# Diode Circuit Problem

by dashkin111
Tags: circuit, diode
 P: 48 1. The problem statement, all variables and given/known data See attachment 2. Relevant equations We are to assume the voltage drop across diodes is .7v (don't use the diode equation!). 3. 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 theres no current, v=-5? Am I approaching this correctly? Attached Thumbnails
PF Patron
 Quote by dashkin111 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 theres no current, v=-5? Am I approaching this correctly?