Engineering Solve Diode Circuit Problem: Homework Equations & Attempt at Solution

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The discussion revolves around solving a diode circuit problem with a voltage drop of 0.7V across the diodes. The initial assumption was that both diodes were "on," leading to a calculated current of -0.57 mA, indicating an incorrect assumption. After revising the assumption to consider the rightmost diode (D2) as "off," a new current of 0.43 mA was calculated. The participant also questioned the voltage across the circuit, concluding that with no current, the voltage would be -5V. The conversation emphasizes the importance of verifying assumptions in circuit analysis.
dashkin111
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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)...
I+ \frac{V_{1}-.7-(-5)}{5k\Omega}=\frac{5-V_{1}}{10k\Omega}

I+ \frac{.7-.7+5}{5k\Omega}=\frac{5-.7}{10k\Omega}

I=-.57mA

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
I = \frac{5-.7}{10k\Omega}

I = .43 mA

Solving for V
Well.. since there's no current, v=-5?
Am I approaching this correctly?
 

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dashkin111 said:
So now I assume the rightmost one is "off" (D2).


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

I = .43 mA

Solving for V
Well.. since there's 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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