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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.
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.
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.
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.
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.
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.
For BJT, points to note are mentioned below.
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.
