# Diode Circuit Problem

• Engineering

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 theres no current, v=-5?

Am I approaching this correctly?

#### Attachments

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Last edited:

Gokul43201
Staff Emeritus
Gold Member
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?
You're on your way. In this second case, what would be the value of V1? Does this agree with your assumption?