Why do we treat R1 and R2 as though they're parallel in this circuit?

Question

I'm working on a homework problem and I need to solve for the base current Ib, and I know to do that I need to find the Thevenin equivalent at the transistor base. My professor solves for it as though R1 and R2 are parallel, but I don't understand why we can do that. Can anyone clarify what's going on? Thank you!

Answer 1

Prof is doing a simplification, reducing two resistors to one. Not much of a simplification, but every bit helps:

^{simulate this circuit – Schematic created using CircuitLab}
You should remember the rules for doing Thevenin equivalent circuits:

• Short voltage sources, open current sources; then solve for Rb resistance seen at the "Vbase" port. In this case, Vcc is replaced with a short (in your mind, not in practice). You should see that this short results in R1 parallel with R2.
• Assume Vbase has nothing connected outside the box... then VBB = voltage across R2 with Vcc active (in this case).

We usually measure all voltages with respect to GND. In this case, Vbase is at a positive potential (above GND) because Vcc is a positive voltage (above GND).

Answer 2

The context of this question is a DC analysis (analysis of the DC bias) as opposed to an small signal AC analysis.

Normally a DC-block capacitor would be in series with Vi, and since you are solving for a DC bias, you set all AC sources to zero and AC block capacitors become open circuits. Therefore, Vi would be open and floating.

Then instead of the base being driven by Vcc, R1, and R2 , it can be thought of as just being driven directly by a voltage source through an output impedance.