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I have proved one formula of

output impedance ro = (Va+Vce)/Ic = Va/Ic'

where Ic is collector current and Ic' is collector current neglecting early effect.

But reading from standard texts other formulas are diverging like for gm{trans-conductance} = Ic/Vt

is formula given in Razavi, Neamen .

But I am getting

gm=β/rb = Ic'/Vt

where rb is base emitter resistance and β is common emitter current gain.

Am I doing some mistake or its rather approximation the books are taking and in later case plz comment if the approximation is rather universally valid.

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    \$\begingroup\$ The Early effect can be modeled in the small signal equivalent circuit as a resistance between collector and emitter. See: allaboutcircuits.com/technical-articles/… Scroll down to "Accounting for the Early Effect" You're trying to make gm include the Early effect, that's over complicating things. It is much easier to leave gm alone and add a resistor to represent the Early effect. As we're dealing with a small signal model, that can be done. \$\endgroup\$ Jun 17, 2019 at 13:40
  • \$\begingroup\$ Try read this electronics.stackexchange.com/questions/299672/… But for a small signal purpose analysis we simply use this \$r_o = \frac{V_A}{I_C}\$ \$\endgroup\$
    – G36
    Jun 17, 2019 at 14:51
  • \$\begingroup\$ @Bimpelrekkie rb=Vt/Ibq by dynamic resistance of diode in base emitter junction in eber-molls model and current through dependent current source is βIbq so transconductance = β/rb which is Ic'/Vt where Ic' is current through dependent current source on collector side and not current flowing in collector terminal so where I am wrong \$\endgroup\$ Jun 17, 2019 at 20:26
  • \$\begingroup\$ Instead of concentrating on all the formulas, draw the small signal model with and without Early effect. Represent the Early effect as a resistor between emitter and collector. You simply cannot include the early effect into the controlled current source for Ic. So forget about the formulas for a moment and THINK what the Early does to the small signal behavior. \$\endgroup\$ Jun 17, 2019 at 20:43

1 Answer 1

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Model those like this

schematic

simulate this circuit – Schematic created using CircuitLab

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