I am reading the book: Problems & Solution of Electronic Devices & Circuits, and on page 119 I got stumped.

They write that :

$$A_V = A_I \cdot \frac{Z_L}{Z_i}= \cdots = \frac{-h_f}{h_i(Y_L+h_o)-h_r h_f}$$

And then they write that for \$Y_L =\infty\$ \$ A_V = \frac{-h_f}{h_i h_o -h_r h_f} \$

Where \$A_V\$ is voltage gain, \$Y_L\$ is the load admittance, and the \$h\$'s are the \$h\$ parameters.

What I don't understand is why we get the last relation when \$Y_L = \infty \$, shouldn't we get \$ A_V =0 \ for \ Y_L = \infty \$.

Perhaps they meant for \$ Z_L =1/Y_L \$, \$Z_L = \infty \$, but that's not how it's written in the book.

Can someone explain?


Of course, for infinite YL (short circuit) the gain is zero. Hence, you are right and the authors of the book book made an error (typo?). Without any external load resistance (ZL infinite), which applies, for example, to a very (nonrealistic) inductance and a corresponding high frequency, the gain would approach a value that is determined by the internal output resistance ro=1/ho of the transistor only. However, that´s a theoretical exercise only.

  • \$\begingroup\$ Why is it only a theoretical exercise?; I mean approximately it can happen, take Z_L as large as you want, then approximately ro= 1/ho. All the anlysis in circuits uses approximations anyway. \$\endgroup\$ – MathematicalPhysicist Nov 20 '14 at 9:18
  • \$\begingroup\$ OK - at first I agree with you that all such analyses are approximations only. My comment (theoretical exercise) was based on the fact that (a) an infinite ohmic resistance would stop the proper functioning of the BJT and (b) an inductor with an impedance much larger than ro is also not very practical. However, you are right: "approximately" it could happen. \$\endgroup\$ – LvW Nov 20 '14 at 9:32

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