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In order to amplify an AC signal, it's required to provide a DC offset in order to block negative values of current to go through the input. Input signals and output signals oscillate centred on the Q point. What would this look like for an AC signal that is not DC-offset? Would the input signal be stretched vertically and then clipped?

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    \$\begingroup\$ Get hold of a free sim tool and simulate. \$\endgroup\$
    – Andy aka
    Nov 6, 2018 at 10:50
  • \$\begingroup\$ I cant find any good software, do you mind telling me pls? \$\endgroup\$
    – user97449
    Nov 6, 2018 at 11:04
  • \$\begingroup\$ This is what the base biasing pair of resistors is for. They create a Thevenin voltage source and Thevenin resistance to the base in order to set the quiescent point. There are almost countless examples illustrating both abstract and concrete situations, one after another, on this site. \$\endgroup\$
    – jonk
    Nov 6, 2018 at 11:12
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    \$\begingroup\$ Micro-cap student edition \$\endgroup\$
    – Andy aka
    Nov 6, 2018 at 11:54
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    \$\begingroup\$ DC bias does not "affect" the Q-point of a BJT - it rather DEFINES the Q-point. \$\endgroup\$
    – LvW
    Jun 21, 2023 at 9:13

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What would this look like for a an AC signal not DC offset?

It sounds like you are describing a class C amplifier and that type of amplifier has a load line response like this: -

enter image description here

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