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Frequency Modulation

While changing the amplitude of a radio signal is the most obvious method to modulate it, it is by no means the only way. It is also possible to change the frequency of a signal to give frequency modulation or FM. Frequency modulation is widely used on frequencies above 30 MHz, and it is particularly well known for its use for VHF FM broadcasting.

Although it may not be quite as straightforward as amplitude modulation, nevertheless frequency modulation, FM, offers some distinct advantages. It is able to provide near interference free reception, and it was for this reason that it was adopted for the VHF sound broadcasts. These transmissions could offer high fidelity audio, and for this reason, frequency modulation is far more popular than the older transmissions on the long, medium and short wave bands.

In addition to its widespread use for high quality audio broadcasts, FM is also sued for a variety of two way radio communication systems. Whether for fixed or mobile radio communication systems, or for use in portable applications, FM is widely used at VHF and above.


What is frequency modulation, FM?

To generate a frequency modulated signal, the frequency of the radio carrier is changed in line with the amplitude of the incoming audio signal.

Frequency Modulation, FM

Frequency Modulation, FM

When the audio signal is modulated onto the radio frequency carrier, the new radio frequency signal moves up and down in frequency. The amount by which the signal moves up and down is important. It is known as the deviation and is normally quoted as the number of kilohertz deviation. As an example the signal may have a deviation of ±3 kHz. In this case the carrier is made to move up and down by 3 kHz.

Broadcast stations in the VHF portion of the frequency spectrum between 88.5 and 108 MHz use large values of deviation, typically ±75 kHz. This is known as wide-band FM (WBFM). These signals are capable of supporting high quality transmissions, but occupy a large amount of bandwidth. Usually 200 kHz is allowed for each wide-band FM transmission. For communications purposes less bandwidth is used. Narrow band FM (NBFM) often uses deviation figures of around ±3 kHz. It is narrow band FM that is typically used for two-way radio communication applications. Having a narrower band it is not able to provide the high quality of the wideband transmissions, but this is not needed for applications such as mobile radio communication.


Advantages of frequency modulation, FM

FM is used for a number of reasons and there are several advantages of frequency modulation. In view of this it is widely used in a number of areas to which it is ideally suited. Some of the advantages of frequency modulation are noted below:

  • Resilience to noise: One particular advantage of frequency modulation is its resilience to signal level variations. The modulation is carried only as variations in frequency. This means that any signal level variations will not affect the audio output, provided that the signal does not fall to a level where the receiver cannot cope. As a result this makes FM ideal for mobile radio communication applications including more general two-way radio communication or portable applications where signal levels are likely to vary considerably. The other advantage of FM is its resilience to noise and interference. It is for this reason that FM is used for high quality broadcast transmissions.


  • Easy to apply modulation at a low power stage of the transmitter: Another advantage of frequency modulation is associated with the transmitters. It is possible to apply the modulation to a low power stage of the transmitter, and it is not necessary to use a linear form of amplification to increase the power level of the signal to its final value.


  • It is possible to use efficient RF amplifiers with frequency modulated signals: It is possible to use non-linear RF amplifiers to amplify FM signals in a transmitter and these are more efficient than the linear ones required for signals with any amplitude variations (e.g. AM and SSB). This means that for a given power output, less battery power is required and this makes the use of FM more viable for portable two-way radio applications.



Summary

Frequency modulation is widely used in many areas of radio technology including broadcasting and areas of two way radio communication. In these applications its particular advantages can be used to good effect. For the future, other forms of digital modulation are becoming more widely used - DAB for radio broadcasting and a number of other formats such as TETRA for two-way radio communication systems. Despite these changes, FM will remain in use for many years to come as there are many advantages of frequency modulation for the areas in which it has gained a significant foothold in recent years.

(Original article on http://www.electronics-radio.com/articles/radio/modulation/frequency_modulation/fm.php)

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Amplitude Modulation

Amplitude modulation or AM as it is often called, is a form of modulation used for radio transmissions for broadcasting and two way radio communication applications. Although one of the earliest used forms of modulation it is still in widespread use today.

The first amplitude modulated signal was transmitted in 1901 by a Canadian engineer named Reginald Fessenden. He took a continuous spark transmission and placed a carbon microphone in the antenna lead. The sound waves impacting on the microphone varied its resistance and in turn this varied the intensity of the transmission. Although very crude, signals were audible over a distance of a few hundred metres, although there was a rasping sound caused by the spark.

With the introduction of continuous sine wave signals, transmissions improved significantly, and AM soon became the standard for voice transmissions. Nowadays, amplitude modulation, AM is used for audio broadcasting on the long medium and short wave bands, and for two way radio communication at VHF for aircraft. However as there now are more efficient and convenient methods of modulating a signal, its use is declining, although it will still be very many years before it is no longer used.


What is amplitude modulation?

In order that a radio signal can carry audio or other information for broadcasting or for two way radio communication, it must be modulated or changed in some way. Although there are a number of ways in which a radio signal may be modulated, one of the easiest, and one of the first methods to be used was to change its amplitude in line with variations of the sound.

The basic concept surrounding what is amplitude modulation, AM, is quite straightforward. The amplitude of the signal is changed in line with the instantaneous intensity of the sound. In this way the radio frequency signal has a representation of the sound wave superimposed in it. In view of the way the basic signal "carries" the sound or modulation, the radio frequency signal is often termed the "carrier".

What is amplitude modulation, AM

Amplitude Modulation, AM

When a carrier is modulated in any way, further signals are created that carry the actual modulation information. It is found that when a carrier is amplitude modulated, further signals are generated above and below the main carrier. To see how this happens, take the example of a carrier on a frequency of 1 MHz which is modulated by a steady tone of 1 kHz.

The process of modulating a carrier is exactly the same as mixing two signals together, and as a result both sum and difference frequencies are produced. Therefore when a tone of 1 kHz is mixed with a carrier of 1 MHz, a "sum" frequency is produced at 1 MHz + 1 kHz, and a difference frequency is produced at 1 MHz - 1 kHz, i.e. 1 kHz above and below the carrier.

If the steady state tones are replaced with audio like that encountered with speech of music, these comprise many different frequencies and an audio spectrum with frequencies over a band of frequencies is seen. When modulated onto the carrier, these spectra are seen above and below the carrier.

It can be seen that if the top frequency that is modulated onto the carrier is 6 kHz, then the top spectra will extend to 6 kHz above and below the signal. In other words the bandwidth occupied by the AM signal is twice the maximum frequency of the signal that is used to modulated the carrier, i.e. it is twice the bandwidth of the audio signal to be carried.


Amplitude demodulation

Amplitude modulation, AM, is one of the most straightforward ways of modulating a radio signal or carrier. The process of demodulation, where the audio signal is removed from the radio carrier in the receiver is also quite simple as well. The easiest method of achieving amplitude demodulation is to use a simple diode detector. This consists of just a handful of components:- a diode, resistor and a capacitor.

AM diode detector

AM Diode Detector

In this circuit, the diode rectifies the signal, allowing only half of the alternating waveform through. The capacitor is used to store the charge and provide a smoothed output from the detector, and also to remove any unwanted radio frequency components. The resistor is used to enable the capacitor to discharge. If it were not there and no other load was present, then the charge on the capacitor would not leak away, and the circuit would reach a peak and remain there.


Advantages of Amplitude Modulation, AM

There are several advantages of amplitude modulation, and some of these reasons have meant that it is still in widespread use today:

  • It is simple to implement


  • it can be demodulated using a circuit consisting of very few components


  • AM receivers are very cheap as no specialised components are needed.


Disadvantages of amplitude modulation

Amplitude modulation is a very basic form of modulation, and although its simplicity is one of its major advantages, other more sophisticated systems provide a number of advantages. Accordingly it is worth looking at some of the disadvantages of amplitude modulation.

  • It is not efficient in terms of its power usage


  • It is not efficient in terms of its use of bandwidth, requiring a bandwidth equal to twice that of the highest audio frequency


  • It is prone to high levels of noise because most noise is amplitude based and obviously AM detectors are sensitive to it.



Summary

AM has advantages of simplicity, but it is not the most efficient mode to use, both in terms of the amount of space or spectrum it takes up, and the way in which it uses the power that is transmitted. This is the reason why it is not widely used these days both for broadcasting and for two way radio communication. Even the long, medium and short wave broadcasts will ultimately change because of the fact that amplitude modulation, AM, is subject to much higher levels of noise than are other modes. For the moment, its simplicity, and its wide usage, mean that it will be difficult to change quickly, and it will be in use for many years to come.

(Original article on http://www.electronics-radio.com/articles/radio/modulation/amplitude_modulation/am.php)

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Quadratic Equations

What is a quadratic equation?

A quadratic equation is an equation that can be written in this form.

ax2+bx+c=0

The a,b, and c here represent real number coefficients. So this means we are talking about an equation that is a constant times the variable squared plus a constant times the variable plus a constant equals zero, where the coefficient a on the variable squared can't be zero, because if it were then it would be a linear equation.

Examples


2x2+3x+1=0, x2+x=2x+3, (x+2)(x+3)=5

All these equations are equivalent to equations of the above form. The first one is already in that form. The second one can be put into it by subtracting 2x+3 from both sides. The third one can be put into it by multiplying out and then subtracting 5 from both sides.

Standard Form


The form

ax2+bx+c=0

is the standard form for a quadratic equation, and for future reference, here the letter a will always mean the coefficient on the square of the variable, and b will be the coefficient on the variable, and c will be the constant term. To get a quadratic into standard form you must remove all parentheses and combine all like terms and add or subtract something from both sides so that the right side will be zero. Once you have your equation in standard form you can identify a,b, and c.

Example

This and many of the other examples below are from my MathHelp collection of problem sets, Quadratic Equations. For more practice and worked out examples for this or any other techniques explained here, click on the MathHelp link at the bottom of the page.

Problem: Write the equation in standard form and identify a,b, and c.

Solution: Multiply the left side out and then subtract the 5 from both sides.

Solving

Now lets talk about solving these equations.

Quadratic equations are harder to solve than linear equations, because once you have them in standard form it is hard to simplify them any further, and in this form there are still two occurrences of the variable, so it's hard to see what we can do to get the variable alone.

So we have to find some clever tricks to get around this problem.

Solving by Factoring

One trick is to solve the equation by factoring. This trick works because of the principle of zero products. The principle of zero products says

If A and B are real numbers and AB=0, then either A=0 or B=0.

This is a very special property that only zero has. For other numbers there are lots of ways to multiply and get them, but not for zero. For zero, the only way to multiply numbers and get it, is if one of the numbers is zero.

The principle of zero products allows us to reduce a complicated equation to simpler equations provided the right side of the equation is zero, and the equation is factored, because we can set each of the factors equal to zero.

  • To solve a quadratic by factoring, first you must make sure it is in standard form. It is especially important that it is set equal to zero, because remember, the principle of zero products only works for zero.
  • Then you must factor the left side.
  • Then you set each of your factors equal to zero and solve the equations you get to find the solutions to your equation.

Example

Problem: Solve the equation.
Solution:
The second line here comes from setting x+1 and x+6 equal to 0 by the principle of zero factors.

What if you can't factor?

But some quadratics are difficult to factor, so for these equations we need other methods. The method of completing the square is a method that will work for any quadratic, but it is a little bit complicated, so I will introduce it slowly and step by step. But first to give you an overview of where we are going, I will show you a simple example of it.

Consider the following equation.

x2-2x-1=0

This looks like a nice simple friendly equation, but we can't solve it by factoring, because we can't find two numbers to multiply and get -1 and add and get -2, so we are going to have to find another method.

But if only that minus sign on the 1 weren't there, then we could factor it really easily, in fact it would be a perfect square. How can we make that minus sign go away?

Well, one thing we could do is add 2 to both sides of the equation, and then the equation would become

x2-2x+1=2
and this factors to
(x-1)2=2

Now, I know what you're saying. You are saying, "But you said that you have to get it set equal to 0 to solve it by factoring, because the principle of zero products only works for 0. What good does it do to have something factored and set equal to 2?"

And if you are saying this to yourself, you are absolutely right. But this isn't just any old factorization. It is aperfect square, and maybe you can do something with a perfect square set equal to 2.

If we could figure out how to solve equations like

(x-1)2=2

that is, perfect squares set equal to numbers, then we could solve an equation like

x2-2x-1=0.

And if we could find a way to add a number to both sides of other quadratics so that we can put them into the form perfect square equals constant, then maybe we could be able to solve them too.

This means that to help us solve quadratic equations, we need to learn two skills.

  1. Solve equations of the form (x+k)2=d, where k and d are numbers.
  2. Find a way of figuring out what number to add to both sides of a quadratic equations so that the left side will become a perfect square.

To work our way up to the task of solving equations of the form

(x+k)2=d

let's first start with the slightly easier task of solving equation of the form

x2=d

How do we solve an equation of the form

x2=d?

If x is greater than 0 then the obvious answer is

but this is not quite right because it only gives you the positive square root of d, and all positive numbers have two square roots, a positive one and a negative one. So to be sure that you are getting all solutions to an equation of this form, your answer must be

If x less than 0 then what happens? What kind of number can you square and get a negative number? If you square a positive number then clearly you get a positive number. But if you square a negative number then you have a product of two negative numbers, so you still get a positive number. So what is left for squaring and getting a negative number? Nothing. So the equation has no solutions.

Now let's look at the more general equation of the form

(x+k)2=d

This is really not much harder since anything you can do with x you should be able to do with x+k. x+k represents a number too. So solve for x+k and then add something to both sides of the equation to get x alone.

Example 1

Problem: Solve the equation.
Solution:

Example 2

Problem: Solve the equation.
Solution: Anything you can do with x, you can do with x+1, and once you find x+1 all you have to do to get x is subtract 1.

Completing the Square

Now to problem number two, that of finding something to add to a quadratic to make it a perfect square.

This is what is meant by completing the square, and the secret to it is to expand out the expression

(x+k)2

and see what makes perfect squares tick. Applying our formula for squaring a binomial, we get

(x+k)2=x2+2kx+k2

The key here is to look at the relationship between the coefficient on x and the constant coefficient. The coefficient on x is 2k and the constant term is k2. This means that if we know the coefficient on x, and we want to know what the constant term has to be for the expression to be a perfect square, then we need to divide the coefficient on x by 2 to get k, and then square to get k2.

So if you have an expression of the form

x2+bx

and you want to find something to add to it to make it a perfect square, then you need to

  1. Divide b by 2 to get k
  2. Square k to get k2.

Example

Problem: Complete the square.
Solution:
The yellow part is the scratch paper. On the scratch paper you first divide the 9 by 2 and then square the result. Don't worry about the minus sign, because it will go away when you square anyway. Then the number you get will be the number you need to add to the expression to make a perfect square out of it. After you do that it is good practice to write it as the square that it is. For that you can use the first line of your scratch paper and match the sign with the sign of the second term of the original expression.

I hope the above has helped you understand the process of completing the square. If not, there is another approach to it that I have written an article about that you might find interesting for further understanding. It is a geometrical approach based on the method that many earlier mathematician used.

Solving by Completing the Square

Now we are ready to use the method of completing the squares to solve quadratic equations. The best way to do this is as follows.
  1. Add something to both sides so that the left side has no constant term.
  2. Figure out what to add to the left side to make it a perfect square, and add that to both sides.
  3. Write the left side as the perfect square that it is and do the arithmetic on the right side.
  4. Solve the equation you get by the methods of equations of the form (x+k)2=d

One thing we left out. So far all of the equations we have solved have had a coefficient of 1 on x2. What do we do if we have a coefficient other than 1 on there?

Well, we don't really have any method of completing the squares to deal with that situation, so the easiest thing to do is just divide both sides by it and put up with the fractions. With completing the squares, fractions are not so bad to deal with because there is no guess work.

Example

Problem: Solve the equation.
Solution:
First we divide both sides of the equation by 2, to get a coefficient of 1 on the first term. 0/2 is still 0. Then we add 1/2 to both sides so that it is easier to figure out what to add to make the left side a perfect square. Then we complete the square in order to figure out what to add to both sides to make the left side a perfect square. The scratch work for this is shown in yellow. The best way to divide a fraction by 2 is to multiply it by 1/2. Then we write the left side as the perfect square that it is, and do the arithmetic on the right side. Now take square roots and add 3/4 to both sides to get the final answer.

The Quadratic Formula

Now that you have learned the method of completing the squares, I will tell you a secret. The methods of completing the squares is such a good method for solving quadratics that it is very seldom used for it. But for solving quadratic equations, it is such a good method that it puts itself out of business.

You see, with such a mechanical method like the method of completing the squares, why not just apply it to the general quadratic equation and solve all quadratics in the world at once, and be done with it, and never have to use algebra to solve a quadratic again.

Problem: Solve the equation:

Solution:

Just do it the same way with the letters as you did with the numbers. First divide both sides by a. Then subtract c/a from both sides to be able to see more easily what to add to the left side to make it a perfect square. Then complete the square on the scratch paper and add what you get to both sides. Write the left side as the perfect square that it is. Instead of doing arithmetic on the right side you have to do a little bit of algebra, using 4a2 as a common denominator. Then take square roots and subtract b/2a from both sides and use 2a a common denominator to get the final answer.

We have just solved all quadratics in the world at once and derived the quadratic formula, which says:

For any real numbers a,b, and c, the solutions to the equation

ax2+bx+c=0

are

Solving by the Quadratic Formula


Since this formula is somewhat long and complicated, it is best to evaluate it in two smaller pieces by first evaluating the thing inside the radical,

b2-4ac

and then put the result into the formula

The quantity

b2-4ac

even has a name. It is called the discriminant. And there is another advantage to computing it first. Since it is what is in the radical, it can't be negative if there are going to be solutions to the equation, because you can't take a square root of a negative number, (unless you use the imaginary numbers, and we're not yet ready for them here) so if the discriminant comes out negative, then you don't have to do any more work, and all you have to do is write "no solution" on your paper and you are done. Sometimes you can determine this quite quickly by estimating, particularly if a and c are very large and b is small.

Example 1

Problem: Solve the equation.
Solution:

Example 2

Problem: Solve the equation.
Solution: No solution. b is -2, which is very small in comparison to a=6 and c=27, so you don't even have to compute the discriminant to see that it is going to be negative.

Imaginary Solutions

This section is for more advanced students who know about imaginary numbers. If you know about imaginary numbers, you don't have to stop when you see the square root of a negative number, because with imaginary numbers you can take the square root of a negative number. To find the square root of any negative number you just take the square root of the corresponding positive number and multiply it by i, the square root of -1. This makes sense at least once you believe in the idea that the square root of -1 is i, because of the multiplication rule for square roots.
(It is customary usually to write the real number after the i when it is a square root so that it is clear that the i is not inside the radical.)

Once you know how to find square roots of negative numbers, you find imaginary solutions to quadratics by the completing the square or the quadratic formula pretty much like you find real ones. For the following examples the instruction is to solve the equation.

Example 1:

x2+2x+5=0

Solution:

Completing the Square:

Explanation:

First we add -5 to both sides to get the constant on the right side of the equation so that it is more clear what we need to add to the left side to make it a perfect square. Then we complete the square. 2/2=1, 12=1, so 1 is the number we add to the left side to get a perfect square. Whatever you do to one side of an equation, you have to do to the other side, so we also add 1 to the right side of the equation. Then we write the left side as the perfect square that it is, and do the arithmetic on the right side. In this equation we get a negative number on the right side, but with imaginary numbers we can deal with that. The two square roots of -4 are 2i and -2i, so x+1 has to be one of them. Then to find out what x is, all we have to do is add -1 to both sides.

Quadratic Formula:

Explanation:

First compute the discriminant and find its square root. The square root of -16 is the square root of 16 times i, 4i. Then we just fit in the square root in its place in the formula. The numerator has a common factor of 2 that we can factor out and cancel with the 2 in the denominator.

Example 2:

x2+x+1=0

Completing the Square:

Explanation:

Again we first subtract 1 from both sides so that it is more clear what we need to add. For the completing the square part, this time the coefficient on x is 1. Half of 1 is 1/2 and (1/2)2=1/4, so 1/4 is the completing the square number. We add it to both sides, because anything you add to one side you have to add to the other side. Then again we get a negative number to take plus and minus square roots of, but we can handle that with imaginary numbers. We just break it up into square roots of -1, 3, and 4. The square root of -1 is i, the square root of 4 is 2, and the square root of 3 is the square root of 3. Then to get the final answer we add -1/2 to both sides.

Quadratic Formula:

Explanation:

First we figure out the discriminant, which comes out to be -3, and then we take its square root. The square root of a negative number is just the square root of the corresponding positive one times i. This time we can't the square root of 3 is an irrational number, so it is better left undone thinking of radical 3 is the name for the exact number that you multiply by itself to get 3. Then after we have found the square root of -3, we can put it in place of the radical in the formula to get our final answer.

Example 3:

3x2+2x+3=0

Completing the Square:

Explanation:

In this one there is a coefficient other than 1 on the x2, namely 3, so first we have to get rid of it by dividing both sides by 3. Then from there it is pretty much like the other examples. Next we add -1 to both sides to see more clearly what we need to add to make the left side a perfect square. The coefficient on x is 2/3. To find half of 2/3 we multiply it by 1/2 and get 1/3. Then to find the completing the square number, we square that and get 1/9, which we add to both sides of the equation. Then we write the left side as the perfect square that it is and do the arithmetic on the right side. And again since this section is about imaginary solutions, we get a negative number on the right side of the equation to take the square root of. Then after taking plus and minus square roots, we add -1/3 to both sides to get the final answer.

Quadratic Formula:

Explanation:

First we find the discriminant, and we get -32. The square root of -32 is the square root of -1 times the square root of 16 times the square root of 2. The square root of -1 is i, the square root of 16 is 4, and the square root of 2 is something nasty and irrational, so it is left as the square root of 2. Then we put this square root in place of the radical in the formula and simplify to get the final answer.

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Factoring Polynomials

Common Factors

The first thing you should always do when factoring is to take out a common factor. This is the simplest technique of factoring, but it is important even when you learn fancier techniques, because you will make your later work much easier if you always look for common factors first. Taking out common factors is using the distributive property backwards. The distributive property says

a(b+c)=ab+ac.

The idea behind taking out a common factor is to look for something like the right side here where there is a common factor, here it would be a, and turn it into something like the left side to factor it.

Example:

4x5+12x4-8x3=4x3(x2+3x-8)

A good trick for finding the largest common factor when you are using this method to factor polynomials is to find the greatest common factor of the numbers and the smaller power of the variable, so here the greatest common factor of the numbers is 4 and the smallest power of x is 3, so we can take out 4x3 as a common factor.

Grouping

Grouping is a fancier technique that is based on taking out common factors. For grouping we split the polynomial in two pieces and take out common factors in each of them. If we get the same thing left over in each piece, then we can take that big thing out as a common factor, and this will factor the polynomial. It is, however, important to note that this only works if we do get the same thing. Simply taking out common factors of pieces of a polynomial is not factoring it. To factor a polynomial we must write the whole polynomial as a product of two polynomials.

Example:

5x3+10x2+3x+6=5x2(x+2)+3(x+2)=(5x2+3)(x+2)

Here we take out a common factor of 5x2 from the first two terms and a common factor of 3 from the second two terms. This would not be a good method of factoring except that this polynomial is kind of special in that what is left when taking out the common factors in both cases is x+2. Then what we do is take out x+2 as a common factor of the two big chunks. The idea is that anything we can do with x we can do with x+2, because it represents a number too. If it helps, blur the (x+2) or think of it as one big strange looking letter.

General Quadratics

To learn to factor second degree polynomials, also called quadratics we first make it simpler by looking at ones where the coefficient on x2 is 1. To figure out how to factor these polynomials we need to look at what happens when you multiply things of the form (x+A)(x+B), like for example (x+3)(x+4). Applying FOIL to this multiplication we get

x2+4x+3x+12=x2+7x+12.

Notice that the coefficient on x comes from adding the inners and outers, so it will always be the sum of the two constants. The constant term will always come from the lasts, so it will be the product of the two constants. So really we could use this as a special trick for multiplying anything of this form. We just add the constants for the middle coefficient and multiply them for the last term. So to multiply

(x+5)(x-2)

we can use this shortcut and get

x2+3x-10

by just adding the 5 and the -2 to get the 3 and multiplying the 5 and -2 to get the -10. This means that in order to go the other way around, that is to factor into this kind of form, what you need to find is two number that multiply to the last term and add to the the middle coefficient. So that means that to factor the above polynomial all you need to do is find two numbers that multiply to -10 and add to 3, and that would be 5 and -2, and that tells us that it will factor into

(x+5)(x-2).

Of course we could cheat this time, because we already knew the answer, but for many polynomials it wouldn't be too hard even if we didn't, but there is a little bit of guess work here, so it may take a little practice to get used to it. One thing you can do if you get stuck is to list all of the ways to factor the last term and then play around with them until you find one that adds up to the right thing. You can even simply try all of them, which might seem like a lot of work, but by doing it you will get a feel for how it works and before too long you won't really have to try all of them. Let's say you wanted to factor

x2+17x+60.

This one is a little bit tricky, because there are a lot of ways to factor 60. To get an idea about what happens with these, lets list them all.

1x60
2x30
3x20
4x15
5x12
6x10

1+60=61, no good.
2+30=32, no good.
3+20=23, no good.
4+15=19, closer, but still no good.
5+12=17, eureka, we've found one.

This means that we can factor the polynomial as

(x+5)(x+12).

One way to avoid simply trying them all is to do some estimating and also notice some patterns. One pattern that is helpful is that when the numbers are farther apart, like 60 and 1, they add to bigger numbers and when they are closer together, like 6 and 10, they add to smaller numbers. There are some other tricks you can play as well. One I find handy sometimes is swapping prime factors. Suppose in this one you guessed 6x10 and say that it was 16, which is just a little too small. Since it is too small, what you want is to factor it so that the factors are a little farther apart. Sometimes you can come up with this by prime factoring the two numbers and then switching some factors.

6=2x3
10=2x5

One way to make these farther apart would be to switch the 3 and the 2 and get

2x2=4
3x5=15,

but that gives us a sum that is too big. The way to do it that does work is to give the 10's 2 to the 6 and get

2x2x3=12
5=5.

It is also important to consider the signs of the coefficients and what happens when you multiply and add plus and minus numbers. The simple case is when both coefficients are plus like the one above where you only need to use positive numbers. The case where the middle coefficient is negative and the last terms is positive is almost as easy. What you need then is two numbers that multiply to a positive number and add to a negative number. The only way to get that is if they are both negative. Then because when you add two negative numbers you add and your answer is negative, you can think of it pretty much the same way you do when they are both positive. The only thing you have to do different is make them both negative instead of both positive. So for example, suppose you want to factor

x2-5x+6.

Then you still just need to find two numbers to multiply to 6 and add to 5, which would be 2 and 3, but since we really want -5, we would use -2 and -3 and factor it

(x-2)(x-3).

If the last term is negative it is a little bit different. Then regardless of what the sign of the middle coefficient is, one of your numbers has to be positive an the other negative to multiply to a negative number. That means that when you add those numbers, in terms of positive number arithmetic, it is a subtraction. When you add a positive and a negative number you subtract and use the sign of the larger one, so you need to think in terms of subtracting to get the middle term, and if the middle coefficient is positive you want the positive one to be bigger, but if it is negative, you want the negative one to be bigger. For example if we wanted to factor

x2+x-12,

we could do it by finding two numbers to multiply to 12 and subtract to 1. That would be 4 and 3. Then since the middle coefficient is 1 and not -1, we would use 4 and -3 and factor the polynomial as

(x+4)(x-3).

On the other hand, if we wanted to factor

x2-x-12,

we could think about it the same way, what two numbers multiply to 12 and subtract to 1 and get 4 and 3, but since this time we want a -1, we would use -4 and 3 and get

(x-4)(x+3).

Next we turn to the more difficult problem of when the first coefficient is not 1. Here the guessing is a little more complicated, because the coefficient you get from the inners and outers will not be just the sum of the two constants, because they will get multiplied by the x coefficients. For example, lets say we want to factor

3x2+11x+10.

We need to make the product of the firsts give us 3x2. In this problem that isn't too difficult, because there is only one way to do it, and that is 3x and x, so we want a factorization of the form

(3x )(x ).

Now, the product of the lasts still has to be 10, but it is not their sum that has to be 11, it is something a bit more complicated, because in the sum of the inners and outers one of the numbers will be multiplied by 3. For example, lets suppose we tried 2 for the first one and 5 for the second one. Then it would be

(3x+2)(x+5).

That wouldn't work, because the sum of the inners and outers wouldn't be 11x. Instead it would be 15x+2x=17x. The first and last terms would be right, but that is not good enough. But if we put the 5 in the first factor and the 2 in the second factor, it would work, since then it would be

(3x+5)(x+2)

and the sum of the inners and outers would be 6x+5x, which is indeed 11x. But how are you to come up with that arrangement other than trial and error? To some extent you can't, you just get better at the trial and error with practice. Again, you could if worse came to worse try all possible combinations. For most of the problems you will be expected to do in algebra classes there won't be that many possible ways to factor the last term anyway. For 10 it is just 1x10 and 2x5, but here this makes a few more possibilities than last time, because the order will matter. But still for this problem there are only 4 possibilities,

(3x+10)(x+1),
(3x+1)(x+10),
(3x+2)(x+5),
(3x+5)(x+2).

In the beginning it won't hurt you to just try them all and see what happens. Again, this can be quite helpful in getting a feel for what happens with this. It sounds like a long process, but after you do it a few times, you will find that you can eliminate a lot of possibilities without fully trying them. There are again some techniques you can use involving estimating sizes. In general you will get larger numbers by using factors that are farther apart and by placing the numbers so that the large numbers multiply large numbers and the small numbers multiply small numbers and you will get smaller numbers by doing the reverse. So for example, here (3x+1)(x+10), where the 10 is multiplied by the 3 and the 1 is multiplied by the 1, gives us the largest possible middle term and (3x+5)(x+2), where the 5 is multiplied by the 1 and the 2 is multiplied by the 3, gives us the smallest possible middle term. Also in these problems you can eliminate some possibilities right away by estimating. If I was approaching this problem I would see right away that I couldn't put the 10 so that it multiplied the 3, because the outers alone would be too big. I would even be pretty suspicious of using the 10 at all, because even 10 times 1 is almost 11, so there wouldn't be too much left for the other term. So I would guess it would be 5 times 2, and putting the 5 where it gets multiplied by 3 gives 15 which is already too big on its own, so there is really only one possibility that has a chance. With a little practice you may find it surprising that you will be able to guess right the first time a good amount of the time, and when your first guess isn't right, usually your second one is. I find when doing these that sometimes the right solution just sort of appears before me magically, but I have been doing these in front of classes for a lot of years. Just the same, with practice you may find that this happens to you more and more often.

More Examples

For these examples the instruction is to factor the polynomial.

Example 1:

2x2+9x-5

Solution:

(2x )(x )

There is only one way to factor 5, 5 and 1, so those are the numbers we are going to use. The only question is where to put them. Here since the last term is minus, one of the numbers must be minus and the other plus. That means that the inners and outers will subtract instead of add. Since 9 is large, particularly since it will be a difference we want to split things unevenly, so the best guess is to place the 5 so that it gets multiplied by the 2 and the 1 so it gets multiplied by 1.

(2x 1)(x 5)

Since we want to get a positive 9, we want the one that gives us the bigger product to be positive and the one that gives us a smaller product to be negative. That gives us

(2x-1)(x+5).

And that should work, because we get 10-1=9. The firsts and lasts will work without problems. Check again to make sure the inners and outers do in fact work. We get

10x-1x=9x,

so it works.

Example 2:

6x2-31x+35

Solution:

This one is a bit harder, because we have a choice for the firsts. We can use 6x and x, or 2x and 3x, and we don't really know which one will work. Sometimes it is possible to use the relative sizes of the middle and last terms to help us guess, but there isn't really any way to know for sure. Mainly we just have to try one, and see if we can find anything for it, and if we can't try the other. For this one, I'm not really sure at all which one it is going to be, so let's try 6x and x first and see what happens.

(6x )(x )

We need to multiply to get 35, so it must be 1 and 35 or 5 and 7. There are some possibilities that we can eliminate right away. Since the last term is plus, the signs of the two numbers will be the same, so for the middle term we will be adding. The only difference the minus on the middle term makes is that both of the numbers will be minus. We can't put the 7 or the 35 so they get multiplied by the 6, since that would be already too big. Even putting the 5 so that it gets multiplied by the 6 doesn't look like a good idea, because that only leaves 1 for the inners, but that is really the only possibilities with this set up, since actually we can't use the 35 at all, since it alone is bigger than 31. If we put the 5 with the x, so that it would get multiplied by the 6, then we would have to put the 7 with the 6x to be multiplied by the x. In case you have gotten lost, that would be

(6x+7)(x+5).

But without even precisely working that out, we can tell that this would be too big, since 7 is bigger than 1. So we have come to the end of the line with 6x and x and must see what we can get with 2x and 3x.

(2x )(3x )

Okay, we have already figured out that there is no way that we can use 35 and 1, because the 35 alone would be too big, so we have to use 5 and 7. Let's just try both ways to see what happens.

(2x-5)(3x-7)

This gives us 14+15=29, close, but that's not good enough.

(2x-7)(3x-5)

This give us 21+10=31, which works!!

Notice that without too much difficulty we eliminated all but two possibilities, by some simple estimating, and even if you had to try the two remaining possibilities simply by trial and error, that wouldn't be that bad. With some practice, maybe you wouldn't even have to do that. I might guess the right answer on this one right away, because of seeing that if I put the 7 to be multiplied by the 3, I will get 21, which is 10 off from what I want to get, and I know at least that 10 is divisible by the other numbers, so there is a chance that they might multiply to 10. After you do these for a while you sort of pick up on patterns like that. Another thing you can do is to save time when you are trying the numbers out, don't bother with the signs, and sort those out after you see that the numbers work, and of course don't bother with the variable either, because you know that will be there. So in this correct arrangement you would simply figure 3x7 for the inners and 2x5 for the outers, and check to see if that adds up to your middle coefficient.

Example 3:

4x2-5x-6

Solution:

For this one again we have to be careful, because we get a choice between 4x and x or 2x and 2x. This time I am going to deliberately choose the wrong one, because I know the answer to this problem, because I cheated by multiplying the factors to get the problem. Let's try 2x and 2x, since I know it is really 4x and x.

(2x )(2x )

Since the last term is minus, we will be looking for the middle term to be the difference of the outers and inners. 6 can be factored 6x1 and 2x3 and we need a difference of 5, but remember, that includes multiplying by the 2's. This time since the coefficients on the x's are the same, it won't matter which order we put them in. 6 and 1 will give us 12-2=10 and 3 and 2 will give us 6-4=2, so neither will work. If you are clever about it you don't have to go through all that to see it, because since both numbers are getting multiplied by 2, you will be getting the difference of two even numbers, which must be even, and 5 is odd. You may not be able to precisely explain it like that, but you still might be able to intuitively look at it and see that the inner outer difference just doesn't look like it could come out to 5. So any way we do it, we can see that 2x and 2x, won't work, so we will have to try the other possibility 4x and x.

(4x )(x )

6 and 1 just looks like it won't work. One way the difference looks too big and the other way it looks too small, so lets try 2 and 3. We are looking for a difference of 5. If we put the 3 so it gets multiplied by the 4, that looks like it will make the difference too large, since the 3x4 part is 12 and the other part is small, so it looks like the other way around is the best choice. Let's try it.

(4x 3)(x 2)

Write the numbers in first without signs, because that is quicker and we can adjust the signs later. This gives us 8-3=5, so that looks good. Now adjust the signs. We want it to be a -5, so we want the negative one to be bigger. Now here we have to be careful, because it is not the bigger of the two numbers that should be negative it is the one that is bigger after multiplying by the x coefficients. The 2 is smaller than the 3, but it is getting multiplied by the 4 for make 8 and the 3 is multiplying by 1 to get 3, and it is these products that we are adding together. So we want to get -8 and +3, so that they add up to -5, so that means that even though the 2 is smaller, it is the 2 that will get the minus sign, so the final answer is

(4x+3)(x-2).

Let's check that to make sure. For the firsts we get 4x2 as planned. For the outers we get -8x. For the inners we get 3x. And for the lasts we get -6. Putting that together we get

4x2-8x+3x-6=4x2-5x-6,

so it works!!

With some practice, even though this method may seem a bit fiddling at first, you should be able to get good enough at it so that you can manage most problems. But in case you have one where both the first and last terms can be factored a lot of different ways and you get stuck, there is another method based on grouping you can use. It is longer, but involves less guesswork. Because it involves less guesswork it can be tempting to give up on the guesswork and only use this method, but I don't think this is a good idea. It is better to save this method for problems where you get really stuck doing it the other way, because otherwise you won't get the practice that it takes to get good at doing it the other way. In other words, if I teach you this other way, you must promise not to abuse it. So read this part only if you can make that promise.

Okay, now that I know I am only talking to those who are ready for this, here is the trick. We multiply the first and last coefficients and then play the game we played for ones with a first coefficient of 1, but with this number instead of just the last term. We take that number and look for two numbers that will multiply to it and add to the middle coefficient. Then we use those numbers to break up the middle term, and if we do that it turns out that the 4 term polynomial we get will always factor by grouping. This might be a bit hard to grasp without examples, so I will show you how to use this method for the above three examples.

Example 1:

2x2+9x-5

Alternate Solution:

Multiply the 2 and the -5 to get -10

Now find two numbers that multiply to -10 and add to 9.

That would be 10 and -1

Now use these numbers to break up the middle term and the polynomial becomes

2x2+10x-x-5.

It doesn't matter what order we put them in. We could use -x+10x as well, and in the end we would come out with the same answer.

Now I claim we can factor this by the grouping method. For the first two terms take out a common factor of 2x and for the second two terms take out a common factor of -1, and we get

2x(x+5)+-1(x+5)

and look, isn't that amazing, we get a common factor of x+5 that we can take out and we get

(2x-1)(x+5),

the same thing we got before.

Example 2:

6x2-31x+35

Alternate Solution:

Multiply 6 times 35 to get 210.

Find two numbers to multiply to get 210 and add to get 31.

Not quite as easy as last time, but -21 and -10 will do.

Now use these numbers to break up the middle term and we get

6x2-21x-10x+35.

Now factor this by grouping. Take out a common factor of 3x from the first two terms and -5 from the second two terms and this becomes

3x(2x-7)-5(2x-7)

Again the leftover parts are both the same, this time 2x-7, so we can take that out as a big common factor and get

(3x-5)(2x-7).


Example 3:

4x2-5x-6

Alternate Solution:

Multiply the 4 and the -6 to get -24.

Then find two numbers that multiply to -24 and add to -5.

-8 and 3 will work.

Use these numbers to break up the middle term and we get

4x2-8x+3x-6.

Now factor this by grouping. Take out a common factor of 4x from the first two terms and a common factor of 3 from the second two and we get

4x(x-2)+3(x-2)

Then take out x-2 as a factor to get the answer of

(4x+3)(x-2).

Just make sure to keep an open mind. It is never a good idea in mathematics class to get stuck on one way of doing something.

Difference of Squares

There is a special formula for factor a difference of squares that comes from applying a shortcut multiplication formula backwards. It turns out that whenever you multiply the sum and difference of the same things, the inners and outers cancel out and you only get two terms, and this gives you the formula

(A+B)(A-B)=A2-B2 .

To use the formula for factoring we need to write it backwards as

A2-B2 =(A+B)(A-B).

So if we see something that is a difference of squares, we can factor it as the sum and difference.

Example:

x2-9=(x+3)(x-3)

Sometimes we can use the formula more than once.

x8-1=(x4+1)(x4-1)=(x4+1)(x2+1)(x2-1)=(x4+1)(x2+1)(x-1)(x+1)

This brings up another point. Factoring is like prime factoring of numbers. Always factor completely, that is keep factoring the factors until nothing more will factor. The ones that can't factor come along for the ride. Sums of squares can't be factored, so they come along for the ride.

Perfect Squares

Sometimes we can also use the squaring formulas,

(A+B)2=A2+2AB+B2
(A-B)2=A2-2AB+B2

in reverse for factoring. If you see a polynomial with three terms and the first and last terms are perfect squares, you can check and see if the middle term is right for this kind of factorization.

Example:

x2+14x+49

The first and last terms are perfect squares. If this were to fit into the pattern of the first of these formulas, it would have to be

(x+7)2

The first and last terms are right for that. Is the middle term right. The middle term should be twice the product of 7 and x, which is 14x, and yes, that is what we have, so this factorization works. For this problem we could have also used the method above for general quadratics, but for some, particularly like ones with fractions, we can get it easier this way.

Sum and Difference of Cubes

There are special formulas that need to be memorized for factoring the sum and difference of cubes.

A3+B3=(A+B)(A2-AB+B2)
A3-B3=(A-B)(A2+AB+B2)

Notice that although it isn't possible to factor a sum of squares, it is possible to factor a sum of cubes. It turns out in general that it is always possible to factor sums of odd powers, but not even powers. The way I always remember these formulas is that the first factor is the original without the cubes and the second factor is almost like the squaring formulas, but without the 2 and the sign on the second term of it is always the opposite of the sign in the original and in the first factor. In other words these formulas almost look like

A3+B3=(A+B)(A-B)2
A3-B3=(A-B)(A+B)2

except that there is no 2 on the middle term. Also when you are factoring, once you have written down the first factor, you have A and B handy, so you can easily read them off from it for writing down the second factor.

Example 1:

x3+8

Solution:

Since the first term is x cubed and the second term is 2 cubed, this is a sum of cubes. For the first factor we just remove the cubes and write down

(x+2).

Then for the second factor, just take these two terms that are written down in the first factor and plug them into the formula for the second factor, which is just like the formula for the square of x-2, except that you don't double for the middle term, the middle term is just the product of the two terms, so we square the x and then take the product of 2 and x and then square the 2 and get

(x+2)(x2-2x+4).

Example 2:

x3-27

Solution:

This one is a difference of cubes, where A is x and B is 3, so for the first factor we get

(x-3).

Now for the second factor we just plug these two numbers into the formula for the second factor. Square the x for the first term, multiply 3 times x for the second term and square the 3 for the last term and we get

(x-3)(x2+3x+9).

Sum and Difference of Higher Powers

There are formulas for sums and differences of higher powers, but mostly people don't bother to learn them. You can either look them up, or work them out by long division. But some higher powers can be done by using the difference of square and sum and difference of cubes formulas, namely any power that is a multiple of 2 or 3. Any power that is a multiple of 2 is a perfect square and any power that is a multiple of 3 is a perfect cubes. This is true because powers to powers multiply. So for example x16 is a perfect square, because it is (x8)2 and x24 is a perfect square and a perfect cube, since it is (x12)2 and (x8)3. So you can factor differences of even powers by using the difference of squares formula and sums and differences of powers that are multiples of 3 by using the sum and difference of cubes formulas. And sometimes you can factor more than once this way. Sometimes when you approach it two different ways it can look like you get different answers. An interesting example of that is when you have a difference of 6th powers where you can treat it either as a difference of squares or a difference of cubes.

Example:

x6-1

Method 1:

Look at it as a difference of squares. (x3)2=x6, 12=1.

x6-1=(x3-1)(x3+1),

but we can go further, because each of these factor can be factored using the sum and difference of cubes formulas.

x3-1=(x-1)(x2+x+1)

and

x3+1=(x+1)(x2-x+1),

so putting this all together, we get

x6-1=(x3-1)(x3+1)=(x-1)(x2+x+1)(x+1)(x2-x+1).

Method 2:

Look at it as a difference of cubes. (x2)3=x6, 13=1.

x6-1=(x2-1)(x4+x2+1)

The first factor can be factor further as a difference of squares, and we get

(x-1)(x+1)(x4+x2+1),

but it looks like that is all we can do, because it is hard to see how to factor the other factor.

But we should get the same answers with both methods, because after all, they are equal, so it must be true that

x4+x2+1=(x2+x+1)(x2-x+1),

but it is not at all clear how you would figure that out if you did it that way. For this reason it is generally better when you have problems like this where there is a choice to look at it first as a difference of squares.

But there is a way you can get the complete factorization the other way, and it involves a fancier factorization that you can use for some other problems as well. I don't know any official name for it, but I call it completing the square from the inside, because it involves adding and subtracting something that will change the middle term so that the first chunk is a perfect square, and then factoring it as a big difference of squares.

Completing the Square from the Inside (Extra for Experts)

This is a method of factoring that we can use when we have a 4th degree polynomial that has no odd terms, and we can't factor simply by treating x2 as the variable. For some polynomials of this sort we can factor them much easier by treating x2 as the variable. For example if we want to factor x4-5x2+6, we can do that. We just ask ourselves what two numbers will multiply to 6 and add to -5, and since that is -2 and -3, we can factor it to (x2-2)(x2-3), just like if it were x2-5x+6. But it turns out that if we have one like that where we can't find numbers to multiply to the last term and add to the middle coefficient, we can often still factor them by using this method of completing the square from the inside.

Example:

x4+x2+1

We can't factor this one like the last one, because we can't find two numbers to multiply to 1 and add to 1. What we do with a problem like this is try to find something to add to the middle term so that the polynomial becomes a perfect square, and we add and subtract it. By doing this we can turn it into a difference of squares. In this case if there was a 2 coefficient on the x2, we would have x4+2x2+1, which would be a perfect square, so what we do is add x2 and subtract it, to get

x4+2x2+1-x2=(x2+1)2-x2

If we couldn't see by guessing what the coefficient would need to be for this we could get it by doing the opposite of what we normally do for completing the square, take the square root of the constant term, and then double that, so here this would be

sqrt(1)=1
2·1=2.

Then to get that coefficient we figure out what multiple of x2 we need to add to the existing term to get it. In this case since we already have 1, we need to add another 1 to get 2.

After we do this we get a big difference of squares, so we can use the difference of squares formula to factor it and get

[(x2+1)+x][(x2+1)-x].

Then getting rid of the extra parentheses and writing things in standard order this becomes

(x2+x+1)(x2-x+1),

just like we got from the other method.

Here are a couple more examples of this method.

Example 1:

x4+2x2+9

Solution:

sqrt(9)=3
2·3=6

So here the middle term needs to be 6. Since it is 2 now, we need to add and subtract 4x2. Then we get

x4+2x2+9=x4+6x2+9-4x2=(x2+3)2-4x2=[(x2+3)+2x][(x2+3)-2x]=(x2+2x+3)(x2-2x+3).

Example 2:

x4-19x2+25

Solution:

Here if the middle coefficient were a -10, then it would be a perfect square, so add and subtract 9x2.

sqrt(25)=5
2·5=10

x4-19x2+25=x4-10x2+25-9x2=(x2-5)2-9x2=[(x2-5)+3x][(x2-5)-3x]=(x2+3x-5)(x2-3x-5).

Tips

  1. First look for a common factor.
  2. Always factor completely. This is like prime factoring.

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Transistors 101

Function

transistorsTransistors amplify current, for example they can be used to amplify the small output current from a logic IC so that it can operate a lamp, relay or other high current device. In many circuits a resistor is used to convert the changing current to a changing voltage, so the transistor is being used to amplify voltage.

A transistor may be used as a switch (either fully on with maximum current, or fully off with no current) and as an amplifier(always partly on).

The amount of current amplification is called the current gain, symbol hFE.
For further information please see the Transistor Circuits page.


Types of transistor

NPN and PNP transistor symbols
Transistor circuit symbols
There are two types of standard transistors, NPN and PNP, with different circuit symbols. The letters refer to the layers of semiconductor material used to make the transistor. Most transistors used today are NPN because this is the easiest type to make from silicon. If you are new to electronics it is best to start by learning how to use NPN transistors.

The leads are labelled base (B), collector (C) and emitter (E).
These terms refer to the internal operation of a transistor but they are not much help in understanding how a transistor is used, so just treat them as labels!

A Darlington pair is two transistors connected together to give a very high current gain.

In addition to standard (bipolar junction) transistors, there are field-effect transistors which are usually referred to as FETs. They have different circuit symbols and properties and they are not (yet) covered by this page.


Transistor leads
Transistor leads for some common case styles.

Connecting

Transistors have three leads which must be connected the correct way round. Please take care with this because a wrongly connected transistor may be damaged instantly when you switch on.

If you are lucky the orientation of the transistor will be clear from the PCB or stripboard layout diagram, otherwise you will need to refer to a supplier's catalogue to identify the leads.

The drawings on the right show the leads for some of the most common case styles.

Please note that transistor lead diagrams show the view from below with the leads towards you. This is the opposite of IC (chip) pin diagrams which show the view from above.

Please see below for a table showing the case styles of some common transistors.


Crocodile clip, photograph © Rapid Electronics
Crocodile clip
Photograph © Rapid Electronics.

Soldering

Transistors can be damaged by heat when soldering so if you are not an expert it is wise to use a heat sink clipped to the lead between the joint and the transistor body. A standard crocodile clip can be used as a heat sink.

Do not confuse this temporary heat sink with the permanent heat sink (described below) which may be required for a power transistor to prevent it overheating during operation.


Heat sink
Heat sink

Photograph © Rapid Electronics

Heat sinks

Waste heat is produced in transistors due to the current flowing through them. Heat sinks are needed for power transistors because they pass large currents. If you find that a transistor is becoming too hot to touch it certainly needs a heat sink! The heat sink helps to dissipate (remove) the heat by transferring it to the surrounding air.

For further information please see the Heat sinks page.


Testing a transistor

Transistors can be damaged by heat when soldering or by misuse in a circuit. If you suspect that a transistor may be damaged there are two easy ways to test it:

testing a transistor
Testing an NPN transistor

1. Testing with a multimeter

Use a multimeter or a simple tester (battery, resistor and LED) to check each pair of leads for conduction. Set a digital multimeter to diode test and an analogue multimeter to a low resistance range.

Test each pair of leads both ways (six tests in total):

  • The base-emitter (BE) junction should behave like a diode and conduct one way only.
  • The base-collector (BC) junction should behave like a diode and conduct one way only.
  • The collector-emitter (CE) should not conduct either way.
The diagram shows how the junctions behave in an NPN transistor. The diodes are reversed in a PNP transistor but the same test procedure can be used.

testing a transistor
A simple switching circuit
to test an NPN transistor

2. Testing in a simple switching circuit

Connect the transistor into the circuit shown on the right which uses the transistor as a switch. The supply voltage is not critical, anything between 5 and 12V is suitable. This circuit can be quickly built on breadboardfor example. Take care to include the 10kohm resistor in the base connection or you will destroy the transistor as you test it!

If the transistor is OK the LED should light when the switch is pressed and not light when the switch is released.

To test a PNP transistor use the same circuit but reverse the LED and the supply voltage.

Some multimeters have a 'transistor test' function which provides a known base current and measures the collector current so as to display the transistor's DC current gain hFE.


Transistor codes

There are three main series of transistor codes used in the UK:
  • Codes beginning with B (or A), for example BC108, BC478
    The first letter B is for silicon, A is for germanium (rarely used now). The second letter indicates the type; for example C means low power audio frequency; D means high power audio frequency; F means low power high frequency. The rest of the code identifies the particular transistor. There is no obvious logic to the numbering system. Sometimes a letter is added to the end (eg BC108C) to identify a special version of the main type, for example a higher current gain or a different case style. If a project specifies a higher gain version (BC108C) it must be used, but if the general code is given (BC108) any transistor with that code is suitable.
  • Codes beginning with TIP, for example TIP31A
    TIP refers to the manufacturer: Texas Instruments Power transistor. The letter at the end identifies versions with different voltage ratings.
  • Codes beginning with 2N, for example 2N3053
    The initial '2N' identifies the part as a transistor and the rest of the code identifies the particular transistor. There is no obvious logic to the numbering system.

Choosing a transistor

Most projects will specify a particular transistor, but if necessary you can usually substitute an equivalent transistor from the wide range available. The most important properties to look for are the maximum collector current IC and the current gain hFE. To make selection easier most suppliers group their transistors in categories determined either by their typical use or maximum power rating.

To make a final choice you will need to consult the tables of technical data which are normally provided in catalogues. They contain a great deal of useful information but they can be difficult to understand if you are not familiar with the abbreviations used. The table below shows the most important technical data for some popular transistors, tables in catalogues and reference books will usually show additional information but this is unlikely to be useful unless you are experienced. The quantities shown in the table are explained below.

NPN transistors
CodeStructureCase
style
IC
max.
VCE
max.
hFE
min.
Ptot
max.
Category
(typical use)
Possible
substitutes
BC107NPNTO18100mA45V110300mWAudio, low powerBC182 BC547
BC108NPNTO18100mA20V110300mWGeneral purpose, low powerBC108C BC183 BC548
BC108CNPNTO18100mA20V420600mWGeneral purpose, low power
BC109NPNTO18200mA20V200300mWAudio (low noise), low powerBC184 BC549
BC182NPNTO92C100mA50V100350mWGeneral purpose, low powerBC107 BC182L
BC182LNPNTO92A100mA50V100350mWGeneral purpose, low powerBC107 BC182
BC547BNPNTO92C100mA45V200500mWAudio, low powerBC107B
BC548BNPNTO92C100mA30V220500mWGeneral purpose, low powerBC108B
BC549BNPNTO92C100mA30V240625mWAudio (low noise), low powerBC109
2N3053NPNTO39700mA40V50500mWGeneral purpose, low powerBFY51
BFY51NPNTO391A30V40800mWGeneral purpose, medium powerBC639
BC639NPNTO92A1A80V40800mWGeneral purpose, medium powerBFY51
TIP29ANPNTO2201A60V4030WGeneral purpose, high power
TIP31ANPNTO2203A60V1040WGeneral purpose, high powerTIP31C TIP41A
TIP31CNPNTO2203A100V1040WGeneral purpose, high powerTIP31A TIP41A
TIP41ANPNTO2206A60V1565WGeneral purpose, high power
2N3055NPNTO315A60V20117WGeneral purpose, high power
Please note: the data in this table was compiled from several sources which are not entirely consistent! Most of the discrepancies are minor, but please consult information from your supplier if you require precise data.
PNP transistors
CodeStructureCase
style
IC
max.
VCE
max.
hFE
min.
Ptot
max.
Category
(typical use)
Possible
substitutes
BC177PNPTO18100mA45V125300mWAudio, low powerBC477
BC178PNPTO18200mA25V120600mWGeneral purpose, low powerBC478
BC179PNPTO18200mA20V180600mWAudio (low noise), low power
BC477PNPTO18150mA80V125360mWAudio, low powerBC177
BC478PNPTO18150mA40V125360mWGeneral purpose, low powerBC178
TIP32APNPTO2203A60V2540WGeneral purpose, high powerTIP32C
TIP32CPNPTO2203A100V1040WGeneral purpose, high powerTIP32A
Please note: the data in this table was compiled from several sources which are not entirely consistent! Most of the discrepancies are minor, but please consult information from your supplier if you require precise data.

StructureThis shows the type of transistor, NPN or PNP. The polarities of the two types are different, so if you are looking for a substitute it must be the same type.
Case styleThere is a diagram showing the leads for some of the most common case styles in the Connecting section above. This information is also available in suppliers' catalogues.
IC max.Maximum collector current.
VCE max.Maximum voltage across the collector-emitter junction.
You can ignore this rating in low voltage circuits.
hFEThis is the current gain (strictly the DC current gain). The guaranteed minimum value is given because the actual value varies from transistor to transistor - even for those of the same type! Note that current gain is just a number so it has no units.
The gain is often quoted at a particular collector current IC which is usually in the middle of the transistor's range, for example '100@20mA' means the gain is at least 100 at 20mA. Sometimes minimum and maximum values are given. Since the gain is roughly constant for various currents but it varies from transistor to transistor this detail is only really of interest to experts.
Why hFE? It is one of a whole series of parameters for transistors, each with their own symbol. There are too many to explain here.
Ptot max.Maximum total power which can be developed in the transistor, note that a heat sink will be required to achieve the maximum rating. This rating is important for transistors operating as amplifiers, the power is roughly IC × VCE. For transistors operating as switches the maximum collector current (IC max.) is more important.
CategoryThis shows the typical use for the transistor, it is a good starting point when looking for a substitute. Catalogues may have separate tables for different categories.
Possible substitutesThese are transistors with similar electrical properties which will be suitable substitutes in most circuits. However, they may have a different case style so you will need to take care when placing them on the circuit board.


Darlington pair

Darlington pairThis is two transistors connected together so that the amplified current from the first is amplified further by the second transistor. This gives the Darlington pair a very high current gain such as 10000. Darlington pairs are sold as complete packages containing the two transistors. They have three leads (B, C and E) which are equivalent to the leads of a standard individual transistor.

You can make up your own Darlington pair from two transistors.
For example:

  • For TR1 use BC548B with hFE1 = 220.
  • For TR2 use BC639 with hFE2 = 40.
The overall gain of this pair is hFE1 × hFE2 = 220 × 40 = 8800.
The pair's maximum collector current IC(max) is the same as TR2.

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