Common Emitter amplifier Bias Circuit Simplification This is a slide from one of my semiconductor devices lectures. I understand how the bias circuit works and how to calculate the resistor values however I do not understand how this simplification works.

I see that the capacitors and Vcc source has been removed however I cant understand how the two circuits are equivalent as wouldnt current I1 (from R1) and Ic (from Rc) be different when connected to the AC source?

Thevenin simplification comes to mind but maybe i'm not on the right track!

Any help appreciated.

Andy

• For AC analysis aka small signal analysis, VCC is considered the same as ground (that's one reason why you need decoupling capacitors : to guarantee that assumption is valid!) – Brian Drummond May 2 '16 at 11:22
• Andy West, what is the source of this very uncommon and confusing "equivalent circuit" (mixture of two concepts).? – LvW May 2 '16 at 13:22
• How do you mean? As I mentioned in the question It is from lecture slides on BJT amplifier configurations from a Semiconductor device lecture series @LvW – TheAndyEngineer May 2 '16 at 14:08
• I'm learning this stuff for the first time so hence the question. I find the lecture style confusing. I've been trying to use the book 'Electroinc Devices' by Floyd however I could not find an example in this manner in the book either – TheAndyEngineer May 2 '16 at 14:12
• A linear small-signal equivalent circuit contains ONLY linear passive elements (resistors) and active elements which are linearized around the dc operating point (controlled sources). In your diagram there ist still a transistor symbol, which needs DC bias. But this bias is not shown (because it is intended to be small-signal linear equivalent diagram). This is a contradiction. Either a complete circuit diagram (first figure) or a linearized equivalent. But your second diagram is a mixture. – LvW May 2 '16 at 14:57