How do I plot the load line for a nonlinear network with an ideal diode?

AI Thread Summary
To plot the load line for a nonlinear network with an ideal diode, the analysis involves considering two states: the diode on and off. When the diode is off, the voltage is expressed as Va = Ia(1000), and when it is on, Va = Ia(500), with a transition occurring at Va = 1V. The resulting graph features two lines with a slope change at Ia = 0.001. The load line is determined by the external circuit, specifically the source and the 1kΩ resistor. Understanding these relationships is crucial for accurately plotting the iA vs. vA characteristics.
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Homework Statement


Plot the iA vs. vA characteristics for the nonlinear network shown in Figure 4.49. Assume the diode is ideal.

The nonlinear network from part (a) is connected as shown in Figure 4.50. Draw the load line on your v i characteristics from part (a), and find iT

physicspicture.jpg


Homework Equations


V=IR

The Attempt at a Solution



For part A I have separated the problem into two states diode on and diode off. When the diode is off the voltage Va (which I have control over) should be Va=Ia(1000)

When the diode is on, then the voltage I have is Va=Ia(500)

This transition occurs at Va= 1V so the graph looks like two lines with a change in slope at Ia=.001.

There is no solutions manual for this so I was wondering if I am on the right track. This makes sense to me since when the diode is on we have a linear circuit as well as when the diode is off.
My other attempt (at the load line) has been futile as I don't really know how to go about doing this[/B]
 
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The loadline is set by the circuit external to the load, i.e., the circuit external to the NLD block. So in Fig 4.50 the loadline is the V-I plot determined by the source and the 1kΩ resistor.
 
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