Transistor, AC amplification

Let's start with R1 and R2. These resistors form a voltage divider. Without any load the voltage at the midpoint were Vb=12k/(12k+150k)*24V = 1.78V. The schematic says 1.67V because some current from R1 is used as base current for the transistor, thus increasing the voltage drop across R1.
So let's assume 1.67V is correct.
The 0.7V you're talkinmgh about is the base-emitter voltage Vbe. However, in this circuit there is a resistor Re from emitter to GND. The emitter current creates a voltage drop across this resistor. The schematic states that the emitter voltage is 1.17V. That makes Vbe=1.67V-1.17V=0.6V That's reasonable since Vbe=0.6...0.8V for a typical transistor (note that this includes your 0.7V).

Re is used to stabilize the circuit. Assume there is no Re: Vbe is temperature dependent with ~-2 mv/degree C. That means that the base current follows temperature and so does the collector current. The whole circuit would be rather temperature sensitive.
Enter Re. If the Vbe should decreas, Ibe will rise. An increase in Ibe means an increase in Ie and thus an increase in the voltage drop across Re. Thus Vbe is decreased, Ibe is reduced etc. The circuit becomes much more stable.

This is called negatve feedback and is often used to stabilize any system (mechanical, electrical, biological etc.).

Sonce the negative feedback will redice the AC gain, too, you will often find this circuit modified by a capacitor in parallel to Re. The capacitor is a short circuit for AC and thus reduces teh negative feedback.


Harald

 

I wrote:

which is completely correct, considering standard mathematical rules.

The second pair of parenthesis introduced here is correct, but not necessary. A second pair of parenthesis would have been necessary had I intended to write

But I didn't mean to.

 

No problem
Your equation is correct anyway.