Calculating RoE for AC Equivalent Transistor Amplifiers

[Jammin_James]

Hey everyone,

In my EET program we're going over AC equivalent transistor amplifiers. We've gone over how to do some of them in class but I can't seem to figure out how to find roe. I have the output characteristic of our 2N2222 transistor and made the DC load line but I'm not sure were to go from there. Am I supposed to take the change in voltage and divide it by the change in Ib (from the Q point)?

Thanks for the help!

 

[The Electrician]

I would think that roe would be the ratio of a small change in collector voltage to the small change in collector accompanying it; Ib is not involved. It should be fairly large, as you can see by looking at the characteristic curves; notice how they are almost horizontal.

 

[Integral]

roe is fish eggs.

Perhaps you mean $\rho$ or rho.

 

[rbj]

roe is fish eggs.

Perhaps you mean $\rho$ or rho.

no, i think he means one of the "h-parameters" called $h_{oe}$ (or the reciprocal of it).

where the quiessent point is (the "Q-point") in the I_C vs. V_CE, the slope of the curve at that point is your $h_{oe}$ parameter. that slope should be pretty small for a decent BJT and is physically a conductance quantity. the reciprocal of that slope is the output resistance of the Norton or Thevenin equivalent of the dependent source that models the transistor.

$$r_{oe} = \frac{1}{h_{oe}}$$

that's what i think the OP means.

 

[Jammin_James]

That's it, thanks rbj.

Sorry about the confusion; I didn't realized the formula function on here.