Diode Circuit Analysis

The general procedure is for diode circuit analysis is:

1. Draw with diode off (removed from circuit). Solve for $V_D$ and any other unknowns like $V_A$ and $V_B$.
2. Draw with diode on (replace with $V_F$ source). Solve for $V_D$ and any other unknowns, like $V_A$ and $V_B$.
3. Use “off” model when $V_D<V_F$ and “on” when $V_D\ge V_F$

Example

Use a basic diode model that does not conduct any current until 0.7V is applied, at which point it conducts any amount of current. Solve for $V_A$ and $V_B$.

Step 1: Off model

First, we model the circuit with the diode off. Since the diode is off, we can basically just draw it as an open circuit:

What’s the diode voltage when it’s off? We simply have:

$$\begin{array}{rl}{V}_{D+}& =V_A\\ V_{D-}& =V_B\\ \therefore V_D& =V_A-V_B\end{array}$$

Solving for $V_A$ and $V_B$ (as voltage dividers):

$$\begin{array}{r}{V}_A=5\cdot \frac{4}{1+4}=4\text{ V}\\ V_B=10\cdot \frac{12}{48+12}=2\text{ V}\end{array}$$

In the off-state, our diode voltage is $V_A-V_B=2\text{ V}$.

Since $2\text{ V}>0.7\text{ V}$ and $V_A$ and $V_B$ don’t change in the problem, we know that the diode is on!

Step 2: On model

Now, we model the circuit with the diode on:

We can solve this using Supernodes:

The equations are:

$$\begin{array}{r}\frac{V_A-5}{1}+\frac{V_A}{4}+\frac{V_B-10}{48}+\frac{V_B}{12}=0\\ ⟹12V_A+V_B=50\end{array}$$

and

$$V_A=V_B+0.7\text{ V}$$

Solving this gives:

$$V_A=3.9\text{ V},V_B=3.2\text{ V}$$