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Maybe it's a stupid question but as the title says, in DC analysis how does the current from Vcc flows? Where is the - and the + of the voltage source? And how does the current flows in AC analysis when we also have Vi involved?

  • 1
    \$\begingroup\$ Maybe this will help electronics.stackexchange.com/questions/301617/… and this electronics.stackexchange.com/questions/310471/… \$\endgroup\$
    – G36
    Mar 7, 2021 at 13:10
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    \$\begingroup\$ It would be nice to know what is your current understanding of the circuit, to know how to start helping you to understand it. \$\endgroup\$
    – Justme
    Mar 7, 2021 at 13:36
  • \$\begingroup\$ Well I do know that in DC analysis that capacitor is like an open switch, so we do not consider Vi at all \$\endgroup\$
    – Elena
    Mar 7, 2021 at 13:45
  • \$\begingroup\$ In DC analysis Rb bias the transistors depens on Rb value, in amplifier it is usualy to point when T2 start to conduct, so thru path Vcc-Rc-Tce2-Re flows some minimal current. \$\endgroup\$
    – user208862
    Mar 7, 2021 at 16:12
  • \$\begingroup\$ After aplying AC signal on Vi the transistors opens more so the current thru path Vcc-Rc-Tce2-Re increases what descreases Voltage at collector of T2. This voltage is usually the output of amplifier (in voltage amplifiers). \$\endgroup\$
    – user208862
    Mar 7, 2021 at 16:20

1 Answer 1


Whenever you see Vcc or +9V or any other voltage flag, it's really just a shortcut way of saying "this is a voltage source relative to ground." That means I can redraw your circuit as:


simulate this circuit – Schematic created using CircuitLab

So the + side of the voltage source is connected to the "V" in your circuit, and the minus side is ground. Vi works the same way. That first capacitor will effectively be a short to AC (if the capacitor is large enough). So current will flow in from Vi in to the base of Q1, which will cause more current to flow through RC and the base of Q2, which will cause even more current to flow through RC and RE. It's a Darlington Pair circuit.

  • \$\begingroup\$ I get it now, thank you so much! \$\endgroup\$
    – Elena
    Mar 7, 2021 at 15:39

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