Diode Circuit Problem

  • Engineering
  • Thread starter dashkin111
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  • #1
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Homework Statement


See attachment


Homework Equations


We are to assume the voltage drop across diodes is .7v (don't use the diode equation!).


The Attempt at a Solution



I first was assuming that both diodes were "on". Then my work is as follows:

V1 is the voltage directly above diode D1 (the left most diode)...



[tex]I+ \frac{V_{1}-.7-(-5)}{5k\Omega}=\frac{5-V_{1}}{10k\Omega}[/tex]

[tex]I+ \frac{.7-.7+5}{5k\Omega}=\frac{5-.7}{10k\Omega}[/tex]

[tex]I=-.57mA[/tex]

Now since I is negative, one of my assumptions is wrong? So now I assume the rightmost one is "off" (D2).


So.. then
Solving for I
[tex]I = \frac{5-.7}{10k\Omega}[/tex]

[tex]I = .43 mA[/tex]

Solving for V
Well.. since theres no current, v=-5?



Am I approaching this correctly?
 

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Answers and Replies

  • #2
Gokul43201
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Science Advisor
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So now I assume the rightmost one is "off" (D2).


So.. then
Solving for I
[tex]I = \frac{5-.7}{10k\Omega}[/tex]

[tex]I = .43 mA[/tex]

Solving for V
Well.. since theres no current, v=-5?



Am I approaching this correctly?
You're on your way. In this second case, what would be the value of V1? Does this agree with your assumption?
 

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