Solve a Diode Problem: Voltage Sources, VDD, Part A

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Discussion Overview

The discussion revolves around a homework problem involving diode circuits, specifically focusing on the behavior of voltage sources and current flow through diodes in a given circuit configuration. Participants explore concepts related to circuit analysis, including assumptions about diodes being on or off, and the implications of voltage sources in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the 10/-20V voltage sources are equivalent to VDD or if they are different, expressing confusion about their role in the circuit.
  • Another participant suggests that if a diode is off, it can be treated as an open circuit, implying no current flows through it.
  • A participant proposes a method for analyzing the circuit by treating the 10V source as grounded and discusses the implications for the loop equation.
  • There is a mention of a simulation result indicating a current of 34mA through D2 when D1 is off, prompting questions about the correctness of this outcome.
  • One participant emphasizes that the 10V supply cannot be ignored as it provides a path to ground, suggesting a need for a more comprehensive loop analysis.
  • Another participant recommends nodal analysis as a potentially effective approach for solving the circuit problem.

Areas of Agreement / Disagreement

Participants express differing views on how to treat the voltage sources and the diodes in the circuit, with no consensus reached on the correct approach or final answer. The discussion includes both agreement on certain analysis methods and disagreement on the implications of diode states.

Contextual Notes

Participants rely on various assumptions about the behavior of diodes and voltage sources, which may not be universally accepted. The discussion also highlights the complexity of circuit analysis, with different interpretations of the circuit configuration leading to varied conclusions.

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Homework Statement


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The Attempt at a Solution


I know this might seem like a stupid question but are the 10/-20V voltage sources like the 5.3V or are they VDD (which is something different?) I'm also stuck on part a. If D2 is off, that means its not conducting, so zero current, but would there still be a voltage drop across it? Then since its -20V on one side does that mean V(D2)=20V gain to the node on top?
 
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asdf12312 said:

The Attempt at a Solution


I know this might seem like a stupid question but are the 10/-20V voltage sources like the 5.3V or are they VDD (which is something different?) I'm also stuck on part a. If D2 is off, that means its not conducting, so zero current, but would there still be a voltage drop across it? Then since its -20V on one side does that mean V(D2)=20V gain to the node on top?

Think of the 10/-20V sources as batteries which connect to the indicated terminals and then to ground.

If you are told to assume that a diode is off, simply remove it from the circuit (it supposedly conducts no current so is equivalent to an open circuit. If you are told to assume that a diode is on, for a first pass analysis replace it with a battery equivalent to the junction forward bias (0.7V in this case).
 
OK, I guess I can treat the 10V as a voltage source that is grounded then. So it won't be part of the loop equation, then I do I(D1)=(5.3+V(D1))/6k where V(D1)=0.7, I(D1)=-1mA. Since there can't be negative current, D1 on/D2 off doesn't work. D1/D2 both off obviously doesn't work either. When I simulated the circuit for D1 off/D2 on I got I(D2)=34mA. is this the answer?
 
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asdf12312 said:
OK, I guess I can treat the 10V as a voltage source that is grounded then. So it won't be part of the loop equation, then I do I(D1)=(5.3+V(D1))/6k where V(D1)=0.7, I(D1)=-1mA. Since there can't be negative current, D1 on/D2 off doesn't work. D1/D2 both off obviously doesn't work either. When I simulated the circuit for D1 off/D2 on I got I(D2)=34mA. is this the answer?

You can't ignore the 10V supply because it has a path through your loop to ground via the 1.93 kΩ resistor. In effect you have a second loop (stick a battery in there for the 10V supply and wire its negative lead to the ground symbol at the bottom to "see" the loop).

You might find that nodal analysis will treat you well here. Only one node to worry about!

Your simulation yields the correct current for ID2 (it's what I've just calculated).
 

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