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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.


  • \$\begingroup\$ 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!) \$\endgroup\$
    – user16324
    May 2, 2016 at 11:22
  • \$\begingroup\$ Andy West, what is the source of this very uncommon and confusing "equivalent circuit" (mixture of two concepts).? \$\endgroup\$
    – LvW
    May 2, 2016 at 13:22
  • \$\begingroup\$ 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 \$\endgroup\$
    – andowt
    May 2, 2016 at 14:08
  • \$\begingroup\$ 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 \$\endgroup\$
    – andowt
    May 2, 2016 at 14:12
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    May 2, 2016 at 14:57

1 Answer 1


The equivalent circuit shows only the components relevant to the signal.

The capacitors are assumed to have an impedance low enough to be considered a short for the signal so they're replaced by a wire.

R1 supplies bias current to the NPN in the equivalent circuit we already know what the bias current is (it is derived in the original circuit) so R1 in the equivalent circuit is there only to show the influence of R1 on the signal (a voltage divider consisting of Rs, R1 and R2).

  • \$\begingroup\$ From what you've said I now understand why the caps have been removed however I'm still unsure on the resistors. Surely the current through R1 will change as its connected to Vac not Vcc? \$\endgroup\$
    – andowt
    May 2, 2016 at 11:24
  • \$\begingroup\$ You're forgetting that the equivalent circuit deals only with the circuit from the signal's point of view. For the signal, the supply rail is the same as ground, on ground and supply there is no signal. You're thinking about DC current, that is only applicable to the original circuit. The equivalent circuit deals with AC signals only or more accurate: derivatives (in the mathematical sense !) of signals only. What is the derivative of a constant Vcc ? \$\endgroup\$ May 2, 2016 at 11:32
  • \$\begingroup\$ To me, the second circuit is not really an equivalent linear small-signal representation. Instead, it is a (uncommon) mixture between relevant small signal components and real non-linear part (transistor). Classical equivalent circuit diagrams use a transconductance block (voltage controlled current source) to show the transistors operation in a certain fixed bias point. \$\endgroup\$
    – LvW
    May 2, 2016 at 13:20
  • \$\begingroup\$ @LvW I fully agree with you, that transistor symbol is out of place. I choose to ignore that since the question was about the circuit around the transistor. To get to the real small signal equivalent, the transistor needs to be replaced by its small signal equivalent model. \$\endgroup\$ May 2, 2016 at 13:24
  • \$\begingroup\$ Yes - and that is why you had the problem of mentioning the DC current in your comment - although it plays no role in a linear small signal representation (it is included in the transconductance block gm). \$\endgroup\$
    – LvW
    May 2, 2016 at 13:39

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