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I am very new here and almost just as new at learning about electronics. I am a musician and sound engineer first and foremost and I just recently have caught the circuits bug. My first big task has been to learn how a transistor amplifier works. I understand the concepts surrounding current and voltage gain, and I think I understand the basics around biasing the base and collector terminals.

What I just can't seem to grasp (and this may just be a very silly question so go easy on me) is how the transistor makes an amplified copy of the base AC signal. From what I can tell, the current never actually flows from the base to the output of a simple Class A transistor circuit. Maybe I just don't understand the concept of current and voltage well enough, but it seems crazy to me that the output has an amplified version of the complex AC input signal, when the current to the output really only flows from the DC bias, down through a resistor, to the output of the circuit and collector! It just seems like magic.

To be clear, I understand mathematically how it all works. I understand that if there is a negative current from the collector to the output junction, there is a gain in voltage. I guess I just don't understand the amplification process intuitively and I would love someone to break it down for me like the noob I am. Maybe I also am struggling with a lack of knowledge of how an AC signal flows through a circuit.

Thank you !!

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  • \$\begingroup\$ When you play with a friend on a see-saw, you can make the other person move, without ever going over to the other side of the see-saw and pushing on them. \$\endgroup\$
    – The Photon
    Commented Apr 9, 2020 at 17:44
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    \$\begingroup\$ Throw a snowball at a snowy mountain that the snowball starts an avalanche. Electrons instead of snow. The transistor has lots of snowy electrons. The ones flowing through base trigger an avalanche at the collector. \$\endgroup\$
    – scorpdaddy
    Commented Apr 9, 2020 at 17:57
  • \$\begingroup\$ If you are ready to take the next step beyond building from kits, but you never intend to become an actual electronics engineer, then I highly recommend this book: amazon.com/Art-Electronics-Paul-Horowitz-ebook/dp/B01BYJO2JU It will teach you just enough theory to be able to understand, and design your own simple transistor circuits. The intended audience was starving grad students in the hard sciences, who don't want to learn electronics design, but who need to invent simple circuits to control and interface with scientific experiments. \$\endgroup\$ Commented Apr 9, 2020 at 18:48

9 Answers 9

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From what I can tell, the current never actually flows from the base to the output of a simple Class A transistor circuit.

Understand that we have chosen that we are going to use the voltage, not the current, as the signal that carries information. It doesn't have to be this way but in most cases it is. Also, understand that current flows, voltage doesn't. Current is not the information carrier therefore the current does not need to be read, therefore current doesn't actually have to flow out of the output to be read. Voltage does not flow but does appears at the output so it can still be read.

Maybe I just don't understand the concept of current and voltage well enough, but it seems crazy to me that the output has an amplified version of the complex AC input signal, when the current to the output really only flows from the DC bias, down through a resistor

You never stated what exactly it was that seemed crazy to you here: whether it was because current doesn't flow out the output (which was roughly answered above) or whether it is because you have something coming from a DC-bias but have a AC output. I will just point out that the current you are referring to as a DC-bias isn't a constant. It does vary the way an AC signal would (that's how it produces the signal after all), but it is offset so these variations occur around a DC-bias.

I guess I just don't understand the amplification process intuitively.

Suppose I decided my signal (the information carrier) was a current, not the voltage. Now, suppose I wanted to convert this current signal into a voltage signal (so that it is in a form where the voltage is now the information carrier).

So I would use a current-to-voltage converter...in other words, a resistor. I run this current through a resistor and use the voltage that appears across it. If I choose a large resistor, the voltage produced is larger. If I choose a smaller resistor, the voltage produced is smaller.

Now, suppose that instead of my original information carrier (the signal) was not a current, but another voltage. What would happen if I somehow converted this voltage to a current, then ran that current through a resistor? It would produce another voltage of course. But what if chose the resistor such that the end voltage was actually a bigger voltage than what I started out with? I just amplified it. Similar idea.

The transistor in here is the "somehow" part that converts a voltage to a current. This would be the case if you use a MOSFET.

If you used a BJT, which converts current to current but have chosen the voltage to be your information carrier, then you have to use a base resistor to convert the voltage to a current.

In both cases, the current produced by the transistor is forced through a resistor to turn it back into a voltage with the resistor being sized to produce a larger voltage than what you started with.

Of course, this isn't nearly everything there is to it since the transistor has an amplification effect too (the current you put into a BJT base is not the equal to the current that is produced across collector-emitter). This allow power amplification, not just the voltage level. THis is contrast to something like a transformer. For example, a step-up transformer increases the voltage from input to output but does not increase the power. This is because it has no power source in addition to the signal source. All the power a transformer receives is from the signal and it can't just add power to the signal from nowhere.

What does power amplification really mean when you have chosen voltage as your output information carrier? What it means is that if you try to draw more current from the output, it won't distort the output voltage too much (it retains its information). Whereas if you try to draw more output current than input current from a step-up transformer when you are using voltage as the output signal, the output voltage will drop like a rock.

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  • \$\begingroup\$ Corrected some letter typos that completely changed the meaning of the word and sentence \$\endgroup\$
    – DKNguyen
    Commented Apr 9, 2020 at 20:21
  • \$\begingroup\$ this was very helpful in my overall understanding of voltage, current, power, and amplification. thank you!! \$\endgroup\$
    – Jake Z
    Commented Apr 13, 2020 at 21:41
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An device-physicists explained the bipolar transistor this way:

when electrons enter the base wires, the electrons proceed on to the base-silicon contact region, and once in the silicon will continue on, urged by small electric fields into the actual base-interaction region, located between the emitter and collector.

In the base-interaction region, charges are released from the emitter (emitted) with intention of annihilating the charges creeping in from the base wires.

BUT MOST OF THE EMITTED CHARGES MISS and overshoot and are gathered up in the collector region.

The ratio of those charges from the emitter that rach the collector (most of them), to the annihilation events, is the Power Gain, the BETA.

To have an amplifier responsive to even the smallest audio signal, the transistor is operated with input DC voltages and output DC currents, that provide low-distortion response. Your distortion will not be zero (the Vin/Out curve is after all an exponential), but for input voltages of a few milliVolts, the distortion will be about 5% or smaller.

Notice we are biasing (operating) the transistor with a significant WASTE of power, so the transistor gain-stage is always ready to respond; in the common-emitter circuit, a +1mV input fluctuation will give you some small (or large, depending on your design) negative-going change on the collector. And a -1mV input fluctuation will give you some small (or large) positive-going change on the collector.

Now you need to learn about load-lines, and you'll see why the polarity inversion occurs.

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The answer of the OP's fundamental question, "How does a transistor amplify?" is simple and surprising: The transistor does not amplify... it attenuates. Then what does "amplification" mean? How does the transistor make "an amplified copy of the base AC signal"?

The idea of the basic transistor amplifying stage (e.g., the so-called common-emitter stage) is extremely simple and intuitive. It can be thought as of a humble R1-R2 voltage divider but with a voltage-controlled transfer ratio R2/(R1 + R2). It is supplied by a powerful constant voltage source (the power supply) with many times higher than the input voltage. Either R1 or R2 (sometimes both) is replaced by a transistor acting as a "voltage-controlled resistor" (indeed, it is non-linear... but this is not so important when explaining the concept of amplification).

Thus the input voltage controls the resistance of R1 or R2, the transfer ratio of the voltage divider and accordingly, the part of the supply voltage appearing at the output of the voltage divider. In this way, it can vary from zero to the supply voltage... and we can think of it as of an "amplified copy of the input voltage".

In this arrangement, the transistor decreases (attenuates) the supply voltage... but since it is many times higher than the input voltage, the ratio between the variations of the attenuated (output) voltage and controlling (input) voltage can be bigger than one. This trick is named "amplification".

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    \$\begingroup\$ Yes - I think, this is a nice and descriptive way for explaining the the general "concept of amplification" for all kinds of voltage amplifiers. \$\endgroup\$
    – LvW
    Commented Apr 10, 2020 at 9:25
  • \$\begingroup\$ A very absurd way of amplifying... A low-energy input source controls a high-energy power source by throwing away some of its energy... but still the rest of energy is greater than the input energy... and we say that the energy is "amplified"... \$\endgroup\$ Commented Apr 10, 2020 at 9:45
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Forget it Jake, it's transistor town.

But if you really want to dive into how a transistor "transists", try here: https://www.allaboutcircuits.com/textbook/semiconductors/chpt-2/bipolar-junction-transistors/

The short answer is, transistors (the bipolar junction kind, or BJT), through the magic of semiconductor physics, amplify current. A BJT wired as an amplifier takes a small change in base-to-emitter current and translates to a large change in collector-to-emitter current. That’s the whole thing, right there: these changes (AC signals) ride on top of the transistor bias (DC currents).

The current gain, or beta (or hFE in the datasheets) for a typical transistor is large enough that its input can be treated as a high-ish impedance, and the output as low impedance.

More about that here: https://www.electronics-notes.com/articles/electronic_components/transistor/current-gain-hfe-beta.php

So a properly-biased BJT base can be AC-coupled to the source, like you would for a line-level amp. The line-in AC voltage couples to the base bias current as variations that follow the input. The current gain of the transistor will amplify these variations to larger ones between collector and emitter.

Finally, understand that base current only flows from base to emitter. Collector current likewise only flows from collector to emitter. So the emitter current is the sum of base-emitter and collector-emitter current. The key is, collector-emitter current is much larger than the base current, so the output current variation is dominated by it.

More here: https://www.electronics-tutorials.ws/transistor/tran_2.html

A related device, the field effect transistor, or FET, has a nearly infinite input impedance. FETs are used for example to amplify electret microphones (including the venerable Telefunken U47), converting mic's high impedance to low impedance suitable for driving down a cable. They work more like vacuum tubes, being voltage controlled.

More here: https://www.electronics-tutorials.ws/amplifier/amp_3.html

And finally, to help visualize these, try playing around with this stuff here: http://www.falstad.com/circuit/

There's demos for the BJT and FET available, and you can edit them and see the effects.

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This comes late however. A transistor amplifier uses a small current, to vary a large current, and gives that large current all the characteristics of the small current. When a small current flows across the base emitter junction of a transistor, it can cause a greater current to flow from collector to emitter, with same characteristic as the small current.

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"From what I can tell, the current never actually flows from the base to the output of a simple Class A transistor circuit"

Jake - I am afraid, your problem is that you can think only in terms of CURRENT. But you should not forget the role of the input VOLTAGE.

In fact, it is the input voltage between the base and the emitter node which controls and determines the current between the collector and the emitter. At the same time, there is a small current into the base (which is a certain and - more or less - fixed part of the colector current). Buth this base current is something like a secondary effect (unwanted, but cannot be avoided). During design/calculation this current can and is , in many cases, neglected.

Between the input voltage (Vbe) and the output current Ic there is an exponential relationship which is known as Shockley`s equation Ic=Io[exp(Vbe/Vt)-1]

Comment: Jake, it is my impression that you want to understand how the BJT works, right?

Therefore this comment. I know that in some answers it will be claimed that the BJT would be a current-controlled device. That means: The base current would control and determine the collector current. But this is wrong - and it cannot be prooved. It is just a claim - that`s all.

This comes from the fact that in some textbooks such a current-control function is mentioned - mostly in textbooks which do not go into the details. They think that the formula Ic=beta*Ib would say anything about cause and effect. But this is not the case. It is just another form of Ib=Ic/beta. Nobody can EXPLAIN to you HOW a small base current Ib could control a much larger collector current Ic. On the other hand, there are many explanations/verifications for voltage control. I can give you serious references.

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  • \$\begingroup\$ The current-controlled view is a very handy thing for calculation, though. \$\endgroup\$
    – Hearth
    Commented Apr 9, 2020 at 20:41
  • \$\begingroup\$ Yes - I agree. For calculation purposes (even for most design tasks) it works. No doubt about it. In electronics, there are some other areas where we use - with success - a model that is not in full accordance with the physical "truth". But - for my opinion - a good engineer should be aware of this difference. \$\endgroup\$
    – LvW
    Commented Apr 10, 2020 at 9:12
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    \$\begingroup\$ Sredni Vashtar...analogsystemsrf has mentioned that both currents (Ib and Ie) do exist. I am really surprised that you consider this as a kind of evidence for current-control. Do you really neglect the primary role of the E-field in the base region? On the other hand - I know your position since a long time...it seems you are not open for some corrections..... \$\endgroup\$
    – LvW
    Commented Apr 14, 2020 at 9:48
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    \$\begingroup\$ Thank you - you are a very polite man. No further comment. Oh sorry - still one recommendation: Consult some good and serious books. \$\endgroup\$
    – LvW
    Commented Apr 15, 2020 at 12:48
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    \$\begingroup\$ Pwerhaps the above linked contribution from wbeaty helps to revise you opinion? \$\endgroup\$
    – LvW
    Commented Apr 19, 2020 at 9:22
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You can use the transistor as a current amplifier with unity gain or a voltage-controlled active resistor load or a current-ratio controlled switch or a current-ratio controlled current-source/sink depending on the configuration.

Basics to your understanding before an explanation may be given:

Necessary conditions for transistor operation

  • proper base-emitter biasing and proper collector-emitter polarity.
    • Bipolar junction transistors require a base current in the proper direction to inject minority carriers into the NPN base layer
    • PNP requires "hole" injection from emitter to base with a positive emitter
    • NPN requires "electron" injection from emitter to base with a negative emitter
    • the B-E jcn must be heavily doped while the B-C jcn. must be very lightly doped.
  • Both base and collector currents join at the emitter terminal with a linear gain of about 100 unless Vce is saturated (<<1V) then it drops rapidly towards 10% of the max linear gain.

  • The base-emitter voltage follows the log curve for a low impedance diode

  • the collector-emitter curve follows a high impedance current source controlled by Vbe
  • the input voltage to output current may more linear by adding an emitter resistor
  • even greater linearity is achieved with negative feedback from collector to base with a known source impedance ratio and excess open-loop voltage gain at some balanced bias voltage.
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It's like a pantograph producing an enlarged identical line drawing while tracing a smaller one.

enter image description here

Courtesy Wikipedia

https://en.wikipedia.org/wiki/Pantograph

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  • \$\begingroup\$ Nice analogy... but it is more appropriate to explain in this intuitive way the operation of the step-up transformer. I hope you will guess why... \$\endgroup\$ Commented Apr 9, 2020 at 20:22
  • \$\begingroup\$ Thank you, Circuit fantasist. I give up! \$\endgroup\$
    – vu2nan
    Commented Apr 10, 2020 at 8:31
  • \$\begingroup\$ @vu2nan, I'll help you a little bit:) Is a lever a mechanical amplifier? \$\endgroup\$ Commented Apr 10, 2020 at 9:27
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    \$\begingroup\$ @Circuit fantasist, I got it now! Thank you very much. \$\endgroup\$
    – vu2nan
    Commented Apr 10, 2020 at 10:27
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These answers have all been very interesting and helpful. I didn't expect this much info from the community on my first post! I guess everybody likes talking about transistors though...

My questions has pretty much been answered. My main takeaway is I really had a fundamental misunderstanding of what amplification is and what it means to "carry a signal".

It was my impression that a current is what literally "carries" the sound wave from one end of a circuit to another because in my mind I conflated current with electrons and, to me, electrons are the only moving physical objects in the circuit. I also had a false impression somewhere deep down that voltage is equivalent to overall loudness. For that reason, it was really the fact that current flowed into the collector of the transistor that was freaking me out. (seeing as I thought current literally carried the sound, this was confusing me).

I'm not sure where I got these ideas but if any of you are teachers and it's helpful for you to know that this is what I thought coming in, there you go.

I now realize that current is simply the rate at which electrons are moving at any given point. I also realize that loudness at the output of the amplifier is dependent on current, voltage, and of course power.

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  • \$\begingroup\$ If you haven't already, try the Falstad applet to help visualize how the BJT works in a circuit. falstad.com/circuit \$\endgroup\$ Commented Apr 14, 2020 at 1:20
  • \$\begingroup\$ this is great!! thank you. \$\endgroup\$
    – Jake Z
    Commented Apr 14, 2020 at 18:25

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