Signal in transistor

Question

In a common emitter NPN transistor, we learn that it works as an amplifier. What we are shown commonly in books and everywhere is that the transistor is amplifying the voltage or current. Right after that they say "This is how a small sound is converted to a large sound."

What does this sound have to do with current or voltage? Do they mean the current or voltage as sound or both?

Moreover,bin the input side of the transistor, apart from the biasing voltage,bwe see that the electric signal that we want to amplify is shown as an AC voltage. Why is the signal converted to voltage and not anything else?

I read this in high school books but I understand how the amplification of voltage or current works but I never really understood how the amplification of current or voltage means amplification of electrical signal or even why the signal we give as input is given by voltage. Please enlighten me.

Answer 1

The transistor is said to have a particular gain, or beta. This might be 10, 100, or 1000.

What this means is, if you put some (tiny) amount of current into the base, then gain times as much current goes in/out of the collector.

If that current is static (DC) then for a transistor with beta of 10, 10µA into the base will cause about 100µA to flow at collector. Hence, it is amplifying.

AC is just DC that changes over time. In the case of a microphone, it may supply 0-10µA to the base. This value is constantly changing. And if a transistor is used with a beta of 100, then 1mA will be (constantly changing) at the collector as well - 100x as much current.

If you routed that 10µA changing signal to a headphone speaker directly, it would be inaudible because that is not enough current to cause the speaker to move. A 1mA signal however, would be clearly heard.

Answer 2

We use electronics to amplify sound because we don't have a really good way of amplifying sound directly.

To amplify you have to add energy to a signal, how would you do this with sound?

Think about a megaphone. There is the old type that are just a cone with a hole in the end that you speak into. You can use one of these to make your voice sound louder, but this is just concentrating the existing sound energy in one direction, it is not adding any additional energy. This is not amplification.

Now think about a modern amplified megaphone. It uses a microphone to generate an electrical signal representing your voice. It then uses an amplifier to add energy to that signal. The energy comes from a battery. This is amplification. The amplified signal can now be applied to a speaker which uses the electrical energy to create sound energy.

So you've taken a 'small sound', converted it to an electrical signal, increased the power of that electrical signal using electronics and a battery, and converted it back to a 'large sound'.

Answer 3

You've already received some useful answers. I'll take a slightly different slant at the question.

First off, I want to emphasize the point that the word Amplification is just a word. We use words to communicate. But nature itself doesn't understand the word. It just does physics stuff. So please don't imagine that the term means anything to nature or reality. It doesn't. It just has meaning for us.

Usually, we use the term (as opposed to the anonym retardation) for any situation where something important to us is made ever so much moreso (a phrase I take from the Homer Price series of kids books.)

For example, ocean waves are amplified as they approach the shore. As the ocean water shallows, the velocity of the arriving wave decreases and the amplitude of the arriving wave grows. Since we don't care as much about the velocity (which is retarding) and do care a lot more about the height (which is being amplified) then we might very well call that wave amplification due to shoaling. No electronics involved at all. And, from a different perspective, no energy was added to the system, either.

So the meaning of the word amplification has to do with what's important to a human in some specific, given situation.

It is a bit of a subjective human term, though.

There's much else to say. Your question is quite broad. But I'll hold short. Just one last thought. There's a lot of thermal noise in the world. And it is important to have signals within an electronic device that are large with respect to the signal-referred thermal noise that all parts possess so that the noise interferes less-so than it might, otherwise.

Answer 4

What does this sound have to do with current or voltage? Do they mean the current or voltage as sound or both?

The most convenient way to amplify sound is electronically. A microphone converts sound waves to a varying voltage or current, then that signal is amplified, then the result is applied to a speaker which converts the varying voltage into sound waves.

This is also how sound is stored and transmitted -- the microphone makes a varying voltage, that variation is captured somehow and put away on a shelf (or on a drive, these days), then the variation is retrieved, turned into a voltage, amplified, and applied to a speaker.

I read this in high school books but I understand how the amplification of voltage or current works but I never really understood how the amplification of current or voltage means amplification of electrical signal.

In it's purest theoretical form, a signal is just a "something" that varies in time. When you're playing with that theory, signals are often just expressed as functions of time, but with names that look like variables, i.e. \$x(t)\$ denotes a signal.

A signal can be impressed upon a voltage. I.e., if you have that pure theoretical signal \$x(t)\$, you might say that your voltage signal is \$v(t) = \left(5 \mathrm V \right) x(t)\$, where multiplying \$x(t)\$ by a constant \$5 \mathrm V\$ makes it into a quantity in the real world.

or even why the signal we give as input is given by voltage. Please enlighten me.

It isn't always. For most electrical circuits, using voltage to carry signals is the easiest and most convenient way. Electronic signals can also be carried as a current. In a more general sense, signals can be carried on almost other quantity. Instances of this are pressure waves in air or water -- i.e. sound waves, or the intensity of light in a fiber-optic cable, or the movement of some mechanical part.

Answer 5

I'll try to enlighten you, but it's not clear to me where your understanding fails. So I apologise for the length of this answer, but I cannot assume what you know and what you don't.

It's not easy to understand what voltage is, or what current is, without a really good understanding of Kirchhoff's laws, and Ohm's law. So that's where I would send you first. However those principles are necessarily mathematical in nature, and I understand why beginners are reluctant to - well - begin there.

For an understanding of voltage and current without first visiting Kirchhoff, perhaps it's sufficient to realise that both are different manifestations of the same underlying phenomenon. When someone says "apply an AC voltage to point X", they mean to somehow cause the potential there to change. (I prefer the term "potential" when the context is "voltage at some point", since that voltage is always measured or quoted relative to the voltage at some other point. The term "voltage", to me at least, implies some difference in potential between two points. This is purely personal preference, and some use the terms "potential" and "voltage" interchangeably)

When somebody says "input", they are almost certainly referring to a point in a circuit where we impose some kind of change, and that change is always going to be described in terms of a fluctuation (or perturbation) of either current or potential there. Both interpretations are equivalent. By somehow forcefully raising or lowering the potential somewhere (I don't say here how that may be achieved), necessarily there will be a corresponding change in current flowing into (or through) that point. Conversely, by injecting or sucking electric current into or out of some point in a circuit, this perturbation will result in a corresponding change in the potential at that point.

By analogy, you only need to imagine a section of pipe, through which water may flow, and see that if you somehow raise the water pressure (potential) at one end of the pipe, while keeping the pressure at the other end unchanged, you will have increased the difference in pressure (voltage) between the two ends, and consequently the rate of flow of water (current) through he pipe must increase. The two phenomena (flow rate and pressure) are inseparable, and are only the measurable manifestations of the same underlying change.

You could argue that any means by which you bring about an increase in water flow rate will also result in a measurable change in the pressure difference between the two ends, and this becomes a chicken-and-egg type scenario.

Which is it? A change in current causes a corresponding change in voltage, or a change in voltage causes current to change? To an engineer, the two are one and the same underlying phenomenon, the only difference is in how we measure that change. For a water system, you could use pressure gauges to measure the change in pressure difference, and infer from this information what the accompanying change in flow rate must have been, or you could use a flow meter to determine that flow rate changed, and infer the corresponding change in pressure difference that must also have accompanied it.

For completeness, I will add that the relationship between change in current and corresponding change in voltage, is called "impedance". In the water pipe analogy, impedance would be a measure of how much the pipe restricts flow. A fatter pipe has less impedance to flow than a narrow pipe.

To address your confusion about how sound is mixed up in all this, it isn't. An "amplifier" consisting of electronic components has no concept of sound, only current and voltage. That is, an electronic amplifier can only respond to perturbations of current/voltage somewhere in a circuit (its "input") and respond with a much greater proportional change in current/voltage somewhere else (the "output").

As it is, an amplifier can't do anything useful for humans, because humans deal with phenomena such as sound, light, heat and so on, "real life" stuff. To make our amplifier useful to humans, we must take some real-life phenomenon, convert it into a current or voltage, have the amplifier amplify that, and then convert the larger current/voltage at its output back into a real-life form.

Devices that perform this "conversion" are called transducers. The transducers useful in the field of sound are the "microphone" and the "loudspeaker". The microphone produces a voltage/current that varies in proportion the position of its diaphragm. In other word, that's a voltage/current waveform that is a facsimile of the vibrations of the microphone diaphragm caused by incident sound waves. The loudspeaker does the exact opposite, by moving a diaphragm in response to voltage across/current through it, resulting in sound waves.

By placing an electronic amplifier between them, you are able to produce large movements of a large diaphragm, in proportion to tiny movements in a tiny diaphragm, and in this sense you have "amplified the sound". The ensemble could be called an "audio amplifier".

You can apply the same principle to any physical phenomenon, such as light or motion or heat, phenomena that humans deal with. All you need is a pair of appropriate transducers to convert between phenomena in the realm of the physical world, and some electrical facsimile which we call "signals", and an amplifier to "process" those signals in the electrical realm.

A defining characteristic of amplifiers is their ability to provide more energy at their output than they receive at their input. In this above example, the output sound is much louder than the input sound, implying that more energy came out of the microphone/amplifier/loudspeaker combination than went in. There's no such thing as a free lunch, though. The physical law "conservation of energy" means that you can never have more energy out than in, for any system, and the natural question that arises should be how does our circuit here manage to achieve this impossible feat?

The answer is that the amplifier draws whatever energy it needs to operate the loudspeaker from an external source of energy, like a battery or mains power supply. Essentially, the amplifier produces a copy of its input at its output, but at a much greater "energy level" (I hate that phrase, because it raise more questions, and is a grossly inadequate description of what's going on, but it will have to do for now), using energy derived from some source like a battery. The energy in the battery had to be supplied ahead of time (by charging it), in order for that energy to be made available, at a later date, to the amplifier.

You alluded to "AC" in your question, and I feel that I should just mention that AC is nothing more than perturbations of current/voltage somewhere in a system. And as I explained before, current and voltage are intimately related, and fundamentally inseparable. It's really just a question of which phenomenon, pressure or flow, voltage or current, more appropriately describes a signal in the context of the application. For various reasons which I won't go into here, voltage usually wins the vote, but allow me one last stab at showing you how the two are only superficially different:

^{simulate this circuit – Schematic created using CircuitLab}

Here, I've connected a resistor R1 (with impedance 100Ω) directly across the output of a microphone MIC1. I am measuring the current \$I\$ flowing around the loop with ammeter AM1, and Kirchhoff's Current Law tells me that the current through the microphone and the current through the resistor must be the same.

Also I measure the voltage across the resistor, which must be the same as the voltage across the microphone, according to Kirchhoff's Voltage Law, as they are connected in parallel.

Lastly, Ohms law tells me the relationship between the two, and illustrates how impedance (resistance, in this case) enters the picture:

$$ \frac{V}{I} = R = 100\Omega $$

Clearly, if you tell me the voltage, I can infer the current, and tell you that value. If you tell me the current, I can infer the voltage.

This is a contrived and trivial example which is unlikely to appear in any real world circuit, but it does illustrate that when somebody says "the input is an AC voltage", whatever currents flow can be derived from what you know about the impedances present there and the voltage applied there. Conversely, when you are told "we inject a current into the input", then with knowledge of the impedances present you can infer the resulting changes in potential.

So when it is stated that something is "converted to a voltage", you immediately know that it's also converted to a current, the value of which will depend on whatever impedance exists between that potential difference. It is incorrect to say "it's converted to a voltage, and not something else".

If however, the microphone produces some voltage, but that voltage is not applied across an impedance of some sort, then no current can flow, by Ohm's law. In this sense it's probably more appropriate to refer to the microphone's output signal in terms of voltage, instead of current. Often voltage is the most appropriate descriptor in some given context. When there's an impedance present to conduct current, though, and that impedance is carefully chosen to get as much energy as possible out of the microphone's puny signal, then it may be more appropriate to talk about about the current produced by the microphone. After all, when we talk about power and energy, this needs information about both current and voltage, as made clear by the power equation \$P = I \times V\$. It's all about context. What is the most appropriate phenomenon to use to describe the nature and behaviour of something?