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Lesson 1: Diode

# Diode as a circuit element

Diodes conduct current in one direction but not the other. We solve a diode circuit graphically by plotting a diode i-v curve and resistor to find the intersection. Written by Willy McAllister.
The diode is our first semiconductor device. The diode's distinctive feature is that it conducts current in one direction, but not the other. We won't go into the details of how a diode does this, or how it's made. Fortunately, you don't have to know how to make a diode before using it in a circuit.

The $i$-$v$ curve of a diode is modeled by this non-linear equation:
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{q\phantom{\rule{0.167em}{0ex}}v/k\text{T}}-1\right)$
• We will define terms like forward bias, reverse bias, and saturation current.
• You will learn some tips for identifying the terminals of a real-world diode.
• We will solve a diode circuit using a graphical method.

## Diode symbol

The schematic symbol for a diode looks like this:
The black arrow ▶ in the symbol points in the direction of the diode's forward current, $i$, the direction where current flow happens. The diode's voltage, $v$, is oriented with the $+$ sign on the end where forward current comes into the diode. We use the sign convention for passive components. The optional curved orange arrow also indicates the voltage polarity.

## Diode $i$‍ -$v$‍  curve

This is a typical $i$-$v$ curve for a silicon diode. A diode is a non-linear device:

## Forward and reverse current

### Forward current

Let's say we place a very small positive voltage, like $+0.2$ volts, across a silicon diode. That puts us on the right side of the $i$-$v$ curve. With this small positive voltage, almost no forward current flows. When the voltage increases up to around $0.6\phantom{\rule{0.167em}{0ex}}\text{V}$ measurable current starts to flow through the diode in the forward direction. As the voltage moves a little above $0.6\phantom{\rule{0.167em}{0ex}}\text{V}$, the current through the diode rises rapidly. The $i$-$v$ curve is nearly vertical at this point (it tips a little to the right).
With a positive voltage on its terminals, we say the diode is forward biased. A diode is forward biased when its voltage is anywhere on the $+$voltage side of the origin. In normal operation, the voltage across a forward biased silicon diode is somewhere between $0.60-0.75\phantom{\rule{0.167em}{0ex}}\text{V}$. If you externally force the voltage higher than $0.75$ volts, the diode current gets very large and it may overheat.

### Reverse current

If you put a negative voltage to a diode, so the $-$ terminal is at a higher voltage than the $+$ terminal, this puts us over on the left side of the $i$-$v$ curve. We say the diode is reverse biased. In the reverse direction, the current is very close to zero, just ever so slightly negative, below the voltage axis.
A reverse biased diode can't hold out forever. When the voltage reaches a high negative value known as the breakdown voltage, ${\text{V}}_{\text{BR}}$, the diode starts to conduct in the reverse direction. At breakdown, the current sharply increases and becomes very high in the negative direction. A breakdown voltage ${\text{V}}_{\text{BR}}$ of $-50\phantom{\rule{0.167em}{0ex}}\text{V}$ is typical of ordinary diodes. Most of the time you don't allow the diode voltage to get near ${\text{V}}_{\text{BR}}$.

## Diode terminals

When you draw diodes, the schematic symbol clearly indicates the direction of forward current flow. You don't really need names for the two terminals. But, if you are handling real diodes to build a circuit, you have to figure out which way to point the diode. Real diodes are so small there isn't room to paint a little diode symbol on them, so you need to identify the terminals some other way.
The two terminals of a diode are the anode and cathode.

### How can I remember the anode and cathode?

For the longest time I could not remember which end of the diode was called the anode and which was the cathode, I looked it up every time. I finally came up with this memory aid. The German word for cathode is Kathode. The big K kind of looks like a diode symbol.
Flip the diode symbol around until it reads like a K. The Kathode is the terminal on the left.

### Identifying the terminals of a real-life diode

Diodes are made on small chips of silicon. They are delivered to you in all sorts of tiny packages. There are a few different ways to indicate which diode terminal is which.
Diode packages like the glass and black plastic cylinders shown above usually have a painted bar near one end. The bar on the package is the bar of the diode symbol, so it indicates the cathode.
This red LED (light emitting diode) has wire leads of different length. The forward current goes into the longer lead (anode). The package may have a bump or tab sticking out on the forward current side.

### Identify the anode and cathode with a meter

A good way to verify the identity of the terminals is using an ohm-meter to figure out the forward current direction. On the resistance setting, $\mathrm{\Omega }$, the meter puts a small voltage on its test leads (this is why an ohm meter needs a battery). You use that small voltage to see which way current flows.
The diode is flipped in each image. If the ohm-meter reads a finite resistance, that means the diode is conducting a small current in the forward direction, and the red $+$ lead from the meter is touching the anode. If the resistance reads O.L (for overload), the diode is not conducting current. That means the red $+$ test lead is touching the cathode.
Your meter might have a diode setting, a little diode symbol. If it does, it will display the forward voltage when the red lead is touching the forward current terminal (the anode) as shown below.

## Diode $i$‍ -$v$‍  equation

The diode $i$-$v$ relationship can be modeled with an equation. This equation is based on the physics underlying the diode action, along with careful measurements on real diodes.
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{q\phantom{\rule{0.167em}{0ex}}v/k\text{T}}-1\right)$
The plot above doesn't look very much like an exponential curve, and the current for negative voltages appears to be $0$. If we expand the current scale a whole bunch $\left(\text{milliamperes}$ $\to$ $\text{picoamperes}\right)$ the exponential shape becomes apparent (the voltage scale is expanded, too). You can see a tiny negative ${\text{I}}_{\text{S}}$ flows when the diode is reverse biased:
${\text{I}}_{\text{S}}$ is the saturation current. This current flows backwards when the diode is reverse biased. A typical value for ${\text{I}}_{\text{S}}$ in silicon is ${10}^{-12}\phantom{\rule{0.167em}{0ex}}\text{A}$, ($1$ picoampere). For germanium diodes, a typical value for ${\text{I}}_{\text{S}}$ is ${10}^{-6}\phantom{\rule{0.167em}{0ex}}\text{A}$, ($1$ microampere).
It is best to think of this diode equation as a model of a diode, rather than as a law. The equation represents an abstract ideal diode. The actual behavior depends on how it is made, its temperature, and how much you care about the fine details.

## Detailed look inside the diode $i$‍ -$v$‍  equation

[This next part takes apart the diode equation in some detail. You don't need this to use a diode in a circuit. It is okay to jump down to the example.]
There are many new parameters in the diode equation. Let's go through them carefully.
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{q\phantom{\rule{0.167em}{0ex}}v/k\text{T}}-1\right)$
$v$ is the voltage across the diode. We find it up top in the exponential term, which explains why current $i$ has an exponential dependence on voltage $v$.
Now lets look at all that other stuff up in the exponent of ${e}^{q\phantom{\rule{0.167em}{0ex}}v/k\text{T}}$.
We know exponents have no dimensions, so the other terms in the exponent have to end up with units of $1/v$.
$q$ is the charge on an electron, in coulombs:
$q=1.602×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}$.
$k$ is Boltzmann's constant, a very important number in physics. The energy of a particle increases with temperature. If you know the temperature of a particle, $k$ tells you how much kinetic energy the particle has just by virtue of being warm. The units of Boltzmann's constant are energy per kelvin.
$k=1.380×{10}^{-23}\phantom{\rule{0.167em}{0ex}}\text{J/K}\phantom{\rule{0.167em}{0ex}}\text{(joules per kelvin)}$
$\text{T}$ is the temperature measured from absolute zero in $\text{kelvin}$ or $\text{K}$. A temperature of absolute zero, or $0\phantom{\rule{0.167em}{0ex}}\text{K}$ is $-273{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}\phantom{\rule{0.167em}{0ex}}\text{(celsius)}$.
If a particle is at $\text{T}=300\phantom{\rule{0.167em}{0ex}}\text{K}$, (room temperature), then it has energy:
$k\text{T}=1.380×{10}^{-23}\phantom{\rule{0.167em}{0ex}}\text{J/K}\cdot 300\phantom{\rule{0.167em}{0ex}}\text{K}=4.14×{10}^{-21}\phantom{\rule{0.167em}{0ex}}\text{J}$
If our particle is an electron, it has a known charge, and we can talk about its energy per charge. Energy per charge might sound familiar. Its other name is voltage.
$\frac{k\text{T}}{q}=\frac{4.14×{10}^{-21}\phantom{\rule{0.167em}{0ex}}\text{J}}{1.602×{10}^{-19}\phantom{\rule{0.167em}{0ex}}\text{C}}=25.8\approx 26\phantom{\rule{0.167em}{0ex}}\text{mV}$
At room temperature (around $300\phantom{\rule{0.167em}{0ex}}\text{K}$), $k\text{T}/q$ is $26$ millivolts. That's the energy of a normal everyday electron. The exponent of the diode equation, $v/26\phantom{\rule{0.167em}{0ex}}\text{mV}$, is comparing the diode voltage to the energy of an ordinary electron.
If you feel like it, you can write the diode equation for room temperature as:
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{v/26\text{mV}}-1\right)$
This non-linear diode $i$-$v$ equation is harder to deal with than the linear $i$-$v$ equations for $\text{R}$, $\text{L}$, and $\text{C}$. There are very few cases where you will be asked to use this equation to find an analytical solution. The usual approach to diode circuits is to perform a graphical solution or to use a circuit simulation program to get an approximate answer.

## Diode circuit example

Let's build a circuit with a diode. This circuit has a green light-emitting diode (LED).
The resistor and diode share the same current, $i$. We want to find $i$ and the voltage that appears across the diode, ${v}_{\text{D}}$.
All elements share the same current, so we'll focus on equations for current.
For the diode, we get the current $i$ in terms of ${v}_{\text{D}}$ from the diode equation at room temperature:
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{{v}_{\text{D}}/26\phantom{\rule{0.167em}{0ex}}\text{mV}}-1\right)$
For the resistor, if we can come up with an equation for $i$ in terms of ${v}_{\text{D}}$, we'll be able to plot that equation on the same $i$-$v$ graph as the diode. Ohm's Law for the resistor is:
$i=\frac{{v}_{\text{R}}}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}$
We know ${v}_{\text{R}}=3\phantom{\rule{0.167em}{0ex}}\text{V}-{v}_{\text{D}}$, so the resistor current becomes:
$i=\frac{3\phantom{\rule{0.167em}{0ex}}\text{V}-{v}_{\text{D}}}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}$
I'll rearrange the equation so it looks like the slope-intercept form of a line:
$i=-\frac{1}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}\phantom{\rule{0.167em}{0ex}}{v}_{\text{D}}+\frac{3\phantom{\rule{0.167em}{0ex}}\text{V}}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}$
$i=-\frac{1}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}\phantom{\rule{0.167em}{0ex}}{v}_{\text{D}}+9\phantom{\rule{0.167em}{0ex}}\text{mA}\phantom{\rule{2em}{0ex}}$
The slope of the resistor load line is $-\frac{1}{330}$.
The $i$-intercept is $9\phantom{\rule{0.167em}{0ex}}\text{mA}$.

### Graphical solution

We now have two simultaneous equations with two unknowns, $i$ and ${v}_{\text{D}}$:
$i=-\frac{1}{330\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }}\phantom{\rule{0.167em}{0ex}}{v}_{\text{D}}+9\phantom{\rule{0.167em}{0ex}}\text{mA}$
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{{v}_{\text{D}}/26\phantom{\rule{0.167em}{0ex}}\text{mV}}-1\right)$
We can solve these two equations by a graphical method, by plotting them on the same scale, and seeing where they cross. At the point where they intersect, the current in the resistor equals the current in the diode.
We get a pretty accurate answer by reading the intersection point off the graph:
${v}_{\text{D}}=0.6\phantom{\rule{0.167em}{0ex}}\text{V}$ and $i=7.2\phantom{\rule{0.167em}{0ex}}\text{mA}$
Reading from the graph is often all the accuracy you need.

### Concept check

problem 1
What is the current when the diode voltage ${v}_{\text{D}}$ is $0$?
$i=$
$\phantom{\rule{0.167em}{0ex}}\text{mA}$

problem 2
What is the voltage when the resistor line touches the $v$-axis? $\left(i=0\right)$
$v=$
$\text{V}$

problem 3
Does the $i$-axis intercept of the resistor line depend on the value of $\text{R}$?

problem 4
Does the $v$-axis intercept of the resistor line depend on the value of $\text{R}$?

### Brighter

Suppose you build this circuit and the LED is not bright enough. The brightness goes up if you increase the current. How might you do that?
Try changing something about the circuit to increase brightness. Then sketch a new graphical solution.
Suppose we go for maximum brightness by leaving out the resistor altogether. Is this a good idea or a bad idea?
Imagine how the resistor line changes as the resistor goes from $200\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ to $0\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$.

### If we had used an LED $i$‍ -$v$‍  curve

If we had done this problem with an $i$-$v$ curve for an LED with a forward voltage of $2\phantom{\rule{0.167em}{0ex}}\text{V}$, the solution would have come out something like this:
A typical LED gives off a nice light with a current of around $20\phantom{\rule{0.167em}{0ex}}\text{mA}$. To get this current you adjust the slope of the resistor load line until the intersection point is at the current you want. Then figure out the resistor value from the slope. The slope of the resistor load line is $-\frac{1}{\text{R}}$.
A standard-value $47\phantom{\rule{0.167em}{0ex}}\mathrm{\Omega }$ resistor comes close to the target $20\phantom{\rule{0.167em}{0ex}}\text{mA}$. The resistor value will come out different if you have a different supply voltage, of course.
Most LED datasheets give you a forward voltage specification, but not an equation for a diode curve as I've shown on the plot above. It is okay to approximate the diode with a straight vertical line at the forward voltage. You will get a pretty good graphical solution.

## Summary

The schematic symbol and terminal names for a diode:
The diode equation is:
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{q\phantom{\rule{0.167em}{0ex}}v/k\text{T}}-1\right)$
${\text{I}}_{\text{S}}$ is the saturation current. For silicon, ${\text{I}}_{\text{S}}={10}^{-12}\phantom{\rule{0.167em}{0ex}}\text{A}$ is a typical value.
In the exponent of the diode equation, the term $k\text{T}/q$ is equivalent to $26\phantom{\rule{0.167em}{0ex}}\text{mV}$ if the diode is at room temperature. $k$ is Boltzmann's constant, $\text{T}$ is the temperature in kelvin, and $q$ is the charge on an electron in coulombs. If the diode is near room temperature, the diode equation can be written as:
$i={\text{I}}_{\text{S}}\phantom{\rule{0.167em}{0ex}}\left(\phantom{\rule{0.167em}{0ex}}{e}^{v/26\phantom{\rule{0.167em}{0ex}}\text{mV}}-1\right)$
We demonstrated a graphical solution for a diode circuit, and it gave a pretty good answer without a lot of work. In general, graphical solutions are a good way to go after any circuit with a non-linear element.

There are many types of diodes, differing in materials and processing, and specialized for different uses. Here are a few:
• Silicon diode - Silicon is the most common material used to make diodes. Silicon has a typical forward voltage of $0.6-0.7\phantom{\rule{0.167em}{0ex}}\text{V}$.
• Germanium diode - Made from a different element. Germanium diodes have a lower forward voltage of $0.25-0.30\phantom{\rule{0.167em}{0ex}}\text{V}$.
• Schottky diode - Made from a silicon-to-metal contact. The forward voltage is lower than regular silicon diodes, in the range of $0.15–0.45\phantom{\rule{0.167em}{0ex}}\text{V}$.
• Zener diode - Intentionally operated in the breakdown region, used as a voltage reference.
• LED (light-emitting diode) - Does what its name says. Otherwise, it acts like a regular silicon diode. LEDs are made by combining materials on either side of Silicon on the periodic table. For example, a yellow LED can be made from gallium arsenide phosphide (GaAsP).
• Photodiode - This diode has a window to let light fall directly on the silicon surface. The current in the diode is proportional to the intensity of light. Solar cells are a form of photodiode.
• Small signal or switching diode - A silicon diode constructed to be very fast going from forward to reverse bias and back. This is done by making the diode physically very small.

## Want to join the conversation?

• Will there be any future semiconductor video, mabye about ICs or transistors?
• GMB,

Perhaps but it will take some time. On its current trajectory this website is still in the early parts of what would be considered Electrical Engineering I01. Traditionally there is another class called linear circuits that come after EE1. Sorry, no IC or transistors there either. We will need to wait and see where KA takes us next.

In the interim may I recommend two books:

"Practical Electronics for Inventors" By Scherz and Monk

"Art of Electronics" by Horowitz and Hill

Regards,

APD
• Can increasing the electric field of an intrinsic semiconductor increase the generation of electrons or conductivity?
• Hello Dzifa,

Sure, as a rule increasing the voltage on a semiconductor will cause more current. It's hard to give a more precise answer:

Yes: forward biased diode

Yes: more leakage current on a reverse biased diode - but keep going and you get all the current plus dead diode...

Sort of: for a transistor - here a transistor acts as constant current source.

Regards,

APD
• From the graphical solution of the diode it is shown that if the voltage is increased then the diode current will not increase that much but at the same time current through the resistor will increase . But diode and resistor are both sharing the same current . Isn't this a contradiction ?
• Yes, the diode and resistor share the same current. The goal of the graphical solution is to find points on both the diode curve and resistor curve where the current is identical in the two components. The currents are the same where the two graphs intersect.

But the first sentence is backwards. The horizontal axis is voltage and the vertical axis is current. If you increase the voltage on a diode a little (above 0.6v) the current will go up A LOT. You can turn that statement around... If you change the current through a diode (move up/down on the vertical axis) the voltage across the diode does not change very much. That's one of the key properties of a diode, above about 0.6v, the voltage drop across the diode doesn't change very much for any reasonable current, it stays in the range 0.6 - 0.9 v.
• Can't you use indium and antimony to make a diode?
• Yes. This would be an example of what is called a III-V semiconductor (named after the column's of the Periodic Table. These semiconductors are used to make thermal imagers.
• Don't you damage a diode if you get past the breakdown voltage?
• Not necessarily. Damage happens if too much heat is generated. If you can control that, the diode survives. A type of diode called the Zener diode is intentionally operated in reverse breakdown.
• Answer to question # 1. If the voltage across the diode is zero, there will be no current. Why is the load line of the resistor used to answer the question, instead of the diode equation line?
• You do use the diode equation:
i = Is (e^(qv/kT)−- 1)
so for v=0,
i = Is (e^0 - 1) = 0
• did you use a 9v battery to light the led light
• Hello Jay,

You certainly can. BUT you must select an appropriate resistor to limit the diode current. Without a proper resistor the diode will be damaged.

Regards,

APD