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2003 Paperback New Book New and in stock. 8/28/2003. *****PLEASE NOTE: This item is shipping from an authorized seller in Europe. In the event that a return is necessary, you
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More About This Textbook
Overview
This book enables those with a limited knowledge of the field to understand both the theory and practice of valve audio amplifier design, such that they can analyse and modify circuits, and build or restore an amplifier. Design principles and construction techniques are provided so readers can devise and build from scratch, designs that actually work.
This is a completely uptodate practical guide for people working with valves. Morgan Jones takes the reader through each step in the process of design, starting with a brief review of electronic fundamentals relevant to valve amplifiers, simple stages, compound stages, linking stages together, and finally, complete designs. Practical aspects, including safety, are addressed throughout.
Editorial Reviews
Booknews
For readers with a limited knowledge of the field, Jones, an exBBC engineer, examines the theory and practice of designing valve audio amplifiers to the level that they can analyze and modify circuits and build or restore an amplifier. He walks through each step in the process of design, beginning with a brief review of electronic fundamentals. The first edition was published in 1995, and the second includes several new topical sections and example applications. Annotation c. Book News, Inc., Portland, OR (booknews.com)From the Publisher
"Jampacked with theory, circuit analysis, and DIY basics, it will walk you through all stages of design so that you can create your own wonders. Jones is an exBBC engineer with a cool writing style and you'll find it a nopain education."HiFi News and Record Review
"Valve Amplifiers is an extremely wellwritten book, containing a wealth of information that all audio designers and builders will find useful."
Glass Audio
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Read an Excerpt
Valve Amplifiers
By Morgan Jones
Newnes
Copyright © 2012 Elsevier Ltd.All right reserved.
ISBN: 9780080966410
Chapter One
Circuit AnalysisIn order to look at the interesting business of designing and building valve amplifiers, we need some knowledge of electronics funmentals. Unfortunately, fundamentals are not terribly interesting, and to cover them fully would consume the entire book. Ruthless pruning is, therefore, necessary to condense what is needed in one chapter.
It is thus with deep sorrow that the author has had to forsaken complex numbers and vectors, whilst the omission of differential calculus is a particularly poignant loss. All that is left is ordinary algebra, and although there are lots of equations, they are timid, miserable creatures and quite defenceless.
If you are comfortable with basic electronic terms and techniques, then please feel free to go directly to Chapter 2, where valves appear.
MATHEMATICAL SYMBOLS
Unavoidably, a number of mathematical symbols are used, some of which you may have forgotten, or perhaps not previously met:
a [equivalent to] b a is totally equivalent to b
a = b a equals b
a [approximately equal to] b a is approximately equal to b
a [varies] b a is proportional to b
a [not equal to] b a is not equal to b
a > b a is greater than b
a < b a is less than b
a ≥ b a is greater than, or equal to, b
a ≤ b a is less than, or equal to, b
As with the = and [not equal to] symbols, the four preceding symbols can have a slash through them to negate their meaning (a [contains as member] b, a is not less than b).
√a the number which when multiplied by itself is equal to a (square root)
a^{n} a multiplied by itself n times. a^{4} = a × a × a × a (a to the power n)
± plus or minus
∞ infinity
° degree, either of temperature (°C), or of an angle (360° in a circle)
 parallel, either parallel lines, or an electrical parallel connection
Δ a small change in the associated value, such as ΔV_{gk}.
ELECTRONS AND DEFINITIONS
Electrons are charged particles. Charged objects are attracted to other charged particles or objects. A practical demonstration of this is to take a balloon, rub it briskly against a jumper and then place the rubbed face against a wall. Let it go. The balloon remains stuck to the wall. This is because we have charged the balloon, and so there is an attractive force between it and the wall. (Although the wall was initially uncharged, placing the balloon on the wall induced a charge.)
Charged objects come in two forms: negative and positive. Unlike charges attract, and like charges repel. Electrons are negative and positrons are positive, but whilst electrons are stable in our universe, positrons encounter an electron almost immediately after production, resulting in mutual annihilation and a pair of 511 keV gamma rays.
If we don't have ready access to positrons, how can we have a positively charged object? Suppose we had an object that was negatively charged, because it had 2,000 electrons clustered on its surface. If we had another, similar, object that only had 1,000 electrons on its surface, then we would say that the first object was more negatively charged than the second, but as we can't count how many electrons we have, we might just as easily have said that the second object was more positively charged than the first. It's just a matter of which way you look at it.
To charge our balloon, we had to do some work and use energy. We had to overcome friction when rubbing the balloon against the woollen jumper. In the process, electrons were moved from one surface to the other. Therefore, one object (the balloon) has acquired an excess of electrons and is negatively charged, whilst the other object (woollen jumper) has lost the same number of electrons and is positively charged.
The balloon would, therefore, stick to the jumper. Or would it? Certainly it will be attracted to the jumper, but what happens when we place the two in contact? The balloon does not stick. This is because the fibres of the jumper were able to touch the whole of the charged area on the balloon, and the electrons were so attracted to the jumper that they moved back onto the jumper, thus neutralising the charge.
At this point, we can discard vague talk of balloons and jumpers because we have just observed electron flow.
An electron is very small, and doesn't have much of a charge, so we need a more practical unit for defining charge. That practical unit is the coulomb (C). We could now say that 1 C of charge had flowed between one point and another, which would be equivalent to saying that approximately 6,240,000,000,000,000,000 electrons had passed, but much handier.
Simply being able to say that a large number of electrons had flowed past a given point is not in itself very helpful. We might say that a billion cars have travelled down a particular section of motorway since it was built, but if you were planning a journey down that motorway, you would want to know the flow of cars per hour through that section.
Similarly in electronics, we are not concerned with the total flow of electrons since the dawn of time, but we do want to know about electron flow at any given instant. Thus, we could define the flow as the number of coulombs of charge that flowed past a point in one second. This is still rather longwinded, and we will abbreviate yet further.
We will call the flow of electrons current, and as the coulomb/second is unwieldy, it will be redefined as a new unit, the ampere (A). Because the ampere is such a useful unit, the definition linking current and charge is usually stated in the following form.
This is still rather unwieldy, so symbols are assigned to the various units: charge has symbol Q, current I and time t.
This is a very useful equation, and we will meet it again when we look at capacitors (which store charge).
Meanwhile, current has been flowing, but why did it flow? If we are going to move electrons from one place to another, we need a force to cause this movement. This force is known as the electro motive force (EMF). Current continues to flow whilst this force is applied, and it flows from a higher potential to a lower potential.
If two points are at the same potential, no current can flow between them. What is important is the potential difference (pd).
A potential difference causes a current to flow between two points. As this is a new property, we need a unit, a symbol and a definition to describe it. We mentioned work being done in charging the balloon, and in its very precise and physical sense, this is how we can define potential difference, but first, we must define work.
This very physical interpretation of work can be understood easily once we realise that it means that one joule of work would be done by moving one kilogramme a distance of one metre in one second. Since charge is directly related to the mass of electrons moved, the physical definition of work can be modified to define the force that causes the movement of charge.
Unsurprisingly, because it causes the motion of electrons, the force is called the ElectroMotive Force, and it is measured in volts.
The concept of work is important because work can be done only by the expenditure of energy, which is, therefore, also expressed in joules.
work done (joules) = energy expended (joules)
W = E
In our specialised sense, doing work means moving charge (electrons) to make currents flow.
Batteries and Lamps
If we want to make a current flow, we need a circuit. A circuit is exactly that a loop or path through which a current can flow, from its starting point all the way round the circuit, to return to its starting point. Break the circuit, and the current ceases to flow.
The simplest circuit that we might imagine is a battery connected to an incandescent lamp via a switch. We open the switch to stop the current flow (open circuit) and close it to light the lamp. Meanwhile, our helpful friend (who has been watching all this) leans over and drops a thick piece of copper across the battery terminals, causing a short circuit.
The lamp goes out. Why?
Ohm's Law
To answer the last question, we need some property that defines how much current flows. That property is resistance, so we need another definition, units and a symbol.
This is actually a simplified statement of Ohm's law, rather than a strict definition of resistance, but we don't need to worry too much about that.
We can rearrange the previous equation to make I or R the subject.
These are incredibly powerful equations and should be committed to memory.
The circuit shown in Figure 1.1 is switched on, and a current of 0.25 A flows. What is the resistance of the lamp?
R = V/I = 240/0.25 = 960 Ω
Now this might seem like a trivial example, since we could easily have measured the resistance of the lamp to 3½ significant figures using our shiny, new, digital multimeter. But could we? The hot resistance of an incandescent lamp is very different from its cold resistance; in the example above, the cold resistance was 80 Ω.
We could now work the other way and ask how much current would flow through an 80 Ω resistor connected to 240 V.
I = V/R = 240/80 = 3 A
Incidentally, this is why incandescent lamps are most likely to fail at switchon. The high initial current that flows before the filament has warmed up and increased its resistance stresses the weakest parts of the filament, they become so hot that they vaporise, and the lamp blows.
Power
In the previous example, we looked at an incandescent lamp and rated it by the current that flowed through it when connected to a 240 V battery. But we all know that lamps are rated in watts, so there must be some connection between the two.
This may not seem to be the most useful of definitions, and, indeed, it is not, but by combining it with some earlier equations:
W = QV
So:
P = QV/t
But:
Q = It
So:
P = IVt/t
We obtain:
P = IV
This is a fundamental equation of equal importance to Ohm's law. Substituting the Ohm's law equations into this yields:
P = V^{2}/R
P = I^{2}/R
We can now use these equations to calculate the power rating of our lamp. Since it drew 0.25 A when fed from 240 V, and had a hot resistance of 960 , we can use any of the three equations.
Using:
P = V^{2}/R
P = 240^{2}/960
P = 60 W
It will probably not have escaped your notice that this lamp looks suspiciously like an AC mains lamp, and that the battery was rather large. We will return to this later.
Kirchhoff's Laws
There are two of these: a current law and a voltage law. They are both very simple and, at the same time, very powerful.
The current law states:
The algebraic sum of the currents flowing into, and out of, a node is equal to zero.
0 = I_{1} + I_{2} + I_{3} ...
What it says in a more relaxed form is that what goes in, comes out. If we have 10 A going into a node, or junction, then that much current must also leave that junction – it might not all come out on one wire, but it must all come out. A conservation of current, if you like (see Figure 1.2).
current flowing into the node: I_{1} = 10 A
currents leaving the node: I_{2} = 4 A
I_{3} = 6 A
total current leaving the node: I_{total} = 10 A
From the point of view of the node, the currents leaving the node are flowing in the opposite direction to the current flowing into the node, so we must give them a minus sign before plugging them into the equation.
0 = I_{1} + I_{2} + I_{3} = 10 A + (4 A) + (6 A) = 10  4  6
This may have seemed pedantic, since it was obvious from the diagram that the incoming currents equalled the outgoing currents, but you may need to find a current when you do not even know the direction in which it is flowing. Using this convention forces the correct answer!
It is vital to make sure that your signs are correct.
The voltage law states:
The algebraic sum of the EMFs and potential differences acting around any loop is equal to zero.
This law draws a very definite distinction between EMFs and potential differences. EMFs are the sources of electrical energy (such as batteries), whereas potential differences are the voltages dropped across components. Another way of stating the law is to say that the algebraic sum of the EMFs must equal the algebraic sum of the potential drops around the loop. Again, you could consider this to be a conservation of voltage (see Figure 1.3).
Resistors in Series and Parallel
If we had a network of resistors, we might want to know what the total resistance was between terminals A and B (see Figure 1.4).
(Continues...)
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