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.

canadiansmith
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Hi I am having trouble figuring out what to do with negative reference voltages in my circuit. I attached my problem(Problem 3) and I know that some current will be drawn out of the ground and I think all of the diodes will be on but I am unsure how to approach this problem.
 

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You pic is upside down
 
Sorry Here it is right side up
 

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OK, you are right about the diodes all being on. You need to show some work here. What have you tried so far? Have you read the forum rules about how to post homework problems?
 
Yes I have read the rules regarding homework. So far I have redrawn the circuit with batteries in place of the diodes each = to 0.7 volts. The 0.7 V batteries all oppose their assigned current ( ie the negative end of the battery is the same side as the cathode on the diode.
So first I tried to figure out each node using KVL starting with Va. But quickly I have figured out this cannot work. So right now I am sort of suck because I am not sure if what I am trying to do is to get each negative potential voltage to zero or if I should somehow find a way to sum up the voltage sources.
 
Oh and the current from the +5 Voltage source across R1 is 1mA
 
phinds said:
OK, you are right about the diodes all being on. You need to show some work here. What have you tried so far? Have you read the forum rules about how to post homework problems?

You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.
 
gneill said:
You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.

Hm ... I worked the whole thing under that assumption and got good and consistent answers. Are you sure I need to redo it?
 
ok so I have Va=7.15V and the current on the left side = 1.43 mA
Vb= 4.3V and current right side = 0.86mA. But how do I know that I can make the assumption to turn D2 off?
 
  • #10
Like with D2 off I get Va>Vb which would mean that my assumption would not turn D2 off.
 
  • #11
Oh wait. Va should be >Vb so the assumption is correct. Thanks for your help. One last question: when choosing diodes to be on or off do you simply take your best guess or is there a more systematic approach?
 
  • #12
canadiansmith said:
Like with D2 off I get Va>Vb which would mean that my assumption would not turn D2 off.

Is Va > Vb always a sufficient condition to have D2 on?
 
  • #13
phinds said:
Is Va > Vb always a sufficient condition to have D2 on?

I believe so. The polarity on the open switch that is used to replace D2 would be such that the diode would be turned off. Assuming my Va and Vb are correct
 
  • #14
canadiansmith said:
I believe so. The polarity on the open switch that is used to replace D2 would be such that the diode would be turned off. Assuming my Va and Vb are correct

Well, you need to think about this some more. What does "on voltage" on a diode mean to you?
 
  • #15
Well its the voltage required to turn on the diode to allow current through it. Usually 0.7V for silicon.
 
  • #16
So is it always turned on if that voltage is greater than zero?
 
  • #17
no. If the voltage is less than the "on" voltage then the diode will still be off.
 
  • #18
gneill said:
You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.

OOPS --- that IS a bad assumption isn't it.

Good catch.
 
  • #19
canadiansmith said:
no. If the voltage is less than the "on" voltage then the diode will still be off.

Good.

Also, note the post I just made about D2 being on.
 
  • #20
OK. So if d2 is on does this change my value for va and vb? I don't think it would.
 
  • #21
canadiansmith said:
OK. So if d2 is on does this change my value for va and vb? I don't think it would.

I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.
 
  • #22
phinds said:
I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.

What? I am confused. What post about D2 being on? Can you please explain.
 
  • #23
canadiansmith said:
What? I am confused. What post about D2 being on? Can you please explain.

Read through the posts again in order. You'll get it.
 
  • #24
oh no. I see what you are saying now. So is everything that i have done wrong now? Like the way I calculated Va and Vb is this still valid.
 
  • #25
Suppose that you remove D2 from the circuit (excise it... open circuit it). What would be the potential VA? What would be the potential VB? What's VA - VB?
 
  • #26
Va-Vb would be 7.15-4.3 = 2.85V
 
  • #27
canadiansmith said:
Va-Vb would be 7.15-4.3 = 2.85V

Where would a 7.15V potential come from? The highest potential source I can see is +5V.
 
  • #28
That would be the potential with respect to ground.
 
  • #29
canadiansmith said:
That would be the potential with respect to ground.

What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.
 
  • #30
gneill said:
What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.

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
 

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