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[HELP]Thevenin Norton with dependent source in Transistor
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[QUOTE="gneill, post: 5133942, member: 293536"] Setting Vin = 0 implies that you are determining the output impedance at the quiescent operating point. If you think about it, for small signal analysis it's really the only practical way to define the output impedance. Alternatively, consider applying the superposition theorem. If there is some input signal Vin and and external current source on the output (you can stick a current source or a voltage source externally, but a current source makes the analysis simpler). Then the voltage at the output will depend upon contributions by both. Showing just the output circuit: [ATTACH=full]84567[/ATTACH] The output impedance will be the ##ΔV_o/ΔI_{ext}## contribution, and will be independent of ##i_b##. [/QUOTE]
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[HELP]Thevenin Norton with dependent source in Transistor
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