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A diode is called a diode because it has two distinct electrodes (i.e. terminals), called the anode and the cathode. A diode is electrically asymmetric because current can flow freely from the anode to the cathode, but not in the other direction. In this way, it functions as a one-way valve for current. The diode's asymmetric behavior can be used to create circuits for signal processing (such as an AC to DC converter).
Figure 1

Figure 2

The I-V characteristic of a diode can be plotted on an oscilloscope in the lab. Its shape is nonlinear due to the diode's electrical asymmetry.
Figure 3

The offset ideal model approximates the nonlinear I-V characteristic in Fig. 3 as a piecewise linear graph with a threshold voltage parameter $V_{\text{on}}$. The typical value for a silicon diode is $V_{\text{on}}=0.7\text{ V}$.
Figure 4

Within each diode state, the diode can be replaced with a simpler linear element. If the diode is OFF, it can be replaced by an open circuit which has $I_D=0$ and can have any value of $V_D$. Otherwise, the diode is ON and it can be replaced by an element with fixed voltage drop $V_D=V_{\text{on}}$ and any value of $I_D$.
Condition | State | Behavior | Replacement Rule |
---|---|---|---|
$\begin{aligned}V_D < V_{\text{on} }\end{aligned}$ | OFF | $\begin{aligned}I_D=0\end{aligned}$ | ![]() |
$\begin{aligned}I_D > 0\end{aligned}$ | ON | $\begin{aligned}V_D=V_{\text{on} }\end{aligned}$ | ![]() |
The flowchart below allows you to solve for unknown voltages and currents in a circuit containing sources, resistors and one or more diodes, which are modeled by the offset ideal model.
Figure 5

Guess whether each diode is OFF or ON. (You should use your intuition as much as possible to make good guesses.)
Replace each diode with a linear element according to the rule in Table 1 and redraw the circuit.
In the resulting linear circuit, solve for all the voltages and currents where the diodes used to be, using KCL, KVL, Ohm's law and/or the node-voltage method.
Check whether the appropriate condition in Table 1 holds for each diode.
If any diode fails its condition, change one or more of your guesses and repeat this procedure from step 2.
Otherwise, all diodes satisfy their conditions and your guesses are correct.
Solve for the desired voltages and currents using the correct guesses for the diode states.
After some practice, you are expected to do problems involving one or two diodes quickly without explicitly redrawing the circuit in step 2.
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A diode like the ones in Fig. 1 is made of semiconductor material (such as crystalline silicon) with impurities added in a process called doping. Different impurities are added to the anode and cathode sides to make the diode electrically asymmetric.
The semiconductor on the anode side is doped with atoms having 3 valence electrons each (such as indium, aluminum or gallium). Relative to pure silicon which has 4 valence electrons, it has a deficit of electrons and these missing electrons are called holes. A semiconductor doped in this way is called p-type, where the "p" stands for the positively-charged holes.
The semiconductor on the cathode side is doped with atoms having 5 valence electrons each (such as phosphorus or arsenic) and, thus, has an excess of electrons relative to pure silicon. This kind of doped semiconductor is called n-type because the excess electrons are negatively charged.
Figure 6

If the bias voltage $V_D>0$, we say that the p-n junction and diode are forward-biased. Note that this condition is slightly different to $V_D>V_{\text{on}}$ for the diode to be ON.
Figure 7

If $V_D < 0$, the p-n junction and diode are said to be reverse-biased. Again, this condition is slightly different to $V_D < V_{\text{on}}$ for the diode to be OFF.
Figure 8

In ECE 340: Solid State Electronic Devices, you will study these physical phenomena of p-n junctions quantitatively and derive a more realistic exponential model of a diode's I-V characteristic (shown below) than the offset ideal model.
\begin{align}
I_D = I_S\left(\text{exp}\left(\frac{q}{kT}V_D\right)-1\right) \label{DIO-IVE}
\end{align}
where
\begin{align}
I_S &= \text{saturation current}\\
q &= \text{charge of an electron} = 1.6 \cdot 10^{-19} \text{ C}\\
k &= \text{Boltzmann constant} = 1.38 \cdot 10^{-23} \text{ JK}^{-1}\\
T &= \text{temperature (K)}
\end{align}
At room temperature ($T=300$ K), equation $\eqref{DIO-IVE}$ simplifies to
\begin{align}
I_D = I_S\left(\text{exp}\left(38.5 V_D\right)-1\right)
\end{align}