than Vd (the diode doesn't start conducting until the drop across R1 exceeds Vd).
Not so much wrong as impractical. There will be two distinct versions of the transformation (current, resistance) which depend upon whether or not the diode is conducting or not. You need to know when to apply which version for a given value of I_in. But since that's essentially what you're trying to find out, life will just get more complicated...

5. Mar 8, 2013

### ElijahRockers

Well. Spring break is here so, at least for the moment, life will be simpler. :P