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Regarding the below liner circuit model, as far as I understand to find the output impedance the input current Ib is set to zero and the output resistance is calculated as Vc/Ic. I think this way of modeling and calculating the output resistance comes from the Thevenin model. So the definition output impedance will be Vc/Ic when input port is open circuit (?).

enter image description here

So if I set Ib to zero, hfr * Ib becomes zero and the only thing left is hoe where Ic = Vc * hoe so can we say that the output impedance is 1/hoe ?

I'm not sure above is correct but in some texts they say the output impedance is infinity theoretically? How can that be estimated? If wrong how is the output impedance is derived following the circuit above?

What is meant by the following? I had to take photo to add the original text: enter image description here

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can we say that the output impedance is 1/hoe ?

The output admittance is simply \$h_{OE}\$. That's what \$h_{OE}\$ is by definition. So yes output impedance is the reciprocal.

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  • \$\begingroup\$ Please see my edit. The motivation to write the question was that statement of the text. I don't understand why neglecting means setting the output admittance hoe to zero. I dont get the point what is the significance of doing that. Is hoe really too small in real? Is that because he approximates it to zero? Isnt hoe the conductance of CE region? Im trying to understand the physical meaning of these in relation to the actual transistor. \$\endgroup\$ – cm64 Nov 22 '18 at 10:37
  • \$\begingroup\$ Given that there will mostly always be a collector resistor that is in parallel with 1/hoe you can often ignore the effect of hoe. \$\endgroup\$ – Andy aka Nov 22 '18 at 12:23
  • \$\begingroup\$ what about Hre? \$\endgroup\$ – analogsystemsrf Nov 22 '18 at 16:21

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