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What I understand the AC analysis of a bjt amplifier is to just remove the DC sources from the circuit and short all the capacitors. My question is if we do so, how R1 and R_c is going to be grounded in AC analysis? Vcc is removed so aren't they just joined by a short or wire? Why they are taking them as grounded in AC circuit?

  • 1
    \$\begingroup\$ DC sources should be replaced by a short circuit, not left open. So VCC is equivalent to ground. \$\endgroup\$ Dec 8, 2020 at 19:29
  • \$\begingroup\$ Thank you, i think now i've got it. It was silly \$\endgroup\$
    – Sayan
    Dec 8, 2020 at 19:33
  • 1
    \$\begingroup\$ Effectively, the DC source does not exist to an ac signal. \$\endgroup\$ Dec 8, 2020 at 20:17


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