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I know there are lots of videos, documents on the internet and I can say that I've already looked most of them. However, I cannot get the insight of the working principle, amplification operation of these amplifiers. I know basics of transistors and their characteristics. However I am stuck with amplifying action.

I guess someone may come up with a nice and neat explatory answer.

Thanks in advance.

muyustan
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    There is such a vast amount of data on transistor gain and common emitter vs. common base designs that I am not sure what you missed. Across many sources you will get the abstract and the technical reason why they have gain. They drown you in working principals, so what are you missing? –  Dec 01 '19 at 22:14
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    Try and write a short explanation of the working principle into your question. That will give us something to work with / against! – Transistor Dec 01 '19 at 22:35
  • Are you asking about common emitter amplifiers specifically, or transistors in general? – Aloha Dec 02 '19 at 07:41
  • @PNDA not specific to CE, I chose it because it is a common topology and actually used to amplify voltage, I thought it would be easier. The answers below were kind of the answer type I was hoping to get. But, still open to new answers. – muyustan Dec 02 '19 at 08:09
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    Here you can read the way I see it https://electronics.stackexchange.com/questions/355899/how-is-possible-that-with-same-ibase-there-is-more-than-one-vce/355955#355955 – G36 Dec 02 '19 at 13:56
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    @muyustan...A short comment from my side to the "mystic" quantity re (as extensively discussed below the contribution from Jonk) : The quantity re is always present and - for my opinion - it should not be supressed (re< – LvW Dec 03 '19 at 09:55

8 Answers8

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Here is a very simplistic intuitive explanation of this most widespread transistor circuit:

Think of the transistor as a "voltage-controlled resistor" (with the proviso that it is non-linear)... something like a rheostat whose slider is driven by the input voltage. The input of this device (where voltage is applied) is the base-emitter junction, and its output (where resistance appears) is the collector-emitter junction. Thus, we may consider the "complex" 3-terminal transistor as composed of two simpler 2-terminal elements. Now it remains to decide where to apply the input voltage and where to take the output voltage from.

If we do not know about the principle of the common ground in circuitry, we are happy people... and we simply apply the input voltage to the base-emitter junction ("+" to the base, "-" to the emitter). To take the output voltage, we make a "voltage divider" by connecting an additional (collector) resistor in series to the transistor output. We supply this network and take the voltage drop across one of them as the output voltage.

However, we soon learn that (for a number of reasons) in circuitry devices are connected by one of their terminals to a common reference point called “ground”... and we begin looking for a solution to the problem.

We are inventive enough to present the input (differential) voltage as a difference between two input single-ended (grounded) voltages applied to the base and emitter... and thus we solve the problem with the common ground. Now we can change the base voltage as an input while keeping the emitter voltage constant... or change the emitter voltage as input by keeping the base voltage constant... we can even change both... but this is another story... With respect to the voltage change, the constant voltage behaves as an (AC) ground; hence the name “common emitter” and “common base”.

In the common emitter stage we keep the emitter voltage constant, particularly zero (grounded emitter) and change the base voltage as an input.

Circuit fantasist
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You can think of it as a current-controlled rheostat. There is a diode between the base and the emitter, and an ammeter in series with the diode that measures IB. There is a rheostat between the collector and the emitter, with another ammeter in series for measuring IC. And then there is the magic: a small demon who continuously reads both currents, and adjusts the rheostat in order to keep the ratio of the currents constant. Here is what it looks like when you open it and look inside:

Demon inside transistor

(Credits: Illustration by Paul Horowitz and Winfield Hill, from The Art of Electronics. T-shirt by Adafruit.)

Depending on the components you put around the transistor, the demon may find himself unable to get a high enough IC, even with the rheostat set to the minimum. We call this regime “saturation”. The regime where the demon manages to control IC is the linear regime, used in amplification.

Edgar Bonet
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    The answers are getting more and more interesting... So, what is saturation? The very idea of the so-called "amplification" is quite absurd. First, the transistor introduces some initial resistance (losses) that decreases the current. Then, it begins decreasing its resistance to increase the current... Finally, the resistance becomes zero… the transistor has nothing more to decrease... and as they say, it "saturates". To continue below the zero point, it needs an additional internal source with negative polarity... and then it will behave as a "negative resistor"... – Circuit fantasist Dec 02 '19 at 13:30
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I'll try and keep it to the absolute minimum needed.

The signal at the base is copied at the emitter, less by about \$700\:\text{mV}\$. So the emitter just "follows" the base around. This voltage is impressed on the emitter resistor, causing a varying emitter current that varies directly with changes in the base voltage. So \$I_\text{E}\approx \frac{V_\text{B}-700\:\text{mV}}{R_\text{E}}\$. Since \$I_\text{C}\approx I_\text{E}\$, it follows that \$V_\text{C}=V_\text{CC}-I_\text{C}\cdot R_\text{C}\$ or:

$$\begin{align*}V_\text{C}&=V_\text{CC}-\frac{V_\text{B}-700\:\text{mV}}{R_\text{E}}\cdot R_\text{C}\\\\ &=V_\text{CC}+700\:\text{mV}\cdot\frac{R_\text{C}}{R_\text{E}}-V_\text{B}\cdot\frac{R_\text{C}}{R_\text{E}}\end{align*}$$

Taking the derivative, we find:

$$\begin{align*}\text{D} \:V_\text{C}&=\text{D} \:\left[V_\text{CC}+700\:\text{mV}\cdot\frac{R_\text{C}}{R_\text{E}}-V_\text{B}\cdot\frac{R_\text{C}}{R_\text{E}}\right]\\\\ &=-\frac{R_\text{C}}{R_\text{E}}\cdot \text{D}\:V_\text{B}\\\\\therefore\\\\ \frac{\text{d}\,V_\text{C}}{\text{d}\,V_\text{B}}&=-\frac{R_\text{C}}{R_\text{E}}\end{align*}$$

So, \$A_v\approx -\frac{R_\text{C}}{R_\text{E}}\$.

With respect to your comment here below, the above does not apply to a grounded emitter, where another highly temperature dependent and highly signal dependent term arrives: \$r_e\$.

Quoting from Paul Horowitz and Winfield Hill's "The Art of Electronics," 3rd edition, page 94, right column:

"The extra voltage gain you get by using \$R_\text{E}=0\$ comes at the expense of other properties of the amplifier. In fact, the grounded emitter amplifier, in spite of its popularity in textbooks, should be avoided except in circuits with over-all negative feedback."

Q.E.D.

jonk
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  • Only to add that in this configuration, none of the transistor terminals is common... and all transistor terminals "move" when the input voltage "moves". It is also not "common collector" because the collector is not directly connected to V+. In fact, the lower end of the emitter resistor is common here. To make it "common-emitter", we should connect a capacitor in parallel to the emitter resistor to "stiff" the voltage across it... or just connect the emitter directly to ground ("grounded emitter"). What is this circuit is another interesting topic to discuss... – Circuit fantasist Dec 02 '19 at 07:31
  • Thanks for the answer but your calculations get indefinite for grounded emitter, i.e RE=0. I think you should revise your answer. – muyustan Dec 02 '19 at 08:06
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    @muyustan A grounded emitter situation is vastly more complicated, since the gain will vary quite substantially with the applied signal leading to very highly distorted outputs. These almost always need to be rectified using global negative feedback networks as part of the added stages that must also surround it. As such, it's beyond the scope of my intended answer. Your question was quite short and certainly didn't convey any need for that situation. You should ***expand*** your question, if you want that situation addressed directly. At a minimum. – jonk Dec 02 '19 at 17:34
  • *"With respect to your comment here below, the above does not apply to a grounded emitter, where another highly temperature dependent and highly signal dependent term arrives: re."* This sentence of yours a little bit sounds like little *re* is not there when RE is present. Which looks not right to me, however I am not sure. – muyustan Dec 02 '19 at 19:34
  • @muyustan Little \$r_e\$ is an AC phenomenon that derives directly from the Shockley equation simplification of the Ebers-Moll model. (You take the derivative to find the slope of the curve at some operating point.) It is *not* a DC phenomenon. So it is present when considering AC gain, for example. But it is not present when computing the DC quiescent operating point. For single stages, which should *never* use \$R_\text{E}=0\$, it's usually the case that \$R_\text{E}\$ is very much larger than \$r_e\$ so that the AC gain is almost independent of the applied signal and distortion is less. – jonk Dec 02 '19 at 19:58
  • @jonk yes, but what I want to say that yout phrase that I quoted sounds like *re* is not a concern whlie dealing circuits containing RE but it becomes a concern when dealing with grounded emitter circuits. Is this the situation you want to imply? edit : I guess you edited your comment while I was writing this one. – muyustan Dec 02 '19 at 20:02
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    @muyustan For a well-managed CE amplifier circuit, the value of \$R_\text{E}\$ will be very much larger than \$r_e\$. Usually, if it is affordable, the quiescent voltage drop across \$R_\text{E}\$ will be at least 10 times the worst expected value for \$V_T\$. I prefer at least 40 times, or more. So yes, it is the case that \$r_e\$ is not usually a concern because it is intentionally swamped out by \$R_\text{E}\$. I've written about this many times here on EESE, with much more equation development (including how to derive \$r_e\$.) You've only to search. – jonk Dec 02 '19 at 20:06
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    Edited: @muyustan To a reasonably good approximation Re = 26/Ie_mA Ohms. eg for Ie = 1 mA re = 26 Ohms, for Ie = 10 mA re = 2.6 Ohms etc. This is true regardless of whether RE is present or not. Obviously for gain dominated by Rl and Re, variations in Ie play havoc with linearity. When RE is present and >> Re then gain ~= Rl/ (RE + Re) is much closer to linear. So, yes, most of your "it sounds like" comments above are about correct. – Russell McMahon Dec 02 '19 at 20:45
  • @RussellMcMahon thanks, are these related to the bypass capacitor used across the RE? If they are so unrelated don't bother to explain here. – muyustan Dec 02 '19 at 20:51
  • @muyustan "these" hardly conveys your question adequately. Being clearer with what you don't understand better helps us help you. Re is internal emitter resistance. RE is external emitter resistor. Work through what I said. Work out why it's true. See how it affects gain and results with and without RE bypassed. – Russell McMahon Dec 02 '19 at 21:01
  • @muyustan Where in your question did you mention a bypass capacitor across \$R_\text{E}\$? Such designs ***also*** *"should be avoided except in circuits with over-all negative feedback."* There are designs which are perfectly fine, stand-alone, where the emitter resistor is divided into two resistors and the capacitor bypasses only one of them. That is commonly used to provide another degree of freedom in setting the voltage gain separately from the quiescent DC biasing point. But I think you are conflating a great many different CE designs into a single mush and may, therefore, be confused. – jonk Dec 02 '19 at 21:15
  • @RussellMcMahon well, I know that bypassing RE provides a higher ac signal gain and meantime it provides stability to the circuit in DC manners(because bypass cap. will not ground emitter for DC). The confusion in my head came after these comments in this question and specificaly on this answer. You say that using RE causes AC gain to become much or less independent of Vbase, however while doing AC analysis we will short the capacitor and thus RE will be again non-existing meaning emitter is again connected to the ground. I cannot make the connections. – muyustan Dec 02 '19 at 21:15
  • @jonk I didn't mention it in my question, I should think it and maybe ask in another question. – muyustan Dec 02 '19 at 21:17
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    @muyustan You should ***NOT*** bypass \$R_\text{E}\$ unless you are designing a multistage system and will be including substantial global NFB. It should be preceded by a pre-amplifier stage and followed by yet another stage, at a minimum. Never stand-alone. For a single stage amplifier, you ***may*** bypass some fraction of \$R_\text{E}\$. But not all of it. So there is always some Ohmic resistance within the AC gain calculation in order to swamp out \$r_e\$. (It also helps to substantially increase the input impedance of the stage, too.) – jonk Dec 02 '19 at 21:24
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    @muyustan Sorry if I confused rather than helped. My point is that Re is inversely current dependant (with a value of close to 26/I Ohms). **Overall gain = Rl/(RE_effective)**, where RE_effective = (Re plus any unbypassed component of RE) (= what Jonk is saying in his last comment above). If you bypass ALL of RE for signal purposes it is the same as having no RE for signal purposes and you get massive non linearities as current varies. | There is no magic here - it all follows very logically from the above **Overall gain = Rl/(RE_effective)**. Make sense of that and it should be clear. – Russell McMahon Dec 03 '19 at 03:28
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    @muyustan Re " ... however while doing AC analysis we will short the capacitor ..." -> In this situation, and in most of life, you do what you do that is appropriate to achieve what you wish to achieve. IF you wish to know the AC results with RE bypassed you may short the capacitor for AC analysis. If you want to know the AC result without a bypass cap in real life you do NOT short the cap for Ac analysis. If you want the real world result for partial bypass, or a small cap or a cap with its own series R or ...., you do what is appropriate for AC analysis. ... ie – Russell McMahon Dec 03 '19 at 05:03
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    ... | You need to avoid jumping into "for this operation we do this standard thing, even if it makes no sense". – Russell McMahon Dec 03 '19 at 05:03
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An even simpler, non-mathematical explanation (that does however involve "magic"):

schematic

simulate this circuit – Schematic created using CircuitLab

V in is applied to the base. The cathode of the base "diode" is always about 0.7V lower than the base. However base current will always be much lower than the current in the load resistor, because RM is a "magic" resistor that adjusts itself to create the current gain of the circuit, which is very roughly the hfe of the transistor.

Basically, it is an amplifier with lots of current gain, unity voltage gain (i.e. no voltage gain), that also shifts the input signal by about 0.7V.

Oh, and it can only source current, never sink it of course.

danmcb
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  • +1 for the original explanation of the emitter follower. It is also interesting that the "magic resistor" (the collector-emitter junction) originally behaves as a current source but here it is forced by the negative feedback to behave as a voltage source. Figuratively speaking, RM and LOAD form a "magic voltage divider" that adjusts its output voltage equal to the input voltage. – Circuit fantasist Dec 02 '19 at 15:07
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Let me try an answer (nearly) without formulas:

The "secret" behind the property of voltage amplification is the fact that the bipolar transistor (BJT) works - similar to the FET - as a voltage controlled current source Ic=f(VBE). This current Ic is driven by the DC source, but it is controlled by the voltage VBE. Any change in VBE (input signal, input voltage swing) will change the current Ic (output current swing).

Now we use a resistor Rc to convert this output current swing into an output voltage swing (output signal). This output signal can be larger than the input signal because we can select a rather large value for Rc (as long as we have a supply voltage of several volts to allow a corresponding DC drop across Rc).

Property of the voltage-current transfer function: delta(VBE)/delta(Ic)=26mV/Ic (Example: 26 Ohm for Ic=1mA)

That means: For all Rc values larger than 26 Ohms we have voltage amplification (assuming Ic=1mA, which would cause a DC drop of 1V across Rc=1kOhm, which is not a problem for a supply voltage above 5volts or so..)

Comment to the very simple "rheostat" models: These models have the disadvantage that any change in Vsupply would also change the current through the rheostat-resistance. However, this is not the case in reality. I even would say that it is a misleading over-simplification to model the collector-emitter path as a (controllable) resistance.

LvW
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  • Is not vbe always constant lets say at 0.7 volts? – muyustan Dec 02 '19 at 10:52
  • Yes, VBE must be DC biased with app. 0.7 volts....but note that I spoke about changes of VBE [delta(Vbe) swings around this bias point]. – LvW Dec 02 '19 at 11:05
  • @LvW, it is not a simple "rheostat" with static (ohmic) resistance; it is a "dynamic rheostat" with current-stable nonlinear resistance. This means it varies not only when the input voltage varies but also when Rc varies. What else could be the "collector-emitter path" other than resistance... non-linear but still resistance that dissipates power? – Circuit fantasist Dec 02 '19 at 12:54
  • Ok - but everybody looking to the model (diagram) sees an ohmic resistor symbol....To me, a valid model must allow to apply all classical rules (Ohm, Kirchhof...) - and the result must approximately realistic. – LvW Dec 02 '19 at 13:19
  • But there is also a man (the "demon" of @EdgarBonet below) that controls the "rheostat" and determines its behavior. The combination of the rheostat and man is the "transistor" in this conceptual arrangement. These are means of intuitively understanding circuits by human beings - rheostats controlled by mans and demons, vessels filled with water and then emptied, springs and levers pulled up and down... – Circuit fantasist Dec 02 '19 at 14:01
  • Well - a model is a model...nothing else. I think, some models can reflect the behavior of real devices rather good - and some other not so good. Some models allow the application of all the known rules from circuit theory and some other need some artificial parts (like "demons"). I think, everybody can decide which models fits best to his brain and can best explain the main functions to him. I always remember the sentence: "Everything should be made as simple as possible, but no simpler " (Einstein) – LvW Dec 02 '19 at 15:41
  • As an example I refer to the T-model for a BJT (which is one of three most documented small-signal BJT-models). Here the transconductance gm is modelled as a resistance re=1/gm. For my understanding (and my way of thinking) this is a very bad model because I know that 1/gm is not a resistance in the classical sense (a two-pole). This model does not help me to understand whats really going on.....because the parameter gm (as part of another, better small-signal model) constitutes the connection between input (Vbe) and output (Ic) in form of the transfer functions slope gm=d(Ic)/d(Vbe). – LvW Dec 02 '19 at 15:57
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schematic

simulate this circuit – Schematic created using CircuitLab

Very simple non-mathematical description (see comments below for all the bits I missed out).

With nothing on the Base (input) you just have the Collector (top) connected to the power rail and the Emitter (bottom) connected to ground.

Recall that the Base-Emitter and Base-Collector junctions are like diodes. In this configuration, the Base-Collector "diode" is reverse-biased so it looks like an open circuit and no current flows. The Output is therefore at the same voltage as the power rail.

When you put a small positive voltage on the input, the Base-Emitter junction becomes forward-biased and current starts to flow.

The injection of charge-carriers into the Base region modifies the characteristics of the Base-Collector junction in such a way that it allows current to flow from the Collector to the Emitter. So what was once an open-circuit (collector -> emitter) becomes a partial conductor whose resistance depends on the current injected at the base.

Current flowing in at the Collector pulls the output level low. The extent of the swing depends on the current at the base (this is the amplification effect).

Of course, a transistor is really a dynamic component so if you have an oscillating signal (e.g. from a guitar pickup) at the input, you will get a similar oscillation in the output voltage. Since this output voltage is coming from the power rail, it can have as much power as you like - enough to drive a speaker, for example.

Oscar Bravo
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  • Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/101764/discussion-on-answer-by-oscar-bravo-how-does-a-common-emitter-amplifier-work-in). Any conclusions reached should be edited back into the question and/or any answer(s). – Dave Tweed Dec 03 '19 at 21:34
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A transistor can amplify because the collector current is very sensitive to changes in the voltage at the base, especially between 0.5 and 0.7 volts of base voltage. For example, a 1% change in voltage at the base can result in a 100% change or more in the current at the collector.

What is the mechanism that makes the collector current so sensitive to the base voltage? This requires an understanding of pn junctions, but in simple terms, as the base voltage increases it reduces the barrier between the collector and emitter which is already quite thin. Eventually the barrier is narrow enough that current can flood across the collector and emitter since the potential between these two is at the voltage set by the power rail. Imagine a very wide dam holding back water and that the height of the dam can be lowered. As we lower the height of the dam it approaches the height of the water at which point a lot of water will start to flow of the lip of the dam. You can image that a 1% change in dam height could result in a 50% change in water flow over the dam.

Often however, what we want at the collector is not a change in current but a change in voltage. To convert changes in current to changes in voltage we add a resistor between the collector and voltage rail. Thus as the collector current increases the voltage drops across the collector and ground. Note this also means that the output signal will be inverted but this doesn't often matter. For example, if you have a sine wave coming into the base with a peak to peak voltage of 5 mV, the output voltage will mimic the sine wave (but 180 degrees out of phase) but with a much larger peak to peak voltage. BUT I said before, amplification occurs between 0.5 and 0.7 volts, so how can 5 mV make a difference? This is what transistor biasing is for, to bring the operating range of the transistor into the range of the input signal.

rhody
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For BJT, points to note are mentioned below.

  1. Emitter current or collector current is exponentially proportional to base-emitter voltage.
  2. Emitter current is directly proportional to base current. (Ic = beta * Ib)
  3. VCE of a BJT should be greater than or equal to VBE to keep transistor in active region (region where we want BJT to operate to get gain).
  4. Trans-conductance of a BJT is Ic/Vt (Vt is thermal voltage). Means, trans-conductance of BJT depends on collector current.
  5. Ie = Ic + Ib

Now, if we want to get gain, our first hope is that gain will not change much as i go on applying input voltage. But, clearly from above points, if i apply INCREMENTAL input voltage across VBE, Ic is will change exponentially wrt VBE. This is not the kind of gain we want. Trick here is to implement negative feedback by putting a resistor at the emitter. In this way, change in VBE causes exponential change in Ic, that Ic flows through emitter as well and built exponential voltage across emitter resistor. Now, as emitter voltage itself has exponential built in it (as Vb is input voltage and Ve is kind of exponentially related to Vb (This is applicable only for an instance of time)), VBE will have negative exponent built in (VBE = VB - VE and VE is positive exponent for an instant) which will bring back the collector current in proportion to input voltage. Thus we have achieved proportional relationship between collector current and VBE.

Now, if we connect a resistor RC to collector side, this current will flow though that resistor which will generate output voltage. This output voltage is proportional to input voltage applied and proportionality constant is nothing but INCREMENTAL COLLECTOR CURRENT * RC.

I hope you understand difference in BIASING and INCREMENTAL picture of circuits.

To feed a thought to ponder, if emitter resistor is not used, VBE to IC relationship is exponential in nature, and thus gain.

Omibuddyy
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  • "VCE of a BJT should be almost equal to VBE to keep transistor in active region". I am looking at CE stage in active mode (amplifies about 100x) and... VCE is 6V, while VBE is 0.72 V. They do not appear to be almost equal. Care to review this part? – Sredni Vashtar Dec 03 '19 at 18:41
  • @SredniVashtar Corrected answer for the same. Thanks for noticing it! – Omibuddyy Dec 03 '19 at 23:52
  • However, there is a case where VCE = VBE... and the transistor is in active mode - the input part of the BJT current mirror (the so-called "active diode"). But this is not a humble CE stage; this is a CE stage with voltage type negative feedback between the collector and base. – Circuit fantasist Dec 05 '19 at 19:20
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    So by humble CE stage you mean no emitter resistor case? – Omibuddyy Dec 05 '19 at 20:10
  • Yes, I meant CE without emitter resistor. Strictly speaking, in the "emitter resistor case", the emitter is not common; the lower end of the resistor is common. So this configuration is not "common emitter; it is sooner "common emitter resistor":) You can make it CE if connect a big capacitor in parallel to the resistor. Your explanation of the "proportional relationship between collector current and VBE" is interesting. Is it you insight or you saw it somewhere? – Circuit fantasist Dec 05 '19 at 23:06
  • So at the end of my answer, i gave a thought to ponder on for OP as to think what happens if emitter resistor is not there. I wanted to answer the question in such a way that i gives insight to negative feedback as well. Because it's not easy for a beginner to understand how negative feedback can be thought of in single transistor circuit! As you mentioned that circuit will become CE if capacitor is put across emitter resistor, is in itself kind of confusing maybe for OP is he doesn't clearly understand small signal and large signal analysis. So i tried to explain that as well in my answer. – Omibuddyy Dec 06 '19 at 03:48
  • The explanation regarding proportional relationship between collector current and VBE is my own insight. I hope i am correct and clear in the answer and hope it helps OP and others! I didn't get upvotes though! So maybe my answer lacks in some aspect. I would like if anyone tells me where! – Omibuddyy Dec 06 '19 at 03:53
  • The original thinking does not inspire enthusiasm in traditionally minded people… IMO we will obtain linearity if cascade an antilog and log converters as it is made in a BJT current mirror - https://en.wikibooks.org/wiki/Circuit_Idea/How_to_Reverse_Current_Direction. In your case there is only an antilog converter connected in the loop. So I think we should find some other explanation of the linearity... – Circuit fantasist Dec 06 '19 at 06:58
  • I understand. I hope this answers helps people. That's all. – Omibuddyy Dec 06 '19 at 07:26