A semiconductor diode is what you get when you place a p-type semiconductor next to (and in contact with) an n-type semiconductor. The doped semiconductors are placed together as shown in Figure 1.
Remember the p-type has extra holes, the n-type has extra electrons.
[Hint - - The way I remember these associations is to think of 'n' as negative, electrons have negative charge and the n-type has more electrons, while 'p' is positive and with extra holes it is more positive because of the missing electrons. This is an example of a memorizing to learn technique.]
Recommended Books on Amazon (affiliate links) | ||
---|---|---|
Join Amazon Student - FREE Two-Day Shipping for College Students |
Figure 1 - Semiconductor Diode and Circuit Symbol
src - wikipedia
Current Convention
Remember that electric current flows in the opposite direction as the electrons.
Diodes are circuit elements that allow current to flow in only one direction. In the opposite direction, current is blocked. This unique behaviour is largely due to the p-n junction that is formed as a result of placing a p-type semiconductor in contact with an n-type semiconductor.
As you know, the p-type semiconductor is missing electrons and the n-type semiconductor has extra electrons. Consequently, when they are placed in contact, some of the electrons on the n-type side diffuse to the p-type side to fill in the holes. This happens quickly during fabrication and by the time we get the diode, the pn-junction is fully formed. We call this pn-junction the depletion or space charge region. Figure 2 shows the result.
Figure 2 - pn-Junction
src - wikipedia
There is a lot going on in this figure, so let's discuss some key points.
1. The width of the depletion region depends on many factors including the materials used and how many impurity atoms are introduced.
2. The depletion region may not be symmetric. This is due to some of the same factors that determine the width of the region.
3. Due to the locations of the electrons and holes, an electric field is produced in the depletion region.
4. Because of the induced electric field, there is small potential voltage, \(\Delta V\), across the diode. Again, the value depends on the materials, the amount of doping and the width of the depletion region. Representative values are \(0.3V\) for germanium and \(0.7V\) for silicon.
The diodes in Figures 1 and 2 are unbiased, meaning that there is no applied differential voltage across them. Now we will discuss what happens when we apply a voltage across the diode terminals.
Applying A Voltage Across A Diode
Figure 3 - Diode Set-Up
adapted from All About Circuits
In order to apply a voltage across a diode, we need to have a way to connect a voltage source and we need to know which end is which. To accomplish this, we have a set-up similar to Figure 3. To connect a voltage source, we use the anode and cathode leads. The anode lead connects to the p-type material of the diode and, similarly, the cathode lead connects to the n-type material of diode. The cathode, n-type, end is marked with a coloured band on the device.
We can apply a voltage across a diode in two possible directions. These are referred to as forward biased and reverse biased. The strength of the voltage causes the diode to operate in 3 possible regions, forward, reverse and breakdown.
Forward Biased Diode
Figure 4 - Forward Biased Diode
[built with DigiKey SchemeIt]
In the forward biased configuration, the positive end of the voltage source is connected to the p-type lead or anode terminal. This can be done either as shown in figure 4 with a resistor between the voltage source and the diode or the resistor may be placed after the diode. We use the resistor to control the current through the diode. Here is a video explaining resistor placement.
video by Build Electronics Circuits |
---|
In an ideal diode, when the diode is forward biased, there should be no resistance and the current should flow freely. Since there is no resistance, there would be no voltage drop either.
In actuality, what happens is that the holes in the p-type region and the electrons in the n-type region are pushed toward the depletion region. This causes the depletion region to reduce in size allowing current to flow more freely. As the voltage is increased, current increases. There is a slight voltage drop across the diode, V_{d}. This value is about 0.7 V for silicon (0.3 V for germanium, 0.2 V for a Schottky diode).
Figure 5 - I-V Characteristic
src - wikipedia - diodes
Looking at Figure 5, we can see that, once the applied voltage matches V_{d}, the current is unimpeded and free to flow through the diode. This is graphed in Figure 5 as the forward region. (Note: The value of V_{d} in this plot is about 0.3-0.4 V.)
Reverse Biased Diode
Figure 6 - Reverse Biased Diode
src - wikipedia - p-n junction
When the diode is placed in reverse biased mode things get interesting. The idea of an ideal diode is to not allow any current to flow in the opposite direction. This is very nearly achieved in an actual diode. In reverse biased mode, the depletion region is widened, thus blocking almost all current flow.
Referring to Figure 5 again, when the voltage is in the opposite direction and kept small, notice that there is a small leakage current, i.e. the current is very small in the negative direction and is referred to as I_{s}, saturation current. Representative values are \(10^{-12}A\) for silicon and \(10^{-6}A\) for germanium.
As long as the reverse biased voltage does not get too high, the diode will block all current except for the saturation current. The diode is said to be operating in the reverse region. However, once the voltage gets high enough, the diode enters the breakdown region. In Figure 5, this is represented at about 50V, however, this value will be different for different types of diodes.
Once the breakdown region is reached, some diodes may be permanently damaged. But if the product of the reverse voltage and reverse current does not exceed the diode's power rating, the diode will not be damaged and will continue to fully function. Some diodes, like the Zener diode , are designed to allow current flow in both directions for various voltages without damage.
Uses For Diodes
Okay, so now that we have an idea how diodes work, what can we use them for? One major application is to put a couple of p-n junction diodes together to build a BJT transistor.
Learn how electronics really work.
Diode Circuits |
---|
Related Topics |
Links |
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
---|---|---|---|
\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
To bookmark this page, log in to your account or set up a free account.
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
| |
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. |