Can I Transform Diode in Parallel for V_out vs. I_in Graph?

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Discussion Overview

The discussion revolves around the challenge of graphing V_out as a function of I_in in a circuit involving a diode in parallel with a resistor. Participants explore the implications of the diode's behavior on the transformation of the circuit and the resulting output graph.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant describes their previous success with Norton-Thevenin transformations but expresses uncertainty about applying this method with the diode in parallel.
  • Another participant suggests modeling the diode as a resistor that changes value based on current direction, proposing two scenarios: one where the diode conducts and one where it does not.
  • A participant questions if V_out can be expressed simply as the voltage drop across the resistor and a voltage source for positive and negative I_in values, respectively.
  • Further clarification is provided regarding the behavior of the diode when I_in is negative and small, noting the diode's conduction threshold.
  • Concerns are raised about the practicality of source transformations due to the diode's behavior, indicating that multiple transformations may be necessary depending on the diode's state.

Areas of Agreement / Disagreement

Participants generally agree on the complexity introduced by the diode in the circuit and the need for careful consideration of its state when applying transformations. However, there is no consensus on the best approach to graphing V_out versus I_in, as various models and interpretations are discussed.

Contextual Notes

Participants note the dependence on the diode's conduction state and the implications for the transformation process, highlighting the need for clarity on when to apply each model. The discussion does not resolve the mathematical steps involved in the transformations.

Who May Find This Useful

This discussion may be useful for students and practitioners working on circuit analysis involving diodes, particularly in understanding the complexities of modeling and graphing output in response to varying input currents.

ElijahRockers
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Homework Statement



Graph V_out as a function of I_in

540x405.png


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?:

540x405.png


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
 
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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.
 
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:
ElijahRockers said:
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?
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).
Also, I'm assuming my attempt at source transforming is wrong, because of the diode, right?

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... :smile:
 
gneill said:
life will just get more complicated... :smile:

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

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