# How does a common emitter amplifier work (intuitively)

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.

• 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?
– user105652
Dec 1 '19 at 22:14
• Try and write a short explanation of the working principle into your question. That will give us something to work with / against! Dec 1 '19 at 22:35
• Are you asking about common emitter amplifiers specifically, or transistors in general? Dec 2 '19 at 7:41
• Here you can read the way I see it electronics.stackexchange.com/questions/355899/…
– G36
Dec 2 '19 at 13:56
• @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<<RE) in our formulas because it is very important to UNDERSTAND the working principle of the BJT. We should realize that re=1/gm - and that the transconductance gm=d(Ic)/d(VBE) constitutes the transfer properties between input and output! In other circuits (common collector, diff. amplifier,..) we need the quantity gm (note: re is NOT a resistance, it is a transresistance"!)
– LvW
Dec 3 '19 at 9:55

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.

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:

(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.

• 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"... Dec 2 '19 at 13:30

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.

• @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 2 '19 at 17:34
• @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 2 '19 at 20:06
• 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. Dec 2 '19 at 20:45
• @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 2 '19 at 21:24
• @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. Dec 3 '19 at 3:28

An even simpler, non-mathematical explanation (that does however involve "magic"):

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.

• +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. Dec 2 '19 at 15:07

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.

• Is not vbe always constant lets say at 0.7 volts? Dec 2 '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 2 '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? Dec 2 '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 2 '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... Dec 2 '19 at 14:01

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.

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). Dec 3 '19 at 21:34

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.

• "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? Dec 3 '19 at 18:41
• @SredniVashtar Corrected answer for the same. Thanks for noticing it! Dec 3 '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. Dec 5 '19 at 19:20
• So by humble CE stage you mean no emitter resistor case? Dec 5 '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? Dec 5 '19 at 23:06