Electric circuit with negative reference voltages and diodes

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Discussion Overview

The discussion revolves around the challenges of analyzing an electric circuit that includes negative reference voltages and diodes. Participants explore the behavior of diodes under various assumptions and the implications of these assumptions on circuit analysis, particularly focusing on voltage calculations and current flow.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to handle negative reference voltages in their circuit analysis.
  • Another participant suggests that all diodes will be forward biased, but later questions this assumption by considering the case where one diode (D2) is not conducting.
  • Participants discuss the calculations for voltages Va and Vb, with one participant asserting that Va should be greater than Vb to keep D2 off, while another challenges this assumption.
  • There is a discussion about the "on voltage" of diodes, with participants clarifying that a diode requires a specific voltage to conduct.
  • One participant calculates Va and Vb based on the circuit configuration, leading to a discussion about the validity of these calculations and the implications for D2's state.
  • Participants explore the relationship between the voltage across resistors and the overall circuit potential, leading to a realization about the behavior of D2 in the circuit.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial assumptions regarding the state of the diodes. There are multiple competing views on how to approach the analysis of the circuit, particularly concerning the assumptions about diode conduction and the calculations for voltages.

Contextual Notes

Participants express uncertainty about the assumptions made regarding diode states and the calculations of voltages, indicating that the analysis is dependent on these assumptions. There are unresolved questions about the implications of negative potentials on the circuit behavior.

Who May Find This Useful

This discussion may be useful for students or individuals working on circuit analysis involving diodes and negative voltages, particularly those seeking to understand the complexities of voltage relationships in such circuits.

  • #31
Is this wrong? If so please help me. I have been working on this all day.
 
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  • #32
canadiansmith said:
Ok I found Va by (R1/R1+R2)*(5-(-100-0.7) = 7.15V since the bottom of the left side is -10. Likewise for Vb i found 0-0.7-(-5) = 4.3 V

Your formula for VA cannot yield 7.15V. It can't! VA must be less than 5V, since he current is flowing down through R1 (so there's a voltage drop across R1), and the top of R1 is at only +5V. It might be convenient to find the current in the branch first, then find the voltage drop on R1. +5V minus that voltage drop will give you VA.

The drop across D3 is 0.7V, so VB can only be 0.7V below ground potential. That makes VB -0.7V.
 
  • #33
ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off
 
  • #34
canadiansmith said:
ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off

Show your calculation for Va if D2 is not conducting.
 
  • #35
gneill said:
Show your calculation for Va if D2 is not conducting.

Exactly what I was going to say :smile:
 
  • #36
va would equal 5-VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the -10 draw and the +5 push. therefore the total current in the left side would be 5+10-.7 =14.3V/R1+R2 = 1.43mA
VR1 = 1.43mA*R1 = 7.15 V
and Va = 5-7.15V = -2.15V is this right?
 
  • #37
canadiansmith said:
va would equal 5-VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the -10 draw and the +5 push. therefore the total current in the left side would be 5+10-.7 =14.3V/R1+R2 = 1.43mA
VR1 = 1.43mA*R1 = 7.15 V
and Va = 5-7.15V = -2.15V is this right?

It is. So in conclusion...?
 
  • #38
In conclusion D2 is in fact off. Thank you very much. I got very confused with the negative potentials. But I realize now that they cannot make the potential at any point higher than the original potential.
 
  • #39
The important thing is that you got there in the end :smile:
 
  • #40
good work ... sounds like you learned something worth learning
 
  • #41
Thank you both very much for your help. I didn't think I would ever see the light at the end of the tunnel. Thanks again
 

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