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

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 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

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

Why are most transistors made of silicon? Why not lead? sure the farther away the valence electrons, the easier it is to acquire holes.

Choosing a material to make transistors is a balancing act. You want to modify the material (add impurities) to get a wide spectrum of semi-conducting properties (ranging from almost insulating to nearly fully conducting). If you choose a really big atom like Lead because it is easy to take away electrons, you won't be able to modify it with impurities to make it more like an insulator. The element Silicon sits in the "just right" spot in the periodic table so it can be manipulated to get a full range of conduction properties. Another neighboring element used for making diodes is Germanium. A lot of high speed transistors are made from a combination of Gallium and Arsenic. All these elements are nestled close to each other in the periodic table.

What is in the center of the glass diodes pictured above?

Hello GMB, The heart of the diode is a PN junction made of highly refined silicon (with a few controlled impurities). This link may help: https://en.wikipedia.org/wiki/P%E2%80%93n_junction You may also want to take a look at Adafruit's tutorials: https://www.youtube.com/watch?v=ap7edIKkykA Regards, APD

So what is the basic difference between an LCD and LED monitor?

Liquid crystal diodes block light. Light emitting diodes emit it.

Main content

Course: Electrical engineering > Unit 4

Lesson 1: Diode

Diode as a circuit element

Google Classroom

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.

Semiconductor materials fall between insulators and conductors. They usually act like insulators, but we can control how much they conduct by changing the way they are made and by applying voltages.

The most well-known and well-understood semiconductor material is Silicon (Si, atomic number

14

). Silicon is by far the most common material used for creating semiconductor devices. More is known about silicon than perhaps any other material on Earth.

Other semiconductor materials include germanium (Ge, atomic number

32

, right below Silicon on the periodic table), and gallium-arsenide, a

1 : 1

ratio of Gallium and Arsenic (atomic numbers

31

and

33

, on either side of germanium).

Our ability to finely control the conducting properties of silicon allows us to create modern marvels like computers and mobile phones and every other complex electronic device. The details of how a semiconductor works are governed by our understanding of quantum mechanics.

Where we're headed

The

i

v

curve of a diode is modeled by this non-linear equation:

i = I_{S} (e^{q v / k T} - 1)

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 V

measurable current starts to flow through the diode in the forward direction. As the voltage moves a little above

0.6 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 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.

There is a tiny bit of current that flows when a diode is reverse biased. We call it the reverse saturation current. A well-made diode has a typical reverse current of

I_{S} = 10^{- 9}

10^{- 12} A

. In most situations, this is close enough to zero to be ignored.

In some cases (an integrated circuit with millions of diodes, or you want to detect really tiny currents), the reverse saturation current becomes important and you give it a bad-sounding name: leakage current.

A reverse biased diode can't hold out forever. When the voltage reaches a high negative value known as the breakdown voltage,

V_{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

V_{BR}

- 50 V

is typical of ordinary diodes. Most of the time you don't allow the diode voltage to get near

V_{BR}

The word bias is an older term. It does not have a single precise definition.

In everyday use, bias can refer to a person's attitude, usually implying unfairness or favoritism. You also might use it in a favorable way, "They show a bias for taking action." Or you might describe a tendency, "The goalkeeper has a bias for jumping to the left on penalty kicks."

In electronics, we use bias in the sense of pulling towards or favoring one side. It only comes up in a few situations. One of them is when talking about diodes as we are doing here. Forward bias means an externally applied voltage is tugging the diode towards the forward conducting side of its

i

v

curve. Reverse bias is the opposite, a negative external voltage pulls the diode into its reverse bias region.

The other place you come across this term: You apply a bias voltage to the terminals of a device (usually a transistor) to position it within a voltage range where it works best. For example, if a transistor works best between

1

and

3

volts, you design a bias voltage centered at

2

volts, right in the middle of its happy range.

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,

Ω

, 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 = I_{S} (e^{q v / k T} - 1)

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

(milliamperes

\to

picoamperes)

the exponential shape becomes apparent (the voltage scale is expanded, too). You can see a tiny negative

I_{S}

flows when the diode is reverse biased:

I_{S}

is the saturation current. This current flows backwards when the diode is reverse biased. A typical value for

I_{S}

in silicon is

10^{- 12} A

, (

1

picoampere). For germanium diodes, a typical value for

I_{S}

10^{- 6} 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 = I_{S} (e^{q v / k T} - 1)

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 v / k 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 \times 10^{- 19} 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.

When we measure the temperature in Celsius or Fahrenheit, we say "degrees celsius" or "degrees fahrenheit". You write temperatures as

23^{\circ} C

73^{\circ} F

, with the little circle

^{\circ}

degree symbol. The units of absolute temperature named after Lord Kelvin are defined to already be degrees. So we say just "kelvin" instead of "degrees kelvin", since that would be redundant. The temperature in kelvin is written without the little degree circle, like this:

- 273^{\circ} C = 0 K

Also, make a note to not confuse big

K

for kelvin with little

k

for Boltzmann's constant.

k = 1.380 \times 10^{- 23} J/K (joules per kelvin)

T

is the temperature measured from absolute zero in

kelvin

K

. A temperature of absolute zero, or

0 K

- 273^{\circ} C (celsius)

If a particle is at

T = 300 K

, (room temperature), then it has energy:

k T = 1.380 \times 10^{- 23} J/K \cdot 300 K = 4.14 \times 10^{- 21} 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 T}{q} = \frac{4.14 \times 10^{- 21} J}{1.602 \times 10^{- 19} C} = 25.8 \approx 26 mV

At room temperature (around

300 K

k T / q

26

millivolts. That's the energy of a normal everyday electron. The exponent of the diode equation,

v / 26 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 = I_{S} (e^{v / 26 mV} - 1)

This non-linear diode

i

v

equation is harder to deal with than the linear

i

v

equations for

R

L

, and

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_{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_{D}

from the diode equation at room temperature:

i = I_{S} (e^{v_{D} / 26 mV} - 1)

For the resistor, if we can come up with an equation for

i

in terms of

v_{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_{R}}{330 Ω}

We know

v_{R} = 3 V - v_{D}

, so the resistor current becomes:

i = \frac{3 V - v_{D}}{330 Ω}

I'll rearrange the equation so it looks like the slope-intercept form of a line:

i = - \frac{1}{330 Ω} v_{D} + \frac{3 V}{330 Ω}

i = - \frac{1}{330 Ω} v_{D} + 9 mA

The slope of the resistor load line is

- \frac{1}{330}

The

i

-intercept is

9 mA

Graphical solution

We now have two simultaneous equations with two unknowns,

i

and

v_{D}

i = - \frac{1}{330 Ω} v_{D} + 9 mA

i = I_{S} (e^{v_{D} / 26 mV} - 1)

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_{D} = 0.6 V

and

i = 7.2 mA

Reading from the graph is often all the accuracy you need.

Concept check

problem 1

What is the current when the diode voltage $v_{D}$ ‍ is $0$ ‍?

i =

mA

problem 2

What is the voltage when the resistor line touches the $v$ ‍-axis? $(i = 0)$ ‍

v =

V

problem 3

Does the $i$ ‍-axis intercept of the resistor line depend on the value of $R$ ‍?

problem 4

Does the $v$ ‍-axis intercept of the resistor line depend on the value of $R$ ‍?

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.

One way to get more diode current is to reduce the resistance. Lower resistance makes the resistor line steeper. If we reduce the resistor to

200 Ω

and replot the line, we get a new solution:

Making the resistor

200 Ω

increases the LED current to

12 mA

, making it brighter.

You could also see what happens if you increase the voltage. The resistor line moves in a different way when the voltage is adjusted. Go ahead give that a try on your own.

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 Ω

0 Ω

If we make the resistor smaller and smaller, the green resistor line becomes steeper and steeper as the intersection point on the

i

axis goes up and up.

The bottom end of the resistor line is stuck at

v = 3 V

. With

0 Ω

resistance, the line points straight up and does not intersect the diode curve until way way up there in current. Two things can happen at this point. The diode burns out from the excess heat, or, if the diode survives, the battery runs down in just a little while. So it turns out to be a bad idea to leave out the resistor.

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 V

, the solution would have come out something like this:

A typical LED gives off a nice light with a current of around

20 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}{R}

A standard-value

47 Ω

resistor comes close to the target

20 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 = I_{S} (e^{q v / k T} - 1)

I_{S}

is the saturation current. For silicon,

I_{S} = 10^{- 12} A

is a typical value.

In the exponent of the diode equation, the term

k T / q

is equivalent to

26 mV

if the diode is at room temperature.

k

is Boltzmann's constant,

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 = I_{S} (e^{v / 26 mV} - 1)

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.

Addendum: Types of diodes

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 $0.6 - 0.7 V$ ‍.
Germanium diode - Made from a different element. Germanium diodes have a lower forward voltage of $0.25 - 0.30 V$ ‍.
Schottky diode - Made from a silicon-to-metal contact. The forward voltage is lower than regular silicon diodes, in the range of $0.15 - 0.45 V$ ‍.
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.

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GMB2003
Posted 8 years ago. Direct link to GMB2003's post “Will there be any future ...”
Will there be any future semiconductor video, mabye about ICs or transistors?
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(8 votes)
Answer
- APDahlen
  Posted 8 years ago. Direct link to APDahlen's post “GMB, Perhaps but it will...”
  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
  Comment on APDahlen's post “GMB, Perhaps but it will...”
  (6 votes)
Dzifa Kwaku
Posted 8 years ago. Direct link to Dzifa Kwaku's post “Can increasing the electr...”
Can increasing the electric field of an intrinsic semiconductor increase the generation of electrons or conductivity?
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(4 votes)
Answer
- APDahlen
  Posted 8 years ago. Direct link to APDahlen's post “Hello Dzifa, Sure, as a ...”
  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
  Comment on APDahlen's post “Hello Dzifa, Sure, as a ...”
  (5 votes)
utshaw
Posted 7 years ago. Direct link to utshaw's post “From the graphical soluti...”
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 ?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Yes, the diode and resist...”
  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.
  Button navigates to signup page
  (3 votes)
Alexander Wu
Posted 7 years ago. Direct link to Alexander Wu's post “Can't you use indium and ...”
Can't you use indium and antimony to make a diode?
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(3 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Yes. This would be an exa...”
  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.
  Button navigates to signup page
  (5 votes)
Alexander Wu
Posted 7 years ago. Direct link to Alexander Wu's post “Don't you damage a diode ...”
Don't you damage a diode if you get past the breakdown voltage?
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(3 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Not necessarily. Damage h...”
  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.
  Comment on Willy McAllister's post “Not necessarily. Damage h...”
  (3 votes)
Michael
Posted 7 years ago. Direct link to Michael's post “Answer to question # 1. ...”
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?
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(3 votes)
Answer
- Tracy Muhlbauer
  Posted 7 years ago. Direct link to Tracy Muhlbauer's post “You do use the diode equa...”
  You do use the diode equation:
  i = Is (e^(qv/kT)−- 1)
  so for v=0,
  i = Is (e^0 - 1) = 0
  Comment on Tracy Muhlbauer's post “You do use the diode equa...”
  (2 votes)
jay charland
Posted 8 years ago. Direct link to jay charland's post “did you use a 9v battery ...”
did you use a 9v battery to light the led light
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(2 votes)
Answer
- APDahlen
  Posted 8 years ago. Direct link to APDahlen's post “Hello Jay, You certainly...”
  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
  Button navigates to signup page
  (4 votes)
Bari Yakubu
Posted 7 years ago. Direct link to Bari Yakubu's post “Why are most transistors ...”
Why are most transistors made of silicon? Why not lead? sure the farther away the valence electrons, the easier it is to acquire holes.
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(2 votes)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Choosing a material to ma...”
  Choosing a material to make transistors is a balancing act. You want to modify the material (add impurities) to get a wide spectrum of semi-conducting properties (ranging from almost insulating to nearly fully conducting). If you choose a really big atom like Lead because it is easy to take away electrons, you won't be able to modify it with impurities to make it more like an insulator. The element Silicon sits in the "just right" spot in the periodic table so it can be manipulated to get a full range of conduction properties. Another neighboring element used for making diodes is Germanium. A lot of high speed transistors are made from a combination of Gallium and Arsenic. All these elements are nestled close to each other in the periodic table.
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  (3 votes)
GMB2003
Posted 8 years ago. Direct link to GMB2003's post “What is in the center of ...”
What is in the center of the glass diodes pictured above?
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(2 votes)
Answer
- APDahlen
  Posted 8 years ago. Direct link to APDahlen's post “Hello GMB, The heart of ...”
  Hello GMB,
  
  The heart of the diode is a PN junction made of highly refined silicon (with a few controlled impurities).
  
  This link may help: https://en.wikipedia.org/wiki/P%E2%80%93n_junction
  
  You may also want to take a look at Adafruit's tutorials: https://www.youtube.com/watch?v=ap7edIKkykA
  
  Regards,
  
  APD
  Comment on APDahlen's post “Hello GMB, The heart of ...”
  (2 votes)
Anjali Mohapatra
Posted 5 years ago. Direct link to Anjali Mohapatra's post “So what is the basic diff...”
So what is the basic difference between an LCD and LED monitor?
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(1 vote)
Answer
- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “Liquid crystal diodes blo...”
  Liquid crystal diodes block light.
  Light emitting diodes emit it.
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  (2 votes)

Electrical engineering