BJT Four resistor bias circuit analysis

bjttransistorsvoltage divider

I am currently studying BJTs but am having trouble analysing the four resistor bias circuit. My textbook gives the following circuit and says we can use voltage divider across R1 and R2 to find VB.
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But I feel like we can't ignore RE when we try to find VB. Shouldn't RE (and the 0.7V drop across VBE) combine with R2 in parallel and then use the voltage divider with that parallel combination and R1?

Best Answer

I suppose, your design starts with a given value for Ic, correct? Hence, you know the corresponding value for the base current Ib (based on a given current gain value). More than that, I assume you have fixed already the values for Rc and Re.

Because of Ie=Ic+Ie you know the voltage at the emitter node as well as on the base node (0.65...07 V larger). With these information you can design the voltage divider at the base. Don´t forget that it is a LOADED divider (load=base node).

During calculation you have to make some assumptions:

1.) Vbe=(0.65...0.7): Choice of the value is of less importance because of Re feedback, which stabilizes the DC operating point,

2.) Resistance niveau of the voltage divider: It is common practice to select resistances which allow a current through the divider chain which is approximately (6...10) times the base current.

(Explanation: You have to find a trade-off. Small resistors give an unwanted small signal input resistance, but allow a good dc stabilizatin - and vice versa).