One simple rule is presented in many sources that I read (for example, Art of Electronics.) This rule tells me that I have to choose resistors R1 and R2 to meet the condition: (R1 || R2) << h21e*Re.
I don't understand the reasons for it. I read that emitter follower are available for low-resistance load (this is not clear for me.)
I have already lost track of how many years I have been thinking about this circuit and how many times I have explained it... and still do. And if someone asks me why I do it, I won't be able to answer them. The OP is long gone; new OPs have their own problems, and this one is unlikely to interest them. The chance to attract the attention of "those who know" and "those who can" is slim to none. So, my only hope remains future visitors to the site (EE archive of questions is a veritable treasury). Oh yes, there is one more reason to do it, and that is CircuitLab with its amazing capabilities...
This rule tells me that I have to choose resistors R1 and R2 to meet the condition: (R1||R2) << h21e*Re. I don't understand the reasons for it.
For you to understand it, Mr. H&H should explain that through the voltage divider R1-R2 an imperfect voltage source is made with output resistance R1||R2 which is loaded by the input resistance h21*Re of an emitter follower. It should then reveal the problem. In practice, however, it turns out that there is none or it is insignificant. There would be a problem and it would be big if the divider directly fed the emitter resistor Re. But this is done through a transistor, which does some "magic" and increases Re a hundredfold. This should be well explained.
I will disclose the idea in my favorite way - building the circuit step by step, and illustrating each step by CircuitLab experiments. I will explore the circuits in DC mode so you can observe the quantities through the measuring instruments and the DC Live Simulation (hovering the mouse over the circuit elements).
First we obtain the input voltage using the voltage divider R1-R2; assume it is half of the supply voltage. We connect a voltmeter Vout to the divider output; its resistance is very high, so the divider is unloaded (open circuit).
simulate this circuit – Schematic created using CircuitLab
Then we connect a low-resistance (1 kΩ) load through an ammeter to the divider's output. It significantly loads the divider - the current approaches 1 mA and the voltage drops below 1 V.
If we connect a diode in series with the load (to make a connection with the next step), the load voltage will drop by about 0.7 V.
The significant loading is undesirable because it changes Vout; we need to somehow reduce the impact of the load to the divider's output. We have two options - either significantly reduce R1 and R2 (to make the divider "harder") or significantly increase the load resistance RL. The former is not desirable because later we will apply an AC input voltage at the midpoint, and the input source will be loaded by the divider. The latter is directly infeasible but we can artificially increase the resistance with a clever idea with the figurative name "bootstrapping".
So, let's connect a transistor as an emitter follower but to see the meaning of this arrangement, let's initially disconnect the collector from Vcc. This configuration is no different from the above, except that the diode is replaced by the base-emitter junction; and the result is the same - a significant loading.
Real circuit: Connect the collector to Vcc, and you will see the magical effect of this connection. The transistor maintains almost the same voltage as the input across Re, but now the load consumes its current not from the divider's output but directly from Vcc. The current consumed from the divider is hundreds of times (h21 or β) less than before when the load was directly connected to the divider.
Equivalent circuit: As though a load with virtual resistance h21.Re is connected to the divider.
Let's finally bring this great bootstrapping idea to perfection in three steps:
Get a voltage source and connect to it a voltmeter through an ammeter. There is no current flowing through this "load" (open circuit).
Then, connect a low-resistance load RL to the voltage source. As a result, a significant current flows through the load.
By current source: Finally, connect a current source to the load and adjust the current so that the voltage across it is equal to the input voltage. As a result, no current flows through the ammeter; all the current is diverted through RL (Vin prevents it to flow through the ammeter). The input voltage source "has the feeling" that there is no load connected (virtual open circuit); it "sees" nothing.
Let's sweep V and IL...
... to see it graphically.
By voltage source in parallel: Of course, we can directly connect a voltage source in parallel to RL, and the result will be the same.
This arrangement was used in the 19th century to measure a voltage by an imperfect voltmeter (with low resistance). Let's try it by setting a 1 kΩ voltmeter resistance (open the voltmeter parameters window and set the resistance).
Let's now sweep V and VL...
... to see it graphically.
By voltage source in series: With the same success, we can connect the voltage source in series to RL. This is the classic bootstrapping arrangement.
With this trick, we can increase the op-amp input resistance up to infinity (to make it "ideal").
If we sweep V and VL...
... we will get the same graphical results as above.
The current drawn by the base is dependent on the transistor parameter hfe (which varies greatly from one part to another of the same type, and also with temoerature etc). The base current will be the emitter current (emitter voltage divided by Re) divided by the hfe.
The source (Thevenin equivalent) resistance looking into the divider is R1||R2.
So they are saying that the voltage drop at the base due to the base current should be small in relation to the supply voltage. That way, hfe can vary over a wide range with little effect on the bias.
A typical choice is a factor of 10, so if hfe is 200 and Re is 1K you would like R1||R2 < 20K.
The explanation behind this requirement is the fact that the transistor acts as a voltage-controlled current source. (I know that in some books it is described in a simplified manner as current-controlled, but that`s not true and cannot explain the mentioned - and other - requirements).
For common-emitter configurations, it is the task of the emitter resistor RE to provide DC stabilization of the operating point. For this purpose an unwanted current increase d(IE) produces a corresponding voltage increase d(VE) at the emitter node which reduces the VBE=VB-VE voltage correspondingly. As a result - the current IE comes back nearly to its original value.
For the shown circuit (emitter follower) the resistor RE does the same job - although it provides - at the same time - the output signal.
However, this current-controlled voltage feedback is effective only if the voltage VB at the base node is kept constant. For this reason, the resistive level of the base voltage divider should be as low as possible (in order to produce a "stiff" bias voltage VB, independent on possible IB variations). As a minimum requirement, it must be much lower than the load resistance for this voltage divider. Because the effective emitter resistance - as seen from the base node - is h21*RE (due to the current IE=h21*IB) we arrive at the requirement R1||R2 much smaller than h21*RE.
However, we have to watch the resulting input impedance (should not be too low) and the power consumption of the whole circuit. Hence, a trade-off is always necessary between "good stabilization" (low value of R1||R2) and the input impedance (R1||R2 not too low). In most cases, this trade-off results in the following rule of thumb for the current I1 through R1: I1~(6...10)IB.
Hence, the resistors R1 and R2 are determined by R1=(Vcc-VB)/I1 and R2=VB/(I1-IB) with VB=0.7V+IE*RE .
Comment: Both equations for R1 and R2 show an interesting and important property of the circuit: We do not know the "exact" value for VBE which gives the desired quiescent current IE. However, it does not matter too much if we assume VBE=0.65V or VBE=0.7V because VBE appears together with IE * RE (normally, several volts) in the equations (VB=VBE+IE*RE). Hence, the influence of the actual VBE value is reduced due to the feedback caused by RE.
